Energy Conservation And The Work Energy Theorem part a
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Energy Conservation And The Work Energy Theorem part b
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Work Done By A Constant Force
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Explanation About the Integral of Work
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How to Calculate the Integral of a non Constant Force
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Deriving Work And Energy Equations
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Calculating Conservative Forces From Potential Energy
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How To Check If A Force Is Conservative
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Calculating Potential Energy From Conservative Forces
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[{"Name":"Energy Conservation And The Work Energy Theorem part a","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation","Duration":"3m 54s","ChapterTopicVideoID":9031,"CourseChapterTopicPlaylistID":5417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.460","Text":"Hello. In this video,"},{"Start":"00:02.460 ","End":"00:07.080","Text":"we\u0027re going to talk about the Conservation of Energy and the Work Energy Theorem."},{"Start":"00:07.080 ","End":"00:08.460","Text":"This is really important."},{"Start":"00:08.460 ","End":"00:13.560","Text":"It can show us in principle how we really work with energy."},{"Start":"00:13.560 ","End":"00:17.415","Text":"Now I\u0027m going to do this a little differently from a pedagogical perspective."},{"Start":"00:17.415 ","End":"00:19.275","Text":"First, I want to show you the formula."},{"Start":"00:19.275 ","End":"00:23.880","Text":"Then later on in the second video for anyone who\u0027s interested and I really do suggest it."},{"Start":"00:23.880 ","End":"00:25.830","Text":"I\u0027ll show you where that formula comes from."},{"Start":"00:25.830 ","End":"00:29.835","Text":"It\u0027s not a very difficult development and I think it can really be a great enrichment,"},{"Start":"00:29.835 ","End":"00:32.040","Text":"but it will be in the second video."},{"Start":"00:32.040 ","End":"00:35.060","Text":"Keeping that in mind, it\u0027s as though we\u0027re starting from"},{"Start":"00:35.060 ","End":"00:38.375","Text":"the end and then we\u0027re going to move back towards the explanation."},{"Start":"00:38.375 ","End":"00:42.910","Text":"But putting the pedagogy aside for a second, let\u0027s get started."},{"Start":"00:42.910 ","End":"00:48.035","Text":"The first formula I want to introduce to you is the formula for kinetic energy."},{"Start":"00:48.035 ","End":"00:51.545","Text":"Kinetic energy is symbolized by E_k."},{"Start":"00:51.545 ","End":"00:57.780","Text":"It\u0027s a set formula which equals 1/2 MV^2."},{"Start":"00:58.090 ","End":"01:01.370","Text":"For now, we\u0027re going to treat this as a definition for"},{"Start":"01:01.370 ","End":"01:03.725","Text":"those who are interested later on in the second video,"},{"Start":"01:03.725 ","End":"01:06.125","Text":"I will explain where this formula comes from."},{"Start":"01:06.125 ","End":"01:09.230","Text":"But if you look at it, it\u0027s important to understand it as is also"},{"Start":"01:09.230 ","End":"01:12.335","Text":"its 1 1/2 the mass times the velocity squared."},{"Start":"01:12.335 ","End":"01:14.635","Text":"Now if we\u0027re squaring the velocity,"},{"Start":"01:14.635 ","End":"01:16.770","Text":"we\u0027re squaring a vector."},{"Start":"01:16.770 ","End":"01:18.275","Text":"If we square vector,"},{"Start":"01:18.275 ","End":"01:19.880","Text":"we get a scalar value."},{"Start":"01:19.880 ","End":"01:24.795","Text":"We get a value with no direction, only magnitude."},{"Start":"01:24.795 ","End":"01:28.430","Text":"Kinetic energy, as well as all other types of energy,"},{"Start":"01:28.430 ","End":"01:31.475","Text":"are always in terms of scalars."},{"Start":"01:31.475 ","End":"01:35.705","Text":"The second energy we\u0027re going to talk about is potential energy."},{"Start":"01:35.705 ","End":"01:38.510","Text":"We talked about in previous lectures."},{"Start":"01:38.510 ","End":"01:40.655","Text":"In the sense that it is also,"},{"Start":"01:40.655 ","End":"01:44.810","Text":"it is equivalent to the work that a certain force will be doing."},{"Start":"01:44.810 ","End":"01:49.805","Text":"It can be symbolized with U or with E_p sometimes as well."},{"Start":"01:49.805 ","End":"01:53.330","Text":"Depending on the force at work,"},{"Start":"01:53.330 ","End":"01:59.494","Text":"it can be equal to different set functions or set equations,"},{"Start":"01:59.494 ","End":"02:03.740","Text":"2 of the most common examples are the gravitational potential energy,"},{"Start":"02:03.740 ","End":"02:08.465","Text":"which is mgh, and the elastic potential energy,"},{"Start":"02:08.465 ","End":"02:12.040","Text":"which equals 1/2(k) delta x^2."},{"Start":"02:12.040 ","End":"02:15.050","Text":"In physics, 1 will almost exclusively use"},{"Start":"02:15.050 ","End":"02:19.370","Text":"the potential energy of gravity and the potential elastic energy in our problems."},{"Start":"02:19.370 ","End":"02:22.970","Text":"But there exists many other types of potential energy as well."},{"Start":"02:22.970 ","End":"02:28.070","Text":"The last energy I want to talk about is total energy or general energy."},{"Start":"02:28.070 ","End":"02:30.260","Text":"It\u0027s basically the sum of all of these energies."},{"Start":"02:30.260 ","End":"02:35.665","Text":"It\u0027s symbolized with E. It can also be called the mechanical energy."},{"Start":"02:35.665 ","End":"02:38.455","Text":"It equals E_k,"},{"Start":"02:38.455 ","End":"02:41.905","Text":"the kinetic energy, plus U the potential energy."},{"Start":"02:41.905 ","End":"02:45.460","Text":"Or more accurately, E_k plus all the"},{"Start":"02:45.460 ","End":"02:49.255","Text":"various types of U that we might have in a given problem."},{"Start":"02:49.255 ","End":"02:50.830","Text":"In 1 particular problem,"},{"Start":"02:50.830 ","End":"02:54.385","Text":"we will probably encounter gravitational potential energy."},{"Start":"02:54.385 ","End":"02:58.780","Text":"There will probably be an elastic potential energy in certain problems as well."},{"Start":"02:58.780 ","End":"03:03.805","Text":"There may be multiple other potential energy associated with the same problem."},{"Start":"03:03.805 ","End":"03:07.330","Text":"Our total energy equals our kinetic energy"},{"Start":"03:07.330 ","End":"03:11.810","Text":"plus all the different types of potential energy in the problem."},{"Start":"03:11.810 ","End":"03:14.665","Text":"When we\u0027re talking about our total energy,"},{"Start":"03:14.665 ","End":"03:17.885","Text":"which we can also call our total mechanical energy."},{"Start":"03:17.885 ","End":"03:20.600","Text":"We always have a constant kinetic energy,"},{"Start":"03:20.600 ","End":"03:22.850","Text":"which is always 1/2mv^2."},{"Start":"03:22.850 ","End":"03:28.590","Text":"We have multiple formulas depending on the different potential energies we\u0027re using."},{"Start":"03:29.030 ","End":"03:31.580","Text":"Our total energy, or again,"},{"Start":"03:31.580 ","End":"03:35.840","Text":"our total mechanical energy,"},{"Start":"03:35.840 ","End":"03:39.980","Text":"is going to be changing depending on what potential energies we\u0027re dealing with."},{"Start":"03:39.980 ","End":"03:42.980","Text":"Now, I want to give a deeper explanation."},{"Start":"03:42.980 ","End":"03:45.440","Text":"The next video for those who are interested."},{"Start":"03:45.440 ","End":"03:47.780","Text":"The larger question still remains,"},{"Start":"03:47.780 ","End":"03:49.145","Text":"how do we use these terms?"},{"Start":"03:49.145 ","End":"03:50.360","Text":"Or maybe more accurately,"},{"Start":"03:50.360 ","End":"03:54.690","Text":"how do we use these terms to solve problems in physics?"}],"ID":9304},{"Watched":false,"Name":"How to Approach the Exercises","Duration":"4m 44s","ChapterTopicVideoID":9032,"CourseChapterTopicPlaylistID":5417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.510","Text":"The first thing I want to do with every problem is"},{"Start":"00:03.510 ","End":"00:06.345","Text":"check if the energy is conserved in the problem."},{"Start":"00:06.345 ","End":"00:11.055","Text":"What does that mean? It means that all of your forces are conservative forces."},{"Start":"00:11.055 ","End":"00:14.235","Text":"If all of your forces are conservative forces,"},{"Start":"00:14.235 ","End":"00:16.080","Text":"what that means is you can solve your problems,"},{"Start":"00:16.080 ","End":"00:18.705","Text":"solve your exercise the way that we solved"},{"Start":"00:18.705 ","End":"00:21.870","Text":"the last exercise by finding the initial force,"},{"Start":"00:21.870 ","End":"00:24.720","Text":"finding the final force, and setting them equal."},{"Start":"00:24.720 ","End":"00:28.440","Text":"I want to take an opportunity to talk about a common mistake as well here."},{"Start":"00:28.440 ","End":"00:32.100","Text":"Some people think that if you have conservative forces,"},{"Start":"00:32.100 ","End":"00:35.700","Text":"it means you only have internal forces and no external forces."},{"Start":"00:35.700 ","End":"00:37.040","Text":"That is not true."},{"Start":"00:37.040 ","End":"00:39.210","Text":"That is absolutely untrue and it will get"},{"Start":"00:39.210 ","End":"00:41.895","Text":"you in a lot of trouble with some of these exercises."},{"Start":"00:41.895 ","End":"00:46.415","Text":"External forces are not related necessarily to the preservation of energy."},{"Start":"00:46.415 ","End":"00:47.900","Text":"They\u0027re related to momentum."},{"Start":"00:47.900 ","End":"00:50.765","Text":"In fact, you can have external forces that"},{"Start":"00:50.765 ","End":"00:54.380","Text":"are conservative forces that do preserve energy."},{"Start":"00:54.380 ","End":"00:57.080","Text":"Well, some of them don\u0027t, some of them still do."},{"Start":"00:57.080 ","End":"01:02.510","Text":"A conservative force does not mean an absence of external forces."},{"Start":"01:02.510 ","End":"01:06.155","Text":"This similar mistake can be made in the opposite way."},{"Start":"01:06.155 ","End":"01:10.535","Text":"You can also have internal forces that are not conservative forces."},{"Start":"01:10.535 ","End":"01:13.430","Text":"This is not a good measure by which to judge if you"},{"Start":"01:13.430 ","End":"01:16.180","Text":"have a problem with preservation of energy."},{"Start":"01:16.180 ","End":"01:17.975","Text":"Let\u0027s return to our method."},{"Start":"01:17.975 ","End":"01:21.035","Text":"We need to check if we have conservative forces."},{"Start":"01:21.035 ","End":"01:23.795","Text":"The easiest way to check if we have conservative forces"},{"Start":"01:23.795 ","End":"01:27.155","Text":"is to see if a given forest has potential energy."},{"Start":"01:27.155 ","End":"01:34.490","Text":"A couple of examples here above are the potential energy of gravity,"},{"Start":"01:34.490 ","End":"01:37.265","Text":"U_g, something can fall,"},{"Start":"01:37.265 ","End":"01:41.060","Text":"or the potential energy that\u0027s elastic potential energy that of a spring,"},{"Start":"01:41.060 ","End":"01:43.565","Text":"because a spring can always spring back."},{"Start":"01:43.565 ","End":"01:46.400","Text":"If you have that potential energy in it,"},{"Start":"01:46.400 ","End":"01:50.395","Text":"your energy is inherently a conservative force."},{"Start":"01:50.395 ","End":"01:55.295","Text":"Now, there are 2 other ways that I can check if I have a conservative force."},{"Start":"01:55.295 ","End":"01:57.110","Text":"The first is the rotor."},{"Start":"01:57.110 ","End":"01:58.190","Text":"We talked about that earlier,"},{"Start":"01:58.190 ","End":"02:00.425","Text":"so I\u0027m not going to get into that too much here."},{"Start":"02:00.425 ","End":"02:07.670","Text":"The second is if we have a force whose net work done is 0,"},{"Start":"02:07.670 ","End":"02:10.010","Text":"then it is also a conservative force."},{"Start":"02:10.010 ","End":"02:12.830","Text":"That means that in going from point A to point B,"},{"Start":"02:12.830 ","End":"02:15.605","Text":"if the work done equals 0,"},{"Start":"02:15.605 ","End":"02:18.980","Text":"then you have a conservative force."},{"Start":"02:18.980 ","End":"02:24.440","Text":"Let\u0027s assume for a second that we have a problem that has conservation of energy."},{"Start":"02:24.440 ","End":"02:27.440","Text":"We\u0027ve checked all of our forces and they\u0027re all conservative forces."},{"Start":"02:27.440 ","End":"02:28.610","Text":"Let\u0027s say for this example,"},{"Start":"02:28.610 ","End":"02:31.175","Text":"we\u0027re using the force of gravity and the force of a spring."},{"Start":"02:31.175 ","End":"02:34.130","Text":"We can then check the box that says that we have conservation of"},{"Start":"02:34.130 ","End":"02:37.790","Text":"energy because we\u0027re dealing with only potential energy in this case."},{"Start":"02:37.790 ","End":"02:39.220","Text":"We move on to our next step."},{"Start":"02:39.220 ","End":"02:42.290","Text":"Now, our next step is we can set our energies equal to each other."},{"Start":"02:42.290 ","End":"02:43.540","Text":"What do I mean by that?"},{"Start":"02:43.540 ","End":"02:46.220","Text":"We can look at our final moment and"},{"Start":"02:46.220 ","End":"02:49.930","Text":"our initial moment in the particular time we\u0027re looking at and set them equal."},{"Start":"02:49.930 ","End":"02:54.845","Text":"Our initial moment, the beginning of the problem is E_i, initial energy."},{"Start":"02:54.845 ","End":"02:58.945","Text":"Our last moment is final energy E_f."},{"Start":"02:58.945 ","End":"03:03.875","Text":"We know that when we\u0027re trying to find the value of the initial energy,"},{"Start":"03:03.875 ","End":"03:07.220","Text":"we can set it equal to the kinetic energy plus the potential energy,"},{"Start":"03:07.220 ","End":"03:13.490","Text":"that\u0027s 1.5mv squared plus whatever the potential energies might be in a given problem."},{"Start":"03:13.490 ","End":"03:16.985","Text":"Then I can take my final energy E_f."},{"Start":"03:16.985 ","End":"03:21.380","Text":"Again, I set that equal to the kinetic energy 1.5mv"},{"Start":"03:21.380 ","End":"03:25.745","Text":"squared plus whatever our potential energy may be at that moment."},{"Start":"03:25.745 ","End":"03:27.895","Text":"Again, it depends on the problem."},{"Start":"03:27.895 ","End":"03:30.720","Text":"Now, when I say I\u0027m setting the 2 equal,"},{"Start":"03:30.720 ","End":"03:35.385","Text":"it literally means that I take an equation where E_i=E_f."},{"Start":"03:35.385 ","End":"03:37.230","Text":"This is the beauty with energy."},{"Start":"03:37.230 ","End":"03:41.285","Text":"If you don\u0027t really care what happened between point E_i and point E_f,"},{"Start":"03:41.285 ","End":"03:43.625","Text":"because as long as there\u0027s conservation of energy,"},{"Start":"03:43.625 ","End":"03:44.930","Text":"we can know that they\u0027re equal."},{"Start":"03:44.930 ","End":"03:50.225","Text":"This makes our job a lot easier because we don\u0027t really care how it got from E_I to E_f."},{"Start":"03:50.225 ","End":"03:54.185","Text":"It could have been a very simple trajectory or very complicated 1 that\u0027s hard to grasp."},{"Start":"03:54.185 ","End":"03:57.820","Text":"What we really care about is those 2 points."},{"Start":"03:57.820 ","End":"04:00.105","Text":"Then we set the 2 equal."},{"Start":"04:00.105 ","End":"04:01.965","Text":"When we subtract things out,"},{"Start":"04:01.965 ","End":"04:05.150","Text":"what we\u0027ll usually get is either are answered directly or"},{"Start":"04:05.150 ","End":"04:11.035","Text":"some data point or value that will solve the problem for us."},{"Start":"04:11.035 ","End":"04:13.965","Text":"To remind you what this looks like,"},{"Start":"04:13.965 ","End":"04:17.885","Text":"E_i breaks down into the initial kinetic energy and initial potential energy,"},{"Start":"04:17.885 ","End":"04:20.650","Text":"which is 1.5mvi squared,"},{"Start":"04:20.650 ","End":"04:22.900","Text":"the initial velocity plus U_i,"},{"Start":"04:22.900 ","End":"04:25.255","Text":"the initial potential energy."},{"Start":"04:25.255 ","End":"04:27.585","Text":"E_f breaks down the same way."},{"Start":"04:27.585 ","End":"04:31.010","Text":"It\u0027s equal to 1.5mvf squared,"},{"Start":"04:31.010 ","End":"04:34.040","Text":"the final velocity, plus U_f,"},{"Start":"04:34.040 ","End":"04:36.380","Text":"whatever the potential energy may be there."},{"Start":"04:36.380 ","End":"04:40.190","Text":"Let\u0027s look at an example like this where we have conservation of energy."},{"Start":"04:40.190 ","End":"04:44.340","Text":"Then we\u0027ll see what happens if we don\u0027t have conservation of energy."}],"ID":9305},{"Watched":false,"Name":"Exercise- One Mass on a Slope Tied to Another Mass Hanging Vertically","Duration":"11m 7s","ChapterTopicVideoID":9035,"CourseChapterTopicPlaylistID":5417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.430","Text":"Hello. In this exercise,"},{"Start":"00:02.430 ","End":"00:05.610","Text":"we\u0027re dealing with a pretty special system and it has a lot of elements."},{"Start":"00:05.610 ","End":"00:08.340","Text":"The first thing we should do is run through those elements."},{"Start":"00:08.340 ","End":"00:10.320","Text":"First, we have the mass m_1."},{"Start":"00:10.320 ","End":"00:14.970","Text":"M_1 is sitting on an incline with an angle of Theta and while it\u0027s on that incline,"},{"Start":"00:14.970 ","End":"00:18.030","Text":"it rests on a spring and the spring constant is k,"},{"Start":"00:18.030 ","End":"00:21.870","Text":"and the spring is compressed to a distance of d."},{"Start":"00:21.870 ","End":"00:27.555","Text":"This mass is then attached to a rope or a string that runs through an ideal pulley,"},{"Start":"00:27.555 ","End":"00:29.940","Text":"meaning we don\u0027t have to worry about friction and on"},{"Start":"00:29.940 ","End":"00:33.030","Text":"the other side of the string it\u0027s connected to a mass m_2."},{"Start":"00:33.030 ","End":"00:38.175","Text":"Now, this mass m_2 is at a height H above the ground."},{"Start":"00:38.175 ","End":"00:43.685","Text":"The question asks, once we release this system from a resting position,"},{"Start":"00:43.685 ","End":"00:47.915","Text":"at what velocity will the mass m_2 hit the ground?"},{"Start":"00:47.915 ","End":"00:50.390","Text":"There are also some given values at"},{"Start":"00:50.390 ","End":"00:53.945","Text":"the end of the problem that will be incorporated later."},{"Start":"00:53.945 ","End":"00:57.040","Text":"For now, it\u0027s time to solve this problem."},{"Start":"00:57.040 ","End":"01:01.235","Text":"The first thing to notice is that m_1 is lighter than m_2."},{"Start":"01:01.235 ","End":"01:05.545","Text":"That way we know that m_2 is in fact going to fall and m_1 is going to rise."},{"Start":"01:05.545 ","End":"01:07.875","Text":"Now, what I need to do is,"},{"Start":"01:07.875 ","End":"01:10.190","Text":"if I want to look at this problem,"},{"Start":"01:10.190 ","End":"01:12.020","Text":"I first need to see if I\u0027ve conservation of"},{"Start":"01:12.020 ","End":"01:15.335","Text":"energy and the way I need to do that is look at all my forces."},{"Start":"01:15.335 ","End":"01:17.360","Text":"What forces do we have in the problem?"},{"Start":"01:17.360 ","End":"01:19.145","Text":"First of all, we have mg,"},{"Start":"01:19.145 ","End":"01:20.690","Text":"the force of gravity."},{"Start":"01:20.690 ","End":"01:23.180","Text":"We know that the force of gravity is"},{"Start":"01:23.180 ","End":"01:27.760","Text":"a conservative force so we can put a checkmark next to that."},{"Start":"01:27.760 ","End":"01:31.225","Text":"Secondly, we have the F_k,"},{"Start":"01:31.225 ","End":"01:34.280","Text":"we\u0027ll call it the force of the spring,"},{"Start":"01:34.280 ","End":"01:36.335","Text":"and we know that, that is a conservative force."},{"Start":"01:36.335 ","End":"01:37.790","Text":"We also have the normal force."},{"Start":"01:37.790 ","End":"01:42.200","Text":"This is the force going perpendicular to the object m_1, in this case."},{"Start":"01:42.200 ","End":"01:47.300","Text":"We know from the past that m_1 is a conservative force because it\u0027s work equals 0,"},{"Start":"01:47.300 ","End":"01:49.055","Text":"the sum work is 0."},{"Start":"01:49.055 ","End":"01:52.685","Text":"Another force that needs to be addressed is T, the tension."},{"Start":"01:52.685 ","End":"01:56.465","Text":"We can\u0027t immediately discern whether it\u0027s a conservative force or not,"},{"Start":"01:56.465 ","End":"01:57.740","Text":"but you\u0027ll see that it is,"},{"Start":"01:57.740 ","End":"01:59.590","Text":"in a second, I\u0027ll explain why."},{"Start":"01:59.590 ","End":"02:03.675","Text":"Ultimately, the work done by T=0."},{"Start":"02:03.675 ","End":"02:05.815","Text":"What\u0027s the reason for this?"},{"Start":"02:05.815 ","End":"02:09.350","Text":"T is the tension on our ideal string holding mass"},{"Start":"02:09.350 ","End":"02:12.730","Text":"1 to mass 2 and going through an ideal pulley."},{"Start":"02:12.730 ","End":"02:17.150","Text":"We know on some basic level that we\u0027re going to have a T over m_2"},{"Start":"02:17.150 ","End":"02:21.455","Text":"and it\u0027s going to be the exact same going upwards as the T on m_1,"},{"Start":"02:21.455 ","End":"02:24.205","Text":"also going upwards towards the pulley."},{"Start":"02:24.205 ","End":"02:27.200","Text":"We know that it\u0027s the same tension operating on"},{"Start":"02:27.200 ","End":"02:30.260","Text":"positive on one side and a negative on the other side."},{"Start":"02:30.260 ","End":"02:32.345","Text":"We can show this mathematically."},{"Start":"02:32.345 ","End":"02:33.890","Text":"W_T of 1,"},{"Start":"02:33.890 ","End":"02:39.160","Text":"the work of T on m_1 is the integral of F or T.dr."},{"Start":"02:39.160 ","End":"02:41.280","Text":"We can say that on m_1,"},{"Start":"02:41.280 ","End":"02:45.240","Text":"it\u0027s going in a positive direction and for m_2 it\u0027s going in a negative direction."},{"Start":"02:45.240 ","End":"02:47.875","Text":"Basically, the displacement is the same,"},{"Start":"02:47.875 ","End":"02:50.740","Text":"but the angle of m_1 would equal 0 degrees,"},{"Start":"02:50.740 ","End":"02:54.205","Text":"the angle of m2 180 degrees, they\u0027re exact opposites."},{"Start":"02:54.205 ","End":"02:57.085","Text":"W_T1 equals negative W_T2,"},{"Start":"02:57.085 ","End":"03:00.160","Text":"and therefore the total work of T,"},{"Start":"03:00.160 ","End":"03:01.885","Text":"the tension equals 0."},{"Start":"03:01.885 ","End":"03:05.020","Text":"You could say, it\u0027s only bringing energy between m_1 and"},{"Start":"03:05.020 ","End":"03:09.850","Text":"m_2 and not affecting the overall energy of the general system."},{"Start":"03:09.850 ","End":"03:13.870","Text":"On a test, you won\u0027t generally need to be as detailed"},{"Start":"03:13.870 ","End":"03:17.665","Text":"and to explain in such depth how T is working."},{"Start":"03:17.665 ","End":"03:21.830","Text":"We can say that, T is a conservative force and move on."},{"Start":"03:21.830 ","End":"03:26.410","Text":"What we can see here is that in our system we do have a conservation of energy,"},{"Start":"03:26.410 ","End":"03:28.355","Text":"all the forces are conservative."},{"Start":"03:28.355 ","End":"03:30.010","Text":"Now, that I have this understanding,"},{"Start":"03:30.010 ","End":"03:32.530","Text":"I can move on to actually solving the problem."},{"Start":"03:32.530 ","End":"03:34.255","Text":"How are you going to solve the problem?"},{"Start":"03:34.255 ","End":"03:35.665","Text":"Like we\u0027ve discussed before,"},{"Start":"03:35.665 ","End":"03:36.955","Text":"you want to set your equation."},{"Start":"03:36.955 ","End":"03:38.870","Text":"First, you need to find your initial energy,"},{"Start":"03:38.870 ","End":"03:43.285","Text":"then you find your final energy and then you\u0027ll set your equation,"},{"Start":"03:43.285 ","End":"03:46.720","Text":"setting the initial energy equal to the final energy."},{"Start":"03:46.720 ","End":"03:49.660","Text":"Now, we can start writing out or energy equations."},{"Start":"03:49.660 ","End":"03:51.820","Text":"If you recall, an energy equation is"},{"Start":"03:51.820 ","End":"03:55.120","Text":"kinetic energy plus all the different potential energies."},{"Start":"03:55.120 ","End":"03:58.195","Text":"In our case, we have gravity and we have elastic potential energy,"},{"Start":"03:58.195 ","End":"04:00.145","Text":"the potential energy of a spring."},{"Start":"04:00.145 ","End":"04:02.125","Text":"For our initial energy,"},{"Start":"04:02.125 ","End":"04:04.020","Text":"first, for the kinetic portion,"},{"Start":"04:04.020 ","End":"04:11.330","Text":"we have 1/2(m_1 v_1^2) plus 1/2(m_2 v_2^2)."},{"Start":"04:11.330 ","End":"04:15.290","Text":"Each object mass 1 and mass 2 needs its own kinetic energy."},{"Start":"04:15.290 ","End":"04:17.075","Text":"They have their own kinetic energy."},{"Start":"04:17.075 ","End":"04:19.700","Text":"Now, it just so happens that at the first moment,"},{"Start":"04:19.700 ","End":"04:22.190","Text":"both of these equal 0 because the velocity on"},{"Start":"04:22.190 ","End":"04:25.120","Text":"each object as we release them from rest is 0."},{"Start":"04:25.120 ","End":"04:27.830","Text":"Now, we can add gravity to the equation."},{"Start":"04:27.830 ","End":"04:31.009","Text":"Again, we need to split this into 2 different masses."},{"Start":"04:31.009 ","End":"04:32.705","Text":"We\u0027ll have for mass 1,"},{"Start":"04:32.705 ","End":"04:38.100","Text":"m_1 gh_1, and for mass 2, m_2 gh_2."},{"Start":"04:38.100 ","End":"04:43.220","Text":"We have to set our h=0 height for each object and we can do this separately."},{"Start":"04:43.220 ","End":"04:44.705","Text":"There can be 2 different points."},{"Start":"04:44.705 ","End":"04:50.945","Text":"For m_1, we get set h0 as some line that goes through the middle of the mass,"},{"Start":"04:50.945 ","End":"04:54.095","Text":"that can be h0 for our first mass."},{"Start":"04:54.095 ","End":"04:56.225","Text":"Form_2, we could set h0,"},{"Start":"04:56.225 ","End":"04:57.590","Text":"we can set it at the boundary object,"},{"Start":"04:57.590 ","End":"05:00.005","Text":"but it\u0027s easier for us to set it at the ground,"},{"Start":"05:00.005 ","End":"05:03.450","Text":"capital H below the mass m2."},{"Start":"05:03.460 ","End":"05:06.275","Text":"Just to reiterate, because it can be confusing,"},{"Start":"05:06.275 ","End":"05:10.010","Text":"we\u0027ve set 2 different initial heights, 2 different h0s."},{"Start":"05:10.010 ","End":"05:11.525","Text":"For mass 1, we have 1,"},{"Start":"05:11.525 ","End":"05:14.080","Text":"for mass 2, we have a different one and this is okay."},{"Start":"05:14.080 ","End":"05:15.690","Text":"Now, for mass 1,"},{"Start":"05:15.690 ","End":"05:17.694","Text":"if we look at our equation,"},{"Start":"05:17.694 ","End":"05:22.265","Text":"the height at the initial moment is 0 so therefore,"},{"Start":"05:22.265 ","End":"05:25.745","Text":"the potential energy of gravity will fall away."},{"Start":"05:25.745 ","End":"05:28.045","Text":"For mass 2,"},{"Start":"05:28.045 ","End":"05:32.570","Text":"we can see that the height at the initial moment is in fact"},{"Start":"05:32.570 ","End":"05:37.475","Text":"large H above the ground so we can replace h_2 with the variable"},{"Start":"05:37.475 ","End":"05:41.750","Text":"large H. I can put a checkmark above"},{"Start":"05:41.750 ","End":"05:46.940","Text":"the potential energy of gravity and move on to the potential energy of the spring."},{"Start":"05:46.940 ","End":"05:51.285","Text":"Now this, I can write it as 1/2(k) times Delta x^2."},{"Start":"05:51.285 ","End":"05:53.070","Text":"In this case, Delta x is d,"},{"Start":"05:53.070 ","End":"05:55.815","Text":"so we write 1/2(k) d^2."},{"Start":"05:55.815 ","End":"05:58.265","Text":"Now, I have my initial energy."},{"Start":"05:58.265 ","End":"06:00.170","Text":"For now, I\u0027m not going to put in the numbers,"},{"Start":"06:00.170 ","End":"06:02.440","Text":"we\u0027ll incorporate those later."},{"Start":"06:02.440 ","End":"06:07.160","Text":"Now, we can move on to the final energy and before writing this out,"},{"Start":"06:07.160 ","End":"06:13.700","Text":"I want to note that because h=3 meters and d=30 centimeters,"},{"Start":"06:13.700 ","End":"06:20.110","Text":"we can assume that the spring will be fully extended at the final moment in the problem."},{"Start":"06:20.110 ","End":"06:21.620","Text":"That\u0027s something to keep in mind,"},{"Start":"06:21.620 ","End":"06:23.615","Text":"but that\u0027s jumping ahead a little bit."},{"Start":"06:23.615 ","End":"06:26.165","Text":"To start back again in an organized way,"},{"Start":"06:26.165 ","End":"06:28.100","Text":"let\u0027s look at the kinetic energy first,"},{"Start":"06:28.100 ","End":"06:29.950","Text":"looking at the velocities of each mass."},{"Start":"06:29.950 ","End":"06:32.615","Text":"Again, we have to separate between the 2 masses,"},{"Start":"06:32.615 ","End":"06:35.215","Text":"each might have its own velocity."},{"Start":"06:35.215 ","End":"06:40.260","Text":"For the first mass, it\u0027s 1/2m_1 v_1^2,"},{"Start":"06:40.260 ","End":"06:41.625","Text":"and for the second object,"},{"Start":"06:41.625 ","End":"06:48.150","Text":"m_2, it\u0027s 1/2m_2 v_2^2 and this is what we\u0027re trying to find."},{"Start":"06:48.150 ","End":"06:52.685","Text":"Now, at this stage, I can also say that, v_1=v_2."},{"Start":"06:52.685 ","End":"06:54.875","Text":"Because they\u0027re connected by the string,"},{"Start":"06:54.875 ","End":"06:57.020","Text":"whatever distance m_2 moves,"},{"Start":"06:57.020 ","End":"07:01.500","Text":"m_1 has to move the identical distance and an identical amount of time."},{"Start":"07:01.660 ","End":"07:05.690","Text":"Now, we can move on to potential energy."},{"Start":"07:05.690 ","End":"07:09.830","Text":"We need to add in first the potential energy of gravity."},{"Start":"07:09.830 ","End":"07:16.750","Text":"We\u0027re going to use the same formula as before, m_1gh_1 and m_2gh_2."},{"Start":"07:16.750 ","End":"07:18.890","Text":"Now, we have to figure out what those hs are."},{"Start":"07:18.890 ","End":"07:20.180","Text":"For the second mass,"},{"Start":"07:20.180 ","End":"07:24.965","Text":"we know that h_2=0 because the object is on the ground."},{"Start":"07:24.965 ","End":"07:27.215","Text":"We can take that and bring it to 0."},{"Start":"07:27.215 ","End":"07:31.295","Text":"For h_1, we need to figure that out and we will calculate it."},{"Start":"07:31.295 ","End":"07:33.020","Text":"But first, I want to say that,"},{"Start":"07:33.020 ","End":"07:37.835","Text":"the elastic potential energy is 0 because we know that our spring is fully extended,"},{"Start":"07:37.835 ","End":"07:40.240","Text":"the spring has no potential energy."},{"Start":"07:40.240 ","End":"07:42.890","Text":"Because Delta x=0,"},{"Start":"07:42.890 ","End":"07:45.880","Text":"our spring has no potential energy."},{"Start":"07:45.880 ","End":"07:48.965","Text":"Now, let\u0027s move on to calculating h_1."},{"Start":"07:48.965 ","End":"07:52.325","Text":"Again, h_1 is the height that mass 1 is"},{"Start":"07:52.325 ","End":"07:55.925","Text":"above the h=0 baseline at the end of the equation."},{"Start":"07:55.925 ","End":"07:58.010","Text":"First, let\u0027s move these 2 objects that it looks"},{"Start":"07:58.010 ","End":"08:00.380","Text":"like it would at the end of this exercise."},{"Start":"08:00.380 ","End":"08:04.550","Text":"Remember, we\u0027re talking about the moment just before m_2 hits the ground."},{"Start":"08:04.550 ","End":"08:06.890","Text":"We can imagine it that way so we don\u0027t have to deal with"},{"Start":"08:06.890 ","End":"08:10.835","Text":"the forces that come into play once m2 hits the ground."},{"Start":"08:10.835 ","End":"08:13.250","Text":"Assuming we\u0027re dealing with that moment,"},{"Start":"08:13.250 ","End":"08:18.140","Text":"and we know that our mass m_1 has moved the same distance as m_2,"},{"Start":"08:18.140 ","End":"08:19.520","Text":"but instead of moving H,"},{"Start":"08:19.520 ","End":"08:21.425","Text":"large H, vertically,"},{"Start":"08:21.425 ","End":"08:24.590","Text":"it\u0027s moved the same distance large H on a diagonal line,"},{"Start":"08:24.590 ","End":"08:30.170","Text":"a diagonal line that\u0027s specifically parallel to the slope on which mass 1 rests."},{"Start":"08:30.170 ","End":"08:34.160","Text":"We know that we can use this baseline h=0 as"},{"Start":"08:34.160 ","End":"08:39.020","Text":"the starting point and we\u0027re trying to find this vertical line which is parallel to h=0,"},{"Start":"08:39.020 ","End":"08:41.120","Text":"which is the height that m_1 is moved."},{"Start":"08:41.120 ","End":"08:44.120","Text":"We can call this h_1 because that\u0027s what we\u0027re trying to find."},{"Start":"08:44.120 ","End":"08:46.880","Text":"Now, we know that this triangle large H,"},{"Start":"08:46.880 ","End":"08:49.970","Text":"h_1 in the baseline has an angle of Theta"},{"Start":"08:49.970 ","End":"08:53.435","Text":"because it\u0027s parallel to the triangle made by the slope,"},{"Start":"08:53.435 ","End":"08:56.330","Text":"the vertical line, and the base of our initial set."},{"Start":"08:56.330 ","End":"08:59.465","Text":"Therefore, using some basic trigonometry,"},{"Start":"08:59.465 ","End":"09:02.435","Text":"I can replace our variable h_1,"},{"Start":"09:02.435 ","End":"09:05.805","Text":"with large H times the sine of Theta."},{"Start":"09:05.805 ","End":"09:07.700","Text":"When I clean up this equation,"},{"Start":"09:07.700 ","End":"09:14.135","Text":"I\u0027ll replace h_1 with large H sine Theta. Let\u0027s do that."},{"Start":"09:14.135 ","End":"09:16.910","Text":"Let\u0027s first write out our final energy."},{"Start":"09:16.910 ","End":"09:20.345","Text":"We have what\u0027s left is 1/2, and we can say,"},{"Start":"09:20.345 ","End":"09:23.870","Text":"m_1 plus v_1 in parentheses times"},{"Start":"09:23.870 ","End":"09:28.370","Text":"v^2 because it\u0027s the same v and we also add on to that what we found,"},{"Start":"09:28.370 ","End":"09:32.075","Text":"m_1 g large H sine Theta,"},{"Start":"09:32.075 ","End":"09:35.485","Text":"and everything else from the final energy drops off."},{"Start":"09:35.485 ","End":"09:38.255","Text":"Now, we can set this equal to our initial energy,"},{"Start":"09:38.255 ","End":"09:40.475","Text":"which looking above, you\u0027ll recall,"},{"Start":"09:40.475 ","End":"09:45.340","Text":"equals m_2 g large H"},{"Start":"09:45.340 ","End":"09:52.804","Text":"plus the elastic energy of the spring, which is 1/2kd^2."},{"Start":"09:52.804 ","End":"09:55.540","Text":"Now, all that\u0027s left is to incorporate"},{"Start":"09:55.540 ","End":"09:59.590","Text":"the given values and solve for v, the missing variable."},{"Start":"09:59.590 ","End":"10:05.250","Text":"Let\u0027s get started, 1/2 of 1 kilogram plus 2 kilograms,"},{"Start":"10:05.250 ","End":"10:07.930","Text":"that\u0027s 3 kilograms times v^2,"},{"Start":"10:07.930 ","End":"10:12.250","Text":"which is what we\u0027re looking for plus 1 kilogram times"},{"Start":"10:12.250 ","End":"10:17.230","Text":"10 times 3 meters and the sine of Theta 30 degrees is"},{"Start":"10:17.230 ","End":"10:22.020","Text":"1/2 equals 2 kilograms times"},{"Start":"10:22.020 ","End":"10:27.720","Text":"10 times 3 meters plus 1/2 k is 100 newtons per meter,"},{"Start":"10:27.720 ","End":"10:30.180","Text":"and d is 30 centimeters."},{"Start":"10:30.180 ","End":"10:33.020","Text":"Now, if you\u0027ll notice, we\u0027ve been working in meters, meters, kilograms,"},{"Start":"10:33.020 ","End":"10:35.675","Text":"and seconds so instead of writing 30,"},{"Start":"10:35.675 ","End":"10:37.580","Text":"d is actually 30 centimeters,"},{"Start":"10:37.580 ","End":"10:41.390","Text":"which is not 30 meters, it\u0027s 0.3 meters."},{"Start":"10:41.390 ","End":"10:42.710","Text":"When we put this in the problem,"},{"Start":"10:42.710 ","End":"10:48.035","Text":"we have to write d as 0.3^3 and when we solve this,"},{"Start":"10:48.035 ","End":"10:50.205","Text":"and we\u0027re looking for v, again,"},{"Start":"10:50.205 ","End":"11:00.690","Text":"we end up with v equaling 5.745 meters per second."},{"Start":"11:00.690 ","End":"11:04.500","Text":"Now, we found the solution to the exercise."}],"ID":9308},{"Watched":false,"Name":"Example - Conservation of Energy","Duration":"5m 38s","ChapterTopicVideoID":12198,"CourseChapterTopicPlaylistID":5417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"Now we have this example."},{"Start":"00:01.935 ","End":"00:04.380","Text":"If this example is too simple for you,"},{"Start":"00:04.380 ","End":"00:07.035","Text":"feel free to move on and do other exercises."},{"Start":"00:07.035 ","End":"00:08.850","Text":"In this exercise though,"},{"Start":"00:08.850 ","End":"00:16.815","Text":"we have a cart that is moving on a frictionless plane and it is above a slope."},{"Start":"00:16.815 ","End":"00:20.040","Text":"The cart is going to go down a slope with the height H,"},{"Start":"00:20.040 ","End":"00:24.780","Text":"and it has an initial velocity given as u_0."},{"Start":"00:24.780 ","End":"00:26.850","Text":"Now the question asked here is,"},{"Start":"00:26.850 ","End":"00:29.100","Text":"what is going to be the final velocity of"},{"Start":"00:29.100 ","End":"00:31.950","Text":"this cart once it reaches the bottom of the slope?"},{"Start":"00:31.950 ","End":"00:35.240","Text":"Let\u0027s run through how we can solve this problem by breaking"},{"Start":"00:35.240 ","End":"00:38.885","Text":"down our energy into its different components and solving them in order."},{"Start":"00:38.885 ","End":"00:41.270","Text":"The first thing we\u0027re gonna do is check if we have"},{"Start":"00:41.270 ","End":"00:43.415","Text":"a conservation of energy, and how do we do that?"},{"Start":"00:43.415 ","End":"00:46.790","Text":"We check if each force is conserved by"},{"Start":"00:46.790 ","End":"00:51.505","Text":"checking whether it has potential energy or whether it\u0027s work equal 0."},{"Start":"00:51.505 ","End":"00:53.445","Text":"Let\u0027s return to our problem."},{"Start":"00:53.445 ","End":"00:55.235","Text":"We have 2 forces we\u0027re dealing with."},{"Start":"00:55.235 ","End":"00:57.860","Text":"We have the force of gravity and the normal force."},{"Start":"00:57.860 ","End":"01:01.850","Text":"Now we know that gravity is a conservative force."},{"Start":"01:01.850 ","End":"01:04.130","Text":"We can already check that off the list."},{"Start":"01:04.130 ","End":"01:06.230","Text":"For the normal force, we have to check."},{"Start":"01:06.230 ","End":"01:10.160","Text":"Now we know that the normal force is going to be perpendicular or"},{"Start":"01:10.160 ","End":"01:14.440","Text":"orthogonal to the path of the cart at any given point."},{"Start":"01:14.440 ","End":"01:16.325","Text":"At the top, it\u0027ll be something like this."},{"Start":"01:16.325 ","End":"01:17.720","Text":"As the cart goes down the slope,"},{"Start":"01:17.720 ","End":"01:20.810","Text":"it\u0027ll be pointing outwards at something akin to a 45-degree angle."},{"Start":"01:20.810 ","End":"01:23.030","Text":"At the bottom it will again be 90 degrees."},{"Start":"01:23.030 ","End":"01:25.970","Text":"Now if we\u0027re calculating the work of a force that is"},{"Start":"01:25.970 ","End":"01:29.405","Text":"perpendicular to our general direction of motion,"},{"Start":"01:29.405 ","End":"01:31.865","Text":"remember that\u0027s the integral of f.dr,"},{"Start":"01:31.865 ","End":"01:34.910","Text":"the scalar multiplication will equal 0."},{"Start":"01:34.910 ","End":"01:37.770","Text":"This is dr is going forward at all times,"},{"Start":"01:37.770 ","End":"01:40.855","Text":"and what we\u0027re trying to measure is going perpendicular to that."},{"Start":"01:40.855 ","End":"01:45.905","Text":"A good rule of thumb is that the work of perpendicular force is always equals 0,"},{"Start":"01:45.905 ","End":"01:47.435","Text":"in this case is no exception."},{"Start":"01:47.435 ","End":"01:50.705","Text":"There are times when the normal force does perform work."},{"Start":"01:50.705 ","End":"01:52.070","Text":"In this case it does not,"},{"Start":"01:52.070 ","End":"01:55.940","Text":"so it can be considered another force that is conserving."},{"Start":"01:55.940 ","End":"02:00.290","Text":"This is great news for us because it means we have 2 conservative forces,"},{"Start":"02:00.290 ","End":"02:02.105","Text":"so we can move forward with our problem."},{"Start":"02:02.105 ","End":"02:03.965","Text":"Now if we look at our sheet,"},{"Start":"02:03.965 ","End":"02:07.550","Text":"what we\u0027ll find is that our next step is to find"},{"Start":"02:07.550 ","End":"02:12.200","Text":"the initial energy and find the final energy and set them equal to each other."},{"Start":"02:12.200 ","End":"02:13.715","Text":"So let\u0027s start with that."},{"Start":"02:13.715 ","End":"02:15.890","Text":"First, we\u0027re going to find our initial energy."},{"Start":"02:15.890 ","End":"02:17.225","Text":"What\u0027s our initial energy?"},{"Start":"02:17.225 ","End":"02:20.720","Text":"Well, as always, it\u0027s a kinetic energy plus a potential energy."},{"Start":"02:20.720 ","End":"02:23.360","Text":"Our kinetic energy is always written the same,"},{"Start":"02:23.360 ","End":"02:28.400","Text":"it is 1/2 mv^2."},{"Start":"02:28.400 ","End":"02:30.680","Text":"Remember this is our kinetic energy, E_k."},{"Start":"02:30.680 ","End":"02:35.360","Text":"In this case our potential energy is going to be gravitational,"},{"Start":"02:35.360 ","End":"02:37.400","Text":"so it\u0027s going to be mgh,"},{"Start":"02:37.400 ","End":"02:38.795","Text":"and this is our potential energy."},{"Start":"02:38.795 ","End":"02:44.720","Text":"Remember, our total energy is always going to be kinetic energy plus potential energy."},{"Start":"02:44.720 ","End":"02:46.250","Text":"In this particular problem,"},{"Start":"02:46.250 ","End":"02:48.259","Text":"our potential energy is only gravitational."},{"Start":"02:48.259 ","End":"02:51.620","Text":"Well, of course, we could have other potential energies in other problems."},{"Start":"02:51.620 ","End":"02:54.650","Text":"Now to finish filling this in for this specific problem,"},{"Start":"02:54.650 ","End":"02:58.010","Text":"we can replace v with u_0 and"},{"Start":"02:58.010 ","End":"03:02.210","Text":"small h with large H. These are the given variables in our problem."},{"Start":"03:02.210 ","End":"03:05.335","Text":"Now we can find our final energy."},{"Start":"03:05.335 ","End":"03:08.010","Text":"Once we find that, we can set the 2 equal."},{"Start":"03:08.010 ","End":"03:10.895","Text":"Our final energy, we\u0027re only going to have kinetic energy."},{"Start":"03:10.895 ","End":"03:16.160","Text":"That\u0027ll be 1/2 mv final as opposed to v initial squared,"},{"Start":"03:16.160 ","End":"03:18.365","Text":"and our potential energy will be 0."},{"Start":"03:18.365 ","End":"03:22.020","Text":"Now vf, v final is the variable we\u0027re looking for."},{"Start":"03:22.020 ","End":"03:26.030","Text":"We\u0027re going to put a question mark above it for now and get back to it,"},{"Start":"03:26.030 ","End":"03:27.890","Text":"and 0 potential energy."},{"Start":"03:27.890 ","End":"03:32.794","Text":"Why? Because our height relative in this problem are large H equals 0 at the end here."},{"Start":"03:32.794 ","End":"03:34.520","Text":"You can measure that from wheels to wheels,"},{"Start":"03:34.520 ","End":"03:35.870","Text":"you can measure that from cart to cart,"},{"Start":"03:35.870 ","End":"03:38.540","Text":"whatever you\u0027re measuring there it\u0027s the same drop in height,"},{"Start":"03:38.540 ","End":"03:42.290","Text":"so at the end height equal 0 and we have no potential energy left."},{"Start":"03:42.290 ","End":"03:45.435","Text":"Now we can set these 2 equations equal to each."},{"Start":"03:45.435 ","End":"03:49.100","Text":"If our initial energy equals our final energy,"},{"Start":"03:49.100 ","End":"03:53.750","Text":"then we can write that 1/2 of mv"},{"Start":"03:53.750 ","End":"04:03.705","Text":"final^2 equals 1/2 mu initial^2 plus mgh."},{"Start":"04:03.705 ","End":"04:05.563","Text":"Our ms can drop out,"},{"Start":"04:05.563 ","End":"04:07.520","Text":"and we can multiply it by 2,"},{"Start":"04:07.520 ","End":"04:09.420","Text":"and then take a square root of both sides,"},{"Start":"04:09.420 ","End":"04:18.330","Text":"and what we\u0027re left with is v f equals the square root of u_0 squared plus 2gh."},{"Start":"04:18.770 ","End":"04:23.855","Text":"That\u0027s your solution. Just for a quick recap of how we got here."},{"Start":"04:23.855 ","End":"04:25.520","Text":"First, we found our 2 forces,"},{"Start":"04:25.520 ","End":"04:27.095","Text":"in this case,"},{"Start":"04:27.095 ","End":"04:29.405","Text":"gravitational force and the normal force."},{"Start":"04:29.405 ","End":"04:31.535","Text":"We know that the gravitational force is"},{"Start":"04:31.535 ","End":"04:36.060","Text":"a conservative force intuitively and from past problems."},{"Start":"04:36.060 ","End":"04:39.980","Text":"We found that the work of the normal force in this case equal 0,"},{"Start":"04:39.980 ","End":"04:42.530","Text":"so it\u0027s also a conservative force."},{"Start":"04:42.530 ","End":"04:46.010","Text":"Then once we knew that we had a preservation of forces in the equation,"},{"Start":"04:46.010 ","End":"04:47.930","Text":"we set everything equal."},{"Start":"04:47.930 ","End":"04:53.375","Text":"Once we did that, we just need to set our initial and final energies to be the same,"},{"Start":"04:53.375 ","End":"04:56.255","Text":"do a little bit of math and find our solution."},{"Start":"04:56.255 ","End":"04:59.225","Text":"Up until now, the problems we\u0027ve done"},{"Start":"04:59.225 ","End":"05:02.890","Text":"had a preservation or conservation of energy in them."},{"Start":"05:02.890 ","End":"05:05.630","Text":"We know the process is what we just describe."},{"Start":"05:05.630 ","End":"05:07.820","Text":"What you\u0027re also going to encounter are problems"},{"Start":"05:07.820 ","End":"05:11.360","Text":"that do not have a conservation of energy,"},{"Start":"05:11.360 ","End":"05:12.620","Text":"and there\u0027s a different process there."},{"Start":"05:12.620 ","End":"05:16.970","Text":"It doesn\u0027t mean you can\u0027t use the body or the equations that you\u0027re dealing with,"},{"Start":"05:16.970 ","End":"05:20.050","Text":"but you do have to use a different device."},{"Start":"05:20.050 ","End":"05:23.360","Text":"If you don\u0027t have a conservation energy,"},{"Start":"05:23.360 ","End":"05:26.950","Text":"you have to use the theorem of work energy."},{"Start":"05:26.950 ","End":"05:28.895","Text":"We\u0027re going to talk about that in the next lecture."},{"Start":"05:28.895 ","End":"05:30.470","Text":"What I would suggest for now is do"},{"Start":"05:30.470 ","End":"05:35.720","Text":"a few more example exercises with conservation of energy before moving on to that."},{"Start":"05:35.720 ","End":"05:39.330","Text":"In the next video, I\u0027ll explain how that works."}],"ID":12673},{"Watched":false,"Name":"Exercise - an Object is Dropped Above a Vertical Spring","Duration":"5m 52s","ChapterTopicVideoID":12199,"CourseChapterTopicPlaylistID":5417,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.295","Text":"Hello. In this video,"},{"Start":"00:02.295 ","End":"00:05.235","Text":"we have a spring that\u0027s attached to the ground."},{"Start":"00:05.235 ","End":"00:06.660","Text":"It\u0027s an ideal spring,"},{"Start":"00:06.660 ","End":"00:09.060","Text":"meaning it doesn\u0027t have any mass,"},{"Start":"00:09.060 ","End":"00:12.494","Text":"and its spring constant is 50 Newtons."},{"Start":"00:12.494 ","End":"00:16.470","Text":"Now, we have a weight suspended above it in air."},{"Start":"00:16.470 ","End":"00:21.240","Text":"The mass is 2 kilograms and the height is 3 meters above the spring."},{"Start":"00:21.240 ","End":"00:26.055","Text":"It\u0027s dropped from a standing position with no speed down onto the spring."},{"Start":"00:26.055 ","End":"00:27.600","Text":"Part A asks me,"},{"Start":"00:27.600 ","End":"00:31.395","Text":"what is the maximal compression of the spring?"},{"Start":"00:31.395 ","End":"00:35.145","Text":"To solve this, I first need to find if I conservation of energy."},{"Start":"00:35.145 ","End":"00:36.450","Text":"I\u0027ll check my forces."},{"Start":"00:36.450 ","End":"00:38.775","Text":"Now, I have the force of a spring in this problem,"},{"Start":"00:38.775 ","End":"00:40.470","Text":"springs have potential energy."},{"Start":"00:40.470 ","End":"00:42.260","Text":"Therefore, it\u0027s a conservative force."},{"Start":"00:42.260 ","End":"00:45.050","Text":"I\u0027m also dealing with the force of gravity."},{"Start":"00:45.050 ","End":"00:50.095","Text":"I know that gravity is a conservative force because it has potential energy as well."},{"Start":"00:50.095 ","End":"00:51.950","Text":"I have conservative forces."},{"Start":"00:51.950 ","End":"00:54.140","Text":"I can now move on to the second part,"},{"Start":"00:54.140 ","End":"00:58.265","Text":"which is finding my initial and final energy and setting them equal."},{"Start":"00:58.265 ","End":"01:01.205","Text":"E_i, my initial energy in this case,"},{"Start":"01:01.205 ","End":"01:04.430","Text":"equals as always, kinetic energy plus potential energy."},{"Start":"01:04.430 ","End":"01:06.995","Text":"In this case, I have 2 types of potential energy,"},{"Start":"01:06.995 ","End":"01:11.220","Text":"I have gravity and I have the force of a spring."},{"Start":"01:11.480 ","End":"01:14.525","Text":"If I want to write out the formula for this,"},{"Start":"01:14.525 ","End":"01:17.945","Text":"E_k is 1/2mv^2,"},{"Start":"01:17.945 ","End":"01:21.655","Text":"U_g is mgh,"},{"Start":"01:21.655 ","End":"01:28.065","Text":"and my elastic energy is 1/2k Delta x^2."},{"Start":"01:28.065 ","End":"01:34.265","Text":"Now I want to fill in what I can for the initial energy using the given pieces of data."},{"Start":"01:34.265 ","End":"01:38.900","Text":"First of all, if we\u0027re calculating my kinetic energy, I know that it\u0027s 0."},{"Start":"01:38.900 ","End":"01:41.345","Text":"Why? Because my body starts at rest."},{"Start":"01:41.345 ","End":"01:43.440","Text":"The object is not moving,"},{"Start":"01:43.440 ","End":"01:45.995","Text":"the weight hasn\u0027t been dropped yet, so that\u0027s 0."},{"Start":"01:45.995 ","End":"01:47.870","Text":"Now if we add in the force of gravity,"},{"Start":"01:47.870 ","End":"01:51.590","Text":"we can replace our little h with our height in this problem."},{"Start":"01:51.590 ","End":"01:55.945","Text":"Let\u0027s set the top of the spring as height equals 0."},{"Start":"01:55.945 ","End":"02:02.700","Text":"That would make the height that it starts at big H. Our force of gravity is mgH."},{"Start":"02:03.530 ","End":"02:08.885","Text":"What we can then do for our spring is we know that\u0027s also equal to 0."},{"Start":"02:08.885 ","End":"02:11.495","Text":"Why? Because the spring is entirely extended,"},{"Start":"02:11.495 ","End":"02:13.750","Text":"it\u0027s not compressed at all."},{"Start":"02:13.750 ","End":"02:16.485","Text":"Now we can find the final energy."},{"Start":"02:16.485 ","End":"02:19.445","Text":"It\u0027s important to know exactly what we\u0027re trying to find."},{"Start":"02:19.445 ","End":"02:21.350","Text":"We\u0027re looking for the moment of"},{"Start":"02:21.350 ","End":"02:24.725","Text":"maximal compression of the spring when the weight is sitting"},{"Start":"02:24.725 ","End":"02:26.990","Text":"on the spring and it can\u0027t go any farther"},{"Start":"02:26.990 ","End":"02:29.990","Text":"down and has yet to start traveling up. It looks like this."},{"Start":"02:29.990 ","End":"02:32.270","Text":"This point, our weight has a velocity of"},{"Start":"02:32.270 ","End":"02:34.760","Text":"0 because if it was still going down any farther,"},{"Start":"02:34.760 ","End":"02:36.095","Text":"if it\u0027s still at more velocity,"},{"Start":"02:36.095 ","End":"02:37.910","Text":"the spring wouldn\u0027t be fully compressed."},{"Start":"02:37.910 ","End":"02:41.270","Text":"Similarly, it hasn\u0027t started going up yet because if it had been going up,"},{"Start":"02:41.270 ","End":"02:44.000","Text":"the spring would not be fully compressed either."},{"Start":"02:44.000 ","End":"02:48.500","Text":"This is a unique trait of springs that the maximal compression is a point"},{"Start":"02:48.500 ","End":"02:52.790","Text":"when any given weight on a spring will have a velocity of 0. This helps us a lot."},{"Start":"02:52.790 ","End":"02:55.555","Text":"It\u0027s good to remember this for other equations as well."},{"Start":"02:55.555 ","End":"02:58.310","Text":"When we look for our final energy,"},{"Start":"02:58.310 ","End":"03:00.365","Text":"we have to remember 1 other thing also,"},{"Start":"03:00.365 ","End":"03:02.870","Text":"which is the height is not 0."},{"Start":"03:02.870 ","End":"03:04.130","Text":"In fact, the height is,"},{"Start":"03:04.130 ","End":"03:05.220","Text":"we can call it Delta x,"},{"Start":"03:05.220 ","End":"03:08.720","Text":"it\u0027s below the height of 0 by an amount of Delta x."},{"Start":"03:08.720 ","End":"03:11.405","Text":"That\u0027s the unknown variable that we\u0027re looking for."},{"Start":"03:11.405 ","End":"03:14.125","Text":"Delta x is the solution to our problem."},{"Start":"03:14.125 ","End":"03:17.495","Text":"Now we can fill in our final energy in the problem."},{"Start":"03:17.495 ","End":"03:21.260","Text":"We know that our kinetic energy equals 0 once again because we just"},{"Start":"03:21.260 ","End":"03:25.170","Text":"talked about how the velocity equals 0. That can be filled in."},{"Start":"03:25.170 ","End":"03:27.080","Text":"If we talk about potential energy,"},{"Start":"03:27.080 ","End":"03:28.580","Text":"first we talk about gravity,"},{"Start":"03:28.580 ","End":"03:30.695","Text":"it\u0027s not exactly equal to 0."},{"Start":"03:30.695 ","End":"03:36.125","Text":"In fact, we have a height of negative Delta x that we just talked about."},{"Start":"03:36.125 ","End":"03:42.800","Text":"Really the term is mg and our h value will be negative Delta x,"},{"Start":"03:42.800 ","End":"03:44.660","Text":"so mg negative Delta x,"},{"Start":"03:44.660 ","End":"03:47.195","Text":"but it\u0027s also a negligible amount, as you\u0027ll see,"},{"Start":"03:47.195 ","End":"03:53.285","Text":"in terms of the height relation between the Delta x and H, 3 meters."},{"Start":"03:53.285 ","End":"03:56.960","Text":"But anyways, we should put this in to be completely correct."},{"Start":"03:56.960 ","End":"03:58.864","Text":"As for the spring,"},{"Start":"03:58.864 ","End":"04:00.800","Text":"we now have kinetic energy and gravity."},{"Start":"04:00.800 ","End":"04:02.120","Text":"We have the spring left."},{"Start":"04:02.120 ","End":"04:09.560","Text":"We know from above that that equals 1/2k times Delta x^2 just like in the formula."},{"Start":"04:09.560 ","End":"04:12.080","Text":"Now that we have our initial and our final energy,"},{"Start":"04:12.080 ","End":"04:16.400","Text":"we can set them equal to each other and use the principle of conservation of energy."},{"Start":"04:16.400 ","End":"04:20.488","Text":"We can say E_i=E_f,"},{"Start":"04:20.488 ","End":"04:22.740","Text":"and we can fill everything in."},{"Start":"04:23.110 ","End":"04:25.700","Text":"E_i, our initial energy,"},{"Start":"04:25.700 ","End":"04:30.150","Text":"equals mgH, and that equals E_f,"},{"Start":"04:30.150 ","End":"04:38.190","Text":"which is mg(negative Delta x) plus 1/2k Delta x^2."},{"Start":"04:38.190 ","End":"04:40.230","Text":"Now I can fill in the values that I know."},{"Start":"04:40.230 ","End":"04:41.695","Text":"My mass is 2,"},{"Start":"04:41.695 ","End":"04:44.705","Text":"I\u0027ll write up my gravity constant is 10,"},{"Start":"04:44.705 ","End":"04:46.895","Text":"and my height is 3."},{"Start":"04:46.895 ","End":"04:51.995","Text":"Again, mg is 2 times 10 and negative Delta x, we don\u0027t know yet."},{"Start":"04:51.995 ","End":"04:54.075","Text":"Of course, that\u0027s what we\u0027re trying to find,"},{"Start":"04:54.075 ","End":"04:57.455","Text":"and 1/2(50) equals 25x^2."},{"Start":"04:57.455 ","End":"04:59.570","Text":"Now we have a quadratic equation."},{"Start":"04:59.570 ","End":"05:01.355","Text":"When we go about solving it,"},{"Start":"05:01.355 ","End":"05:05.525","Text":"we can isolate Delta x and in the end,"},{"Start":"05:05.525 ","End":"05:09.780","Text":"we find that Delta x equals about 2 meters."},{"Start":"05:09.780 ","End":"05:13.170","Text":"It actually turns out that it\u0027s good we didn\u0027t throw out"},{"Start":"05:13.170 ","End":"05:17.375","Text":"the Delta x from our gravity equation because it wasn\u0027t really negligible."},{"Start":"05:17.375 ","End":"05:23.320","Text":"Actually, what we can deduce from that is that the spring is probably pretty weak."},{"Start":"05:23.320 ","End":"05:25.230","Text":"Here you have your solution,"},{"Start":"05:25.230 ","End":"05:27.050","Text":"2 meters equals Delta x,"},{"Start":"05:27.050 ","End":"05:30.785","Text":"which is your maximal compression of the spring."},{"Start":"05:30.785 ","End":"05:34.940","Text":"Just keep in mind again that we really just took our 2 points and it was"},{"Start":"05:34.940 ","End":"05:39.440","Text":"before we drop the object and after it has reached maximal compression,"},{"Start":"05:39.440 ","End":"05:41.120","Text":"at that moment of maximum compression."},{"Start":"05:41.120 ","End":"05:44.375","Text":"You don\u0027t need to break it down into when the mass hits the spring."},{"Start":"05:44.375 ","End":"05:47.210","Text":"Then the second equation there, it\u0027s an extra step."},{"Start":"05:47.210 ","End":"05:50.180","Text":"This is a method to solve the problem."},{"Start":"05:50.180 ","End":"05:53.250","Text":"Now let\u0027s move on to Part b."}],"ID":12674}],"Thumbnail":null,"ID":5417},{"Name":"Energy Conservation And The Work Energy Theorem part b","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Review of Part A","Duration":"1m 13s","ChapterTopicVideoID":9038,"CourseChapterTopicPlaylistID":5418,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"Now we\u0027re starting Part 2 of our unit about"},{"Start":"00:03.510 ","End":"00:07.410","Text":"energy conservation and the theorem of work energy."},{"Start":"00:07.410 ","End":"00:09.690","Text":"But before we move on to Part 2,"},{"Start":"00:09.690 ","End":"00:16.845","Text":"I want to do a quick review of part A. Firstly,"},{"Start":"00:16.845 ","End":"00:19.350","Text":"everybody has a total energy."},{"Start":"00:19.350 ","End":"00:21.645","Text":"The total energy is symbolized with E,"},{"Start":"00:21.645 ","End":"00:23.160","Text":"and it\u0027s made up of 2 parts,"},{"Start":"00:23.160 ","End":"00:25.185","Text":"kinetic energy, E_k,"},{"Start":"00:25.185 ","End":"00:27.735","Text":"and potential energy, U."},{"Start":"00:27.735 ","End":"00:32.443","Text":"Kinetic energy is always 1/2mv^2"},{"Start":"00:32.443 ","End":"00:37.405","Text":"and potential energy depends on the question and depends on the forces at work."},{"Start":"00:37.405 ","End":"00:43.345","Text":"With every exercise you want to ask whether there\u0027s conservation of energy."},{"Start":"00:43.345 ","End":"00:45.785","Text":"The way to find that out is to find out whether"},{"Start":"00:45.785 ","End":"00:49.025","Text":"all of the forces are conservative forces."},{"Start":"00:49.025 ","End":"00:52.670","Text":"If we find that all the forces are conservative,"},{"Start":"00:52.670 ","End":"00:57.185","Text":"then what we need to do is set an equation where the initial energy,"},{"Start":"00:57.185 ","End":"00:58.910","Text":"the energy at the first moment,"},{"Start":"00:58.910 ","End":"01:02.465","Text":"is equal to the energy at the last moment, the final energy."},{"Start":"01:02.465 ","End":"01:05.375","Text":"This is symbolized by E_i and E_f."},{"Start":"01:05.375 ","End":"01:08.915","Text":"Once you have that, you calculate the kinetic energy at each moment,"},{"Start":"01:08.915 ","End":"01:11.000","Text":"the potential energy at each moment,"},{"Start":"01:11.000 ","End":"01:13.980","Text":"and set them equal, and you can solve."}],"ID":9311},{"Watched":false,"Name":"Part B","Duration":"2m 25s","ChapterTopicVideoID":9039,"CourseChapterTopicPlaylistID":5418,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.590","Text":"Up until now we\u0027ve solved problems with the conservation of energy."},{"Start":"00:04.590 ","End":"00:06.600","Text":"We find that all the forces are conservative."},{"Start":"00:06.600 ","End":"00:09.630","Text":"We said E_i the initial energy equal to E_f,"},{"Start":"00:09.630 ","End":"00:11.580","Text":"the final energy and we solve."},{"Start":"00:11.580 ","End":"00:13.470","Text":"The question is what happens when you have"},{"Start":"00:13.470 ","End":"00:17.340","Text":"non-conservative forces in your problem and you don\u0027t have conservation energy."},{"Start":"00:17.340 ","End":"00:22.065","Text":"In those cases, you want to use the work-energy theorem. You can see up here."},{"Start":"00:22.065 ","End":"00:25.335","Text":"The work-energy theorem is an equation of sorts."},{"Start":"00:25.335 ","End":"00:26.955","Text":"You can see it down here."},{"Start":"00:26.955 ","End":"00:29.835","Text":"We\u0027re going to put it in red because it\u0027s important and it\u0027s"},{"Start":"00:29.835 ","End":"00:33.450","Text":"W_nc equals Delta E. What does that mean?"},{"Start":"00:33.450 ","End":"00:37.890","Text":"Nc is an abbreviation of non-conservative."},{"Start":"00:37.890 ","End":"00:39.560","Text":"We\u0027re going to find the work of"},{"Start":"00:39.560 ","End":"00:43.145","Text":"the non-conservative forces and that\u0027s going to equal Delta E,"},{"Start":"00:43.145 ","End":"00:45.260","Text":"the change in the overall energy,"},{"Start":"00:45.260 ","End":"00:47.470","Text":"the change in total energy."},{"Start":"00:47.470 ","End":"00:49.890","Text":"Let\u0027s break this down a little bit."},{"Start":"00:49.890 ","End":"00:51.770","Text":"W_nc, the work of"},{"Start":"00:51.770 ","End":"00:55.580","Text":"the non-conservative forces means finding every non-conservative force,"},{"Start":"00:55.580 ","End":"00:58.325","Text":"calculating the work for each one, and summing them."},{"Start":"00:58.325 ","End":"01:00.005","Text":"That equals Delta E,"},{"Start":"01:00.005 ","End":"01:01.760","Text":"or the change in total energy."},{"Start":"01:01.760 ","End":"01:05.210","Text":"If you recall, total energy is comprised of 2 parts."},{"Start":"01:05.210 ","End":"01:10.505","Text":"Total energy is the kinetic energy plus the potential energy."},{"Start":"01:10.505 ","End":"01:12.215","Text":"When talking about Delta E,"},{"Start":"01:12.215 ","End":"01:13.820","Text":"we\u0027re talking about the difference between"},{"Start":"01:13.820 ","End":"01:16.265","Text":"the energy in the initial moment and the final moment."},{"Start":"01:16.265 ","End":"01:17.795","Text":"Whenever we use Delta,"},{"Start":"01:17.795 ","End":"01:19.340","Text":"in this case, n and others,"},{"Start":"01:19.340 ","End":"01:23.330","Text":"we\u0027re going to subtract the initial moment from the final moment to find that change."},{"Start":"01:23.330 ","End":"01:25.995","Text":"Delta E equals E_f,"},{"Start":"01:25.995 ","End":"01:28.955","Text":"the final energy minus E_i the initial energy."},{"Start":"01:28.955 ","End":"01:31.850","Text":"If you recall, in our problems with conservation of energy,"},{"Start":"01:31.850 ","End":"01:34.070","Text":"we said E_i equal to E_f."},{"Start":"01:34.070 ","End":"01:36.625","Text":"With the work-energy theorem,"},{"Start":"01:36.625 ","End":"01:40.965","Text":"we said Delta E equal to E_f minus E_i."},{"Start":"01:40.965 ","End":"01:48.140","Text":"What this means is that E_f minus E_i equals the work of the non-conservative forces."},{"Start":"01:48.140 ","End":"01:54.530","Text":"Another thing to pay attention to here is that if you notice if W_nc equals 0,"},{"Start":"01:54.530 ","End":"01:57.230","Text":"let\u0027s say in a given example."},{"Start":"01:57.230 ","End":"02:01.675","Text":"We know from our formula that equals E_f minus E_i."},{"Start":"02:01.675 ","End":"02:07.910","Text":"Then we can simplify this whole equation and we\u0027ll actually find that in this case,"},{"Start":"02:07.910 ","End":"02:10.880","Text":"E_f equals E_i or E_i equals E_f,"},{"Start":"02:10.880 ","End":"02:14.870","Text":"which brings us back to our equation for the conservation of energy."},{"Start":"02:14.870 ","End":"02:17.660","Text":"What this means is the work-energy theorem"},{"Start":"02:17.660 ","End":"02:21.590","Text":"accounts for situations where we have conservation of energy."},{"Start":"02:21.590 ","End":"02:25.680","Text":"Let\u0027s look at an example to see how this works."}],"ID":9312},{"Watched":false,"Name":"Example","Duration":"10m 51s","ChapterTopicVideoID":12200,"CourseChapterTopicPlaylistID":5418,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.270","Text":"Let\u0027s take the same example we used before."},{"Start":"00:03.270 ","End":"00:06.000","Text":"Where we have a cart that\u0027s going down a slope,"},{"Start":"00:06.000 ","End":"00:08.985","Text":"and it starts with a speed of u_0,"},{"Start":"00:08.985 ","End":"00:15.765","Text":"and the slope has a height of H. What we need to do is find the final velocity."},{"Start":"00:15.765 ","End":"00:19.215","Text":"As opposed to in the initial problem where there is no friction,"},{"Start":"00:19.215 ","End":"00:23.505","Text":"let\u0027s assume that in this problem 1 of the wheels of the cart fell off, like that."},{"Start":"00:23.505 ","End":"00:26.355","Text":"Now in this situation, we would definitely have friction."},{"Start":"00:26.355 ","End":"00:27.840","Text":"But to give it even more friction,"},{"Start":"00:27.840 ","End":"00:29.700","Text":"let\u0027s say the cart has no wheels."},{"Start":"00:29.700 ","End":"00:32.530","Text":"Let\u0027s say it looks something like this."},{"Start":"00:32.900 ","End":"00:35.940","Text":"Now we can add the friction."},{"Start":"00:35.940 ","End":"00:37.380","Text":"Let\u0027s say it\u0027s a known quantity."},{"Start":"00:37.380 ","End":"00:39.900","Text":"We can add it to our given data on the bottom,"},{"Start":"00:39.900 ","End":"00:45.370","Text":"Mu k. What we\u0027re trying to find is the same thing,"},{"Start":"00:45.370 ","End":"00:49.225","Text":"we\u0027re trying to find at the end once the cart reaches the bottom,"},{"Start":"00:49.225 ","End":"00:51.130","Text":"what will be the final velocity."},{"Start":"00:51.130 ","End":"00:52.570","Text":"When the cart is here,"},{"Start":"00:52.570 ","End":"00:54.520","Text":"once it\u0027s gone to the entire slope,"},{"Start":"00:54.520 ","End":"00:55.900","Text":"what velocity will it have."},{"Start":"00:55.900 ","End":"00:57.610","Text":"We can call that V_f."},{"Start":"00:57.610 ","End":"00:59.950","Text":"Now how do I solve this problem?"},{"Start":"00:59.950 ","End":"01:03.235","Text":"The first thing I want to check is if I have conservation of energy."},{"Start":"01:03.235 ","End":"01:08.380","Text":"Now we know that because we have friction we\u0027re not going to have conservation of energy."},{"Start":"01:08.380 ","End":"01:10.840","Text":"I can\u0027t set E_i,"},{"Start":"01:10.840 ","End":"01:13.525","Text":"initial energy, equal to E_f, final energy."},{"Start":"01:13.525 ","End":"01:14.710","Text":"That won\u0027t work here."},{"Start":"01:14.710 ","End":"01:20.125","Text":"But what I can do is compute the work of the non-conservative forces."},{"Start":"01:20.125 ","End":"01:22.890","Text":"In this case, it\u0027s the work of friction."},{"Start":"01:22.890 ","End":"01:25.110","Text":"In a lot of cases, it will be like that."},{"Start":"01:25.110 ","End":"01:27.980","Text":"That\u0027s going to equal the change in the total energy."},{"Start":"01:27.980 ","End":"01:29.420","Text":"Or put in other words,"},{"Start":"01:29.420 ","End":"01:30.575","Text":"if you recall from before,"},{"Start":"01:30.575 ","End":"01:32.735","Text":"that equals E_f, final energy,"},{"Start":"01:32.735 ","End":"01:35.515","Text":"minus E_i, the initial energy."},{"Start":"01:35.515 ","End":"01:40.235","Text":"Set the work of the force of friction equal to E_f minus E_i."},{"Start":"01:40.235 ","End":"01:42.635","Text":"But first what I want to do is write out E_f,"},{"Start":"01:42.635 ","End":"01:45.565","Text":"write out E_i, and subtract them."},{"Start":"01:45.565 ","End":"01:49.350","Text":"Let\u0027s start with the E_i and E_f,"},{"Start":"01:49.350 ","End":"01:52.290","Text":"and then we\u0027ll move on to W_f_k."},{"Start":"01:52.290 ","End":"01:54.840","Text":"The initial energy, E_i,"},{"Start":"01:54.840 ","End":"01:57.945","Text":"equals 1/2 of mv^2."},{"Start":"01:57.945 ","End":"01:59.550","Text":"In our given data,"},{"Start":"01:59.550 ","End":"02:00.850","Text":"we know that v,"},{"Start":"02:00.850 ","End":"02:02.390","Text":"the initial velocity,"},{"Start":"02:02.390 ","End":"02:04.400","Text":"that is, equals u_0."},{"Start":"02:04.400 ","End":"02:06.080","Text":"We can fill that in, instead of as v,"},{"Start":"02:06.080 ","End":"02:09.575","Text":"we can write it as 1/2mU_0^2."},{"Start":"02:09.575 ","End":"02:14.165","Text":"The other energies it has the potential energy of gravity, mgh."},{"Start":"02:14.165 ","End":"02:15.875","Text":"Now the final energy,"},{"Start":"02:15.875 ","End":"02:21.205","Text":"E_f, equals 1/2 mV_f^2."},{"Start":"02:21.205 ","End":"02:24.500","Text":"Remember, V_f is what we\u0027re looking for."},{"Start":"02:24.500 ","End":"02:28.105","Text":"Now we also have the potential energy of gravity there."},{"Start":"02:28.105 ","End":"02:31.310","Text":"But we know that at this point the height h=0,"},{"Start":"02:31.310 ","End":"02:33.070","Text":"so we can write that as 0."},{"Start":"02:33.070 ","End":"02:38.090","Text":"Now let\u0027s move on to calculating w. The work of the non-conservative force,"},{"Start":"02:38.090 ","End":"02:42.755","Text":"in this case, the work of the force of friction. What does that equal?"},{"Start":"02:42.755 ","End":"02:48.870","Text":"Well, what we have to do is we have to take an integral of the force f_k."},{"Start":"02:48.870 ","End":"02:54.605","Text":"d_r. The first thing we have to do is find the force of friction."},{"Start":"02:54.605 ","End":"02:56.090","Text":"Let\u0027s assume for a second,"},{"Start":"02:56.090 ","End":"02:57.635","Text":"for the sake of simplicity,"},{"Start":"02:57.635 ","End":"03:02.470","Text":"that our object m starts really on the edge of the slope here."},{"Start":"03:02.470 ","End":"03:04.515","Text":"It only has friction on the slope,"},{"Start":"03:04.515 ","End":"03:06.830","Text":"and the moment we\u0027re measuring as our final velocity is really"},{"Start":"03:06.830 ","End":"03:09.275","Text":"the exact moment that it comes off the slope,"},{"Start":"03:09.275 ","End":"03:10.370","Text":"really at the edge there."},{"Start":"03:10.370 ","End":"03:14.120","Text":"So we don\u0027t have to worry about any friction on the flat portions,"},{"Start":"03:14.120 ","End":"03:15.785","Text":"but only on the slope."},{"Start":"03:15.785 ","End":"03:19.055","Text":"We are going to need a couple of other things to figure this out."},{"Start":"03:19.055 ","End":"03:22.430","Text":"First of all, we need to know the angle that this slope is at."},{"Start":"03:22.430 ","End":"03:26.030","Text":"For the sake of the problem let\u0027s call this angle Theta."},{"Start":"03:26.030 ","End":"03:28.610","Text":"Now what we want to calculate is,"},{"Start":"03:28.610 ","End":"03:32.140","Text":"let\u0027s assume the mass is in the middle of the slope here, on its way down."},{"Start":"03:32.140 ","End":"03:34.445","Text":"We need to calculate the force of friction."},{"Start":"03:34.445 ","End":"03:35.960","Text":"Now we know that the friction,"},{"Start":"03:35.960 ","End":"03:38.210","Text":"when our object is in the middle here,"},{"Start":"03:38.210 ","End":"03:40.490","Text":"is going to be working upwards,"},{"Start":"03:40.490 ","End":"03:41.660","Text":"towards the top of the slope."},{"Start":"03:41.660 ","End":"03:45.370","Text":"The way that we calculate that is the force of friction,"},{"Start":"03:45.370 ","End":"03:49.055","Text":"f_k, equals Mu k times N, the normal."},{"Start":"03:49.055 ","End":"03:51.290","Text":"Now the normal, as we know,"},{"Start":"03:51.290 ","End":"03:54.485","Text":"is perpendicular to the direction of the object,"},{"Start":"03:54.485 ","End":"03:57.595","Text":"so it\u0027s also perpendicular to the slope itself."},{"Start":"03:57.595 ","End":"04:00.320","Text":"We\u0027re going to set an axis, the y-axis,"},{"Start":"04:00.320 ","End":"04:02.690","Text":"perpendicular or orthogonal to the slope,"},{"Start":"04:02.690 ","End":"04:06.620","Text":"and the x-axis parallel to the slope, like so."},{"Start":"04:06.620 ","End":"04:10.535","Text":"Let\u0027s add in the force of gravity here, mg."},{"Start":"04:10.535 ","End":"04:14.045","Text":"The way we calculate the forces on the y-axis is"},{"Start":"04:14.045 ","End":"04:21.300","Text":"Sigma f_y equals N minus mg cosine Theta."},{"Start":"04:21.300 ","End":"04:24.400","Text":"It\u0027s good to remember here that the force of"},{"Start":"04:24.400 ","End":"04:28.465","Text":"gravity that is tangential to the slope is sine Theta."},{"Start":"04:28.465 ","End":"04:33.040","Text":"The force that is perpendicular to the slope is cosine Theta."},{"Start":"04:33.040 ","End":"04:37.915","Text":"We can set this to 0 because we know the object doesn\u0027t move along the y-axis."},{"Start":"04:37.915 ","End":"04:43.225","Text":"We can then do is set N equal to mg cosine Theta."},{"Start":"04:43.225 ","End":"04:47.075","Text":"We can put this into our equation above for the force of friction."},{"Start":"04:47.075 ","End":"04:49.720","Text":"We can now write this out in a new way."},{"Start":"04:49.720 ","End":"04:52.000","Text":"F_k, the force of friction,"},{"Start":"04:52.000 ","End":"04:57.770","Text":"equals Mu k times mg cosine Theta."},{"Start":"04:57.770 ","End":"05:00.310","Text":"For now, we\u0027re going to ignore the fact that"},{"Start":"05:00.310 ","End":"05:04.120","Text":"the force of friction is negative and just find the absolute value."},{"Start":"05:04.120 ","End":"05:09.860","Text":"The second thing to realize here is that the force of friction is actually a constant."},{"Start":"05:09.860 ","End":"05:11.715","Text":"Mu k is constant,"},{"Start":"05:11.715 ","End":"05:14.725","Text":"mg is constant, and the angle Theta is constant."},{"Start":"05:14.725 ","End":"05:17.680","Text":"If this is a constant, we can take it out of our integral"},{"Start":"05:17.680 ","End":"05:20.650","Text":"above when calculating the work of f_k,"},{"Start":"05:20.650 ","End":"05:23.245","Text":"and we can solve it a different way."},{"Start":"05:23.245 ","End":"05:25.000","Text":"We solve it as follows."},{"Start":"05:25.000 ","End":"05:33.955","Text":"The absolute value of f_k times the total displacement,"},{"Start":"05:33.955 ","End":"05:36.475","Text":"Delta r, the absolute value of that,"},{"Start":"05:36.475 ","End":"05:38.590","Text":"times the cosine of Alpha,"},{"Start":"05:38.590 ","End":"05:40.435","Text":"the angle between the 2 of them."},{"Start":"05:40.435 ","End":"05:45.380","Text":"We know that the absolute value of the force of friction is listed below."},{"Start":"05:45.380 ","End":"05:48.665","Text":"The magnitude of the displacement we\u0027ll find in a moment."},{"Start":"05:48.665 ","End":"05:51.905","Text":"Then we just have to find cosine of Alpha, the angle between them."},{"Start":"05:51.905 ","End":"05:55.220","Text":"The magnitude of the displacement is going to be"},{"Start":"05:55.220 ","End":"05:56.750","Text":"a vector that goes from the top of"},{"Start":"05:56.750 ","End":"05:59.345","Text":"the slope to the bottom of the slope, in a straight line."},{"Start":"05:59.345 ","End":"06:02.000","Text":"We\u0027re not including anything before the slope or after,"},{"Start":"06:02.000 ","End":"06:03.890","Text":"just the whole line across the slope."},{"Start":"06:03.890 ","End":"06:06.080","Text":"The vector is this red line here."},{"Start":"06:06.080 ","End":"06:08.345","Text":"That is the displacement."},{"Start":"06:08.345 ","End":"06:11.375","Text":"In the same way that we talked before with e,"},{"Start":"06:11.375 ","End":"06:15.545","Text":"Delta r is the change between the initial moment and the final moment."},{"Start":"06:15.545 ","End":"06:20.005","Text":"Delta r equals r_f minus r_i."},{"Start":"06:20.005 ","End":"06:23.015","Text":"If you subtract the final vector from the initial vector,"},{"Start":"06:23.015 ","End":"06:27.710","Text":"you can think of that as another way to find Delta r. Now we see here that Delta r,"},{"Start":"06:27.710 ","End":"06:30.905","Text":"the displacement, is the entire diagonal line"},{"Start":"06:30.905 ","End":"06:34.145","Text":"across the right angle triangle with H as the height,"},{"Start":"06:34.145 ","End":"06:35.435","Text":"capital H, that is."},{"Start":"06:35.435 ","End":"06:36.920","Text":"It\u0027s a right-angle triangle."},{"Start":"06:36.920 ","End":"06:41.540","Text":"Now we can do is use some trigonometry to solve for the magnitude of"},{"Start":"06:41.540 ","End":"06:48.850","Text":"Delta r. H divided by the magnitude of Delta r equals sine of Theta."},{"Start":"06:48.850 ","End":"06:52.760","Text":"Another way to express the magnitude of the displacement,"},{"Start":"06:52.760 ","End":"06:55.100","Text":"that is the magnitude of Delta r,"},{"Start":"06:55.100 ","End":"07:03.180","Text":"is magnitude of Delta r of the displacement equals H divided by sine of Theta."},{"Start":"07:03.840 ","End":"07:08.410","Text":"Now we know the magnitude of Delta r, the displacement."},{"Start":"07:08.410 ","End":"07:12.550","Text":"We know from before the magnitude of the force of friction."},{"Start":"07:12.550 ","End":"07:15.070","Text":"We need to find cosine of the angle Alpha,"},{"Start":"07:15.070 ","End":"07:16.670","Text":"the angle between the 2."},{"Start":"07:16.670 ","End":"07:19.885","Text":"Delta r goes downwards along this diagonal,"},{"Start":"07:19.885 ","End":"07:22.600","Text":"and friction goes upwards along the same diagonal."},{"Start":"07:22.600 ","End":"07:26.050","Text":"The angle between the 2 of them is 180 degrees."},{"Start":"07:26.050 ","End":"07:31.150","Text":"Now, this is important because the cosine of 180 is negative 1,"},{"Start":"07:31.150 ","End":"07:33.835","Text":"which means that we\u0027re dealing with a negative force here."},{"Start":"07:33.835 ","End":"07:35.770","Text":"In general, kinetic friction,"},{"Start":"07:35.770 ","End":"07:38.098","Text":"the force of friction, is a negative force,"},{"Start":"07:38.098 ","End":"07:39.160","Text":"it drags us backwards,"},{"Start":"07:39.160 ","End":"07:43.285","Text":"it pushes us away from the general direction that an object is going to move."},{"Start":"07:43.285 ","End":"07:46.480","Text":"Now we\u0027re at a point where we can start really solving,"},{"Start":"07:46.480 ","End":"07:48.730","Text":"filling things in, setting things equal."},{"Start":"07:48.730 ","End":"07:52.475","Text":"What we\u0027re going to do is on the side here we\u0027ll write everything out."},{"Start":"07:52.475 ","End":"07:58.190","Text":"First, the work of friction is the absolute value of f_k,"},{"Start":"07:58.190 ","End":"08:05.055","Text":"which is Mu k times mg cosine of Theta."},{"Start":"08:05.055 ","End":"08:09.530","Text":"That\u0027s multiplied by the absolute value of Delta r,"},{"Start":"08:09.530 ","End":"08:14.085","Text":"which is H over sine Theta and that\u0027s multiplied by negative 1."},{"Start":"08:14.085 ","End":"08:16.025","Text":"We\u0027ll put a negative sign at the beginning here."},{"Start":"08:16.025 ","End":"08:19.520","Text":"This equals E_f minus E_i."},{"Start":"08:19.520 ","End":"08:22.265","Text":"Let\u0027s first put in E_f, and then we\u0027ll find E_i."},{"Start":"08:22.265 ","End":"08:28.840","Text":"E_f equals 1/2 mV_f^2."},{"Start":"08:28.840 ","End":"08:31.515","Text":"We\u0027ll subtract from that E_i,"},{"Start":"08:31.515 ","End":"08:38.840","Text":"which is 1/2mU _0^2 minus mgH,"},{"Start":"08:38.840 ","End":"08:40.625","Text":"because we know the height at the beginning is large"},{"Start":"08:40.625 ","End":"08:44.615","Text":"H. This is really all I need to solve."},{"Start":"08:44.615 ","End":"08:46.510","Text":"Everything is given except for V_f,"},{"Start":"08:46.510 ","End":"08:48.155","Text":"so I can solve for V_f."},{"Start":"08:48.155 ","End":"08:51.800","Text":"The first thing I\u0027ll do is take out all the m\u0027s because I can simplify that."},{"Start":"08:51.800 ","End":"08:57.060","Text":"Remember, I\u0027ve kept this minus sign in here because it is a negative force."},{"Start":"08:57.080 ","End":"09:00.015","Text":"When I solve for V_f,"},{"Start":"09:00.015 ","End":"09:02.060","Text":"what I\u0027m going to get is going to be a little complex,"},{"Start":"09:02.060 ","End":"09:03.770","Text":"but I\u0027ll write it out for you anyways."},{"Start":"09:03.770 ","End":"09:10.110","Text":"V_f equals the square root of u_0^2"},{"Start":"09:10.110 ","End":"09:17.730","Text":"plus 2gH"},{"Start":"09:17.730 ","End":"09:21.045","Text":"times 1"},{"Start":"09:21.045 ","End":"09:26.530","Text":"minus Mu k cotangent Theta."},{"Start":"09:26.740 ","End":"09:29.510","Text":"You should probably simplify this further,"},{"Start":"09:29.510 ","End":"09:31.220","Text":"but I\u0027ll leave that to you."},{"Start":"09:31.220 ","End":"09:34.400","Text":"For now, what I\u0027d like to do is summarize what we did here."},{"Start":"09:34.400 ","End":"09:38.945","Text":"First I checked all my forces and found that it did not have a conservation of energy."},{"Start":"09:38.945 ","End":"09:41.990","Text":"Instead of setting my initial and final forces equal,"},{"Start":"09:41.990 ","End":"09:45.570","Text":"I set.W_f_k equal to the change in"},{"Start":"09:45.570 ","End":"09:50.225","Text":"E. First I found my E_i and my E_f, and wrote them out."},{"Start":"09:50.225 ","End":"09:52.580","Text":"Then I solved for.W_f_k."},{"Start":"09:52.580 ","End":"09:56.165","Text":"That is the work of the force of friction, kinetic friction."},{"Start":"09:56.165 ","End":"10:02.555","Text":"That equals absolute value f_k times the magnitude of r times the cosine of Alpha."},{"Start":"10:02.555 ","End":"10:06.860","Text":"I know I can do this instead of doing an integral because f_k is a constant."},{"Start":"10:06.860 ","End":"10:09.950","Text":"I only know that because I solve for f_k using"},{"Start":"10:09.950 ","End":"10:13.495","Text":"the forces along the y-axis and found it to be constant."},{"Start":"10:13.495 ","End":"10:16.640","Text":"From there I had all my values and I could solve."},{"Start":"10:16.640 ","End":"10:18.755","Text":"Now 1 more thing I\u0027d like to mention,"},{"Start":"10:18.755 ","End":"10:22.835","Text":"is you don\u0027t have to think of the equation in the terms we wrote it out as."},{"Start":"10:22.835 ","End":"10:26.375","Text":"You can also think of it as E initial plus"},{"Start":"10:26.375 ","End":"10:30.800","Text":"the work of the non-conservative forces equals E final."},{"Start":"10:30.800 ","End":"10:33.740","Text":"Both are the same, algebraically they\u0027re the same."},{"Start":"10:33.740 ","End":"10:36.860","Text":"You should find which is most comfortable for you to work with."},{"Start":"10:36.860 ","End":"10:38.539","Text":"I\u0027m talking about this here,"},{"Start":"10:38.539 ","End":"10:41.510","Text":"and I\u0027ll also put it in the lecture slide so that you"},{"Start":"10:41.510 ","End":"10:44.840","Text":"can think about which is easier for you to use, which you prefer."},{"Start":"10:44.840 ","End":"10:51.660","Text":"As you can see, I put here E_i plus W_NC equals E_F."}],"ID":12675}],"Thumbnail":null,"ID":5418},{"Name":"Work Done By A Constant Force","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Work- Formula","Duration":"2m 18s","ChapterTopicVideoID":9041,"CourseChapterTopicPlaylistID":5419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lecture,"},{"Start":"00:01.890 ","End":"00:03.300","Text":"we\u0027re going to talk about work,"},{"Start":"00:03.300 ","End":"00:06.820","Text":"especially in terms of how it\u0027s exerted on a constant force."},{"Start":"00:06.820 ","End":"00:10.035","Text":"Work is symbolized by the letter w,"},{"Start":"00:10.035 ","End":"00:14.175","Text":"and it equals the integral of F dot dr. That\u0027s"},{"Start":"00:14.175 ","End":"00:20.100","Text":"a force scalar multiplication with the trajectory itself."},{"Start":"00:20.100 ","End":"00:23.000","Text":"Basically what we\u0027re doing is taking an integral of"},{"Start":"00:23.000 ","End":"00:26.709","Text":"force times the length of the trajectory."},{"Start":"00:26.709 ","End":"00:29.855","Text":"When do we use this term work?"},{"Start":"00:29.855 ","End":"00:34.250","Text":"It\u0027s really a way to solve problems mathematically."},{"Start":"00:34.250 ","End":"00:36.110","Text":"You can give it all sorts of meanings,"},{"Start":"00:36.110 ","End":"00:38.690","Text":"and we can talk about that and we will talk about that later,"},{"Start":"00:38.690 ","End":"00:40.190","Text":"but the best way to think about it as"},{"Start":"00:40.190 ","End":"00:44.125","Text":"a mathematical tool to help us in certain situations."},{"Start":"00:44.125 ","End":"00:47.105","Text":"For now, I\u0027m not going to deal with this integral itself,"},{"Start":"00:47.105 ","End":"00:51.370","Text":"but talk about a situation when we have a constant force."},{"Start":"00:51.370 ","End":"00:53.280","Text":"If my force is constant,"},{"Start":"00:53.280 ","End":"00:55.040","Text":"and only if my force is constant,"},{"Start":"00:55.040 ","End":"00:59.900","Text":"I can take f out of the integral and I end up with the following formula."},{"Start":"00:59.900 ","End":"01:03.845","Text":"The magnitude of F of the force times the displacement,"},{"Start":"01:03.845 ","End":"01:07.940","Text":"which is the shortest distance between the initial and final points of my trajectory,"},{"Start":"01:07.940 ","End":"01:09.544","Text":"times the cos(Alpha),"},{"Start":"01:09.544 ","End":"01:11.585","Text":"the angle between the 2 of them."},{"Start":"01:11.585 ","End":"01:14.870","Text":"This works because if the force is constant,"},{"Start":"01:14.870 ","End":"01:16.265","Text":"I can take it out of the integral,"},{"Start":"01:16.265 ","End":"01:21.120","Text":"and the integral of dr will give me the displacement."},{"Start":"01:21.560 ","End":"01:24.425","Text":"From there it\u0027s a scalar multiplication"},{"Start":"01:24.425 ","End":"01:27.080","Text":"multiplied by the cosine of the angle between them."},{"Start":"01:27.080 ","End":"01:29.915","Text":"For a bit of a deeper explanation,"},{"Start":"01:29.915 ","End":"01:32.120","Text":"the displacement is the difference between"},{"Start":"01:32.120 ","End":"01:35.635","Text":"the final point and the initial point on my vectors."},{"Start":"01:35.635 ","End":"01:38.330","Text":"For example, if I had"},{"Start":"01:38.330 ","End":"01:42.860","Text":"the following starting point and the following initial and final point,"},{"Start":"01:42.860 ","End":"01:46.830","Text":"and this is my coordinates,"},{"Start":"01:46.830 ","End":"01:51.080","Text":"then from my initial point ri to my final point,"},{"Start":"01:51.080 ","End":"01:53.089","Text":"maybe that was the path it took,"},{"Start":"01:53.089 ","End":"01:56.870","Text":"the displacement would be the shortest distance between the 2 of them."},{"Start":"01:56.870 ","End":"01:59.240","Text":"It would be this line in red."},{"Start":"01:59.240 ","End":"02:04.730","Text":"Keep in mind, it doesn\u0027t matter how I got from my initial point to my final point."},{"Start":"02:04.730 ","End":"02:07.460","Text":"It could have been in a straight line along the displacement."},{"Start":"02:07.460 ","End":"02:10.535","Text":"It could have been the line we drew here of the 2 vectors."},{"Start":"02:10.535 ","End":"02:13.700","Text":"It could have been its own path that squiggles around like this."},{"Start":"02:13.700 ","End":"02:15.245","Text":"It makes no difference."},{"Start":"02:15.245 ","End":"02:17.730","Text":"Let\u0027s look at an example."}],"ID":9314},{"Watched":false,"Name":"Example","Duration":"3m 34s","ChapterTopicVideoID":9042,"CourseChapterTopicPlaylistID":5419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"In this example, I\u0027m given a mass M,"},{"Start":"00:02.790 ","End":"00:06.270","Text":"which is sitting on a surface that has some friction,"},{"Start":"00:06.270 ","End":"00:10.305","Text":"which we usually symbolize with these lines coming out of the bottom of the surface."},{"Start":"00:10.305 ","End":"00:14.235","Text":"Now I\u0027m told that the friction coefficient is Mu_"},{"Start":"00:14.235 ","End":"00:18.140","Text":"k and that the mass is moving"},{"Start":"00:18.140 ","End":"00:22.415","Text":"along the surface at some velocity of arbitrary velocity V,"},{"Start":"00:22.415 ","End":"00:26.225","Text":"and we\u0027re supposed to calculate the work done by fk,"},{"Start":"00:26.225 ","End":"00:28.600","Text":"the force of kinetic friction,"},{"Start":"00:28.600 ","End":"00:31.425","Text":"so what is fk equal to?"},{"Start":"00:31.425 ","End":"00:37.430","Text":"We know that it\u0027s equal to Mu_k times n. If you recall N,"},{"Start":"00:37.430 ","End":"00:39.335","Text":"the normal force is on the y-axis,"},{"Start":"00:39.335 ","End":"00:40.385","Text":"so in this case,"},{"Start":"00:40.385 ","End":"00:47.200","Text":"it\u0027s equal to mg. We can write that the magnitude of fk is equal to Mu_k mg."},{"Start":"00:47.200 ","End":"00:51.450","Text":"If my force is equal to Mu_k mg, it\u0027s constant,"},{"Start":"00:51.450 ","End":"00:53.420","Text":"so I can now skip doing the integral and"},{"Start":"00:53.420 ","End":"00:55.595","Text":"go straight to the formula we just talked about,"},{"Start":"00:55.595 ","End":"00:58.250","Text":"which is the magnitude of the force times"},{"Start":"00:58.250 ","End":"01:02.330","Text":"the displacement of the object times the cosine of the angle between them."},{"Start":"01:02.330 ","End":"01:05.015","Text":"We need to figure out what is the displacement."},{"Start":"01:05.015 ","End":"01:06.545","Text":"Now that has to be given to us."},{"Start":"01:06.545 ","End":"01:11.420","Text":"Let\u0027s assume we\u0027re moving from this initial point to that endpoint along this path."},{"Start":"01:11.420 ","End":"01:13.595","Text":"That means that our displacement if it\u0027s,"},{"Start":"01:13.595 ","End":"01:14.870","Text":"let\u0027s say 2 meters,"},{"Start":"01:14.870 ","End":"01:17.180","Text":"would be a vector from"},{"Start":"01:17.180 ","End":"01:21.310","Text":"the initial point to the endpoint with a length or magnitude of 2 meters."},{"Start":"01:21.310 ","End":"01:24.770","Text":"With that given we know that the magnitude is 2 meters,"},{"Start":"01:24.770 ","End":"01:27.170","Text":"the magnitude of our displacement,"},{"Start":"01:27.170 ","End":"01:30.755","Text":"so that means work W equals the force"},{"Start":"01:30.755 ","End":"01:38.035","Text":"Mu_k mg times 2 meters times the cosine of the angle between the 2."},{"Start":"01:38.035 ","End":"01:41.810","Text":"Notice that before when I was finding the magnitude of my kinetic friction,"},{"Start":"01:41.810 ","End":"01:43.460","Text":"I didn\u0027t care whether it was positive or"},{"Start":"01:43.460 ","End":"01:45.905","Text":"negative because the magnitude is an absolute value."},{"Start":"01:45.905 ","End":"01:49.595","Text":"Now though, when I need to find the angle between the 2 vectors,"},{"Start":"01:49.595 ","End":"01:52.180","Text":"the direction very much matters to me."},{"Start":"01:52.180 ","End":"01:55.460","Text":"I know in this case that my friction will be pulling me to the left,"},{"Start":"01:55.460 ","End":"01:56.870","Text":"will be pulling me backwards,"},{"Start":"01:56.870 ","End":"01:59.555","Text":"and that my displacement is going towards the right,"},{"Start":"01:59.555 ","End":"02:02.870","Text":"meaning that these 2 vectors are facing in exact opposite directions,"},{"Start":"02:02.870 ","End":"02:06.300","Text":"so the angle between them is 180 degrees."},{"Start":"02:06.300 ","End":"02:08.765","Text":"That means that if I fill in my equation,"},{"Start":"02:08.765 ","End":"02:15.590","Text":"I\u0027ll have W equals Mu_k mg times 2 meters times the height cosine of 180 degrees."},{"Start":"02:15.590 ","End":"02:18.170","Text":"The cosine of 180 degrees is negative 1,"},{"Start":"02:18.170 ","End":"02:24.715","Text":"so my answer is W equals negative Mu_k mg times 2."},{"Start":"02:24.715 ","End":"02:30.245","Text":"It\u0027s important to note here that the negative value for the work is"},{"Start":"02:30.245 ","End":"02:32.840","Text":"very much intuitive and makes sense because we"},{"Start":"02:32.840 ","End":"02:35.500","Text":"have a friction pulling our object the opposite way,"},{"Start":"02:35.500 ","End":"02:38.210","Text":"and that this will very often happen."},{"Start":"02:38.210 ","End":"02:41.930","Text":"You\u0027re going to have friction that\u0027s going the opposite direction of the displacement,"},{"Start":"02:41.930 ","End":"02:44.555","Text":"and you\u0027ll have a negative value for your work."},{"Start":"02:44.555 ","End":"02:46.475","Text":"This is the formula you\u0027re going to use often."},{"Start":"02:46.475 ","End":"02:49.895","Text":"You\u0027re going to see this rather frequently, especially with friction."},{"Start":"02:49.895 ","End":"02:51.845","Text":"Let\u0027s take a second example."},{"Start":"02:51.845 ","End":"02:56.545","Text":"Let\u0027s say we have a force capital F that\u0027s going up and towards the right."},{"Start":"02:56.545 ","End":"02:59.105","Text":"Like so you can see it in blue."},{"Start":"02:59.105 ","End":"03:03.200","Text":"If I know that the force F has a given value"},{"Start":"03:03.200 ","End":"03:07.670","Text":"and that it\u0027s at an angle of 30 degrees with my displacement vector."},{"Start":"03:07.670 ","End":"03:12.020","Text":"Then I can calculate the work of the force F by saying"},{"Start":"03:12.020 ","End":"03:16.250","Text":"that it\u0027s the magnitude of F times 2 meters,"},{"Start":"03:16.250 ","End":"03:18.965","Text":"that is, times the cosine of 30 degrees."},{"Start":"03:18.965 ","End":"03:22.280","Text":"Now the cosine of 30 degrees is root 3 over 2."},{"Start":"03:22.280 ","End":"03:27.500","Text":"Root 3 over 2 times 2 this means the magnitude of the force F times root 3."},{"Start":"03:27.500 ","End":"03:30.770","Text":"Keep in mind that these are all values they\u0027re scalars."},{"Start":"03:30.770 ","End":"03:32.255","Text":"There is no direction involved."},{"Start":"03:32.255 ","End":"03:34.500","Text":"It\u0027s only a magnitude."}],"ID":9315},{"Watched":false,"Name":"Sum of Work","Duration":"1m 35s","ChapterTopicVideoID":9043,"CourseChapterTopicPlaylistID":5419,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.055","Text":"Hello. I wanted to talk about 1 basic property of work."},{"Start":"00:05.055 ","End":"00:06.600","Text":"If you recall in the last problem,"},{"Start":"00:06.600 ","End":"00:10.635","Text":"we calculated the work of the force F and the work of the force f_k."},{"Start":"00:10.635 ","End":"00:13.620","Text":"What I can do is I can calculate the work of"},{"Start":"00:13.620 ","End":"00:16.470","Text":"each and every force in a problem and sum them,"},{"Start":"00:16.470 ","End":"00:19.525","Text":"put them together to find the sum of the forces."},{"Start":"00:19.525 ","End":"00:22.050","Text":"There\u0027s a basic property that can help me here,"},{"Start":"00:22.050 ","End":"00:25.395","Text":"which is the work of the sum of forces,"},{"Start":"00:25.395 ","End":"00:28.290","Text":"is equal to the sum of the work."},{"Start":"00:28.290 ","End":"00:30.810","Text":"Meaning that if I take each individual force and add it up,"},{"Start":"00:30.810 ","End":"00:36.375","Text":"it\u0027s the same as taking the work of all forces together."},{"Start":"00:36.375 ","End":"00:37.800","Text":"Returning to our example,"},{"Start":"00:37.800 ","End":"00:39.120","Text":"there are 2 ways we could do this."},{"Start":"00:39.120 ","End":"00:40.470","Text":"First, what we just did,"},{"Start":"00:40.470 ","End":"00:42.845","Text":"we could get each individual force,"},{"Start":"00:42.845 ","End":"00:44.660","Text":"find the work of that and add those."},{"Start":"00:44.660 ","End":"00:48.200","Text":"In our example we would add WF,"},{"Start":"00:48.200 ","End":"00:50.700","Text":"which is root 3 F,"},{"Start":"00:50.700 ","End":"00:55.825","Text":"and add to that negative Mu_k M_g times 2."},{"Start":"00:55.825 ","End":"00:57.935","Text":"The other way to do that is to 1st,"},{"Start":"00:57.935 ","End":"01:01.010","Text":"add the vectors of the different forces and come up with"},{"Start":"01:01.010 ","End":"01:04.940","Text":"some summed vector from the addition of the 2 vectors."},{"Start":"01:04.940 ","End":"01:08.330","Text":"It would probably point in our case in this direction,"},{"Start":"01:08.330 ","End":"01:10.151","Text":"up in a little bit to the left."},{"Start":"01:10.151 ","End":"01:11.990","Text":"It would have a given magnitude."},{"Start":"01:11.990 ","End":"01:13.220","Text":"We could calculate it if we wanted,"},{"Start":"01:13.220 ","End":"01:15.920","Text":"and that should give us the same answer."},{"Start":"01:15.920 ","End":"01:20.900","Text":"Both methods work, usually we\u0027ll find the work of each force and summarize that way."},{"Start":"01:20.900 ","End":"01:22.820","Text":"Just to give a quick recap."},{"Start":"01:22.820 ","End":"01:27.080","Text":"This is the end of our discussion of work on a constant force."},{"Start":"01:27.080 ","End":"01:30.290","Text":"What you do is take the magnitude of the force times"},{"Start":"01:30.290 ","End":"01:35.010","Text":"the length of the displacement times the cosine of the angle between the 2."}],"ID":9316}],"Thumbnail":null,"ID":5419},{"Name":"Explanation About the Integral of Work","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Basic Explanation","Duration":"3m 58s","ChapterTopicVideoID":9044,"CourseChapterTopicPlaylistID":5420,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.505","Text":"Hello. In this video,"},{"Start":"00:02.505 ","End":"00:05.325","Text":"I want to look closer at the integral of the work"},{"Start":"00:05.325 ","End":"00:09.045","Text":"from A to B and see what\u0027s really going on here."},{"Start":"00:09.045 ","End":"00:12.930","Text":"The first thing I want to do is draw 2 axes an x and"},{"Start":"00:12.930 ","End":"00:16.815","Text":"y-axis and on the x and y-axis, I\u0027m going to have 2 points."},{"Start":"00:16.815 ","End":"00:19.290","Text":"The first A will be here,"},{"Start":"00:19.290 ","End":"00:24.300","Text":"and the second and B will draw farther out over here."},{"Start":"00:24.300 ","End":"00:29.100","Text":"Now I\u0027m going to have some object that\u0027s trajectory somehow gets from A to B."},{"Start":"00:29.100 ","End":"00:31.395","Text":"It can be a squiggly line like this."},{"Start":"00:31.395 ","End":"00:33.120","Text":"What does work do?"},{"Start":"00:33.120 ","End":"00:34.410","Text":"What does work measure?"},{"Start":"00:34.410 ","End":"00:39.030","Text":"Work is if we were to break this line down into all of its little pieces,"},{"Start":"00:39.030 ","End":"00:41.150","Text":"imagine each of the dotted lines,"},{"Start":"00:41.150 ","End":"00:43.250","Text":"its own small segment."},{"Start":"00:43.250 ","End":"00:46.925","Text":"Then we can think of each one as a point of dr."},{"Start":"00:46.925 ","End":"00:53.470","Text":"The direction of each dr segment is always tangential to the trajectory."},{"Start":"00:53.470 ","End":"00:58.700","Text":"Each of these tangential lines in the direction of the trajectory,"},{"Start":"00:58.700 ","End":"01:00.470","Text":"we can do each one in red,"},{"Start":"01:00.470 ","End":"01:02.000","Text":"make up the entire line."},{"Start":"01:02.000 ","End":"01:04.145","Text":"When we do the work function,"},{"Start":"01:04.145 ","End":"01:08.330","Text":"we\u0027re breaking up our trajectory into a ton of small little lines,"},{"Start":"01:08.330 ","End":"01:11.440","Text":"each tangential to the trajectory of the object."},{"Start":"01:11.440 ","End":"01:15.365","Text":"I can now look at any given point along this line."},{"Start":"01:15.365 ","End":"01:18.245","Text":"Let\u0027s say we choose a random point in the middle here,"},{"Start":"01:18.245 ","End":"01:22.025","Text":"and the next step is to find the force at that moment."},{"Start":"01:22.025 ","End":"01:24.080","Text":"To see what the force is at the moment,"},{"Start":"01:24.080 ","End":"01:27.155","Text":"let\u0027s say here it\u0027s a vector in that direction,"},{"Start":"01:27.155 ","End":"01:29.465","Text":"sets the force, and before doing the integral,"},{"Start":"01:29.465 ","End":"01:33.295","Text":"you do scalar multiplication with dr."},{"Start":"01:33.295 ","End":"01:36.260","Text":"When I do this scalar multiplication,"},{"Start":"01:36.260 ","End":"01:40.685","Text":"what I\u0027m doing is multiplying the element of the force times"},{"Start":"01:40.685 ","End":"01:45.545","Text":"the dr times the cosine of the angle between the 2 of them."},{"Start":"01:45.545 ","End":"01:50.585","Text":"Another way to look at it is I\u0027m taking the element in the direction of dr,"},{"Start":"01:50.585 ","End":"01:53.150","Text":"of the force, and multiplying it by dr."},{"Start":"01:53.150 ","End":"01:54.740","Text":"If you see here,"},{"Start":"01:54.740 ","End":"02:00.380","Text":"we can imagine that dr is the x-axis and there\u0027s a perpendicular y-axis."},{"Start":"02:00.380 ","End":"02:03.695","Text":"I\u0027m going to ignore the element that\u0027s on the y-axis"},{"Start":"02:03.695 ","End":"02:07.055","Text":"and take the element that drops down to the x-axis,"},{"Start":"02:07.055 ","End":"02:09.985","Text":"that which is in the same direction as dr."},{"Start":"02:09.985 ","End":"02:14.315","Text":"I\u0027m ignoring the y element of the vector F and I take the x element,"},{"Start":"02:14.315 ","End":"02:17.340","Text":"the element that\u0027s parallel to the trajectory."},{"Start":"02:17.340 ","End":"02:23.330","Text":"I\u0027m multiplying that by dr and afterwards I take an integral of this."},{"Start":"02:23.330 ","End":"02:26.870","Text":"What I\u0027m really doing is on any given point of this trajectory,"},{"Start":"02:26.870 ","End":"02:31.115","Text":"I\u0027m taking only the element of the force that\u0027s in the direction of the trajectory,"},{"Start":"02:31.115 ","End":"02:34.915","Text":"multiplying it by some infinitesimal element of dr."},{"Start":"02:34.915 ","End":"02:36.480","Text":"I can choose another point,"},{"Start":"02:36.480 ","End":"02:38.570","Text":"say here and if the F vector,"},{"Start":"02:38.570 ","End":"02:40.775","Text":"the force vector goes in that direction,"},{"Start":"02:40.775 ","End":"02:42.740","Text":"all I care about is the element of"},{"Start":"02:42.740 ","End":"02:45.350","Text":"the force vector that\u0027s in the direction of the trajectory."},{"Start":"02:45.350 ","End":"02:48.125","Text":"I multiply that by a small bit of length,"},{"Start":"02:48.125 ","End":"02:51.320","Text":"the dr and that\u0027s really what\u0027s happening when I do work."},{"Start":"02:51.320 ","End":"02:53.390","Text":"Isn\u0027t any given point I\u0027m taking"},{"Start":"02:53.390 ","End":"02:57.800","Text":"the element of the force that goes in the direction of the trajectory."},{"Start":"02:57.800 ","End":"03:03.949","Text":"What\u0027s interesting here is what work really does is it only measures"},{"Start":"03:03.949 ","End":"03:06.200","Text":"the element of the force going in the direction of"},{"Start":"03:06.200 ","End":"03:10.910","Text":"the trajectory and that\u0027s important because it measures the change in velocity."},{"Start":"03:10.910 ","End":"03:13.460","Text":"Now, if we have an object on this trajectory,"},{"Start":"03:13.460 ","End":"03:16.100","Text":"the only thing we\u0027re looking at is again,"},{"Start":"03:16.100 ","End":"03:18.680","Text":"the element of force and the direction of"},{"Start":"03:18.680 ","End":"03:21.575","Text":"the trajectory and that\u0027s what\u0027s going to change its velocity,"},{"Start":"03:21.575 ","End":"03:23.060","Text":"increase or decrease it."},{"Start":"03:23.060 ","End":"03:25.370","Text":"We ignore altogether the normal force,"},{"Start":"03:25.370 ","End":"03:30.755","Text":"the force going perpendicular to the object because we don\u0027t care about the direction."},{"Start":"03:30.755 ","End":"03:32.510","Text":"Work doesn\u0027t account for direction,"},{"Start":"03:32.510 ","End":"03:34.085","Text":"it doesn\u0027t care about direction."},{"Start":"03:34.085 ","End":"03:41.150","Text":"We only look at the tangential element of the force tangential to the trajectory."},{"Start":"03:41.150 ","End":"03:44.900","Text":"We only find how our velocity is changing."},{"Start":"03:44.900 ","End":"03:50.735","Text":"Put simply work finds how our force influences the velocity of an object,"},{"Start":"03:50.735 ","End":"03:53.420","Text":"not the direction, and just to reiterate,"},{"Start":"03:53.420 ","End":"03:58.620","Text":"that\u0027s why we only look at the element that is tangential to the trajectory."}],"ID":9317},{"Watched":false,"Name":"What is the Vector dr","Duration":"5m 12s","ChapterTopicVideoID":9045,"CourseChapterTopicPlaylistID":5420,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.649","Text":"I wanted to explain in more detail something I\u0027d mentioned earlier which is dr,"},{"Start":"00:04.649 ","End":"00:10.500","Text":"that line which we\u0027d said is always tangential to the trajectory of our object."},{"Start":"00:10.500 ","End":"00:12.705","Text":"To explain it, let\u0027s do the following."},{"Start":"00:12.705 ","End":"00:14.903","Text":"We have our same trajectory from A to B"},{"Start":"00:14.903 ","End":"00:18.960","Text":"and we have an object resting somewhere along it, and let\u0027s put it here."},{"Start":"00:18.960 ","End":"00:24.720","Text":"Right there, okay, That\u0027ll be point C. We want to have the position vector at point"},{"Start":"00:24.720 ","End":"00:30.480","Text":"C. The position vector r goes from the origin to the point where the object is resting."},{"Start":"00:30.480 ","End":"00:33.675","Text":"That\u0027s our new vector r,"},{"Start":"00:33.675 ","End":"00:35.035","Text":"the position vector,"},{"Start":"00:35.035 ","End":"00:40.005","Text":"and let\u0027s say it\u0027s at time t along our trajectory."},{"Start":"00:40.005 ","End":"00:49.700","Text":"Now, if I want to have a new position vector at a new point let\u0027s say dt seconds later,"},{"Start":"00:49.700 ","End":"00:52.162","Text":"dt is just some random amount of time later,"},{"Start":"00:52.162 ","End":"00:54.805","Text":"that will be rt plus dt."},{"Start":"00:54.805 ","End":"00:57.785","Text":"That\u0027ll go to the new position of the object."},{"Start":"00:57.785 ","End":"01:00.690","Text":"It\u0027s moved to here, let\u0027s say in that amount of time."},{"Start":"01:00.690 ","End":"01:07.000","Text":"Rt plus dt will go from the origin to that new point right there."},{"Start":"01:07.000 ","End":"01:11.105","Text":"Let\u0027s say I want to find the difference between the 2 points,"},{"Start":"01:11.105 ","End":"01:13.175","Text":"rt and rt plus dt,"},{"Start":"01:13.175 ","End":"01:17.940","Text":"that would be dr and what I\u0027m looking for is the change."},{"Start":"01:17.940 ","End":"01:20.900","Text":"If you recall from earlier talks we\u0027ve said that when you\u0027re talking about"},{"Start":"01:20.900 ","End":"01:23.880","Text":"change you want to take the final minus the initial,"},{"Start":"01:23.880 ","End":"01:28.160","Text":"so rt plus dt minus rt."},{"Start":"01:28.160 ","End":"01:35.600","Text":"When you do the subtraction what you\u0027ll find is taking rt plus dt minus rt,"},{"Start":"01:35.600 ","End":"01:39.125","Text":"and I\u0027ll write this out in blue here so it\u0027s a little more visible,"},{"Start":"01:39.125 ","End":"01:42.905","Text":"we\u0027ll give you a vector of dr that looks like this."},{"Start":"01:42.905 ","End":"01:44.180","Text":"In fact, as you can see,"},{"Start":"01:44.180 ","End":"01:46.445","Text":"it\u0027s tangential to the trajectory here."},{"Start":"01:46.445 ","End":"01:52.280","Text":"Again, rt plus dr will also give us rt plus dt."},{"Start":"01:52.280 ","End":"01:56.120","Text":"You can write this anyway but this is dr,"},{"Start":"01:56.120 ","End":"01:59.995","Text":"and that\u0027s how it ends up being tangential to the trajectory."},{"Start":"01:59.995 ","End":"02:03.455","Text":"Now this line looks tangential although it\u0027s actually not,"},{"Start":"02:03.455 ","End":"02:08.735","Text":"but if you took dt down to 0 and had an infinitesimally small dr it would"},{"Start":"02:08.735 ","End":"02:15.100","Text":"actually be truly tangential to the trajectory of our object and that\u0027s how you get dr."},{"Start":"02:15.100 ","End":"02:19.700","Text":"If we look at dr from a mathematical perspective,"},{"Start":"02:19.700 ","End":"02:22.285","Text":"rt plus dt minus rt,"},{"Start":"02:22.285 ","End":"02:26.185","Text":"we can see that it\u0027s actually a differential."},{"Start":"02:26.185 ","End":"02:28.130","Text":"We\u0027re taking a differential of"},{"Start":"02:28.130 ","End":"02:33.215","Text":"the vector r. Let\u0027s take this a step back and we can write the vector r,"},{"Start":"02:33.215 ","End":"02:36.800","Text":"the position vector in a different way."},{"Start":"02:36.800 ","End":"02:39.845","Text":"We can write it as x in the direction of x,"},{"Start":"02:39.845 ","End":"02:44.180","Text":"that\u0027s x-hat plus y in the direction of y,"},{"Start":"02:44.180 ","End":"02:49.075","Text":"that\u0027s y-hat, plus z in the direction of z, that z-hat."},{"Start":"02:49.075 ","End":"02:56.135","Text":"Another way we can write that at least in Cartesian coordinates is r=x, y, z."},{"Start":"02:56.135 ","End":"03:00.485","Text":"Now we know that both of these are true ways to represent"},{"Start":"03:00.485 ","End":"03:05.026","Text":"the position vector coming from the origin no matter what that vector is."},{"Start":"03:05.026 ","End":"03:06.880","Text":"It\u0027s a general formula."},{"Start":"03:06.880 ","End":"03:08.880","Text":"We can take that a step further,"},{"Start":"03:08.880 ","End":"03:10.760","Text":"and if we\u0027re writing the differential we can write"},{"Start":"03:10.760 ","End":"03:12.920","Text":"the differential in a general formula that"},{"Start":"03:12.920 ","End":"03:17.615","Text":"applies to all situations regardless of what\u0027s going on by saying that dr,"},{"Start":"03:17.615 ","End":"03:24.825","Text":"the differential of the vector r equals dx, x-hat,"},{"Start":"03:24.825 ","End":"03:26.136","Text":"that\u0027s the direction of x,"},{"Start":"03:26.136 ","End":"03:28.715","Text":"plus dy in the direction of y,"},{"Start":"03:28.715 ","End":"03:34.440","Text":"y-hat, plus dz in the direction of z, z-hat."},{"Start":"03:34.790 ","End":"03:37.955","Text":"This is a general formula for dr."},{"Start":"03:37.955 ","End":"03:40.415","Text":"It\u0027s always true, so dr always equal"},{"Start":"03:40.415 ","End":"03:43.850","Text":"dx in the direction of x plus dy in the direction of y,"},{"Start":"03:43.850 ","End":"03:46.060","Text":"and dz in the direction of z."},{"Start":"03:46.060 ","End":"03:48.305","Text":"Because this is generally true,"},{"Start":"03:48.305 ","End":"03:52.430","Text":"we can apply in all situations including the situation above with work."},{"Start":"03:52.430 ","End":"03:58.100","Text":"We can write our work formula instead of as it is now in the following format."},{"Start":"03:58.100 ","End":"04:07.195","Text":"We can write it as F_xdx plus F_ydy plus F_zdz all as an integral from A-B."},{"Start":"04:07.195 ","End":"04:10.190","Text":"These two are identical they mean the exact same thing,"},{"Start":"04:10.190 ","End":"04:13.310","Text":"they equal the exact same thing and we can show that by showing"},{"Start":"04:13.310 ","End":"04:16.505","Text":"the vector F. Now vectors have a general formula."},{"Start":"04:16.505 ","End":"04:20.810","Text":"In this case for F it would be F_x x-hat plus F_y,"},{"Start":"04:20.810 ","End":"04:24.825","Text":"y-hat plus F_z, z-hat."},{"Start":"04:24.825 ","End":"04:30.715","Text":"If we took a scalar multiplication of the vector f by dr,"},{"Start":"04:30.715 ","End":"04:32.750","Text":"that is the differential of r,"},{"Start":"04:32.750 ","End":"04:34.025","Text":"our position vector,"},{"Start":"04:34.025 ","End":"04:35.780","Text":"we multiply x by x,"},{"Start":"04:35.780 ","End":"04:38.315","Text":"y by y and z by z."},{"Start":"04:38.315 ","End":"04:41.150","Text":"Of course, the x and y multiplied by each other would"},{"Start":"04:41.150 ","End":"04:44.990","Text":"0 out and we end up with the formula that\u0027s on our left."},{"Start":"04:44.990 ","End":"04:47.510","Text":"You end up with F_xdx,"},{"Start":"04:47.510 ","End":"04:50.515","Text":"F_ydy, and F_zdz."},{"Start":"04:50.515 ","End":"04:52.410","Text":"This is a formula you should remember,"},{"Start":"04:52.410 ","End":"04:54.067","Text":"so we\u0027ll put it in red."},{"Start":"04:54.067 ","End":"04:57.800","Text":"You should remember that both are acceptable,"},{"Start":"04:57.800 ","End":"05:01.300","Text":"but the last way we wrote it is sometimes more popular."},{"Start":"05:01.300 ","End":"05:04.370","Text":"The next question of course is how do we use this formula?"},{"Start":"05:04.370 ","End":"05:07.160","Text":"What do we do with it, and what function does it perform?"},{"Start":"05:07.160 ","End":"05:10.050","Text":"For that, we\u0027ll look at the next video."}],"ID":9318}],"Thumbnail":null,"ID":5420},{"Name":"How to Calculate the Integral of a non Constant Force","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation and Formula","Duration":"1m 20s","ChapterTopicVideoID":9214,"CourseChapterTopicPlaylistID":5421,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"Hello. In this video,"},{"Start":"00:02.070 ","End":"00:06.210","Text":"we\u0027re going to calculate the work of a non-constant force."},{"Start":"00:06.210 ","End":"00:10.395","Text":"What we\u0027re really doing is this is truly finding the integral of work."},{"Start":"00:10.395 ","End":"00:13.110","Text":"To remind you, work from 1 point to another,"},{"Start":"00:13.110 ","End":"00:18.210","Text":"in our example we\u0027ll do from A to B is going to equal the integral from"},{"Start":"00:18.210 ","End":"00:25.905","Text":"point A to point B of the force F.dr."},{"Start":"00:25.905 ","End":"00:29.655","Text":"This is the formula for work between the points A and B."},{"Start":"00:29.655 ","End":"00:32.310","Text":"You have to remember that between these two points there are"},{"Start":"00:32.310 ","End":"00:35.220","Text":"an infinite number of trajectories we could follow,"},{"Start":"00:35.220 ","End":"00:37.095","Text":"so we need to know what our trajectory is."},{"Start":"00:37.095 ","End":"00:38.925","Text":"It could be straight like so,"},{"Start":"00:38.925 ","End":"00:41.235","Text":"it could be a little more curved and indirect."},{"Start":"00:41.235 ","End":"00:42.480","Text":"It could really be anything,"},{"Start":"00:42.480 ","End":"00:44.510","Text":"but the point is we need to know what"},{"Start":"00:44.510 ","End":"00:47.500","Text":"our trajectory is if we\u0027re going to calculate the work,"},{"Start":"00:47.500 ","End":"00:49.565","Text":"so this is the second thing we need to know."},{"Start":"00:49.565 ","End":"00:52.205","Text":"What trajectory are we following?"},{"Start":"00:52.205 ","End":"00:55.460","Text":"I can take this formula and rewrite it in"},{"Start":"00:55.460 ","End":"00:58.265","Text":"a different way that can be a little useful for us."},{"Start":"00:58.265 ","End":"01:04.770","Text":"The work from points A to B equals the integral from point A to B"},{"Start":"01:04.770 ","End":"01:12.790","Text":"of Fx dx plus Fy dy plus Fz dz."},{"Start":"01:12.940 ","End":"01:16.820","Text":"This is essentially the product of F.dr."},{"Start":"01:16.820 ","End":"01:21.180","Text":"Dr is the differential of the position vector."}],"ID":9484}],"Thumbnail":null,"ID":5421},{"Name":"Deriving Work And Energy Equations","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Part 1- The Work of the Net Force Equals the Sum of the Work of Each Force Individually","Duration":"3m 53s","ChapterTopicVideoID":9046,"CourseChapterTopicPlaylistID":5422,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.070","Text":"In this set of lectures,"},{"Start":"00:02.070 ","End":"00:04.650","Text":"we\u0027re going to try to understand how we get to"},{"Start":"00:04.650 ","End":"00:08.174","Text":"the equation for the work of the non-conservative forces."},{"Start":"00:08.174 ","End":"00:11.970","Text":"If you recall, the work of the non-conservative forces,"},{"Start":"00:11.970 ","End":"00:16.110","Text":"W_NC equals the change in total energy,"},{"Start":"00:16.110 ","End":"00:20.355","Text":"Delta E. This is the equation that we\u0027re going to try to come to in the end."},{"Start":"00:20.355 ","End":"00:22.845","Text":"Well, this is only background information."},{"Start":"00:22.845 ","End":"00:25.170","Text":"I think it\u0027s critical that you listened because it can be really"},{"Start":"00:25.170 ","End":"00:29.055","Text":"helpful for you to understand what\u0027s going on behind the scene."},{"Start":"00:29.055 ","End":"00:31.205","Text":"In this part of the lecture,"},{"Start":"00:31.205 ","End":"00:34.850","Text":"we\u0027re going to prove that the work of the net force equals the sum of"},{"Start":"00:34.850 ","End":"00:38.640","Text":"the work of each force individually. What does that mean?"},{"Start":"00:38.640 ","End":"00:42.440","Text":"Well, on the 1 hand, we have the work of the net force,"},{"Start":"00:42.440 ","End":"00:44.164","Text":"that is the sum of all forces."},{"Start":"00:44.164 ","End":"00:47.794","Text":"We take that, and then we calculate the work for the sum of all forces,"},{"Start":"00:47.794 ","End":"00:53.630","Text":"that Sigma F. That equals the sum of the work of each force individually."},{"Start":"00:53.630 ","End":"00:55.235","Text":"We\u0027ll call that Fi."},{"Start":"00:55.235 ","End":"01:01.595","Text":"Let\u0027s assume that you have some object with 2 forces acting on it, F_1 and F_2."},{"Start":"01:01.595 ","End":"01:03.275","Text":"F_1 goes to the right,"},{"Start":"01:03.275 ","End":"01:06.650","Text":"and F_2 goes upwards and towards the left."},{"Start":"01:06.650 ","End":"01:10.790","Text":"Now we can find the work of these 2 forces in 2 different ways."},{"Start":"01:10.790 ","End":"01:16.120","Text":"The first is to find the work of F_1 and add that to the work of F_2,"},{"Start":"01:16.120 ","End":"01:18.665","Text":"and that equals the right side of our equation above"},{"Start":"01:18.665 ","End":"01:21.530","Text":"the sum of the work of each force individually."},{"Start":"01:21.530 ","End":"01:23.690","Text":"We take the work of each force individually,"},{"Start":"01:23.690 ","End":"01:25.265","Text":"and then we find the sum."},{"Start":"01:25.265 ","End":"01:29.450","Text":"The other way we can do it is find the addition of these 2 forces."},{"Start":"01:29.450 ","End":"01:30.920","Text":"First, it\u0027d be somewhere in the middle,"},{"Start":"01:30.920 ","End":"01:32.615","Text":"probably something like this."},{"Start":"01:32.615 ","End":"01:36.470","Text":"We can call that F_T for F total because it\u0027s the sum of both forces,"},{"Start":"01:36.470 ","End":"01:38.090","Text":"it\u0027s the total force."},{"Start":"01:38.090 ","End":"01:42.560","Text":"Now that we have F_T or the sum of forces or the net force,"},{"Start":"01:42.560 ","End":"01:44.315","Text":"we can get rid of F_1 and F_2."},{"Start":"01:44.315 ","End":"01:47.810","Text":"We don\u0027t need them anymore because we can perform the operation to"},{"Start":"01:47.810 ","End":"01:51.638","Text":"calculate the work of F_T or the sum or net force,"},{"Start":"01:51.638 ","End":"01:54.060","Text":"and it will have the exact same solution as if we"},{"Start":"01:54.060 ","End":"01:57.645","Text":"found the sum of each force individually."},{"Start":"01:57.645 ","End":"02:01.715","Text":"You can take the integral of the net force.dr,"},{"Start":"02:01.715 ","End":"02:05.430","Text":"or you can find the work of each force individually and add them up,"},{"Start":"02:05.430 ","End":"02:07.280","Text":"in either way they\u0027ll equal the same thing."},{"Start":"02:07.280 ","End":"02:10.160","Text":"It doesn\u0027t matter which way you do the operation."},{"Start":"02:10.160 ","End":"02:13.730","Text":"Now I\u0027d like to explore this in a little more depths."},{"Start":"02:13.730 ","End":"02:16.925","Text":"For those of you who feel that you intuitively understand this,"},{"Start":"02:16.925 ","End":"02:19.820","Text":"you don\u0027t necessarily need to watch the rest of the video."},{"Start":"02:19.820 ","End":"02:22.955","Text":"But I feel that it can still be a useful exercise."},{"Start":"02:22.955 ","End":"02:26.555","Text":"When we talk about the work of the net force,"},{"Start":"02:26.555 ","End":"02:29.030","Text":"what we need to do is perform the work operation,"},{"Start":"02:29.030 ","End":"02:33.935","Text":"which is to take the integral of the net force.dr."},{"Start":"02:33.935 ","End":"02:35.915","Text":"Now, when we say the net force,"},{"Start":"02:35.915 ","End":"02:38.305","Text":"it just means the sum of all forces."},{"Start":"02:38.305 ","End":"02:43.300","Text":"We can open this up and think of it as each force individually within the integral,"},{"Start":"02:43.300 ","End":"02:47.585","Text":"we take the integral of F_1 plus F_2."},{"Start":"02:47.585 ","End":"02:50.615","Text":"Each of these are vectors, plus F_3,"},{"Start":"02:50.615 ","End":"02:52.115","Text":"and so on and so forth,"},{"Start":"02:52.115 ","End":"02:54.580","Text":"until you reach the final force you\u0027re calculating."},{"Start":"02:54.580 ","End":"02:59.840","Text":"You can take all of those and multiply them by dr, scalar multiplication."},{"Start":"02:59.840 ","End":"03:02.765","Text":"With scalar multiplication using commutative property,"},{"Start":"03:02.765 ","End":"03:04.865","Text":"we can open up the parentheses."},{"Start":"03:04.865 ","End":"03:10.280","Text":"We can think of this as F_1 times dr or.dr plus the"},{"Start":"03:10.280 ","End":"03:15.875","Text":"integral of F_2.dr plus the integral of F_3,"},{"Start":"03:15.875 ","End":"03:20.210","Text":"etc, until we reach the final force that we\u0027re calculating."},{"Start":"03:20.210 ","End":"03:24.640","Text":"Each of these integrals is the formula for the work of an individual force."},{"Start":"03:24.640 ","End":"03:29.060","Text":"We have above the work of force 1 plus the work of force 2 etc,"},{"Start":"03:29.060 ","End":"03:30.785","Text":"until we get to our last force."},{"Start":"03:30.785 ","End":"03:34.685","Text":"Meaning that we have above the sum of the work of each force individually,"},{"Start":"03:34.685 ","End":"03:36.800","Text":"which is the right side of our equation before."},{"Start":"03:36.800 ","End":"03:39.650","Text":"So now we\u0027ve actually proven that the work of"},{"Start":"03:39.650 ","End":"03:44.165","Text":"the net force equals the sum of the work of each force individually."},{"Start":"03:44.165 ","End":"03:47.105","Text":"We\u0027re going to need to use this later. Let\u0027s hold on to it."},{"Start":"03:47.105 ","End":"03:50.885","Text":"Remember this, and we\u0027ll come back to it when we get later on in our formula."},{"Start":"03:50.885 ","End":"03:53.730","Text":"For now, let\u0027s move on to Part B."}],"ID":9319},{"Watched":false,"Name":"Part 2- The Work of the Net Force Equals the Change in Kinetic Energy","Duration":"10m 2s","ChapterTopicVideoID":9047,"CourseChapterTopicPlaylistID":5422,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.760","Text":"In this part of the video,"},{"Start":"00:02.760 ","End":"00:05.850","Text":"I want to show how the work of the net force"},{"Start":"00:05.850 ","End":"00:10.005","Text":"equals the change in kinetic energy in a given object."},{"Start":"00:10.005 ","End":"00:13.020","Text":"Let\u0027s assume that we have some object,"},{"Start":"00:13.020 ","End":"00:15.360","Text":"could be that same box from before,"},{"Start":"00:15.360 ","End":"00:16.920","Text":"and we have some number of forces,"},{"Start":"00:16.920 ","End":"00:18.622","Text":"1, 2, 10, a hundred,"},{"Start":"00:18.622 ","End":"00:20.190","Text":"it doesn\u0027t matter, but the sum of"},{"Start":"00:20.190 ","End":"00:23.295","Text":"those forces is pointing in the direction of the vector arrow."},{"Start":"00:23.295 ","End":"00:25.470","Text":"So the net force is that vector arrow."},{"Start":"00:25.470 ","End":"00:31.380","Text":"Now the work of the net force equals the change in kinetic energy, Delta E_k."},{"Start":"00:31.380 ","End":"00:35.865","Text":"Right now we\u0027re only talking about kinetic energy, not total energy."},{"Start":"00:35.865 ","End":"00:37.590","Text":"Total energy we\u0027ll talk about later,"},{"Start":"00:37.590 ","End":"00:40.425","Text":"but for now, we\u0027re only focusing on kinetic energy."},{"Start":"00:40.425 ","End":"00:44.660","Text":"This formula in itself can be remembered and put on your formula sheet if you\u0027d like."},{"Start":"00:44.660 ","End":"00:47.810","Text":"I personally prefer the formula we used in the last part,"},{"Start":"00:47.810 ","End":"00:52.085","Text":"that the work of the net force equals the change in total energy."},{"Start":"00:52.085 ","End":"00:54.050","Text":"But some people prefer this."},{"Start":"00:54.050 ","End":"00:58.520","Text":"First, I want to take an example where the net force is constant."},{"Start":"00:58.520 ","End":"01:03.950","Text":"We can write out the work of the net force in the classic way using an integral."},{"Start":"01:03.950 ","End":"01:11.845","Text":"The work of the net force equals the integral of the net force multiplied scalar by dr."},{"Start":"01:11.845 ","End":"01:13.250","Text":"But because it\u0027s constant,"},{"Start":"01:13.250 ","End":"01:15.185","Text":"we don\u0027t need the integral."},{"Start":"01:15.185 ","End":"01:17.525","Text":"We can write this as the net force,"},{"Start":"01:17.525 ","End":"01:19.640","Text":"scalar multiplied by Delta r,"},{"Start":"01:19.640 ","End":"01:21.140","Text":"which is the displacement."},{"Start":"01:21.140 ","End":"01:22.670","Text":"Now if you recall,"},{"Start":"01:22.670 ","End":"01:31.615","Text":"The net force can also be written using Newton\u0027s law as ma, mass times acceleration."},{"Start":"01:31.615 ","End":"01:41.359","Text":"We can replace that with ma dot Delta r. For the sake of simplicity,"},{"Start":"01:41.359 ","End":"01:46.055","Text":"let\u0027s assume that our object is moving in the direction of F. It starts here."},{"Start":"01:46.055 ","End":"01:48.650","Text":"Let\u0027s say that\u0027s the origin and it moves out in"},{"Start":"01:48.650 ","End":"01:51.680","Text":"the direction of F towards this point over here."},{"Start":"01:51.680 ","End":"01:59.090","Text":"We can draw the object over here so that the displacement will be something like that,"},{"Start":"01:59.090 ","End":"02:01.355","Text":"in the direction of F. Now,"},{"Start":"02:01.355 ","End":"02:05.005","Text":"we can do is we don\u0027t need to worry about the scalar multiplication."},{"Start":"02:05.005 ","End":"02:07.410","Text":"In a situation where this isn\u0027t the case,"},{"Start":"02:07.410 ","End":"02:10.565","Text":"let\u0027s say the initial velocity goes in a different direction,"},{"Start":"02:10.565 ","End":"02:13.460","Text":"then we\u0027re not going to worry about it right now but"},{"Start":"02:13.460 ","End":"02:17.315","Text":"essentially that velocity will continue in a constant form."},{"Start":"02:17.315 ","End":"02:22.180","Text":"Now we can change the scalar multiplication into a classic multiplication."},{"Start":"02:22.180 ","End":"02:25.385","Text":"We have ma times Delta r, the displacement."},{"Start":"02:25.385 ","End":"02:27.380","Text":"Now we can make this our x-axis."},{"Start":"02:27.380 ","End":"02:28.600","Text":"We can choose our axes."},{"Start":"02:28.600 ","End":"02:30.295","Text":"We can say this is the x-axis."},{"Start":"02:30.295 ","End":"02:32.230","Text":"This is the x-axis,"},{"Start":"02:32.230 ","End":"02:34.955","Text":"we can write out the whole formula as"},{"Start":"02:34.955 ","End":"02:41.075","Text":"ma_x times Delta x because we\u0027re moving along the x-axis."},{"Start":"02:41.075 ","End":"02:45.710","Text":"Now, this looks very similar to an equation from kinematics."},{"Start":"02:45.710 ","End":"02:49.835","Text":"The equation itself is v,"},{"Start":"02:49.835 ","End":"02:58.670","Text":"your velocity squared equals the initial velocity v_0^2 plus 2a Delta x."},{"Start":"02:58.670 ","End":"03:02.315","Text":"This formula works for each axis on its own."},{"Start":"03:02.315 ","End":"03:06.365","Text":"We can assume we\u0027re talking here about the x-axis and put an x\u0027s here."},{"Start":"03:06.365 ","End":"03:10.685","Text":"Now if we isolate a Delta x, we can replace it."},{"Start":"03:10.685 ","End":"03:14.000","Text":"We have on one side a Delta x and that"},{"Start":"03:14.000 ","End":"03:22.405","Text":"equals v_x^2 over 2 minus v initial x over 2."},{"Start":"03:22.405 ","End":"03:26.960","Text":"If we put that into our equation above ma_x Delta x."},{"Start":"03:26.960 ","End":"03:28.445","Text":"We can take our 1/2,"},{"Start":"03:28.445 ","End":"03:29.780","Text":"we can take r over 2,"},{"Start":"03:29.780 ","End":"03:37.385","Text":"and put on the side we have 1/2 times m v^2 minus v initial squared."},{"Start":"03:37.385 ","End":"03:43.055","Text":"Remember that v is the final velocity and v_0 is the initial velocity."},{"Start":"03:43.055 ","End":"03:47.960","Text":"When we have 1/2 m v^2 minus v initial squared,"},{"Start":"03:47.960 ","End":"03:51.140","Text":"what we really have is the change in kinetic energy."},{"Start":"03:51.140 ","End":"03:52.795","Text":"Because if you remember,"},{"Start":"03:52.795 ","End":"03:55.685","Text":"1/2 m v^2 equals kinetic energy."},{"Start":"03:55.685 ","End":"03:59.120","Text":"We\u0027re saying the final kinetic energy minus"},{"Start":"03:59.120 ","End":"04:02.870","Text":"the initial kinetic energy equals our change in kinetic energy."},{"Start":"04:02.870 ","End":"04:05.735","Text":"Now, this is of course on the x-axis only."},{"Start":"04:05.735 ","End":"04:07.205","Text":"But as we said before,"},{"Start":"04:07.205 ","End":"04:08.915","Text":"we\u0027re moving along the x-axis."},{"Start":"04:08.915 ","End":"04:13.165","Text":"There is no initial velocity and there is no final velocity in terms of y."},{"Start":"04:13.165 ","End":"04:14.480","Text":"Your y will drop out."},{"Start":"04:14.480 ","End":"04:16.070","Text":"There will be no velocity at all."},{"Start":"04:16.070 ","End":"04:18.275","Text":"You\u0027re left with only your x velocity."},{"Start":"04:18.275 ","End":"04:23.320","Text":"Therefore, what we have written here really equals the change in kinetic energy."},{"Start":"04:23.320 ","End":"04:28.805","Text":"By changing how we expressed our terms and using a small trick of Newton\u0027s second law,"},{"Start":"04:28.805 ","End":"04:34.580","Text":"we were able to prove that the work of the net force equals the change in kinetic energy."},{"Start":"04:34.580 ","End":"04:35.945","Text":"Now, this is a basic explanation."},{"Start":"04:35.945 ","End":"04:39.920","Text":"If anyone is interested in a more deep explanation of how this works,"},{"Start":"04:39.920 ","End":"04:41.480","Text":"please listen to the rest of the video."},{"Start":"04:41.480 ","End":"04:43.280","Text":"But for those who feel that this is enough,"},{"Start":"04:43.280 ","End":"04:45.725","Text":"you can stop here and skip the ending."},{"Start":"04:45.725 ","End":"04:47.420","Text":"I suggest listening to the ending,"},{"Start":"04:47.420 ","End":"04:49.235","Text":"but of course, it\u0027s your decision."},{"Start":"04:49.235 ","End":"04:50.990","Text":"On the right-hand side, I\u0027ll give"},{"Start":"04:50.990 ","End":"04:53.945","Text":"a more detailed explanation that might be a bit more exact."},{"Start":"04:53.945 ","End":"04:57.590","Text":"Let\u0027s assume we\u0027re finding the work of a non-constant force."},{"Start":"04:57.590 ","End":"05:05.570","Text":"The net force equals the integral of F or the net force dot dr."},{"Start":"05:05.570 ","End":"05:07.955","Text":"But of course, the net force is a force in every way,"},{"Start":"05:07.955 ","End":"05:09.530","Text":"so we can break it down into x,"},{"Start":"05:09.530 ","End":"05:10.860","Text":"y, and z components."},{"Start":"05:10.860 ","End":"05:20.835","Text":"We can write this as the integral of F_x dx plus the net force of y,"},{"Start":"05:20.835 ","End":"05:25.720","Text":"dy plus the net force of z, dz."},{"Start":"05:27.370 ","End":"05:30.755","Text":"Now we\u0027ll do the same trick we did before."},{"Start":"05:30.755 ","End":"05:32.555","Text":"Using Newton\u0027s second law."},{"Start":"05:32.555 ","End":"05:35.785","Text":"We\u0027ll replace each force with ma."},{"Start":"05:35.785 ","End":"05:44.840","Text":"We\u0027ll take out all the forces and be left with the following: the integral of ma_x,"},{"Start":"05:44.840 ","End":"05:48.605","Text":"dx plus ma_y,"},{"Start":"05:48.605 ","End":"05:53.495","Text":"dy plus ma_z, dz."},{"Start":"05:53.495 ","End":"05:55.670","Text":"Now we can do another trick."},{"Start":"05:55.670 ","End":"06:01.470","Text":"You know that acceleration equals dv over dt."},{"Start":"06:01.470 ","End":"06:06.800","Text":"So a_x equals dv_x over dt and the same for y and the same for z."},{"Start":"06:06.800 ","End":"06:10.370","Text":"We can replace all of our a\u0027s with dv over dt with"},{"Start":"06:10.370 ","End":"06:14.505","Text":"the respective sign and we can take our m to the outside."},{"Start":"06:14.505 ","End":"06:21.875","Text":"We have m times the integral of dv_x over dt times dx plus"},{"Start":"06:21.875 ","End":"06:32.010","Text":"dv_y over dt times dy plus dv_z over dt times dz."},{"Start":"06:32.270 ","End":"06:36.350","Text":"Now we\u0027re going to use a trick that some mathematicians are frowned upon,"},{"Start":"06:36.350 ","End":"06:39.520","Text":"but it\u0027s totally correct in terms of mathematic principles."},{"Start":"06:39.520 ","End":"06:41.790","Text":"We\u0027re going to talk about dv_x over"},{"Start":"06:41.790 ","End":"06:45.500","Text":"dt as though it\u0027s a normal quantity, not a differential."},{"Start":"06:45.500 ","End":"06:48.410","Text":"We\u0027re going to shift things around with dv_x,"},{"Start":"06:48.410 ","End":"06:51.275","Text":"dv_y, dv_z and it\u0027ll help us out."},{"Start":"06:51.275 ","End":"07:00.285","Text":"We can write this whole thing as m times the integral of dv_x times dx over dt."},{"Start":"07:00.285 ","End":"07:01.970","Text":"This is mathematically correct."},{"Start":"07:01.970 ","End":"07:08.330","Text":"We\u0027ve just moved things around a little plus dv_y times dy,"},{"Start":"07:08.330 ","End":"07:17.430","Text":"dt plus dv_z times dz over dt."},{"Start":"07:17.770 ","End":"07:20.225","Text":"Now that I\u0027ve rearranged things,"},{"Start":"07:20.225 ","End":"07:27.570","Text":"I can show that dx over dt equals v_x and dy over dt is v_y,"},{"Start":"07:27.570 ","End":"07:30.720","Text":"dz over dt is v_z."},{"Start":"07:30.720 ","End":"07:34.010","Text":"Even though this is an instance of division,"},{"Start":"07:34.010 ","End":"07:35.915","Text":"a division of something so small,"},{"Start":"07:35.915 ","End":"07:37.954","Text":"it turns into a derivative."},{"Start":"07:37.954 ","End":"07:42.590","Text":"We can write this as m and the integral of"},{"Start":"07:42.590 ","End":"07:51.640","Text":"v_x dv_x plus v_y dv_ y plus v_z dv_z."},{"Start":"07:51.640 ","End":"07:54.990","Text":"You can see that the v for each component is"},{"Start":"07:54.990 ","End":"07:57.780","Text":"paired with the dv for the same components: you have v_x,"},{"Start":"07:57.780 ","End":"08:01.770","Text":"with dv_x, v_y with dv_y, v_z for dv_z."},{"Start":"08:01.770 ","End":"08:05.225","Text":"Now when I perform the integral on v_x,"},{"Start":"08:05.225 ","End":"08:06.650","Text":"dv_x or v_y,"},{"Start":"08:06.650 ","End":"08:09.545","Text":"dv_y v_z, dv_z, etc."},{"Start":"08:09.545 ","End":"08:13.324","Text":"What I\u0027m left with is m on the outside."},{"Start":"08:13.324 ","End":"08:16.635","Text":"I\u0027m going to keep that on the outside. On the inside."},{"Start":"08:16.635 ","End":"08:21.530","Text":"I have v_x^2 over"},{"Start":"08:21.530 ","End":"08:29.390","Text":"2 and of course within the limits of v_x initial to v_x final."},{"Start":"08:29.390 ","End":"08:34.420","Text":"Plus the same for v_y, v_y^2 over 2."},{"Start":"08:34.420 ","End":"08:40.350","Text":"From the same limits, v_y initial to v_y final plus"},{"Start":"08:40.350 ","End":"08:49.575","Text":"v_z^2 over 2 from the limit of v_z initial to v_z final."},{"Start":"08:49.575 ","End":"08:54.335","Text":"You might already see that this is similar to the results on the left side."},{"Start":"08:54.335 ","End":"08:57.410","Text":"What we need to do is take our 1/2,"},{"Start":"08:57.410 ","End":"08:58.550","Text":"that\u0027s throughout our problem,"},{"Start":"08:58.550 ","End":"09:02.105","Text":"and take and put on the side with our m and put in our limits,"},{"Start":"09:02.105 ","End":"09:03.935","Text":"and we\u0027ll get the solution that we want."},{"Start":"09:03.935 ","End":"09:06.470","Text":"If we take our 1/2 out and put on the side,"},{"Start":"09:06.470 ","End":"09:14.985","Text":"we have 1/2 m times v final squared plus y final squared"},{"Start":"09:14.985 ","End":"09:20.520","Text":"plus v_z final squared minus"},{"Start":"09:20.520 ","End":"09:27.600","Text":"v_x initial squared minus v_y initial squared minus v_z initial squared."},{"Start":"09:27.600 ","End":"09:31.270","Text":"What you can see here is we really have the exact same thing we got before,"},{"Start":"09:31.270 ","End":"09:33.965","Text":"just with a more complicated process to get to it."},{"Start":"09:33.965 ","End":"09:38.630","Text":"Our end result is 1/2 of mv final squared,"},{"Start":"09:38.630 ","End":"09:40.790","Text":"which is our final kinetic energy,"},{"Start":"09:40.790 ","End":"09:45.350","Text":"minus 1/2 mv initial squared."},{"Start":"09:45.350 ","End":"09:47.300","Text":"That\u0027s our initial kinetic energy."},{"Start":"09:47.300 ","End":"09:52.130","Text":"We have the final kinetic energy minus the initial kinetic energy and that,"},{"Start":"09:52.130 ","End":"09:53.315","Text":"like we said before,"},{"Start":"09:53.315 ","End":"09:55.785","Text":"equals the difference in kinetic energy."},{"Start":"09:55.785 ","End":"09:58.730","Text":"Now we\u0027ve proven both ways that the work of"},{"Start":"09:58.730 ","End":"10:03.030","Text":"the net force equals the change in kinetic energy."}],"ID":9320},{"Watched":false,"Name":"Part 3- Reaching the Formula","Duration":"7m 23s","ChapterTopicVideoID":9048,"CourseChapterTopicPlaylistID":5422,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.400","Text":"In this part of the lecture,"},{"Start":"00:02.400 ","End":"00:04.065","Text":"I want to reach the formula,"},{"Start":"00:04.065 ","End":"00:06.735","Text":"the final formula for the work energy theorem."},{"Start":"00:06.735 ","End":"00:09.450","Text":"If you recall, that is that the work of"},{"Start":"00:09.450 ","End":"00:13.740","Text":"the non-conservative forces equals the change in total energy."},{"Start":"00:13.740 ","End":"00:18.075","Text":"We\u0027re going to do this using the things that we found in our last few lectures."},{"Start":"00:18.075 ","End":"00:20.655","Text":"For those who saw, for those who didn\u0027t,"},{"Start":"00:20.655 ","End":"00:23.790","Text":"here\u0027s a little bit of a review of the conclusions we came to."},{"Start":"00:23.790 ","End":"00:26.070","Text":"First, we said that the work of"},{"Start":"00:26.070 ","End":"00:31.725","Text":"the net force equals the sum of the work of each force individually."},{"Start":"00:31.725 ","End":"00:34.350","Text":"We could have 1 force, we could have 2 forces."},{"Start":"00:34.350 ","End":"00:35.910","Text":"We could really have any amount of forces,"},{"Start":"00:35.910 ","End":"00:37.410","Text":"but we can sum them all up,"},{"Start":"00:37.410 ","End":"00:41.265","Text":"and will equal the same thing as the work of the net force."},{"Start":"00:41.265 ","End":"00:48.050","Text":"Secondly, we found that the work of the net force equals the change in kinetic energy."},{"Start":"00:48.050 ","End":"00:49.385","Text":"Only kinetic energy."},{"Start":"00:49.385 ","End":"00:51.170","Text":"We\u0027re not talking about total energy."},{"Start":"00:51.170 ","End":"00:53.675","Text":"This is important and we\u0027ll incorporate it here as well."},{"Start":"00:53.675 ","End":"00:56.030","Text":"The last thing is something we talked about when"},{"Start":"00:56.030 ","End":"00:58.400","Text":"talking about the conservation of energy earlier."},{"Start":"00:58.400 ","End":"01:04.490","Text":"That was that the work of the conservative forces."},{"Start":"01:04.490 ","End":"01:08.315","Text":"Those are the forces that have conservation of energy."},{"Start":"01:08.315 ","End":"01:13.385","Text":"That equals the opposite of the change in potential energy."},{"Start":"01:13.385 ","End":"01:16.300","Text":"That is negative change in potential energy."},{"Start":"01:16.300 ","End":"01:20.930","Text":"Once again, the work of the conservative forces, and I\u0027ll write this here,"},{"Start":"01:20.930 ","End":"01:26.690","Text":"conservative equals negative Delta U."},{"Start":"01:26.690 ","End":"01:30.540","Text":"If any of these assumptions is not clear to you,"},{"Start":"01:30.540 ","End":"01:31.816","Text":"or you don\u0027t agree with them,"},{"Start":"01:31.816 ","End":"01:35.690","Text":"for now assume that they\u0027re true and watch this video then you can go back to"},{"Start":"01:35.690 ","End":"01:37.700","Text":"the other videos to make sure that you"},{"Start":"01:37.700 ","End":"01:41.730","Text":"understand the material on all this background information."},{"Start":"01:41.750 ","End":"01:47.030","Text":"Each of these conclusions is a tool that I\u0027m going to use to find my formula,"},{"Start":"01:47.030 ","End":"01:48.545","Text":"to reach the formula."},{"Start":"01:48.545 ","End":"01:53.555","Text":"What I need to do is finally reach W and C equals Delta"},{"Start":"01:53.555 ","End":"01:58.790","Text":"E. So the first step I\u0027m going to take is set out the formula below that."},{"Start":"01:58.790 ","End":"02:05.440","Text":"W sigma F equals Delta E_k."},{"Start":"02:05.440 ","End":"02:11.140","Text":"That is that the work of the net force equals the change in kinetic energy."},{"Start":"02:11.140 ","End":"02:14.555","Text":"Now I can replace W sigma F,"},{"Start":"02:14.555 ","End":"02:16.910","Text":"that is the work of the net force using"},{"Start":"02:16.910 ","End":"02:21.250","Text":"my second formula with the some of the work of each force individually."},{"Start":"02:21.250 ","End":"02:25.335","Text":"It would be the work of force 1 plus the work of force 2,"},{"Start":"02:25.335 ","End":"02:27.315","Text":"plus the work of force 3,"},{"Start":"02:27.315 ","End":"02:32.275","Text":"etc., equals Delta E_k."},{"Start":"02:32.275 ","End":"02:35.380","Text":"That is the change in kinetic energy."},{"Start":"02:35.380 ","End":"02:38.750","Text":"Again, I\u0027ll calculate the work of each force,"},{"Start":"02:38.750 ","End":"02:41.120","Text":"take the sum of all of those calculations,"},{"Start":"02:41.120 ","End":"02:43.640","Text":"and that will be equal to Delta Ek."},{"Start":"02:43.640 ","End":"02:46.610","Text":"Now if I\u0027m dealing with the work of all of my forces,"},{"Start":"02:46.610 ","End":"02:50.845","Text":"I know that some of my forces will be conservative and some will not."},{"Start":"02:50.845 ","End":"02:54.680","Text":"Let\u0027s say we split our forces here arbitrarily in this case,"},{"Start":"02:54.680 ","End":"02:59.570","Text":"but it works for the example into the conservative forces will be F_1 and F_2."},{"Start":"02:59.570 ","End":"03:07.010","Text":"These are C for conservative and F_3 through however many forces I have until F_n,"},{"Start":"03:07.010 ","End":"03:10.655","Text":"let\u0027s say our non-conservative NC."},{"Start":"03:10.655 ","End":"03:14.690","Text":"Now what I need to do here is remember my last formula,"},{"Start":"03:14.690 ","End":"03:16.880","Text":"the fourth formula from above."},{"Start":"03:16.880 ","End":"03:20.480","Text":"If you recall, the forces that are"},{"Start":"03:20.480 ","End":"03:24.620","Text":"conservative equal the opposite of the change in potential energy,"},{"Start":"03:24.620 ","End":"03:26.420","Text":"that is negative Delta U."},{"Start":"03:26.420 ","End":"03:28.550","Text":"On the right-hand side here,"},{"Start":"03:28.550 ","End":"03:31.040","Text":"I\u0027m going to rewrite this in a slightly different way."},{"Start":"03:31.040 ","End":"03:38.980","Text":"I\u0027m going to write this as negative Delta u_1 minus Delta u_2."},{"Start":"03:38.980 ","End":"03:41.195","Text":"These are from my forces that are conservative,"},{"Start":"03:41.195 ","End":"03:45.500","Text":"plus W_F_3 etc.,"},{"Start":"03:45.500 ","End":"03:47.810","Text":"for all of my non-conservative forces."},{"Start":"03:47.810 ","End":"03:51.135","Text":"This is equal to Delta E_k,"},{"Start":"03:51.135 ","End":"03:52.580","Text":"the change in kinetic energy."},{"Start":"03:52.580 ","End":"03:55.235","Text":"All I have done is replaced by conservative forces"},{"Start":"03:55.235 ","End":"03:58.490","Text":"with the change in potential energy that they represent."},{"Start":"03:58.490 ","End":"03:59.750","Text":"Now I\u0027m going to take all of"},{"Start":"03:59.750 ","End":"04:03.575","Text":"those potential energies and put them on the right side of the problem."},{"Start":"04:03.575 ","End":"04:06.035","Text":"I\u0027m left with W_F_3 etc.,"},{"Start":"04:06.035 ","End":"04:11.540","Text":"on my non-conservative forces equal Delta E_k plus Delta U."},{"Start":"04:11.540 ","End":"04:15.605","Text":"To remind you, Delta E_k equals"},{"Start":"04:15.605 ","End":"04:23.270","Text":"the final kinetic energy Ek_f minus Ek_i the initial kinetic energy."},{"Start":"04:23.270 ","End":"04:26.809","Text":"And of course, Delta U equals"},{"Start":"04:26.809 ","End":"04:32.430","Text":"the final potential energy minus the initial potential energy UF minus UI."},{"Start":"04:32.430 ","End":"04:35.750","Text":"I can rewrite this in a slightly different way than that order,"},{"Start":"04:35.750 ","End":"04:37.460","Text":"but I\u0027ll reorder it as follows."},{"Start":"04:37.460 ","End":"04:38.885","Text":"Take all my final things,"},{"Start":"04:38.885 ","End":"04:48.265","Text":"Ek_f plus U_f and subtract from them the initial things, Ek_i and U_i."},{"Start":"04:48.265 ","End":"04:53.810","Text":"I can put this in parentheses and say Ek_i plus U_i in parentheses and subtract it."},{"Start":"04:53.810 ","End":"04:58.865","Text":"Now we have is the work of the non-conservative forces equals everything on the right."},{"Start":"04:58.865 ","End":"05:01.595","Text":"The potential and kinetic energy of the final moment"},{"Start":"05:01.595 ","End":"05:04.345","Text":"minus the potential and kinetic energy of the initial moment."},{"Start":"05:04.345 ","End":"05:08.735","Text":"Now we can start talking about total energy, total mechanical energy."},{"Start":"05:08.735 ","End":"05:11.240","Text":"If you recall, the formula for"},{"Start":"05:11.240 ","End":"05:15.380","Text":"total mechanical energy is kinetic energy plus potential energy."},{"Start":"05:15.380 ","End":"05:19.400","Text":"You see here that we have 2 sets of kinetic plus potential energy."},{"Start":"05:19.400 ","End":"05:21.170","Text":"Let\u0027s write out this formula."},{"Start":"05:21.170 ","End":"05:25.460","Text":"Total energy equals kinetic energy plus potential energy."},{"Start":"05:25.460 ","End":"05:33.100","Text":"What we have here is the final total energy minus the initial total energy."},{"Start":"05:33.100 ","End":"05:36.590","Text":"If you see final minus initial,"},{"Start":"05:36.590 ","End":"05:39.590","Text":"that\u0027s a hint to you that we\u0027re talking about a change of some sort."},{"Start":"05:39.590 ","End":"05:41.990","Text":"This equals the change in total energy."},{"Start":"05:41.990 ","End":"05:48.870","Text":"We can now do is rewrite this whole equation as the work of F_3 and onwards."},{"Start":"05:48.870 ","End":"05:51.215","Text":"That is the non-conservative forces"},{"Start":"05:51.215 ","End":"05:55.955","Text":"equals the change in total energy. Let\u0027s write that out."},{"Start":"05:55.955 ","End":"05:59.720","Text":"W_NC that is the non-conservative work,"},{"Start":"05:59.720 ","End":"06:03.005","Text":"equals Delta E_t,"},{"Start":"06:03.005 ","End":"06:05.660","Text":"the change in total energy."},{"Start":"06:05.660 ","End":"06:08.630","Text":"We have a change in total energy because"},{"Start":"06:08.630 ","End":"06:11.545","Text":"the final kinetic energy minus the initial kinetic energy,"},{"Start":"06:11.545 ","End":"06:15.430","Text":"and the final potential energy minus the initial potential energy."},{"Start":"06:15.430 ","End":"06:18.110","Text":"These are the 2 elements of total energy."},{"Start":"06:18.110 ","End":"06:21.685","Text":"What does the work energy theorem mean for us?"},{"Start":"06:21.685 ","End":"06:25.855","Text":"Well, it\u0027s a way of switching back and forth between these 2 red formulas,"},{"Start":"06:25.855 ","End":"06:27.790","Text":"the formulas in red boxes."},{"Start":"06:27.790 ","End":"06:32.710","Text":"The work of the net force is a good way of showing us the change in kinetic energy."},{"Start":"06:32.710 ","End":"06:35.560","Text":"But we can also then separate our forces into"},{"Start":"06:35.560 ","End":"06:38.200","Text":"conservative and non-conservative forces and take"},{"Start":"06:38.200 ","End":"06:41.725","Text":"the conservative forces that equal negative Delta U"},{"Start":"06:41.725 ","End":"06:46.105","Text":"and move them from the left side of the equation to the right side of the equation."},{"Start":"06:46.105 ","End":"06:48.280","Text":"Now we have our potential energy and"},{"Start":"06:48.280 ","End":"06:50.875","Text":"our kinetic energy on the right side of the equation,"},{"Start":"06:50.875 ","End":"06:54.080","Text":"and we can put them together to find Delta E_t,"},{"Start":"06:54.080 ","End":"06:56.020","Text":"the change in total energy."},{"Start":"06:56.020 ","End":"06:58.550","Text":"Once we have the change in total energy,"},{"Start":"06:58.550 ","End":"07:01.715","Text":"we can calculate just the non-conservative forces."},{"Start":"07:01.715 ","End":"07:03.860","Text":"There\u0027s your solution, there\u0027s your explanation,"},{"Start":"07:03.860 ","End":"07:05.645","Text":"and that\u0027s how we reached the formula."},{"Start":"07:05.645 ","End":"07:07.985","Text":"For those who understand the mathematics,"},{"Start":"07:07.985 ","End":"07:09.800","Text":"I think it\u0027s a really good way to get"},{"Start":"07:09.800 ","End":"07:12.370","Text":"a better understanding of what\u0027s happening in general."},{"Start":"07:12.370 ","End":"07:13.760","Text":"If you\u0027re up for it,"},{"Start":"07:13.760 ","End":"07:17.690","Text":"I think you should try to develop the formulas to this point on your own."},{"Start":"07:17.690 ","End":"07:20.690","Text":"It can be a very good exercise that will increase your understanding."},{"Start":"07:20.690 ","End":"07:24.120","Text":"That\u0027s the end of the explanation. Thanks for listening."}],"ID":9321}],"Thumbnail":null,"ID":5422},{"Name":"Calculating Conservative Forces From Potential Energy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Conservative Force and Potential Energy","Duration":"3m 21s","ChapterTopicVideoID":9049,"CourseChapterTopicPlaylistID":5423,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.285","Text":"Hello. Now we\u0027re going to talk about conservative forces and potential energy."},{"Start":"00:06.285 ","End":"00:08.295","Text":"What is a conservative force?"},{"Start":"00:08.295 ","End":"00:10.935","Text":"Well, conservative force has 3 main characteristics."},{"Start":"00:10.935 ","End":"00:17.985","Text":"First, the force of the work is not dependent on the trajectory."},{"Start":"00:17.985 ","End":"00:22.545","Text":"That means that if we\u0027re going from point A to point B over here,"},{"Start":"00:22.545 ","End":"00:25.560","Text":"it doesn\u0027t matter how we get from point A to point B."},{"Start":"00:25.560 ","End":"00:30.315","Text":"We can go straight or we can go in some whirling path like so."},{"Start":"00:30.315 ","End":"00:34.860","Text":"The calculation of the work will not change,"},{"Start":"00:34.860 ","End":"00:36.770","Text":"will always be the same."},{"Start":"00:36.770 ","End":"00:38.990","Text":"When I\u0027m talking about the calculation of the work,"},{"Start":"00:38.990 ","End":"00:40.520","Text":"what I mean is that W,"},{"Start":"00:40.520 ","End":"00:45.535","Text":"the work equals the integral on the force times dr,"},{"Start":"00:45.535 ","End":"00:48.510","Text":"and of course, this is from point A to point B so"},{"Start":"00:48.510 ","End":"00:51.930","Text":"I\u0027m talking about the work from point A to point B."},{"Start":"00:51.930 ","End":"00:54.680","Text":"Our first characteristic is that it\u0027s a force where"},{"Start":"00:54.680 ","End":"00:57.919","Text":"the work is not dependent on the trajectory."},{"Start":"00:57.919 ","End":"01:00.950","Text":"The second characteristic is that it\u0027s a force where"},{"Start":"01:00.950 ","End":"01:04.670","Text":"the work of a trajectory that goes back to the initial point."},{"Start":"01:04.670 ","End":"01:07.390","Text":"From A to A would equal 0."},{"Start":"01:07.390 ","End":"01:12.990","Text":"In this case, where we have a trajectory going from A back to A, the work equals 0."},{"Start":"01:13.730 ","End":"01:18.950","Text":"These are your first 2 characteristics of a conservative force."},{"Start":"01:18.950 ","End":"01:20.660","Text":"Our last characteristic of"},{"Start":"01:20.660 ","End":"01:24.545","Text":"a conservative force is that it\u0027s a force with potential energy."},{"Start":"01:24.545 ","End":"01:28.805","Text":"I\u0027m not going to explain yet how we find the potential energy of a conservative force."},{"Start":"01:28.805 ","End":"01:31.430","Text":"But let\u0027s talk about it in terms of an example."},{"Start":"01:31.430 ","End":"01:34.805","Text":"Let\u0027s say that you have a F=mg,"},{"Start":"01:34.805 ","End":"01:39.410","Text":"that is the force of gravity times mass."},{"Start":"01:39.410 ","End":"01:42.155","Text":"Now if we\u0027re talking about our potential energy,"},{"Start":"01:42.155 ","End":"01:43.490","Text":"which we write as U,"},{"Start":"01:43.490 ","End":"01:50.270","Text":"U_ g=mg times h. H is the height above the ground."},{"Start":"01:50.270 ","End":"01:55.565","Text":"You may be asking why this is important and it basically helps us with a shortcut."},{"Start":"01:55.565 ","End":"01:59.090","Text":"A conservative force is defined by,"},{"Start":"01:59.090 ","End":"02:01.070","Text":"again these 3 characteristics."},{"Start":"02:01.070 ","End":"02:03.500","Text":"One of which is that it\u0027s only defined by"},{"Start":"02:03.500 ","End":"02:06.430","Text":"its initial point and its final point or terminal point."},{"Start":"02:06.430 ","End":"02:09.530","Text":"Meaning we don\u0027t care necessarily about the trajectory itself."},{"Start":"02:09.530 ","End":"02:17.180","Text":"Instead of taking a derivative of the force that is to find our work,"},{"Start":"02:17.180 ","End":"02:21.065","Text":"what we can do is instead think of the work from A to B"},{"Start":"02:21.065 ","End":"02:25.580","Text":"in terms of the negative change in the potential energy."},{"Start":"02:25.580 ","End":"02:27.290","Text":"How do we calculate that?"},{"Start":"02:27.290 ","End":"02:31.340","Text":"It\u0027s negative UB, which is the potential energy at B,"},{"Start":"02:31.340 ","End":"02:33.380","Text":"the terminal point, minus UA,"},{"Start":"02:33.380 ","End":"02:35.680","Text":"the potential energy at the initial point."},{"Start":"02:35.680 ","End":"02:37.400","Text":"In the case of conservative force,"},{"Start":"02:37.400 ","End":"02:39.920","Text":"I don\u0027t have to calculate my work in the classic way,"},{"Start":"02:39.920 ","End":"02:42.650","Text":"I can use this formula, negative UB,"},{"Start":"02:42.650 ","End":"02:46.550","Text":"meaning my potential energy at the final point minus UA,"},{"Start":"02:46.550 ","End":"02:50.750","Text":"my potential energy at the initial point equals w equals the work."},{"Start":"02:50.750 ","End":"02:54.080","Text":"This is one of our formulas that we should save so let\u0027s put"},{"Start":"02:54.080 ","End":"02:57.500","Text":"it in a red box and you should put this on your formula sheet."},{"Start":"02:57.500 ","End":"03:02.540","Text":"Now, 1 more conservative force that\u0027s important to know is the elastic force."},{"Start":"03:02.540 ","End":"03:06.650","Text":"This is the force of a spring and it\u0027s defined as negative"},{"Start":"03:06.650 ","End":"03:11.045","Text":"k Delta x. Delta x is the elongation of the spring,"},{"Start":"03:11.045 ","End":"03:21.550","Text":"how far it can extend and the potential energy is defined as 1.5k Delta x^2."}],"ID":9322},{"Watched":false,"Name":"The Force as a Gradient of Potential Energy","Duration":"3m 24s","ChapterTopicVideoID":9050,"CourseChapterTopicPlaylistID":5423,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.875","Text":"What\u0027s the relationship between force and potential energy?"},{"Start":"00:04.875 ","End":"00:08.010","Text":"Well, there\u0027s a formula that will help us understand this."},{"Start":"00:08.010 ","End":"00:10.575","Text":"The formula will go in our formula sheet."},{"Start":"00:10.575 ","End":"00:12.040","Text":"We\u0027ll put it in red here."},{"Start":"00:12.040 ","End":"00:13.565","Text":"It is as follows."},{"Start":"00:13.565 ","End":"00:20.540","Text":"F, the force, equals negative gradient times u. I\u0027ll explain that in a second,"},{"Start":"00:20.540 ","End":"00:24.785","Text":"but for now just let\u0027s remember the formula and put it into our formula sheet."},{"Start":"00:24.785 ","End":"00:29.360","Text":"If you recall, a gradient is a directional derivative."},{"Start":"00:29.360 ","End":"00:32.420","Text":"We take partial derivatives for each of the components x,"},{"Start":"00:32.420 ","End":"00:34.205","Text":"y, and z, and it looks like this."},{"Start":"00:34.205 ","End":"00:40.970","Text":"F equals the negative partial derivative of u/x, that\u0027s the x-component,"},{"Start":"00:40.970 ","End":"00:44.210","Text":"x-hat minus the partial derivative of"},{"Start":"00:44.210 ","End":"00:50.470","Text":"u/y y hat minus the partial derivative of u/z z-hat."},{"Start":"00:50.470 ","End":"00:54.340","Text":"That will give us each component of our force."},{"Start":"00:54.340 ","End":"00:56.960","Text":"Of course, this is for Cartesian coordinates."},{"Start":"00:56.960 ","End":"01:01.415","Text":"There\u0027s a slightly different formula for cylindrical and spherical coordinates,"},{"Start":"01:01.415 ","End":"01:03.140","Text":"which is already on your formula sheet,"},{"Start":"01:03.140 ","End":"01:04.715","Text":"but we\u0027ll deal with that later."},{"Start":"01:04.715 ","End":"01:08.690","Text":"Generally, what you see here is that we\u0027re getting each variable,"},{"Start":"01:08.690 ","End":"01:10.415","Text":"x, y, and z isolated."},{"Start":"01:10.415 ","End":"01:12.680","Text":"If we want just the x variable, for example,"},{"Start":"01:12.680 ","End":"01:15.305","Text":"we do the partial derivative of u/x,"},{"Start":"01:15.305 ","End":"01:18.890","Text":"and that would give us our x variable for Fx."},{"Start":"01:18.890 ","End":"01:22.880","Text":"If we wanted Fy, we would do partial derivative of u/y."},{"Start":"01:22.880 ","End":"01:25.030","Text":"Most of the time,"},{"Start":"01:25.030 ","End":"01:28.330","Text":"my potential energy will only be dependent on 1 variable,"},{"Start":"01:28.330 ","End":"01:30.860","Text":"so then we can just do a derivative of"},{"Start":"01:30.860 ","End":"01:34.770","Text":"that 1 variable instead of doing partial derivatives of each element."},{"Start":"01:34.770 ","End":"01:39.335","Text":"For example, if my potential energy was dependent on x,"},{"Start":"01:39.335 ","End":"01:42.785","Text":"then I would just do a derivative of x to find my answer."},{"Start":"01:42.785 ","End":"01:45.680","Text":"We can look at the example of the potential energy of"},{"Start":"01:45.680 ","End":"01:50.130","Text":"gravity to give us a better understanding of this."},{"Start":"01:50.380 ","End":"01:54.650","Text":"U=mgh, and we\u0027ll set h=y because gravity works vertically."},{"Start":"01:54.650 ","End":"02:00.450","Text":"We have some mass resting y above another surface and it could fall."},{"Start":"02:00.450 ","End":"02:01.970","Text":"The force is equal to"},{"Start":"02:01.970 ","End":"02:06.980","Text":"the negative derivative of u/y, because we don\u0027t have to deal with x,"},{"Start":"02:06.980 ","End":"02:08.000","Text":"we don\u0027t have to deal with z,"},{"Start":"02:08.000 ","End":"02:10.220","Text":"it\u0027s just the derivative of u/y,"},{"Start":"02:10.220 ","End":"02:15.150","Text":"of course, negative, so that equals negative mgy-hat."},{"Start":"02:15.150 ","End":"02:17.930","Text":"What you end up with is potential downwards energy."},{"Start":"02:17.930 ","End":"02:20.600","Text":"You can see the same thing for the potential elastic energy,"},{"Start":"02:20.600 ","End":"02:24.125","Text":"if you recall the formula is 1/2k times Delta x^2."},{"Start":"02:24.125 ","End":"02:27.170","Text":"Delta x is really some point x minus another"},{"Start":"02:27.170 ","End":"02:30.965","Text":"initial point x where your spring is entirely limp."},{"Start":"02:30.965 ","End":"02:33.200","Text":"If this is our spring,"},{"Start":"02:33.200 ","End":"02:37.550","Text":"there\u0027s some point x_0 where the spring is totally limp,"},{"Start":"02:37.550 ","End":"02:41.195","Text":"and there\u0027s another point x where it is now."},{"Start":"02:41.195 ","End":"02:44.545","Text":"That distance is how much it can elongate."},{"Start":"02:44.545 ","End":"02:47.070","Text":"When we do the derivative,"},{"Start":"02:47.070 ","End":"02:49.070","Text":"it\u0027s only in terms of x, we don\u0027t have to deal with y,"},{"Start":"02:49.070 ","End":"02:50.450","Text":"we don\u0027t have to deal with z."},{"Start":"02:50.450 ","End":"02:55.190","Text":"What we\u0027re left with is the force equals the negative derivative of u/x."},{"Start":"02:55.190 ","End":"03:00.020","Text":"What that equals, once our 1/2 and power of 2 fall out when we do the procedure,"},{"Start":"03:00.020 ","End":"03:03.645","Text":"is negative k times x minus x_0,"},{"Start":"03:03.645 ","End":"03:05.070","Text":"which is the change in x."},{"Start":"03:05.070 ","End":"03:07.235","Text":"We get negative k Delta x,"},{"Start":"03:07.235 ","End":"03:09.365","Text":"which is our formula on the left."},{"Start":"03:09.365 ","End":"03:11.930","Text":"Just to recap, we\u0027ve proven here is we can take"},{"Start":"03:11.930 ","End":"03:15.530","Text":"the negative gradient of the potential energy to find our force."},{"Start":"03:15.530 ","End":"03:17.090","Text":"Of course, in most situations,"},{"Start":"03:17.090 ","End":"03:21.715","Text":"that just means a simple derivative in the direction of x or y or whatever it may be."},{"Start":"03:21.715 ","End":"03:24.430","Text":"That\u0027s the short explanation for now."}],"ID":9323}],"Thumbnail":null,"ID":5423},{"Name":"How To Check If A Force Is Conservative","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation","Duration":"3m 22s","ChapterTopicVideoID":9051,"CourseChapterTopicPlaylistID":5424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.695","Text":"Hello. In this video,"},{"Start":"00:01.695 ","End":"00:06.645","Text":"I\u0027m going to discuss how we check if a force in a given equation is a conservative force."},{"Start":"00:06.645 ","End":"00:10.590","Text":"Now the way you do it is with the following mathematical equation."},{"Start":"00:10.590 ","End":"00:12.435","Text":"You take your curl equation,"},{"Start":"00:12.435 ","End":"00:15.675","Text":"which if you recall is the nabla times the force,"},{"Start":"00:15.675 ","End":"00:17.685","Text":"and set it equal to 0."},{"Start":"00:17.685 ","End":"00:23.205","Text":"If you\u0027re curl or rotor rotation equal 0 then you have a conservative force."},{"Start":"00:23.205 ","End":"00:27.990","Text":"Setting the rotor or curl function equal to 0 is"},{"Start":"00:27.990 ","End":"00:32.610","Text":"by far the most common way of checking if we have a conservative force."},{"Start":"00:32.610 ","End":"00:36.195","Text":"That has to do with the definition of a conservative force."},{"Start":"00:36.195 ","End":"00:39.930","Text":"If you recall 3 possible characteristics of a conservative force."},{"Start":"00:39.930 ","End":"00:43.770","Text":"The 1st was that in a trajectory from A to B,"},{"Start":"00:43.770 ","End":"00:46.155","Text":"the work is not depended on the trajectory."},{"Start":"00:46.155 ","End":"00:50.420","Text":"The 2nd was that in a closed trajectory, the work equals 0."},{"Start":"00:50.420 ","End":"00:55.445","Text":"Now neither of these is a good way to check if we have a conservative force."},{"Start":"00:55.445 ","End":"01:00.410","Text":"Because even if we check 1 trajectory or 100 trajectories there\u0027s still"},{"Start":"01:00.410 ","End":"01:03.230","Text":"infinite possibilities out there and we can\u0027t be 100 percent"},{"Start":"01:03.230 ","End":"01:07.130","Text":"certain that every single trajectory equals 0."},{"Start":"01:07.130 ","End":"01:10.490","Text":"The 2nd way we can determine whether we have"},{"Start":"01:10.490 ","End":"01:14.150","Text":"a conservative force is if we know that there\u0027s potential energy."},{"Start":"01:14.150 ","End":"01:17.360","Text":"If you recall that\u0027s the 3rd characteristic of a conservative force,"},{"Start":"01:17.360 ","End":"01:19.280","Text":"is a force of potential energy."},{"Start":"01:19.280 ","End":"01:21.590","Text":"Now, if we can prove that we have"},{"Start":"01:21.590 ","End":"01:24.370","Text":"potential energy and we\u0027ll talk about that in the next lecture."},{"Start":"01:24.370 ","End":"01:27.065","Text":"Or if we\u0027re given that there\u0027s potential energy,"},{"Start":"01:27.065 ","End":"01:31.520","Text":"we can therefore determine that we have a conservative force."},{"Start":"01:31.520 ","End":"01:35.125","Text":"But for now, let\u0027s focus on this mathematical method."},{"Start":"01:35.125 ","End":"01:39.395","Text":"How are we going to do this mathematical operation?"},{"Start":"01:39.395 ","End":"01:42.230","Text":"What I want to do is actually paste here"},{"Start":"01:42.230 ","End":"01:45.320","Text":"a set of formulas from Wikipedia that students generally use."},{"Start":"01:45.320 ","End":"01:48.530","Text":"This might be helpful for you in the future."},{"Start":"01:48.530 ","End":"01:52.640","Text":"Here\u0027s the table that may be a little intimidating, but bear with me."},{"Start":"01:52.640 ","End":"01:56.420","Text":"You can tell that we have a few different operations here,"},{"Start":"01:56.420 ","End":"01:59.770","Text":"and we have them described in different coordinate systems."},{"Start":"01:59.770 ","End":"02:04.190","Text":"While we may sometimes use spherical coordinates and cylindrical coordinates for now,"},{"Start":"02:04.190 ","End":"02:07.295","Text":"let\u0027s focus on Cartesian coordinates it\u0027s what will use most."},{"Start":"02:07.295 ","End":"02:09.950","Text":"You can see that A is our vector field,"},{"Start":"02:09.950 ","End":"02:11.510","Text":"meaning it has x,"},{"Start":"02:11.510 ","End":"02:13.070","Text":"y, and z dimensions."},{"Start":"02:13.070 ","End":"02:19.310","Text":"In Cartesian coordinates or in cylindrical coordinates it could be in terms of r,"},{"Start":"02:19.310 ","End":"02:21.140","Text":"c, and z."},{"Start":"02:21.140 ","End":"02:23.240","Text":"You see the gradient function,"},{"Start":"02:23.240 ","End":"02:24.965","Text":"we\u0027re not interested in that at the moment."},{"Start":"02:24.965 ","End":"02:26.600","Text":"We see the divergence function,"},{"Start":"02:26.600 ","End":"02:29.105","Text":"we\u0027re not interested in that at the moment either."},{"Start":"02:29.105 ","End":"02:33.350","Text":"Below you have the curl function and that\u0027s what we\u0027re going to talk about now."},{"Start":"02:33.350 ","End":"02:37.020","Text":"Just to reiterate curl and rotor mean the same thing."},{"Start":"02:37.020 ","End":"02:41.390","Text":"If you see 1 or the other they\u0027re interchangeable don\u0027t be confused by that."},{"Start":"02:41.390 ","End":"02:44.195","Text":"Now, in our example we used F,"},{"Start":"02:44.195 ","End":"02:45.595","Text":"here they\u0027re using A."},{"Start":"02:45.595 ","End":"02:47.640","Text":"They mean the same thing essentially."},{"Start":"02:47.640 ","End":"02:49.940","Text":"What we\u0027re going to do is multiply these in"},{"Start":"02:49.940 ","End":"02:53.180","Text":"partial differentials by our different coordinates to get our answers."},{"Start":"02:53.180 ","End":"02:55.495","Text":"For the x coordinate it\u0027s going to be partial d, a,"},{"Start":"02:55.495 ","End":"02:58.280","Text":"z over partial y,"},{"Start":"02:58.280 ","End":"03:00.665","Text":"minus partial d,"},{"Start":"03:00.665 ","End":"03:02.570","Text":"a, y over partial z."},{"Start":"03:02.570 ","End":"03:07.255","Text":"You can see how it continues with the y and with the z coordinates."},{"Start":"03:07.255 ","End":"03:12.530","Text":"Just a quick note on this sheet you\u0027ll see in cylindrical coordinates, 1 over rho."},{"Start":"03:12.530 ","End":"03:14.060","Text":"Just a reminder in our course,"},{"Start":"03:14.060 ","End":"03:16.490","Text":"we\u0027re calling that r. We\u0027re calling that variable r,"},{"Start":"03:16.490 ","End":"03:18.860","Text":"so it will be 1 over r. Let\u0027s"},{"Start":"03:18.860 ","End":"03:22.500","Text":"look at an example and see how we can solve a problem like this."}],"ID":9324},{"Watched":false,"Name":"Example","Duration":"1m 45s","ChapterTopicVideoID":9052,"CourseChapterTopicPlaylistID":5424,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.290","Text":"In this example, we\u0027re given the force F,"},{"Start":"00:04.290 ","End":"00:08.460","Text":"which is 2xy in the direction of x plus"},{"Start":"00:08.460 ","End":"00:13.620","Text":"(x^2+z) in the direction of y plus y in the direction of z."},{"Start":"00:13.620 ","End":"00:17.205","Text":"We need to determine if it\u0027s a conservative force."},{"Start":"00:17.205 ","End":"00:20.610","Text":"The way we do that is with our Curl function,"},{"Start":"00:20.610 ","End":"00:22.050","Text":"as you can see listed here,"},{"Start":"00:22.050 ","End":"00:24.435","Text":"and we just need to do the operation."},{"Start":"00:24.435 ","End":"00:26.145","Text":"So let\u0027s jump right into it."},{"Start":"00:26.145 ","End":"00:28.710","Text":"For our x element, we\u0027ll do the following."},{"Start":"00:28.710 ","End":"00:31.260","Text":"We will take dA_z/dy,"},{"Start":"00:31.260 ","End":"00:35.478","Text":"and that accounts for the y in our z element that equals 1."},{"Start":"00:35.478 ","End":"00:39.000","Text":"We\u0027re subtract from that dA_y/dz,"},{"Start":"00:39.000 ","End":"00:41.516","Text":"that is the z\u0027s in our y-direction."},{"Start":"00:41.516 ","End":"00:43.865","Text":"That also equals 1, that zeroes out."},{"Start":"00:43.865 ","End":"00:46.670","Text":"Now, this is a good time to mention that every element has to"},{"Start":"00:46.670 ","End":"00:50.465","Text":"individually even out because each of these coordinates,"},{"Start":"00:50.465 ","End":"00:53.000","Text":"x, y, and z, represents a different direction."},{"Start":"00:53.000 ","End":"00:55.040","Text":"If 1 direction equals more than 0,"},{"Start":"00:55.040 ","End":"00:57.245","Text":"then the whole equation doesn\u0027t equal 0."},{"Start":"00:57.245 ","End":"01:01.430","Text":"Moving on to our y portion, we add that in,"},{"Start":"01:01.430 ","End":"01:03.415","Text":"as you can see here,"},{"Start":"01:03.415 ","End":"01:09.180","Text":"and we\u0027re going to do dA_x/dz, that equals 0."},{"Start":"01:09.180 ","End":"01:13.785","Text":"Similarly, dA_z/dx equals 0."},{"Start":"01:13.785 ","End":"01:16.370","Text":"So that obviously will zero out,"},{"Start":"01:16.370 ","End":"01:18.170","Text":"and we\u0027re 2/3 of our way there."},{"Start":"01:18.170 ","End":"01:19.880","Text":"Now we add our z element in,"},{"Start":"01:19.880 ","End":"01:24.260","Text":"that\u0027s dA_y/dx and that equals 2x."},{"Start":"01:24.260 ","End":"01:27.395","Text":"We\u0027re going to subtract from that dA_x/dy,"},{"Start":"01:27.395 ","End":"01:29.695","Text":"that also equals 2x,"},{"Start":"01:29.695 ","End":"01:31.865","Text":"and that zeroes out as well."},{"Start":"01:31.865 ","End":"01:33.965","Text":"We can do our full calculation,"},{"Start":"01:33.965 ","End":"01:36.365","Text":"x, y, and z each equal 0."},{"Start":"01:36.365 ","End":"01:40.089","Text":"We can determine that this is in fact a conservative force."},{"Start":"01:40.089 ","End":"01:41.255","Text":"That\u0027s it."},{"Start":"01:41.255 ","End":"01:45.360","Text":"That\u0027s an example to go with your explanation. It\u0027s that simple."}],"ID":9325}],"Thumbnail":null,"ID":5424},{"Name":"Calculating Potential Energy From Conservative Forces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation","Duration":"2m 32s","ChapterTopicVideoID":9036,"CourseChapterTopicPlaylistID":5425,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9036.jpeg","UploadDate":"2018-01-17T11:32:18.5570000","DurationForVideoObject":"PT2M32S","Description":null,"MetaTitle":"Explanation: Video + Workbook | Proprep","MetaDescription":"Work and Energy - 10. Calculating Potential Energy from Conservative Forces. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/work-and-energy/10.-calculating-potential-energy-from-conservative-forces/vid9309","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.475","Text":"Hello. In this video,"},{"Start":"00:02.475 ","End":"00:06.645","Text":"I want to find the potential energy of a conservative force."},{"Start":"00:06.645 ","End":"00:10.245","Text":"This is something you may be familiar with from mathematics classes."},{"Start":"00:10.245 ","End":"00:17.020","Text":"In math class, they\u0027ll often call it finding the indefinite integral of the force."},{"Start":"00:17.120 ","End":"00:22.335","Text":"First things first, you have to make sure that you in fact have a conservative force."},{"Start":"00:22.335 ","End":"00:25.050","Text":"If you don\u0027t have a conservative force, this is irrelevant."},{"Start":"00:25.050 ","End":"00:28.620","Text":"The way to check if you have a conservative force is to take"},{"Start":"00:28.620 ","End":"00:32.400","Text":"the rotor of the force and check if that equals 0."},{"Start":"00:32.400 ","End":"00:35.090","Text":"In general, exercises dealing with"},{"Start":"00:35.090 ","End":"00:39.080","Text":"the potential energy of conservative forces will first ask you to verify this."},{"Start":"00:39.080 ","End":"00:41.360","Text":"It\u0027s important to remember how to do this."},{"Start":"00:41.360 ","End":"00:44.585","Text":"Once you know that you have a conservative force,"},{"Start":"00:44.585 ","End":"00:49.640","Text":"you can then try to find what the potential energy of that conservative force is."},{"Start":"00:49.640 ","End":"00:52.265","Text":"Remember, potential energy is signified with U,"},{"Start":"00:52.265 ","End":"00:55.025","Text":"so to find out what U equals."},{"Start":"00:55.025 ","End":"00:58.160","Text":"The whole point of this lecture is to solve for U."},{"Start":"00:58.160 ","End":"01:00.455","Text":"It\u0027s a little more complicated than it looks."},{"Start":"01:00.455 ","End":"01:02.015","Text":"To be honest, it seems simple,"},{"Start":"01:02.015 ","End":"01:04.160","Text":"but there\u0027s a few steps that can be complicated,"},{"Start":"01:04.160 ","End":"01:06.125","Text":"so please pay close attention."},{"Start":"01:06.125 ","End":"01:09.620","Text":"The first thing to note is that the relationship between"},{"Start":"01:09.620 ","End":"01:13.190","Text":"the force and the potential energy is such that"},{"Start":"01:13.190 ","End":"01:14.960","Text":"the force has to equal"},{"Start":"01:14.960 ","End":"01:20.875","Text":"the negative gradient of the potential energy, negative gradient U."},{"Start":"01:20.875 ","End":"01:23.780","Text":"Now what we want to do is extrapolate out"},{"Start":"01:23.780 ","End":"01:26.225","Text":"the gradient and write it in terms of each component,"},{"Start":"01:26.225 ","End":"01:28.055","Text":"x, y, and z."},{"Start":"01:28.055 ","End":"01:30.095","Text":"We can write it as follows."},{"Start":"01:30.095 ","End":"01:36.990","Text":"F equals negative du/dx in the direction of x,"},{"Start":"01:36.990 ","End":"01:41.280","Text":"plus du/dy in the direction of y,"},{"Start":"01:41.280 ","End":"01:45.615","Text":"and du/dz in the direction of z."},{"Start":"01:45.615 ","End":"01:49.565","Text":"In fact, we\u0027re now taking something that was a scalar multiplication,"},{"Start":"01:49.565 ","End":"01:52.910","Text":"something without direction, and we\u0027ve given it direction in terms of x,"},{"Start":"01:52.910 ","End":"01:55.270","Text":"y, and z, Cartesian coordinates."},{"Start":"01:55.270 ","End":"01:57.500","Text":"Now we can take each of these x,"},{"Start":"01:57.500 ","End":"01:59.990","Text":"y, and z and write in terms of the force."},{"Start":"01:59.990 ","End":"02:03.185","Text":"There\u0027s a force of x, force of y, and a force of z."},{"Start":"02:03.185 ","End":"02:08.240","Text":"The force of x equals du/d/x."},{"Start":"02:08.240 ","End":"02:12.585","Text":"The force of z equals negative du/dy,"},{"Start":"02:12.585 ","End":"02:14.930","Text":"and of course, the negative applies to the x as well."},{"Start":"02:14.930 ","End":"02:21.430","Text":"F_z equals negative du/dz."},{"Start":"02:21.430 ","End":"02:25.130","Text":"Now we can see the relationship between the force and the energy."},{"Start":"02:25.130 ","End":"02:28.190","Text":"From here, what I\u0027ll do is use an example to show how"},{"Start":"02:28.190 ","End":"02:32.310","Text":"we find the potential energy of a conservative force."}],"ID":9309},{"Watched":false,"Name":"Example","Duration":"7m 39s","ChapterTopicVideoID":9037,"CourseChapterTopicPlaylistID":5425,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.620","Text":"In our example, you\u0027re asked to find the potential energy of the force"},{"Start":"00:04.620 ","End":"00:09.165","Text":"F equals negative 2xy in the direction of x,"},{"Start":"00:09.165 ","End":"00:14.415","Text":"plus 2 minus (x^2) in the direction of y."},{"Start":"00:14.415 ","End":"00:18.285","Text":"You\u0027re given that the energy at 0,0,"},{"Start":"00:18.285 ","End":"00:20.160","Text":"the origin equals 0."},{"Start":"00:20.160 ","End":"00:23.205","Text":"For now, we\u0027re going to forget about this given piece of data,"},{"Start":"00:23.205 ","End":"00:24.645","Text":"we\u0027ll come back to it later."},{"Start":"00:24.645 ","End":"00:28.830","Text":"But for now, let\u0027s first try to find the potential energy of the force."},{"Start":"00:28.830 ","End":"00:34.185","Text":"What I\u0027m looking for is some force that\u0027s going to fill these requirements below that"},{"Start":"00:34.185 ","End":"00:39.285","Text":"F_x equals negative dU dx F_y equals negative dU dy."},{"Start":"00:39.285 ","End":"00:42.530","Text":"In this case, there\u0027s no F_ z because we\u0027re dealing with a simpler problem."},{"Start":"00:42.530 ","End":"00:43.670","Text":"But if there were,"},{"Start":"00:43.670 ","End":"00:45.815","Text":"F_z would have to follow that as well."},{"Start":"00:45.815 ","End":"00:49.910","Text":"This is really almost impossible to do unless you\u0027re fulfilling"},{"Start":"00:49.910 ","End":"00:54.410","Text":"this first requirement that the gradient of the force equals 0."},{"Start":"00:54.410 ","End":"00:56.690","Text":"In that case, you will be able to find something,"},{"Start":"00:56.690 ","End":"00:59.340","Text":"otherwise it\u0027s near impossible."},{"Start":"01:00.290 ","End":"01:04.010","Text":"Enough talking. Let\u0027s get down to brass tacks."},{"Start":"01:04.010 ","End":"01:11.185","Text":"We can finally do this example by filling in our x with F_x and solving for U."},{"Start":"01:11.185 ","End":"01:15.535","Text":"It\u0027ll give us some understanding of what U equals and we can work from there."},{"Start":"01:15.535 ","End":"01:19.085","Text":"If we write down from our function, the x portion,"},{"Start":"01:19.085 ","End":"01:23.450","Text":"that which goes in the direction of x equals negative 2xy."},{"Start":"01:23.450 ","End":"01:29.660","Text":"According to this, negative 2xy equals negative dU dx."},{"Start":"01:29.660 ","End":"01:31.040","Text":"We can simplify this,"},{"Start":"01:31.040 ","End":"01:33.110","Text":"first the negatives will drop out."},{"Start":"01:33.110 ","End":"01:35.795","Text":"If we simplify that U is isolated,"},{"Start":"01:35.795 ","End":"01:43.450","Text":"we get U equals the integral of 2xy times dx."},{"Start":"01:43.450 ","End":"01:48.205","Text":"When we do this, we find that U equals x^2y."},{"Start":"01:48.205 ","End":"01:51.470","Text":"Now, because we\u0027re doing an indefinite integral,"},{"Start":"01:51.470 ","End":"01:55.400","Text":"we have to add some constant C. C can"},{"Start":"01:55.400 ","End":"01:59.540","Text":"also be dependent on the variable y because we did a partial derivative,"},{"Start":"01:59.540 ","End":"02:04.385","Text":"the C in the y will both drop out if we do a partial derivative of x."},{"Start":"02:04.385 ","End":"02:07.940","Text":"In our example, if we did a partial derivative of x,"},{"Start":"02:07.940 ","End":"02:12.830","Text":"we\u0027d find that we end up with negative 2xy or really 2xy."},{"Start":"02:12.830 ","End":"02:17.270","Text":"Because Cy doesn\u0027t have any x\u0027s in it, it would fall out."},{"Start":"02:17.270 ","End":"02:19.835","Text":"We won\u0027t solve for y right now."},{"Start":"02:19.835 ","End":"02:22.595","Text":"But what\u0027s important is that this requirement is met,"},{"Start":"02:22.595 ","End":"02:25.595","Text":"F_x equals negative dU dx."},{"Start":"02:25.595 ","End":"02:26.825","Text":"Now that we have that,"},{"Start":"02:26.825 ","End":"02:30.455","Text":"we should move on to F_y equals negative dU dy."},{"Start":"02:30.455 ","End":"02:33.020","Text":"We\u0027re not going to perform the same operation."},{"Start":"02:33.020 ","End":"02:34.760","Text":"We\u0027re not going to take F_y,"},{"Start":"02:34.760 ","End":"02:38.135","Text":"set it equal to negative dU dy and do an integral,"},{"Start":"02:38.135 ","End":"02:41.915","Text":"in fact, we\u0027re going to take the results that we found and go backwards."},{"Start":"02:41.915 ","End":"02:45.440","Text":"We have here is x^2y plus some C,"},{"Start":"02:45.440 ","End":"02:50.185","Text":"perhaps multiplied by y equals U."},{"Start":"02:50.185 ","End":"02:53.105","Text":"If we have that, we can take dU, dy from that,"},{"Start":"02:53.105 ","End":"02:54.755","Text":"a partial derivative of y,"},{"Start":"02:54.755 ","End":"02:57.485","Text":"and that will equal F_y. Let\u0027s write this out."},{"Start":"02:57.485 ","End":"03:01.375","Text":"Negative dU dy equals F_y."},{"Start":"03:01.375 ","End":"03:06.555","Text":"That means that negative dU dy is the partial derivative accounting for y."},{"Start":"03:06.555 ","End":"03:12.330","Text":"We end up with the result of x^2 for the x^2y plus some C tag y,"},{"Start":"03:12.330 ","End":"03:13.890","Text":"which we\u0027ll determine soon."},{"Start":"03:13.890 ","End":"03:16.605","Text":"That must equal F_y."},{"Start":"03:16.605 ","End":"03:23.850","Text":"F_y, we know from our function is 2 minus x^2 in the direction of y."},{"Start":"03:23.850 ","End":"03:26.045","Text":"From here we can simplify a little bit."},{"Start":"03:26.045 ","End":"03:29.555","Text":"You notice that we have negative x^2 on both sides, that\u0027ll drop out."},{"Start":"03:29.555 ","End":"03:33.635","Text":"In fact, this happens relatively frequently as you\u0027ll see in later examples."},{"Start":"03:33.635 ","End":"03:37.190","Text":"What we\u0027re left with is C tag y equals"},{"Start":"03:37.190 ","End":"03:40.880","Text":"negative 2 if we multiply everything by negative 1."},{"Start":"03:40.880 ","End":"03:46.975","Text":"Now, the C tag y is actually the derivative of some function with y."},{"Start":"03:46.975 ","End":"03:52.970","Text":"What we can say is this equals the integral of negative 2dy."},{"Start":"03:52.970 ","End":"03:54.985","Text":"When we perform this,"},{"Start":"03:54.985 ","End":"03:59.120","Text":"we end up with is negative 2y plus C tag."},{"Start":"03:59.120 ","End":"04:00.545","Text":"This is a new C tag."},{"Start":"04:00.545 ","End":"04:02.045","Text":"This is not the old C tag,"},{"Start":"04:02.045 ","End":"04:05.120","Text":"but this can\u0027t be dependent on x because that would"},{"Start":"04:05.120 ","End":"04:08.645","Text":"mean that the former function on our left is also dependent on x."},{"Start":"04:08.645 ","End":"04:12.350","Text":"This could be dependent on z if we had a z element."},{"Start":"04:12.350 ","End":"04:14.510","Text":"But in this particular example, we don\u0027t."},{"Start":"04:14.510 ","End":"04:16.980","Text":"I\u0027m sure you\u0027ll see that in the future though."},{"Start":"04:17.140 ","End":"04:20.285","Text":"Now that you\u0027ve found a new Cy,"},{"Start":"04:20.285 ","End":"04:23.785","Text":"you can put it into our new understanding of U above."},{"Start":"04:23.785 ","End":"04:27.550","Text":"We have the skeleton of U and it\u0027s evolving as we go through the process."},{"Start":"04:27.550 ","End":"04:30.770","Text":"We can now do is plug in the new Cy and we\u0027re moving a step"},{"Start":"04:30.770 ","End":"04:34.835","Text":"closer to having a true understanding of u, the potential energy."},{"Start":"04:34.835 ","End":"04:41.805","Text":"We can write this out, U equals x^2y plus negative 2y."},{"Start":"04:41.805 ","End":"04:48.540","Text":"We can just write that as negative or minus 2y plus C tag."},{"Start":"04:48.710 ","End":"04:52.850","Text":"This U actually fulfills all of our requirements."},{"Start":"04:52.850 ","End":"04:57.080","Text":"First of all, F_x equals negative dU dx."},{"Start":"04:57.080 ","End":"05:00.740","Text":"Second of all, F_y equals negative dU, dy."},{"Start":"05:00.740 ","End":"05:05.170","Text":"It fulfills those. All we have left to do is fill in the constant."},{"Start":"05:05.170 ","End":"05:08.210","Text":"This should be familiar with other instances of potential energy."},{"Start":"05:08.210 ","End":"05:10.595","Text":"For example, when we\u0027re dealing with mgh,"},{"Start":"05:10.595 ","End":"05:12.575","Text":"we need to choose h, a constant,"},{"Start":"05:12.575 ","End":"05:14.420","Text":"at whatever point is going to be 0."},{"Start":"05:14.420 ","End":"05:16.445","Text":"We need to find what h(0) is."},{"Start":"05:16.445 ","End":"05:18.760","Text":"Here we need to do something similar."},{"Start":"05:18.760 ","End":"05:21.180","Text":"Now we need to account for this constant."},{"Start":"05:21.180 ","End":"05:24.695","Text":"This is where we put in the piece of given data we have from before."},{"Start":"05:24.695 ","End":"05:27.580","Text":"At 0, 0 U=0."},{"Start":"05:27.580 ","End":"05:30.365","Text":"That means that when x=0 and y=0,"},{"Start":"05:30.365 ","End":"05:32.230","Text":"our potential energy equals 0."},{"Start":"05:32.230 ","End":"05:34.065","Text":"We can plug this in and solve."},{"Start":"05:34.065 ","End":"05:36.500","Text":"U 0,0, that means x=0,"},{"Start":"05:36.500 ","End":"05:40.330","Text":"y=0 would be 0 minus 0 plus C tag."},{"Start":"05:40.330 ","End":"05:42.570","Text":"We\u0027re told that this equals 0."},{"Start":"05:42.570 ","End":"05:46.660","Text":"Here we now know that C tag equals 0."},{"Start":"05:46.760 ","End":"05:54.355","Text":"If C=0, we can drop it from the equation and we\u0027re left with U=x^2y minus 2y."},{"Start":"05:54.355 ","End":"05:56.140","Text":"That\u0027s the solution to the problem."},{"Start":"05:56.140 ","End":"05:58.765","Text":"Let\u0027s do a quick summary of the technique here."},{"Start":"05:58.765 ","End":"06:00.235","Text":"I took 1 of my elements."},{"Start":"06:00.235 ","End":"06:02.020","Text":"It could have been y or it could have been x,"},{"Start":"06:02.020 ","End":"06:05.240","Text":"but I took F_x and I did an integral of it."},{"Start":"06:05.240 ","End":"06:11.875","Text":"What I ended up with was a solution to which I added a constant C that\u0027s dependent on y."},{"Start":"06:11.875 ","End":"06:16.615","Text":"Now I take this solution and do a partial derivative based on y of it."},{"Start":"06:16.615 ","End":"06:18.205","Text":"The solution I get from that,"},{"Start":"06:18.205 ","End":"06:20.080","Text":"I set equal to F_y."},{"Start":"06:20.080 ","End":"06:25.435","Text":"This will give me a new value for C tag y. I can replace that."},{"Start":"06:25.435 ","End":"06:27.790","Text":"This solved for us with 2 variables,"},{"Start":"06:27.790 ","End":"06:31.540","Text":"but let\u0027s assume we add a third variable z that we also had to account for."},{"Start":"06:31.540 ","End":"06:38.700","Text":"Well, this initial Cy in yellow would have been dependent not only on y but also on z."},{"Start":"06:38.700 ","End":"06:40.475","Text":"It could have been C depending on y,"},{"Start":"06:40.475 ","End":"06:42.140","Text":"it could have been C dependent on z."},{"Start":"06:42.140 ","End":"06:43.850","Text":"The next step would have been the same."},{"Start":"06:43.850 ","End":"06:46.190","Text":"We do a partial derivative of y,"},{"Start":"06:46.190 ","End":"06:48.605","Text":"so we can eliminate the y factor."},{"Start":"06:48.605 ","End":"06:52.595","Text":"Then what we\u0027d end up with is some solution in terms of,"},{"Start":"06:52.595 ","End":"06:55.850","Text":"let\u0027s say, it\u0027s still negative 2y plus Cz."},{"Start":"06:55.850 ","End":"06:59.960","Text":"This is only dependent on z because we\u0027ve already eliminated the x factor,"},{"Start":"06:59.960 ","End":"07:01.970","Text":"we\u0027ve already eliminated the y factor."},{"Start":"07:01.970 ","End":"07:05.150","Text":"This last C can only be dependent on z."},{"Start":"07:05.150 ","End":"07:07.714","Text":"Then we would do the second step, again,"},{"Start":"07:07.714 ","End":"07:10.940","Text":"doing a partial derivative of z now and"},{"Start":"07:10.940 ","End":"07:14.390","Text":"setting that equal to F_z as opposed to F_y or F_x."},{"Start":"07:14.390 ","End":"07:20.435","Text":"The solution there would give us some long solution plus a Cz element, as you see here."},{"Start":"07:20.435 ","End":"07:23.315","Text":"On the end, we\u0027d have a C that\u0027s some constant."},{"Start":"07:23.315 ","End":"07:27.080","Text":"We could solve for the constant the same way by using our given data point."},{"Start":"07:27.080 ","End":"07:31.445","Text":"What we\u0027d have is the same thing with a z element on the end."},{"Start":"07:31.445 ","End":"07:35.060","Text":"Ultimately, if you understand how this worked with x and y,"},{"Start":"07:35.060 ","End":"07:39.510","Text":"you\u0027ll understand very easily how it works with z in future examples."}],"ID":9310}],"Thumbnail":null,"ID":5425}]

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