WEBVTT physics/high-school-physics/selhorst-jones 00:00:00.000 --> 00:00:05.000 Hi, welcome back to educator.com, today we are going to be talking about gravitational potential energy. 00:00:05.000 --> 00:00:10.000 In our last section, we mentioned that there is a huge variety of forms that energy can take. 00:00:10.000 --> 00:00:13.000 Then we explored specifically the idea of kinetic energy. 00:00:13.000 --> 00:00:17.000 Now we are going to investigate the idea of gravity giving energy to objects. 00:00:17.000 --> 00:00:22.000 When you lift an object up above the ground, you put gravitational potential energy into it. 00:00:22.000 --> 00:00:24.000 Why do we call it potential energy? 00:00:24.000 --> 00:00:27.000 Potential energy is stored energy that can be later released. 00:00:27.000 --> 00:00:30.000 So, it is something that will allow us to do work later. 00:00:30.000 --> 00:00:32.000 But, that is kind of true of kinetic energy, right? 00:00:32.000 --> 00:00:39.000 The very least, it is able to store, in a way, store the energy as kinetic energy, and then release it as frictional work. 00:00:39.000 --> 00:00:43.000 So, what is the difference between potential energy and kinetic energy? 00:00:43.000 --> 00:00:52.000 If we really want to get rigorous about this, potential energy is energy associated with position/arrangement of a system of objects, and the forces interacting between them. 00:00:52.000 --> 00:01:02.000 That is a little hard to use in our work though, so it is more than enough for us to think of it as storing energy for later use, something that we can hold on to indefinitely, or at least for a while. 00:01:02.000 --> 00:01:22.000 We have got a pretty good understanding intuitively, once again we can appeal to our intuition, and we do not have to worry about having a really rigorous understanding until later Advanced Physics stuff, so for now, we can think about potential energy as just something that, just we are storing energy for a while, and also, we will know about it because we are going to talk about each kind of potential energy very specifically in length. 00:01:22.000 --> 00:01:26.000 Let us start off by thinking about how potential energy works. 00:01:26.000 --> 00:01:36.000 If we have got an object dropped from ever increasing heights, the higher the height it is dropped from , the faster it would be moving when it gets to the ground. 00:01:36.000 --> 00:01:44.000 This means that a higher height implies higher speed at the ground. 00:01:44.000 --> 00:01:50.000 Higher speed means more energy, that means that more height must be more energy. 00:01:50.000 --> 00:01:57.000 If we have got a little mass with a little distance, that is going to be less energy than that same little mass, and a much larger distance. 00:01:57.000 --> 00:02:03.000 Bigger the distance, the greater the energy. 00:02:03.000 --> 00:02:07.000 If we want to talk about things specifically, quantitatively, how much energy does it have? 00:02:07.000 --> 00:02:12.000 If we drop an object of mass 'm' from a height 'h', how much work does the force of gravity do on it? 00:02:12.000 --> 00:02:17.000 The object is going to be pulled down, by a force mg, right? 00:02:17.000 --> 00:02:25.000 The force of gravity is mg, mg always operates down, it is always pulling directly down, just like our height is directly down. 00:02:25.000 --> 00:02:48.000 So, the motion down is going to be straight down, that means, work = force × distance × cos θ = mgh, since they are parallel, cos θ knocked down. 00:02:48.000 --> 00:02:53.000 So, the amount of work is just going to be mgh. 00:02:53.000 --> 00:03:00.000 If gravity does a work of mgh on our box, that is how much energy we must have stored in it by lifting it to that height. 00:03:00.000 --> 00:03:10.000 Therefore, the energy of gravity = mgh. 00:03:10.000 --> 00:03:31.000 Now, think about the fact that, no matter what path we take to get an object up to, say the top of this pillar, if we go directly up, and put it here, or we go like this, and then put it here, once it is still and sitting up there, it is going to have the exact same energy either direction. 00:03:31.000 --> 00:03:36.000 No matter how we get it there, it winds up having the same energy, once they get to the same height. 00:03:36.000 --> 00:03:38.000 The only important thing is the height that it has. 00:03:38.000 --> 00:03:41.000 It matters if it is moving, but once it still, it is just the height. 00:03:41.000 --> 00:03:44.000 It is completely based on the height. 00:03:44.000 --> 00:03:49.000 So, mgh gives us potential gravitational energy. 00:03:49.000 --> 00:03:51.000 Two important things to notice about the formula that we just made: 00:03:51.000 --> 00:03:54.000 It is up to us to set what that base height means. 00:03:54.000 --> 00:03:59.000 For that pillow that we just talked about, it is up to us to know that, that is the zero. 00:03:59.000 --> 00:04:17.000 We could have the zero have up here, and it does not matter, it is all based on a relative idea, so it is comparing the two heights, if we want to talk about what the energy of the ball is here, what the energy of the ball is here, it does not matter if they both have negative, it is still going to have a positive difference, because it would have gone up to get up here. 00:04:17.000 --> 00:04:28.000 Whereas, if it is down here, it is going to be positive, it does not matter, it is positive either way, because this one is less negative from the point of that zero, and this one is more positive from the point of that zero. 00:04:28.000 --> 00:04:34.000 But either way, it is the relative difference, it is the change in 'h', not just the starting height on its own. 00:04:34.000 --> 00:04:42.000 Being at the top of Mt. Everest, and going down 10 feet, is not more energy that starting at the top of Empire State Building, and going all the way to the ground floor. 00:04:42.000 --> 00:04:51.000 There is way more energy in that tall drop from the top of the building down, even though Mt. Everest starts off higher from the centre of the Earth. 00:04:51.000 --> 00:04:55.000 Formula also relies on the idea that 'g' is constant. 00:04:55.000 --> 00:04:58.000 We can do this, because 'g' is reasonably constant near the surface of the Earth. 00:04:58.000 --> 00:05:15.000 We can trust that we can have 9.8, really close to 9.8, remember, if you are on a really high location, depending on where you are on the Earth, specific loads of, like mineral deposits, so there is going to be minor changes in 'g', but more the most part, gravity is really constant on the surface of Earth, we can trust it to be 9.8. 00:05:15.000 --> 00:05:25.000 We can use this formula, but if we went into space, or a different planet, the formula would change as 'g' changes, and if we were to go really far from Earth, we would not even be able to use this anymore. 00:05:25.000 --> 00:05:32.000 It is going to rely on the fact that we can rely on the fairly constant gravity of Earth, within a near area from the surface of the Earth. 00:05:32.000 --> 00:05:52.000 That is what we are going to trust on, whenever we are working on problems, we trust on the fact that we know what the gravitational constant is, and we know that it is going to be fairly, with the acceleration of gravity is not going to change much, whatever distance we go above the Earth, as long as it is a reasonable distance compared to the size of the Earth. 00:05:52.000 --> 00:05:57.000 Finally, to be able to use anything in our energy formulae usefully, we are going to have the fact that energy is conserved. 00:05:57.000 --> 00:06:07.000 We keep talking about hits, and we are going to keep talking about it, energy cannot be created or destroyed, it is always going to be conserved, it is only transferred. 00:06:07.000 --> 00:06:10.000 This means that energy in a system stays constant, unless it is transferred out of the system. 00:06:10.000 --> 00:06:16.000 The energy in the system at the start + the work = energy of the system at the end. 00:06:16.000 --> 00:06:26.000 Positive work means energy is put into the system, work is put into the system, and the energy increases, negative work means energy is taken out, say friction. 00:06:26.000 --> 00:06:32.000 That is what work is, work that is positive, is energy going in, work that is negative, is energy coming out. 00:06:32.000 --> 00:06:34.000 We are ready to tackle the examples. 00:06:34.000 --> 00:06:40.000 If we have a rock dropped from a height of 20 m, and we ignore air resistance, how fast will it be traveling 10 m above the ground? 00:06:40.000 --> 00:06:43.000 Then how fast we will be traveling just before the impact? 00:06:43.000 --> 00:06:51.000 Say we have got this rock, and there is 20 m, and there is the 10 m that it falls first. 00:06:51.000 --> 00:07:00.000 From here to here, what is the change in height?, the change in height = 10, and the change in height here = another 10. 00:07:00.000 --> 00:07:02.000 How fast is it traveling 10 m above the ground? 00:07:02.000 --> 00:07:11.000 At beginning, it starts off at rest, so energy at the beginning, kinetic energy (K.E.) = 0, so there is no energy in it other than potential gravitational energy. 00:07:11.000 --> 00:07:21.000 No work happens, because it does not lose anything to friction, it does not lose anything to air resistance, it is just direct transfer between gravitational potential energy moving directly into K.E. 00:07:21.000 --> 00:07:24.000 For the first one, there is two ways of approaching this one: 00:07:24.000 --> 00:07:38.000 We could look at this as either being, mgΔh = (1/2)mv^2 (we are talking about speed, since this is single dimension.) 00:07:38.000 --> 00:07:58.000 mgΔh, we do not know 'm', that might be a problem, but that we can take it for a bit, we do know 'g' and Δh, and we have got 'm' on both sides, so we can strike out our m's, we do not have to worry about that. 00:07:58.000 --> 00:08:06.000 Remember, gravity works on everything, no matter what the mass is, it has a uniform acceleration, barring air resistance. 00:08:06.000 --> 00:08:19.000 A uniform acceleration caused by gravity happens in our potential gravity to kinetic transform, because of the fact that, that 'm' constant shows up on both sides of the equation. 00:08:19.000 --> 00:08:32.000 That means, gΔh = (1/2)v^2, sqrt(2gΔh) = v. 00:08:32.000 --> 00:08:54.000 Plug in numbers, sqrt(2 × 9.8 × 10) = v = 14 m/s, at 10 m above the ground. 00:08:54.000 --> 00:09:30.000 One other way that we could have done this though, is we could have said, Esystem(beginning) + work = Esystem(end), mghi = mghend + (1/2)mv^2, since work = 0, and also it has got some speed at the end. 00:09:30.000 --> 00:09:43.000 If we move this over, we got, mghi - mghend, which is why we wind up pulling it out, and we look at, mgΔh. 00:09:43.000 --> 00:09:48.000 It is the relative change that tells us how much energy is put into the system from gravitational potential. 00:09:48.000 --> 00:09:52.000 That equals our (1/2)mv^2, and that is why we did not have to put on both sides. 00:09:52.000 --> 00:10:02.000 We could though, some case it is going to be useful to do that, we want to keep that tool in our tool box, but for this case, we are just looking at the change in height, gives us how much energy gets put in. 00:10:02.000 --> 00:10:38.000 If we want to consider what happens when it moves it to all 20, we have got, mgΔh = (1/2)mv^2, 2gΔh = v, 2×9.8×20 = velocity = 19.8 m/s. 00:10:38.000 --> 00:10:42.000 That is the velocity at 0 m above the ground. 00:10:42.000 --> 00:10:59.000 The things are, when we are at 10 m above the ground, we got 14 m/s, and we are at 0 m above the ground, when we are just touching the ground, just the instant before the impact, just that split second before it lands on the ground, 19.8 m/s. 00:10:59.000 --> 00:11:04.000 A 100 g is thrown from a height of 1.4 m above the ground at a speed of 10 m/s. 00:11:04.000 --> 00:11:08.000 At a height of 2.5 m, it has a speed of 7.9 m/s. 00:11:08.000 --> 00:11:10.000 How much of its energy has been lost to air drag? 00:11:10.000 --> 00:11:19.000 This is a case where it is actually going to be useful for us to have height on both sides, I think it is easier that way, you might not, but you can definitely do the other way if you felt like it. 00:11:19.000 --> 00:11:21.000 In this case, we will do my way though. 00:11:21.000 --> 00:11:44.000 Energysystem(beginning) + work = Energysystem(end), we want to figure out what the work is, how much work does the air drag do, and we should get a negative number, because the air drag is going to sap the energy out of it, it is going to suck energy out. 00:11:44.000 --> 00:11:51.000 Esystem(beginning) has two things, it has a height above the ground, and it also has an initial speed. 00:11:51.000 --> 00:12:31.000 What is its initial height?, mghinitial + (1/2)mv^2 + work = mghfinal + (1/2)mvfinal^2 (again, v is speed, note that it could be a vector as well, the important thing is to understand what you are writing, a useful diagram could be really helpful here.) 00:12:31.000 --> 00:12:57.000 This frisbee, it starts off flying at some speed, and with some height, hinitial, and then later in time, it is at higher height, but less speed, and we want to figure out how much work is done between those two moments. 00:12:57.000 --> 00:12:59.000 Those are our two snap shots. 00:12:59.000 --> 00:13:42.000 At the beginning, we have got, in S.I., 0.1 kg, (sometimes we can use other unit system, but if we use 100 g instead of 0.1 kg, we would not be working with joules anymore, we would be working with something else, and we will completely screw up our ability to do this problem properly, because we have got m/s which is S.I., and we have got grams in there, which is not S.I., and things are going to get really funny, it is going to get bad.) 00:13:42.000 --> 00:14:00.000 So, vinitial = 10, then, hinitial = 1.4, hfinal = 2.5, and vfinal = 7.9. 00:14:00.000 --> 00:14:20.000 Plug everything in, we get, 0.1 kg×9.8 m/s/s×1.4 m + (1/2)×0.1×10^2 + work = 0.1 kg×9.8 m/s/s × 2.5 m + (1/2)×0.1 kg × (7.9)^2. 00:14:20.000 --> 00:15:19.000 We cannot make it easier by canceling 'm', because the work does not involve 'm', but it is important to be able to think about what you are doing and catch them when they happen. 00:15:19.000 --> 00:15:31.000 Start calculating everything out, eventually, work = -0.802 J. 00:15:31.000 --> 00:15:37.000 We plug everything in, move things around, put in a calculator, and we get the fact that we get -0.802 J at the end. 00:15:37.000 --> 00:15:52.000 That is what our answer is, it loses that much energy to air drag, which is not very much, but remember, there is not all that much energy in a very light weight frisbee being thrown, not that fast. 00:15:52.000 --> 00:15:57.000 Example 3: Block of unknown mass is slid down a frictionless slide of height 5 m. 00:15:57.000 --> 00:16:08.000 Then it slides along a table where it has a friction coefficient of μk = 0.3, we only have to know the kinetic friction, because it is already starting on motion. 00:16:08.000 --> 00:16:17.000 It starts on a frictionless incline, but very still, it starts to slide down it, and how far does the block slide on the table? 00:16:17.000 --> 00:16:21.000 What is the initial energy in it? 00:16:21.000 --> 00:16:34.000 Ei + work = Eend, so how much energy does it have at the end? 00:16:34.000 --> 00:16:36.000 It is still, so it has no energy. 00:16:36.000 --> 00:16:41.000 The work is going to have to suck out, is going to have to cancel out all the initial energy. 00:16:41.000 --> 00:16:46.000 What energy does it have in the very beginning?, is it moving in the very beginning? 00:16:46.000 --> 00:16:50.000 No!, it just starts, and then slides on a frictionless slide. 00:16:50.000 --> 00:16:58.000 So, how much energy does it have?, it has mgh, its starting height worth of energy. 00:16:58.000 --> 00:17:06.000 We have got, mgh = -W, work here is by friction. 00:17:06.000 --> 00:17:12.000 How much is friction?, force of friction = μk×FN. 00:17:12.000 --> 00:17:19.000 Remember, we do not have to worry about it once it is on the slide, because there is no friction on the slide, so we only have to worry about the flat plane. 00:17:19.000 --> 00:17:35.000 On the flat plane, the only thing pulling down on it is mg, and mg is also equal to FN, so, friction = μkmg. 00:17:35.000 --> 00:18:13.000 We go back to this, we get, mgh = -fdcosθ, θ = 180 degrees, because its motion is this way, but friction is pulling this way, so a 180 degrees. 00:18:13.000 --> 00:18:36.000 So, mgh = -fdcos(180) = fd (cos(180) = -1). 00:18:36.000 --> 00:18:52.000 So, mgh = μk×mg×d. 00:18:52.000 --> 00:18:55.000 Look!, mg gets canceled on both sides. 00:18:55.000 --> 00:19:27.000 So, h = μk×d, d = h/μk = (5 m)/(0.3) = 16.7 m, is the distance that it slides. 00:19:27.000 --> 00:19:41.000 All the energy that it starts with, is just going to be its gravitational potential energy, that is all the energy in it, so it slides down the slide, once it gets to the bottom, it then starts to be zapped by friction. 00:19:41.000 --> 00:20:08.000 So, the distance, starting here, once it gets to the friction table, is going to 16.7 m, because we know that, Ei + W = Eend, (gravitational potential energy) + [(friction force)×(distance it slides)] = 0, since still in the end. 00:20:08.000 --> 00:20:26.000 Of course it might be still higher, but we know are looking at the change in height from start end to the bottom, and we said our base height as being the table's height, that is a clever thing we did, we did not even point out while we were working on it. 00:20:26.000 --> 00:20:40.000 One thing I would like to point out: When we worked on this problem, for almost all of it, we did not put in our numbers, we do not need to put in numbers until the very end. 00:20:40.000 --> 00:21:13.000 It is so much easier to be able to work with μk and sub in at the very end, that is a very useful trick, when you have got a lot of non-sense going around, you only need to substitute in the values until you get to the very end, because you might get the chance to eliminate some of them, it is easier to write letters, than complicated sets of numbers, because you might wind up making a simple error when you are trying to put down five 4 digit numbers, as opposed to 5 letters. 00:21:13.000 --> 00:21:26.000 It is easier to watch what you are doing and have an understanding of what is going on, have an intuitive understanding of what you are doing, what this represents mathematically, by moving things with letters than just these meaningless numbers that you might not understand as well. 00:21:26.000 --> 00:21:33.000 It is really helpful to be able to move things around as letters, and at the very end substitute in. 00:21:33.000 --> 00:21:43.000 Final one: An acrobat of mass, m= 55 kg, swings from a trapeze of length 15 m, starts at rest, and descends 5 m during the sway. 00:21:43.000 --> 00:21:49.000 The trapeze rope has been sabotaged by an evil villain, and it will snap at 800 N of tension. 00:21:49.000 --> 00:21:52.000 Will our acrobat make it across safely? 00:21:52.000 --> 00:21:56.000 This is a question that is important to his life. 00:21:56.000 --> 00:22:06.000 If it is going to snap, where would be the point it would experience the most tension, when is the most tension going to happen? 00:22:06.000 --> 00:22:35.000 If we are able to figure out that it is going to not snap at the point it is experiencing the most tension, then we know he will make it across safely, because the entire time the swing occurs, the acrobat is able to swing across without having to worry about the tension snapping it, because if the tension is maximum at one point, then we know that if we can figure out that one point's maximum tension, it will tell us whether or not it snaps. 00:22:35.000 --> 00:22:40.000 Because everything is going to be at least less than that. 00:22:40.000 --> 00:22:45.000 If everything is less than that, all we figure out is what the maximum tension put on it is. 00:22:45.000 --> 00:22:47.000 Where will the maximum tension occur? 00:22:47.000 --> 00:22:52.000 Now we have to think about, where do we have to get tension from! 00:22:52.000 --> 00:22:55.000 The tension has to keep it in a circle. 00:22:55.000 --> 00:23:22.000 So, the tension is going to be connected to keeping it in a circle, so force centripetal for it to be able to go in a circle, forcecentripetal = m×v^2/r (again, v is speed) 00:23:22.000 --> 00:23:31.000 At what point on the circle will he be traveling the fastest? 00:23:31.000 --> 00:23:41.000 It is going to be the point, where he has converted all of his initial gravitational potential into motion, which is going to be the lowest point on the circle. 00:23:41.000 --> 00:23:45.000 That is going to have the largest 'v'. 00:23:45.000 --> 00:23:55.000 The largest 'v' occurs at the bottom, also the largest 'v' is when there is going to be the most fight between the tension of the rope, and the gravity pulling him down. 00:23:55.000 --> 00:24:07.000 So, mg is going to be pulling down, and the rope is going to have to also give the centripetal force necessary to keep it in the circle, and also beat the gravity force that is trying to pull him down. 00:24:07.000 --> 00:24:14.000 These two things combined will cause the tension to be highest at the bottom, so that is the point that we have to look at. 00:24:14.000 --> 00:24:33.000 We know that, Ei + W = Eend, so the moment we have to look at, the snap shot, is going to be the moment of most tension, that is the point we want to check out and figure out if he is going to survive. 00:24:33.000 --> 00:24:38.000 If he is going to survive, we are going to have to look at that moment of maximum tension. 00:24:38.000 --> 00:24:41.000 So we are going to make that ending snap shot be, when it is pointing directly down. 00:24:41.000 --> 00:24:51.000 There is no work, since there is no friction, everything is smooth. 00:24:51.000 --> 00:25:05.000 So, mgh = (1/2)mv^2, m's cancel out, so we do not have to worry about the mass. 00:25:05.000 --> 00:25:28.000 So, gh = (1/2)v^2, v = sqrt(gh) = 9.9 m/s, is the 'v' at the bottom, that is the speed our acrobat would be traveling if the rope were definitely safe. 00:25:28.000 --> 00:25:45.000 If the rope snaps, it will snap somewhere between here, and here; after this it will wind up being safe, so all we have to do is to test the most extreme point, and we will be able to figure out if it snaps before it gets to the bottom. 00:25:45.000 --> 00:25:53.000 If the bottom tension is greater than 800 N, we know the rope snaps, somewhere on the way to getting to the most extreme tension. 00:25:53.000 --> 00:25:57.000 We do not know precisely where yet, we could figure it out, that would be a more difficult question. 00:25:57.000 --> 00:26:04.000 We know that, vbottom = 9.9 m/s, what is the centripetal force necessary to keep him in a circle? 00:26:04.000 --> 00:26:17.000 That is what it has to be, on the acrobat, forcecentripetal = m×v^2/r, and we know m and r. 00:26:17.000 --> 00:26:37.000 (55 kg)×(9.9)^2/(15 m) = 359 N, is the necessary centripetal force, the necessary sum of forces to be able to keep him in that circle. 00:26:37.000 --> 00:26:42.000 The centripetal force is pulling this way. 00:26:42.000 --> 00:26:49.000 We have got tension, is pulling this way, and has to also be able to be some of mg. 00:26:49.000 --> 00:26:57.000 We know that, T - mg = 359 N = centripetal force. 00:26:57.000 --> 00:27:08.000 T = 359 + mg, at the moment of maximum tension that that rope would experience if it were a safe healthy rope. 00:27:08.000 --> 00:27:18.000 Remember, if it is not a safe healthy rope, it might snap before we even get there, so the actual rope might not even ever experience that tension, because the most it is going to experience is 800 N before it snaps. 00:27:18.000 --> 00:27:21.000 We do not know if he will make it yet, we got to finish this problem. 00:27:21.000 --> 00:27:50.000 359 + mg = 359 N +(55 kg) × (9.8 m/s/s) = 898 N, SNAP!, he does not make it! 00:27:50.000 --> 00:28:04.000 But, on the bright side, there is a net underneath it, let us draw a net, the net catches him, and he works later, and he finds the villain who is responsible for it, and he puts him to justice. 00:28:04.000 --> 00:28:10.000 Hope you enjoyed this, hope you will come back, and we will learn about more about energy, and have an even stronger understanding of how energy works.