For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Energy: Gravitational Potential

- We can think of
*potential energy*as energy stored for future use. This isn't perfectly rigorous, but it gives us a good understanding for now. - The amount of gravitational potential energy is based off of the mass of the object, the gravity involved, and the height of the object:
E _{gravity}= mgh. - We have to set the "base" height. Remember, as usual, we're the ones who have to impose a coordinate system, so it's up to us to determine what we consider the starting height.
- Because of this, the important thing isn't the "absolute" height, but instead the relative height between the start and end heights: ∆h.
- This formula relies on the fact that g is a constant near the surface of Earth (or whatever gravitational body we're dealing with). If g were to vary over the height traveled, we would need a different formula.
- By the conservation of energy, we can look at the entire energy of the system at the start and end:

[Remember, positive work puts energy into the system, while negative work takes it out.]E _{sys, start}+ W = E_{sys, end}. - It's up to decide when we want to take our start and end "snapshots." Carefully choosing what moments we want to compare is key to solving problems.

### Energy: Gravitational Potential

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Why Is It Called Potential Energy?
- Introduction to Gravitational Potential Energy
- Gravitational Potential Energy
- Gravitational Potential Energy: Derivation
- Gravitational Potential Energy: Formulas
- Gravitational Potential Energy: Notes
- Conservation of Energy
- Example 1: Speed of a Falling Rock
- Example 2: Energy Lost to Air Drag
- Example 3: Distance of a Sliding Block
- Example 4: Swinging Acrobat

- Intro 0:00
- Why Is It Called Potential Energy? 0:21
- Why Is It Called Potential Energy?
- Introduction to Gravitational Potential Energy 1:20
- Consider an Object Dropped from Ever-Increasing heights
- Gravitational Potential Energy 2:02
- Gravitational Potential Energy: Derivation
- Gravitational Potential Energy: Formulas
- Gravitational Potential Energy: Notes
- Conservation of Energy 5:50
- Conservation of Energy and Formula
- Example 1: Speed of a Falling Rock 6:31
- Example 2: Energy Lost to Air Drag 10:58
- Example 3: Distance of a Sliding Block 15:51
- Example 4: Swinging Acrobat 21:32

### High School Physics Online Course

### Transcription: Energy: Gravitational Potential

*Hi, welcome back to educator.com, today we are going to be talking about gravitational potential energy.*0000

*In our last section, we mentioned that there is a huge variety of forms that energy can take.*0005

*Then we explored specifically the idea of kinetic energy.*0010

*Now we are going to investigate the idea of gravity giving energy to objects.*0013

*When you lift an object up above the ground, you put gravitational potential energy into it.*0017

*Why do we call it potential energy?*0022

*Potential energy is stored energy that can be later released.*0024

*So, it is something that will allow us to do work later.*0027

*But, that is kind of true of kinetic energy, right?*0030

*The very least, it is able to store, in a way, store the energy as kinetic energy, and then release it as frictional work.*0032

*So, what is the difference between potential energy and kinetic energy?*0039

*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.*0043

*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.*0052

*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.*0062

*Let us start off by thinking about how potential energy works.*0082

*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.*0086

*This means that a higher height implies higher speed at the ground.*0096

*Higher speed means more energy, that means that more height must be more energy.*0104

*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.*0110

*Bigger the distance, the greater the energy.*0117

*If we want to talk about things specifically, quantitatively, how much energy does it have?*0123

*If we drop an object of mass 'm' from a height 'h', how much work does the force of gravity do on it?*0127

*The object is going to be pulled down, by a force mg, right?*0132

*The force of gravity is mg, mg always operates down, it is always pulling directly down, just like our height is directly down.*0137

*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.*0145

*So, the amount of work is just going to be mgh.*0168

*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.*0173

*Therefore, the energy of gravity = mgh.*0180

*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.*0190

*No matter how we get it there, it winds up having the same energy, once they get to the same height.*0211

*The only important thing is the height that it has.*0216

*It matters if it is moving, but once it still, it is just the height.*0218

*It is completely based on the height.*0221

*So, mgh gives us potential gravitational energy.*0224

*Two important things to notice about the formula that we just made:*0229

*It is up to us to set what that base height means.*0231

*For that pillow that we just talked about, it is up to us to know that, that is the zero.*0234

*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.*0239

*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.*0257

*But either way, it is the relative difference, it is the change in 'h', not just the starting height on its own.*0268

*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.*0274

*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.*0282

*Formula also relies on the idea that 'g' is constant.*0291

*We can do this, because 'g' is reasonably constant near the surface of the Earth.*0295

*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.*0298

*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.*0315

*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.*0325

*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.*0332

*Finally, to be able to use anything in our energy formulae usefully, we are going to have the fact that energy is conserved.*0352

*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.*0357

*This means that energy in a system stays constant, unless it is transferred out of the system.*0367

*The energy in the system at the start + the work = energy of the system at the end.*0370

*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.*0376

*That is what work is, work that is positive, is energy going in, work that is negative, is energy coming out.*0386

*We are ready to tackle the examples.*0392

*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?*0394

*Then how fast we will be traveling just before the impact?*0400

*Say we have got this rock, and there is 20 m, and there is the 10 m that it falls first.*0403

*From here to here, what is the change in height?, the change in height = 10, and the change in height here = another 10.*0411

*How fast is it traveling 10 m above the ground?*0420

*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.*0422

*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.*0431

*For the first one, there is two ways of approaching this one:*0441

*We could look at this as either being, mgΔh = (1/2)mv ^{2} (we are talking about speed, since this is single dimension.)*0444

*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.*0458

*Remember, gravity works on everything, no matter what the mass is, it has a uniform acceleration, barring air resistance.*0478

*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.*0486

*That means, gΔh = (1/2)v ^{2}, sqrt(2gΔh) = v.*0499

*Plug in numbers, sqrt(2 × 9.8 × 10) = v = 14 m/s, at 10 m above the ground.*0512

*One other way that we could have done this though, is we could have said, E _{system}(beginning) + work = E_{system}(end), mgh_{i} = mgh_{end} + (1/2)mv^{2}, since work = 0, and also it has got some speed at the end.*0534

*If we move this over, we got, mgh _{i} - mgh_{end}, which is why we wind up pulling it out, and we look at, mgΔh.*0570

*It is the relative change that tells us how much energy is put into the system from gravitational potential.*0583

*That equals our (1/2)mv ^{2}, and that is why we did not have to put on both sides.*0588

*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.*0592

*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.*0602

*That is the velocity at 0 m above the ground.*0638

*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.*0642

*A 100 g is thrown from a height of 1.4 m above the ground at a speed of 10 m/s.*0659

*At a height of 2.5 m, it has a speed of 7.9 m/s.*0664

*How much of its energy has been lost to air drag?*0668

*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.*0670

*In this case, we will do my way though.*0679

*Energy _{system}(beginning) + work = Energy_{system}(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.*0681

*E _{system}(beginning) has two things, it has a height above the ground, and it also has an initial speed.*0704

*What is its initial height?, mgh _{initial} + (1/2)mv^{2} + work = mgh_{final} + (1/2)mv_{final}^{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.)*0711

*This frisbee, it starts off flying at some speed, and with some height, h _{initial}, 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.*0751

*Those are our two snap shots.*0777

*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.)*0779

*So, v _{initial} = 10, then, h_{initial} = 1.4, h_{final} = 2.5, and v_{final} = 7.9.*0822

*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}. *0840

*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.*0860

*Start calculating everything out, eventually, work = -0.802 J.*0919

*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.*0931

*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.*0937

*Example 3: Block of unknown mass is slid down a frictionless slide of height 5 m.*0952

*Then it slides along a table where it has a friction coefficient of μ _{k} = 0.3, we only have to know the kinetic friction, because it is already starting on motion.*0957

*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?*0968

*What is the initial energy in it?*0977

*E _{i} + work = E_{end}, so how much energy does it have at the end?*0981

*It is still, so it has no energy.*0994

*The work is going to have to suck out, is going to have to cancel out all the initial energy.*0996

*What energy does it have in the very beginning?, is it moving in the very beginning?*1001

*No!, it just starts, and then slides on a frictionless slide.*1006

*So, how much energy does it have?, it has mgh, its starting height worth of energy.*1010

*We have got, mgh = -W, work here is by friction.*1018

*How much is friction?, force of friction = μ _{k}×F_{N}.*1026

*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.*1032

*On the flat plane, the only thing pulling down on it is mg, and mg is also equal to F _{N}, so, friction = μ_{k}mg.*1039

*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.*1055

*So, mgh = -fdcos(180) = fd (cos(180) = -1).*1093

*So, mgh = μ _{k}×mg×d.*1116

*Look!, mg gets canceled on both sides.*1132

*So, h = μ _{k}×d, d = h/μ_{k} = (5 m)/(0.3) = 16.7 m, is the distance that it slides.*1135

*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.*1167

*So, the distance, starting here, once it gets to the friction table, is going to 16.7 m, because we know that, E _{i} + W = E_{end}, (gravitational potential energy) + [(friction force)×(distance it slides)] = 0, since still in the end.*1181

*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.*1208

*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.*1226

*It is so much easier to be able to work with μ _{k} 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.*1240

*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.*1273

*It is really helpful to be able to move things around as letters, and at the very end substitute in.*1286

*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.*1293

*The trapeze rope has been sabotaged by an evil villain, and it will snap at 800 N of tension.*1303

*Will our acrobat make it across safely?*1309

*This is a question that is important to his life.*1312

*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?*1316

*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.*1326

*Because everything is going to be at least less than that.*1355

*If everything is less than that, all we figure out is what the maximum tension put on it is.*1360

*Where will the maximum tension occur?*1365

*Now we have to think about, where do we have to get tension from!*1367

*The tension has to keep it in a circle.*1372

*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, force _{centripetal} = m×v^{2}/r (again, v is speed)*1375

*At what point on the circle will he be traveling the fastest?*1402

*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.*1411

*That is going to have the largest 'v'.*1421

*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.*1425

*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.*1435

*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.*1447

*We know that, E _{i} + W = E_{end}, 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.*1454

*If he is going to survive, we are going to have to look at that moment of maximum tension.*1473

*So we are going to make that ending snap shot be, when it is pointing directly down.*1478

*There is no work, since there is no friction, everything is smooth.*1481

*So, mgh = (1/2)mv ^{2}, m's cancel out, so we do not have to worry about the mass.*1491

*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.*1505

*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.*1528

*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.*1545

*We do not know precisely where yet, we could figure it out, that would be a more difficult question.*1553

*We know that, v _{bottom} = 9.9 m/s, what is the centripetal force necessary to keep him in a circle?*1557

*That is what it has to be, on the acrobat, force _{centripetal} = m×v^{2}/r, and we know m and r.*1564

*(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.*1577

*The centripetal force is pulling this way.*1597

*We have got tension, is pulling this way, and has to also be able to be some of mg.*1602

*We know that, T - mg = 359 N = centripetal force.*1609

*T = 359 + mg, at the moment of maximum tension that that rope would experience if it were a safe healthy rope.*1617

*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.*1628

*We do not know if he will make it yet, we got to finish this problem.*1638

*359 + mg = 359 N +(55 kg) × (9.8 m/s/s) = 898 N, SNAP!, he does not make it!*1641

*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.*1670

*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.*1684

2 answers

Last reply by: javier chichil

Wed Oct 9, 2013 1:29 PM

Post by javier chichil on October 8, 2013

hi Vincent:

Hope this question finds you well.

quick question, in minute 27:00, there is this formula T-mg= Fc.

From the diagram shown, i would have thought that it would be T+Fc= mg,

My idea came to mind because of the directions of the forces, T and Fc going up, and mg going down.

could you please comment on this?

thanks

1 answer

Last reply by: ahmed raza

Mon Nov 5, 2012 12:10 AM

Post by Diana Zafra on November 2, 2012

These examples are awesome. It makes physics more interesting when the examples are fun. Thanks!