High School Physics 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)mv² (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)v², 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)mv², 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)mv², 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)mv², 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)mv² + work = mgh_final + (1/2)mv_final² (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)×0.1×10² + 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_kmg.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×v²/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)mv², m's cancel out, so we do not have to worry about the mass.1491

So, gh = (1/2)v², 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×v²/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