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Hello, and welcome to this lesson about calculating energy changes.

It's from the physics unit Energy of Moving Objects, and my name is Mr. Fairhurst.

In this lesson, you're going to find out about how you can use a principle of conservation of energy to calculate changes of energy of moving objects, and in particular, you're going to think about how energy is transferred between the gravitational store and the kinetic store.

These are the key words that you're going to come across during the lesson.

Conservation of energy is the principle that energy can not be created or destroyed, but it can only be transferred from one store to another store, and when it's transferred, some energy's usually dissipated.

It's transferred into the surroundings by warming them up slightly.

The energy that an object has in its gravitational store is what we call its gravitational potential energy, and the energy an object has in its kinetic store is what we call its kinetic energy.

These are the definitions of those keywords.

Any point during the lesson you want to come back and have a look at these and remind yourself of what the definitions are, just pause the video and come back to this slide.

The lesson's divided into three parts.

In the first part, we're going to consider how energy is transferred between stores, in particular, between the kinetic store and the gravitational store.

We're then going to look at how the conservation of energy affects this transfer of energy, and then, in the final part of the lesson, we're going to use all of those ideas in order to calculate the amount of energy transferred in a range of different situations.

Okay, then let's make a start with the first part.

We'll start off by thinking about how energy can be transferred from one store to another store, and we'll do that by considering this toy cannon.

Inside the cannon is a spring that, when it's compressed, the toy cannon has got energy in the elastic store, and when the spring is released, the cannon fires a small ball up into the air.

Now when that spring is compressed and before it's released, the cannon has got a lot of energy in the elastic store, and it's got a little bit of energy in the gravitational store because the ball is raised a little bit up into the air, and then when the spring is released, the ball is shot up into the air and energy is transferred between the stores.

All of the energy, or most of the energy, in the elastic store is transferred into making the ball move, so it's got energy now in the kinetic store and the ball has also risen up into the air, so some energy has also been transferred into the gravitational store.

Now I've got a couple of questions I'd like you to have a go at.

This first one, what energy transfer takes place between firing the toy cannon and the moment the ball reaches its maximum height when it's shot straight up into the air? Just pause the video whilst you choose your answer and then start again once you're ready.

Okay, so how did you get on? The ball shot straight up.

It starts with energy in the elastic store inside the cannon and at it's maximum height, it comes to a momentary stop before it starts falling back down again, so it's only got energy in its gravitational store, so the correct answer is B.

Well done if you've got that one.

This is the second question.

Again, have a look at the question, pause the video whilst you think about it and choose your answer, and start again once you're ready.

Right then, this time you were asked what happens when the cannon is fired horizontally, and at the moment the ball has just left the cannon before it starts to fall.

All of the energy in the elastic store has now been transferred to making the ball move, and it hasn't changed its height yet, so no energy has been transferred into the gravitational store, so this time, the correct answer is A, energy has been transferred from the elastic store just into the kinetic store.

So well done if you've got that answer correct.

The point of this lesson is that we want to be able to calculate the amount of energy transferred between stores, and we're gonna start from this principle that the total amount of energy after firing the cannon, or after an energy transfer, is always the same as a total amount of energy at the start.

Now in this case, if we add up before the cannon was fired the energy in the elastic store and the gravitational store, and compare that to the energy in the different stores afterwards, then what we find is we get the same total amount of energy both times.

This is the principle of conservation of energy.

We can't destroy or create energy, we can only transfer it from one store into another store, or from a few stores into other stores.

Now, if we were to ignore the energy that's dissipated when energy is transferred, it will be quite straightforward to calculate the amount of energy transferred, but energy is dissipated.

When this toy cannon is fired, there's some friction between the ball and the cannon that heats up the barrel of the cannon, and as the ball moves through the air, there's air resistance between the ball and the air, and this makes it really difficult and challenging to calculate the amount of energy transferred in this instance to the gravitational store when that ball reaches its maximum height.

And that's because we don't know exactly how much friction there was, we don't know how much air resistance there was, so we can't easily calculate that, but what we can do is we can use the principle of conservation of energy to calculate the maximum height that the ball could have reached if there was no air resistance and friction involved at all, if we ignore those.

So that's what we're to do in a lot of these physics calculations.

We're going to ignore the energy that's dissipated so that we can work out the maximum transfer of energy, assuming that these were non-existent, or very, very low.

Have a look at this question and see what you think.

What is the maximum amount of energy transferred to the kinetic store as they gannet dives into the water, assuming that we can ignore the air resistance that's involved in this case.

Pause the video as you think about this one, and start again once you're ready.

Okay, then so what do you think? The correct answer is exactly 98 joules of energy.

If the gannet starts off with 98 joules of energy in the gravitational store, as it dives down and reaches the surface of the water, it's got no energy left in the gravitational store, and if none of that energy has been dissipated, then all of that energy will be transferred into the kinetic store, into making the gannet move faster and faster.

Gannet will enter the water at an incredibly high speed, so well done if you've got that right.

What I'd like to do now is this task, I want you to think about a heavy pendulum that's swung from a height of two metres, 2.

00 metres in the air, it swings down, speeds up, and then goes back again to a maximum height where it pauses.

What I'd like you to do is to have a go at those questions and then once you've got the answers, start the video again and we'll check your answers.

Okay, so how did you get on? The first question was where will the pendulum have the maximum speed? Well, as it's moving down and the energy's been transferred from the gravitational store into the kinetic store, it's speeding up, and then as it rises again from the lowest position, energy's been transferred back from the kinetic store and into the gravitational store, so it's losing speed, so its maximum speed will be at the bottom of the swing.

Question two says what is the maximum height it will swing up to? Well, we're assuming here that no energy has been transferred to the surroundings, no energy has been dissipated.

So all the energy in the gravitational store at the start was transferred to the energy in the kinetic store at the bottom, and it's been transferred back to the energy in the gravitational store at the top.

So it's got the same energy in the gravitational store at the end that it had at the start.

In other words, it's at the same height it was before, 2.

00 metres.

Question three was to explain why it can't reach the maximum height, why is it not going to raise up again to exactly two metres? And that's because in real life, there's air resistance, there's friction where the string is attached, the pendulum is attached to the pivot at the top, so that's gonna cause rubbing and it will slow the pendulum down, and it will dissipate energy into the thermal store of the surroundings rather than the gravitational store again.

So that essentially is the answer.

Well done if you've got a good description of that.

We'll now go into part two of the lesson, where we think a little bit more carefully about the conservation of energy and what that tells us about the quantity, the amount of energy in different stores.

To do that, we're going to think about a roller coaster.

I don't know if you've ever been on a roller coaster, but you'll notice that as you get pulled to the top, there's some sort of mechanism that cranks you to the very top, and at the top, there's that moment when the rollercoaster is released from the mechanism just before it starts to fall, and then it freewheels all the way along the track.

It just coasts along, hence the name a rollercoaster.

Now, what we can calculate about the rollercoaster is the amount of gravitational potential energy it has at the start, and we can calculate it using this equation, the gravitational potential energy equals the mass of the rollercoaster time the gravitational field strength times by its height above the ground, so we can calculate the energy it's got at the start before it starts moving and coasting down the track.

Now, as it rolls along the track, friction and air resistors will cause some energy to dissipate into the surroundings, and when it reaches the top of the next rise, the rollercoaster will have energy in the gravitational store and the kinetic store, and there'll be some energy in the thermal store of the surroundings as well.

Now, bearing that in mind and thinking about a real rollercoaster, what I'd like you to do is to answer this question.

Just pause the video whilst you do so and start again once you're ready.

Okay, so how did you get on? We can imagine the rollercoaster starting at the high point there on the left, rolling along, some energy is dissipated, and it will climb up the right hand side and it will stop when it reaches its maximum height.

When it's run out of energy in the kinetic store, it's all been transferred back into the gravitational store and the correct answer here is B.

It won't be quite as high as it started from because some of the energy has been transferred into the surroundings, it's been dissipated.

So well done if you've got that answer.

What I'd now like to do is to think of the same question, but this time, for an ideal rollercoaster, one in which no energy is dissipated, what height will it reach now if there's no dissipation of energy? Pause the video and start again once you're ready.

Okay, so how did you get on? The correct answer here is C, the same height as when it started because no energy's been transferred into the surroundings, so it's got the same total amount of energy at the end as it had at the start, and where it stops at C, so therefore, it's got all of its energy in the gravitational store.

So well done if you've got that one.

That means that for an ideal rollercoaster in which no energy is dissipated, all of the energy in the gravitational store at the top is transferred to the kinetic store at the bottom, and that idea helps us to calculate the amount of kinetic energy it's got at the bottom, and the amount of kinetic energy is exactly equal to the amount of gravitational potential energy added at the top, and the gravitational potential energy is equal to the mass of the rollercoaster times the gravitational field strength times the height that it falls from the start.

What I'd like you to do is to have a quick look at this question and see what you think.

Pause the video whilst you do it and start again once you're ready.

Okay, so how did you get on? A ball was calculated to have 80 joules of gravitational potential energy and is dropped.

How much energy did it have in its kinetic store when it hits the floor? And the correct answer is exactly 80 joules.

We're of course assuming that there's no air resistance and so that no energy has been dissipated.

So well done if you got that right.

I'd now like you to think about the same ball as it hits the ground, but this time, rather than hitting the ground, it's hitting a trampoline.

How much energy will be in the elastic store when the ball first stops moving? Pause the video and start again once you're ready.

Okay, so this time, as the ball hits the trampoline, it stretches the elastic of the trampoline and the energy from the kinetic store is transferred into the elastic store, and again, because no energy dissipates, exactly 80 joules of energy is transferred into the elastic store.

So again, well done if you've got that right.

What I'd now like you to do is to have a go at these four questions, and where it's appropriate, show your working out.

Pause the video and start again once you've got all your answers ready.

Okay, so how did you get on? Let's start with the first question.

Calculate the gravitational potential energy of the ball at the top of its flight.

The ball's got a mass of 0.

2 kilogrammes, and it's shot up into the air with a catapult and it reaches a maximum height of 15.

0 metres.

So to work out its gravitational potential energy, we can use the equation gravitational potential energy equals the mass times the gravitational field strength, which here is 10 newtons per kilogramme, times by the change in height.

And if we do the sums, we get 30 joules of energy.

So well done if you've got that one right.

The next questions all say state.

That means that you don't need to do a calculation.

So question two here, state how much energy was in the kinetic store the moment the ball left the catapult? Well, if no energy is dissipated, it's got the same amount of energy as it has when it reaches the top height, and it's got no energy left in the kinetic store.

So the correct answer there was 30 joules.

The amount of energy it had in the elastic store just before the catapult was released.

Again, that is all of the energy, which is 30 joules, and state again how much energy was in the kinetic store the moment the ball hit the ground.

So it's reached the top, it had 30 joules of energy, it then starts speeding up and hits the ground, having lost all of that gravitational potential energy and transferred it into kinetic energy, so the answer is again 30 joules.

So well done if you've got all of those answers correct.

In the final part of the lesson, we're going to use the equation for calculating the kinetic energy of an object and the equation for calculating the gravitational potential energy of an object together with the principle of conservation of energy in order to calculate energy transfers.

And these are the ideas that we're going to use to do that.

First of all, for an object that changes height, we know that we can calculate the energy transfer to or from the gravitational store using the gravitational potential energy equation, which is the gravitational potential energy equals the mass times of gravitational field strength times by the change in height.

We also know for an object that changes speed, that we can calculate the energy transferred to or from the kinetic store using the equation for kinetic energy of the object, which is the kinetic energy is equal to a half times the mass of the object times its speed squared.

And if we assume that there's no energy dissipated, then when an object changes height or speed, we know that the change in gravitational potential energy is equal in size to the change in kinetic energy, so those two equations are equal to each other and energy is conserved.

So let's put those together and have a think about what happens.

Have a go at this question and see what you think.

Pause the video whilst you have a go and start again once you're ready.

Okay, how did you get on? Which equation or equations need to be used to calculate the amount of energy in the kinetic store when the ball hits the ground? Well, we can calculate from the information we're given the gravitational potential energy of the ball before it was released.

We've got its height, its mass, and the gravitational field strength, and we know that when it hits the ground, all of the energy in its gravitational store will have been transferred to the kinetic store.

So we know the energy in the kinetic store is equal to the energy we calculate for the gravitational potential energy equation.

So we just need equation A.

So well done if you've got that right.

Let's have a look at this question.

A rollercoaster has a mass of 850 kilogrammes.

How much energy will it have in its kinetic store after falling 40 metres from rest? Well, we've got a mass, we've got a height, we've got the gravitational field strength, so we can calculate the gravitational potential energy before it was released, and we know that that's going to be equal to the kinetic energy that's been transferred to the kinetic store as it's falling.

So we can start by stating that the change in gravitational potential energy is equal to the change in kinetic energy, and then calculate the gravitational potential energy using this equation.

If we substitute in the values for mass, gravitational field strength, and the height, we get the energy equals 850 kilogrammes times 10 newtons per kilogramme times 40 metres, which equals 340,000 joules, or 340 kilojoules.

Now we need to just finish off the answer by stating that that is the amount of energy in the kinetic store.

What I'd now like you to do is to have a go at this question.

Pause the video whilst you have a go and start again once you've worked out your answer.

Okay, so how did you get on? A cyclist has got a mass of 75 kilogrammes.

How much energy will she have in the kinetic store after freewheeling down a 23-meter hill from rest? Well, if we start off with this relationship, the change in gravitational potential energy equals a change in kinetic energy, this tells us that we're going to, first of all, calculate the gravitational potential energy that she starts with, and equate that with a change in kinetic energy.

So let's calculate the gravitational potential energy using this equation, substitute the values from the question and calculate the amount of energy, and we get 17,250 joules, or 17 kilojoules rounded up to two significant figures, of energy that she starts with in the gravitational store.

And all of that energy is transferred into her kinetic store as she freewheels down the hill, so our final answer is 17 kilojoules in the kinetic store.

So very well done if you've got that answer.

Now, let's have a look at this question.

This is slightly different.

A ball of mass 0.

5 kilogrammes is kicked up into the air with a speed of eight metres per second.

How high will it reach? At first it seems quite challenging, but let's think about it from the way we've been doing it before.

The mass is 0.

5 kilogrammes, it's got a speed of eight metres per second.

So can use those values to calculate its kinetic energy, and we know that the kinetic energy it starts with will be transferred into the gravitational store, and it will be completely transferred once the ball reaches the maximum height.

So we can start with the same statement we've had before, the change in gravitational potential energy is equal to the change in kinetic energy, but this time, we start by calculating the kinetic energy, and by substituting the values into this equation, we get the kinetic energy equals a half times 0.

5 kilogrammes times eight metres per second times eight metres per second, and that gives us 16 joules of energy in the kinetic store at the moment the ball was kicked, and that's going to be equal to the energy in the gravitational store at the moment the ball reaches its highest point.

So we can now use the equation for gravitational potential energy with this value for energy and the other values from the question, and as you see, the only missing value is the height, so we can use this equation to calculate the height, which is equal to 3.

2 metres.

Now, here's the question for you to have a go at.

Pause the video whilst you do that and start again once you've got your answer.

Okay, so how did you get on? A toy rocket with a mass of 1.

2 kilogrammes is shot up into the air with a speed of 45 metres per second.

How high will it reach? Well, again, we're going to use the idea of the change in gravitational potential energy being equal to the change in kinetic energy, and start by calculating its kinetic energy using this equation.

We can substitute in the values and work out the answer, which is 1,215 joules of kinetic energy at the moment it was launched.

And that is going to be equal to the gravitational potential energy at its highest point, so we can use the equation for gravitational potential energy, substitute in the values, and use those to calculate the height that it reaches, which is 101.

25 metres, which rounded to two significant figures is 100 metres.

So well done if you've got that answer.

This question is a little bit more challenging.

At what speed is a 0.

5 kilogramme ball kicked up at if it reaches a height of 10 metres where the gravitational field strength is 9.

8 newtons per kilogramme? Well, we're given the mass and we're given the height the ball is kicked up to, so we can calculate the gravitational potential energy of the ball when it reaches that maximum height.

But we're asked at what speed was it kicked up at? Well, we know that the gravitational potential energy when it reaches its maximum height is equal to the kinetic energy it has at the moment it's kicked up from the ground.

So we can use this relationship, and first of all, calculate the gravitational potential energy in order to substitute that value into the kinetic energy equation and to calculate the speed.

So let's do that.

There's the gravitational potential energy equation, and we can put in the values from the equation, so that's equal to 0.

5 kilogrammes times 9.

8 newtons per kilogramme times 10 metres, and if we calculate that out, we get 49 joules.

And that's the energy of the ball at its highest point, which is equal to the energy it has, the kinetic energy, when it's kicked up initially.

So what we can now do is to substitute that value into the kinetic energy equation along with the mass of of the ball, so we get 49 joules equals a half times 0.

5 kilogrammes times the speed squared, and we can rearrange that equation to calculate the speed squared, which is 196, and if we square root both sides, we get the speed, which is 14 metres per second.

What I'd like you to do is to have a go at this question, which is very similar, pause the video whilst you do so and start again once you're ready to go on.

Okay, so how did you get on? You're asked to find the speed that a toy rocket of mass 1.

8 kilogrammes is launched at if it reaches a height of 150 metres.

So we know that the gravitational potential energy it has at 150 metres is equal to the kinetic energy it has when it's launched, so we need to use this relationship again and start off by calculating the gravitational potential energy it has at its maximum height.

So we can put in the values we're given from the equation and then we can calculate, that's how, and find that it's gravitational potential energy at 150 metres is 2,646 joules, and that's equal to the kinetic energy it has at the moment it's launched.

So knowing that, we can start with this equation for the kinetic energy, and then substitute the values in and calculate an answer for the speed squared, which is 2,940, and if we square both sides, we get an answer for the speed, which is 54 metres per second.

So very well done if you got that right.

I'd now like you to have a go at these questions.

Don't forget to show all your working out.

Pause the video and start again once you've got all of your answers.

Okay, how did you get on? Let's have a look at your answers.

Question one, a skydiver has a mass of of 68 kilogrammes.

How much energy will she have in the kinetic store after falling 50 metres from rest? Well, we've got a height, we've got a mass, we've got the gravitational field strength so it can calculate the gravitational potential energy she had at the start, and that's going to be equal to the kinetic energy she has after she's fallen 50 metres.

So we'll use this equation to calculate the gravitational potential energy, substitute in the values, calculate them, it's 34,000 joules or 34 kilojoules, and we need you just to state at the end that that is the amount of energy in the kinetic store.

So well done if you've got that answer.

For question two, a cyclist of mass 84 kilogramme stops pedalling at the bottom of the hill.

If he goes up the hill and reaches the height of 4.

05 metres, how much kinetic energy did he have at the bottom of the hill before he started slowing down? Well, again, we've got the relationship, the change in gravitational potential energy is equal to the change in kinetic energy.

We can calculate this time the gravitational potential energy he has when he comes to a stop at his highest point.

And we can use the same equation for gravitational potential energy again, substitute the values in, and when he comes to rest at the highest point, we find he's got an energy of 3,405 joules, or to two significant figures, that is 3.

4 kilojoules, and that is going to be equal to the kinetic energy he had before he started slowing down.

So to get the mark, we simply need to state that point.

So very well done if you've got that right.

Question three, a stunt double of mass 74 kilogrammes is shot vertically out of a cannon at a speed of 22 metres per second.

What height will they reach? Well, for this question, we've got a mass and we've got a speed, so we can calculate the kinetic energy that the stunt double had the moment they left the cannon, and we can say that that is equal to the gravitational potential energy when they reach the top.

And we can use the gravitational potential energy equation then to calculate the height they reached.

So we'll start with the kinetic energy equation, substitute the value we've got in there, so we've got the kinetic energy is a half times 74 kilogrammes times 22 metres per second times 22 metres per second, and that gives us 17,903 joules of kinetic energy, which is going to be equal to the gravitational potential energy when the stunt double reaches their highest point.

So we can now substitute the values into the equation for gravitational potential energy.

The energy is 17,903 joules, and that is equal to 74 kilogrammes times 10 newtons per kilogramme times the height.

We can use that to calculate the height, which comes out as 24.

2 metres, which rounds to two significant figures to be 24 metres.

So well done if you've got that right.

Question four, a volleyball of mass 280 grammes, or 0.

280 kilogrammes, is hit up into the air and reaches a height of 3.

80 metres.

How fast was it moving the moment after it was hit? Well, the gravitational potential energy at its maximum height is equal to the kinetic energy it has at the moment it was hit, so we need to use this relationship.

We start off by calculating the gravitational potential energy at its maximum height, and we can use the values in the equation, substitute it in to work that out, remembering, of course, to use the mass in kilogrammes.

If we work that out, we get the gravitational potential energy of 10.

64 joules at its maximum height, which is equal to the kinetic energy it has at the moment it was hit, so we can use the kinetic energy equation, substitute the value I've just worked out in and the mass of the ball, and use that to calculate the speed squared, and the speed squared is 76.

If we square root both sides, then we find the answer is 8.

7178 metres per second, and I've rounded that up to two significant figures, 8.

7 metres per second.

Now, you might point out that in the question, both the values given to you were to three significant figures, so shouldn't the answer also be to three significant figures? And my answer to you is no, it shouldn't, because the gravitational field strength, 10 newtons per kilogramme, was only given to two significant figures, and we have to go with our answer to the value that we used in the calculation that had the fewest number of significant figures.

So in this case, 8.

7 metres per second to two significant figures is the correct answer.

Well done if you got that, and extremely well done if you've got all four questions right.

So well done for reaching the end of the lesson.

This is a short summary slide of the key learning points from the lesson.

For an object that changes height, we found that the energy transferred to or from the gravitational store is equal to the mass times the gravitational field strength times the change of height.

We found that for an object that changes speed, the energy transferred to or from the kinetic store is equal to half the mass times the speed squared.

And we also found that if no energy is dissipated as an object changes height and speed, then the change in gravitational potential energy is equal to the change in kinetic energy.

So very well done for reaching the end of the lesson again.

I do hope to see you next time.

Goodbye.