# Video: Mechanical Energy Conversion

A ball with an initial velocity of 10 m/s rolls along a curved surface, as shown in the diagram. The mass of the ball is 100 g. Assume that the only energy conversions that take place are between the kinetic energy and the gravitational potential energy of the ball and calculate the speed of the ball at different positions to the nearest meter per second. Find the magnitude of 𝑣₁. Find the magnitude of 𝑣₂. Find the magnitude of 𝑣₃. Find the magnitude of 𝑣₄.

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### Video Transcript

A ball with an initial velocity of 10 meters per second rolls along a curved surface as shown in the diagram. The mass of the ball is 100 grams. Assume that the only energy conversions that take place are between the kinetic energy and the gravitational potential energy of the ball. And calculate the speed of the ball at different positions to the nearest meter per second. Find the magnitude of 𝑣 one. Find the magnitude of 𝑣 two. Find the magnitude of 𝑣 three. Find the magnitude of 𝑣 four.

Okay, so in this question, we need to find out the magnitudes of 𝑣 one, 𝑣 two, 𝑣 three, and 𝑣 four, which are the velocities of the ball at different points along the curve, the curve of course being the red line here. So one of the important things that we’ve been told is that the only conversions in energy that take place are between the kinetic energy and the gravitational potential energy of the ball.

In other words, we only need to deal with those two types of energy. We don’t need to worry about friction between the ball and the surface, for example, or anything like that. So that makes life a lot easier for us.

Now another piece of information we’ve been given is that the mass of the ball is 100 grams. Let’s call this mass 𝑚. And we’ll say that that is equal to 100 grams. However, this is not the most useful piece of information for us in this form. And the reason for that is that we first need to convert it to standard units.

Now the standard unit of mass is the kilogram. So we need to convert this mass from grams to kilograms. To do this, we can recall that one kilogram is equivalent to 1000 grams. So what we can do here is to divide both sides of the equation by 10. And this way, on the left-hand side, we’ll be left with 0.1 kilograms. And on the right, we’ll be left with 100 grams, which is exactly what we needed to convert into kilograms. So we can say that this mass, the mass of the ball, which is 100 grams, is instead equal to 0.1 kilograms.

And at this point, we can move on because we’ve now converted it to standard units. Okay, so at this point, let’s consider the ball when it’s first at this position here. Now at this position, we know the height of the ball above the ground. We know that this height is 25 meters. And we know the speed of the ball. It’s 10 meters per second.

Therefore, with this information, we can work out both the gravitational potential energy of the ball in this position and its kinetic energy. We’ll see whether this is useful in a second. But first let’s recall that the gravitational potential energy of an object, GPE, is given by multiplying the mass of the object 𝑚 by the gravitational field strength of the Earth 𝑔 by the height above the ground ℎ.

And in this case, the height above the ground, like we said earlier, is 25 meters. So at this position, which we’ll call position zero, the ball has a gravitational potential energy, which we’ll call GPE sub zero.

And this gravitational potential energy at position zero is equal to the mass of the ball, which is 0.1 kilograms, multiplied by the gravitational field strength of the Earth, which we can recall is 9.8 meters per second squared, multiplied by the height above the ground, which is 25 meters. And the good thing about this expression is that we’re using the standard units for mass, which is kilograms, the standard units for gravitational field strength, which is meters per second squared, and the standard units for the height above the ground, which is meters.

Therefore, the answer that we get for the gravitational potential energy is also going to be in its standard unit, which is the joule. So when we evaluate the right-hand side of the equation, we find that this evaluates to 24.5 joules. And this is the gravitational potential energy of the ball in position zero.

We can, therefore, move on to working out the kinetic energy of the ball. And to do this, we’ll recall that the kinetic energy of an object, KE, is given by multiplying half by the mass of that object 𝑚 by the velocity of the object 𝑣 squared. So in position zero, the kinetic energy, which we’ll call KE sub zero, is equal to half multiplied by the mass of the ball, which once again is 0.1 kilograms, multiplied by the velocity of the ball, which happens to be 10 meters per second squared. And this is the velocity that we’re using, the velocity in position zero.

Also, yet again, we’ve used the standard unit for mass, which is kilograms, and the standard unit for velocity, which is meters per second. So the kinetic energy we find is also going to be in joules. So evaluating the right-hand side, we find that the kinetic energy of the ball in position zero is five joules. And at this point, we know both the gravitational potential energy of the ball and the kinetic energy of the ball at position zero.

Now since the question tells us that these are the only two types of energy that we need to worry about, we can therefore work out the total energy of the ball at position zero. So the total energy of the ball at position zero, which we’ll call 𝐸 sub zero, is equal to the gravitational potential energy at position zero plus the kinetic energy at position zero. And this happens to be 24.5 joules, which is the gravitational potential energy, plus five joules, which is the kinetic energy. And so the total energy of the ball in this position is 29.5 joules.

Now why is this relevant? Well, it’s because we can use the law of conservation of energy. What conservation of energy tells us is that energy is neither created nor destroyed. In fact, it can only be converted from one form to another. But if energy cannot be created or destroyed, then the total energy of the ball must remain the same throughout its journey. In other words, the total energy of the ball in position zero is the same as the total energy of the ball in position one and position two and position three and position four and anywhere else along the curve.

Now let’s think about this carefully. We’re not saying that the gravitational potential energy stays the same throughout or the kinetic energy stays the same throughout. In fact, the gravitational potential energy and kinetic energy individually change all the time. But what we’re saying is that the total energy or the sum of the gravitational potential energy and the kinetic energy has to stay the same throughout.

So in this case, what happens is, say, for example, we start here at position zero. And the ball is moving along and it starts going towards position one. Well, in this case, it’s moving downwards along the slope. So it’s losing height. And we see that it goes from 25 meters to 15 meters above the ground. So it’s losing gravitational potential energy. But whatever gravitational potential energy loses, it gains as kinetic energy. And this is how the total energy of the ball stays constant. And this is something that we’ll be able to exploit in order to work out the values of 𝑣 one, 𝑣 two, 𝑣 three, and 𝑣 four.

Now at this point, we don’t need to know the gravitational potential energy and kinetic energy at position zero. Instead, what we do need to know is the total energy of the ball. So let’s put a little box around it. And let’s look at position one first of all.

Now as we’ve already said, the total energy of the ball stays the same. So we can write down an equation that tells us that the gravitational potential energy of the ball in position one this time plus the kinetic energy of the ball in position one is equal to the energy, the total energy, in position one.

But as we’ve already said, the total energy stays the same throughout its journey. So 𝐸 sub one is the same as 𝐸 sub zero. So GPE sub one plus KE sub one is equal to 𝐸 sub zero. And we can substitute in the expressions for the gravitational potential energy and the kinetic energy. So 𝑚𝑔 multiplied by the height at position one, which we’ll call ℎ sub one, plus half 𝑚 multiplied by the velocity at position one, which we’ll call 𝑣 sub one squared, is equal to 𝐸 sub zero.

Now in this equation, we already know the value of 𝑚 as well as 𝑔 as well as ℎ one. We’ve been given ℎ one here. And we know 𝑚 once again. We don’t know what 𝑣 one is. But we do know what 𝐸 zero is. In other words, there’s only one unknown in this equation. That’s 𝑣 one. So we can rearrange to find out what 𝑣 one is.

Firstly, we’ll write down the value of 𝐸 sub zero over here to give us a little bit more space to work with. And then what we’ll do is start to rearrange. Firstly, we can subtract the value of 𝑚𝑔ℎ sub one from both sides of the equation. That way, 𝑚𝑔ℎ sub one cancels on the left. And what we’re left with is that half multiplied by 𝑚 multiplied by 𝑣 one squared is equal to 𝐸 naught minus 𝑚𝑔ℎ sub one.

Then what we can do is to multiply both sides of the equation by two over 𝑚. This way, the half cancels with the two in the numerator. And the mass on the left-hand side also cancels. So we’re only left with 𝑣 one squared on the left-hand side. And then at this point, all we need to do is to take the square root of both sides. On the left, that just leaves us with 𝑣 one. And on the right, we’re left with an expression to help us find 𝑣 one.

Now it’s important to note that this expression is in terms of ℎ one, the height at position one. And this is really useful because we can then apply this to position two, three, and four, simply by changing the values one here and here to two or three or four. For now though, let’s keep it as 𝑣 one because that’s what we’re trying to find out first.

So 𝑣 one is equal to square root of two over 𝑚 multiplied by 𝐸 naught minus 𝑚𝑔ℎ one. Let’s plug in all the values on the right-hand side. 𝑣 one is equal to two divided by the mass, which is 0.1, the stuff inside the parentheses 29.5 joules, which is 𝐸 naught minus 𝑚 times 𝑔 times ℎ one. And at this point, we can evaluate this to find that 𝑣 one is equal to 17.20 meters per second.

However, we need to give our answer to the nearest meter per second. So we need to round to this value just before the decimal. Now the value after the decimal is a two. So the seven is going to stay as a seven. It’s not going to round up. Hence, to the nearest meter per second, the value of 𝑣 one is 17. And this is our answer to the first part of the question.

Let’s now move on to finding 𝑣 sub two. And we can do this by replacing the one in this equation with two. Now we can see that the value of ℎ two, the height at position two, is 10 meters. So we plug that value in and keep everything else the same. And we find that 𝑣 two ends up being 19.84 meters per second.

But once again, we need to get our answer to the nearest meter per second. So we’re looking at rounding this value here. Now the number after it is an eight. Eight is larger than five. So this nine is going to round up. In other words, this 19 is going to become a 20. So to the nearest meter per second, our answer is 20. And this is our answer to the second part of the question.

Now we can repeat this process for 𝑣 three to give a value of 14.07 meters per second, which, to the nearest meter per second, is simply 14, and repeat it once again for 𝑣 four, which happens to be 24.28 meters per second. Now once again, to the nearest meter per second, this rounds to be 24. And at this point, we found the velocity of the ball in all four positions along the curve. So we’ve reached the end of our question.