# Lesson Video: Conservation of Energy

In this video we learn the principle of Conservation of Energy, as well as when and how this principle is applied, and that energy can be both potential and kinetic.

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### Video Transcript

In this video, we’re going to talk about the conservation of energy. We’ll learn what this principle says, what its limits are. And we’ll get some practical experience working with energy conservation.

To start out, imagine that you and a friend are busy shooting a music video. Standing on a single-story height by a driveway where a car is parked, you see it would make a great shot if you jumped off this height onto the roof of the car and then from there onto the driveway. The only possible issue you see, as this is a low-budget video production, is that you might damage the car and incur expenses.

To better understand whether getting this shot is a good idea, we’ll want to know more about the conservation of energy. The conservation of energy means that, within an isolated system, the total energy at any one time equals the total energy at any other time. Written as an equation, we can say that the initial system energy is equal to the final system energy. This principle is simple to remember, as long as we keep in mind the limits under which it applies.

The main thing to keep in mind is that energy conservation requires an isolated system, one where no energy enters or leaves. This idea of a system can best be explained through an example. Let’s say you hold a ball in your hand and drop it from some height above ground. As we normally think about this scenario, in our system, we might include the ball, your hand, and the earth. And in that case, we could show that the gravitational potential energy of the ball when it’s in your hand is equal to the kinetic energy of the ball just as it hits the ground. In other words, we could show that the energy of the ball is conserved. But what if our system was not isolated?

Imagine that there was a vertically downward moving wind that, as soon as the ball was released, helped push it towards the ground. In that case, the ball would speed up more than its initial potential energy will lead us to think. When it hit the ground, its kinetic energy would exceed its initial potential energy. It would seem as though energy is not conserved. But that’s only because we violated a condition of conservation of energy. Our system, in the case of the wind coming in, is not an isolated system. We’ve added energy.

When we look at the expression for energy conservation, we realize this covers all kinds of energy. Broadly speaking, energy comes in two main types: kinetic or energy of motion and potential energy. We can expand our energy conservation expression to include these two main types. So we can say that the initial kinetic and potential energy of a system together is equal to the final kinetic and potential energy of that system. And here initial and final are any two times in the evolution of the system. Let’s get some practice applying this conservation principle through a few examples.

A wagon sits at the top of a hill. The wagon is given a push that increases its speed negligibly but is just sufficient to set the wagon in motion down a straight slope. The wagon rolls 53.9 meters down the slope, which is inclined 16.5 degrees below the horizontal, and reaches the bottom of the hill. If friction is negligible, what speed is the wagon moving when it reaches the bottom of the hill?

We can call the speed of the wagon when it reaches the bottom of the hill 𝑣 sub 𝑓. And we’ll start on our solution by drawing a diagram of the situation. A wagon sits perched at the top of a long flat hill whose length we’ve called 𝐿. It’s 53.9 meters. The hill is inclined at an angle below the horizontal — we’ve called 𝜃 — given as 16.5 degrees. We’re told the wagon is given a very slight push to set it in motion and then rolls down the hill. And when it reaches the bottom of the hill, we want to solve for its speed, 𝑣 sub 𝑓.

In this scenario, our system consists of the wagon and the hill. And since energy is neither added to nor taken away from this system, we can say that it’s conserved. Applying the general statement of energy conservation to this scenario, we’ll use our freedom to decide when the initial moment and when the final moments are. Choosing the initial moment to be just when the cart is pushed at the top of the hill, and the final moment when it just reaches the bottom of the hill.

We can expand this conservation expression to say that the kinetic energy plus potential energy initially is equal to the kinetic energy plus potential energy finally. At the outset, the wagon is not in motion. So its initial kinetic energy is zero. And if we choose the height at the bottom of the hill to be our reference level of zero, we can say that the final potential energy of the cart is also zero, since it’s at zero height. So the wagon’s initial potential energy is equal to its final kinetic energy.

The wagon’s potential energy is entirely gravitational. And we recall that gravitational potential energy is equal to 𝑚 times 𝑔 times ℎ. Applying that relationship, we write 𝑚 times 𝑔 times the initial height of the wagon equals KE sub 𝑓. We also recall that an object’s kinetic energy equals half its mass times its speed squared. We can write this in for our KE sub 𝑓 expression.

Looking at this equality, we see that the mass of the wagon appears on both sides. So it cancels out. We want to solve for the final speed of the wagon. So we rearrange this equation and see the wagon’s final speed is equal to the square root of two times 𝑔 times ℎ sub 𝑖. The acceleration due to gravity, 𝑔, we take to be exactly 9.8 meters per second squared.

Looking at our diagram, we see that the height of the hill, ℎ sub 𝑖, relates to the hill length, 𝐿, and the angle of inclination, 𝜃. In particular, we can write that ℎ sub 𝑖 is equal to 𝐿 times the sin of 𝜃. So we substitute 𝐿 sin 𝜃 in for ℎ sub 𝑖 in our expression for 𝑣 sub 𝑓.

Looking at this expression, we’re given the length 𝐿, the angle 𝜃, and 𝑔 is a known constant. So we’re ready to plug in and solve for 𝑣 sub 𝑓. Entering these values on our calculator, we find that, to three significant figures, 𝑣 sub 𝑓 is 17.3 meters per second. That’s the speed of the wagon when it reaches the bottom of the hill.

Now let’s look at an example where, in our initial condition, we have both potential and kinetic energy.

A hiker being chased by a bear runs to the edge of a cliff and jumps off. The hiker is running horizontally at 4.5 meters per second when he reaches the cliff edge. The hiker’s speed when he hits the ground is 16 meters per second. What vertical downward distance did the hiker fall? If the hiker stepped forward off the cliff edge with negligible horizontal velocity, what speed would they hit the ground at?

So we want to solve for the distance the hiker falls. And we also wanna solve for the speed the hiker would have if the hiker left the cliff edge with negligible horizontal velocity. We can call this distance ℎ. And we can label the speed they hit the ground at 𝑣 sub 𝑔. We’ll start by drawing a diagram of the situation.

In the first scenario, our hiker is running away from the bear at a speed of 4.5 meters per second in the horizontal direction, before leaping off the edge of the cliff. We know that, under these conditions, the hiker’s overall speed when he hits the ground is 16 meters per second. And we want to use that information to solve for the height of the cliff, ℎ.

If we approach this from an energy perspective, we can say that the hiker’s energy, kinetic plus potential, just as he leaves the edge of the cliff is equal to his energy just as he lands. Those are our initial and final moments. In this system of the hiker and the cliff, energy is conserved. So we can write that 𝐸 sub 𝑖 is equal to 𝐸 sub 𝑓 and expand that expression to include kinetic and potential energies.

If we set our zero value for height at the base of the cliff, that means that when the hiker lands, he has no potential energy due to gravity. So that term goes to zero. It turns out that’s the only term we can eliminate because, initially, as the hiker leaves the cliff’s edge, he has both gravitational potential energy and kinetic energy due to his motion.

Recalling the equation forms of gravitational potential energy, 𝑚 times 𝑔 times ℎ, and kinetic energy, one-half 𝑚𝑣 squared, we can expand our energy balance equation. So it reads one-half the hiker’s mass times 𝑣 sub 𝑥 squared, the horizontal speed as the hiker leaves the cliff. Plus the hiker’s mass times 𝑔 times the height of the cliff is equal to one-half the hiker’s mass times 𝑣 squared. Where 𝑣 is the hiker’s overall speed when he lands. We see that the mass of the hiker appears in every term. So we can cancel it out. And since we want to solve for the height of the cliff, ℎ, we rearrange this equation to do so. And we find that ℎ is equal to 𝑣 squared minus 𝑣𝑥 squared all over two 𝑔, where we treat the acceleration due to gravity as exactly 9.8 meters per second squared.

Since we know 𝑣 and 𝑣 sub 𝑥, we’re ready to plug in and solve for ℎ. When we do and enter these terms on our calculator, we find that, to two significant figures, ℎ is 12 meters. That’s the height of the cliff from which the hiker jumps.

Next, we picture a slightly different scenario where, in this one, instead of the hiker running off the cliff’s edge at speed, the hiker jumps off from rest. We want to solve for the speed of the hiker when he hits the ground, 𝑣 sub 𝑔. And once again, our initial point will be when the hiker jumps. And the final point will be when the hiker lands.

Just like before, in this system, energy is conserved. But unlike before, the initial kinetic energy of our hiker is zero. In this case, our final potential energy will again be zero since we let our height zero reference point be at the ground level. Our equation, therefore, reduces to PE sub 𝑖 equals KE sub 𝑓, where the potential energy is entirely due to gravitation.

Referring once again to our gravitational potential energy and kinetic energy expressions, we write that 𝑚𝑔ℎ is equal to one-half 𝑚𝑣 sub 𝑔 squared. Once again, the mass of the hiker cancels out from both sides. And rearranging to solve for 𝑣 sub 𝑔, we find it’s equal to the square root of two times 𝑔 times ℎ, the height of the cliff. We know 𝑔 and we’ve solved for ℎ in the earlier part of the problem. So we’re ready to plug in and solve for 𝑣 sub 𝑔. When we enter these terms on our calculator, to two significant figures, it’s 15 meters per second. That’s the hiker’s speed when he reaches the ground after jumping from rest.

Let’s summarize what we’ve learned about the conservation of energy. We’ve seen that energy conservation means that, within an isolated system, energy is neither created nor destroyed. Written as an equation, we’ve said that the initial energy in a closed system is equal to the final energy in that system. We’ve also seen that system energy can consist of both kinetic or moving energy as well as potential energy. This led us to expand our expression for energy conservation to say that initial kinetic plus potential energy is equal to final kinetic plus potential energy. And lastly, we’ve seen that an isolated system is one that neither gains nor loses energy. It’s within this sort of system that energy is conserved.