In this video, we’ll learn how to convert different types of mechanical energy to and from each other and also how to recognize when mechanical energy is dissipated. In the process, we’ll learn what these terms, energy conversion and energy conservation, mean. Since we’re talking about mechanical energy, let’s define that term. The mechanical energy of an object is the sum of its potential and kinetic energies. In a system that consists only of mechanical energy, if the system is closed, then that energy is conserved.
Imagine, for example, we have a system consisting of a ball and a ramp. When the ball is at rest at the top of the ramp, all of the system’s mechanical energy is gravitational potential energy. It equals the mass of the ball times the acceleration due to gravity times the height of the ball above some reference level. Now, if we give this ball a nudge so that it rolls down the slope, we know that it will pick up speed until it’s falling at a maximum speed at the bottom of the ramp we’ll call 𝑣. At this instant in time, we can see that the ball no longer has any gravitational potential energy. But all its energy is now kinetic. It’s equal to one-half 𝑚 times 𝑣 squared. And if we assume that no energy is dissipated as the ball rolls down this incline, then we can say that its mechanical energy is conserved and therefore that 𝑚 times 𝑔 times ℎ must be equal to one-half 𝑚𝑣 squared.
What we’re seeing is an example of energy conversion. The ball’s energy, which began as gravitational potential energy, is converted to kinetic energy. And because no energy was dissipated in the process, energy was also conserved. This is why we were able to write that 𝑚 times 𝑔 times ℎ equals one-half 𝑚𝑣 squared. Now that we’ve seen a system where mechanical energy is conserved, let’s look at one where it isn’t.
Say that we have a mass 𝑚 on the end of a vertical spring with spring constant 𝑘. We’ll consider this mass at two different instants in time as it occupies two different equilibrium positions. First, we’ll say the spring is compressed so that the mass is at a height we’ll call zero. And then the spring is released and comes to rest at its natural length. Considering the mechanical energy of the system, when the spring is compressed, we’ll call that energy 𝐸 sub 𝑖. We know this is equal to the system’s initial potential energy plus its initial kinetic energy. At this instant in time, our mass and spring are not in motion and, therefore, the kinetic energy of our system is zero.
We see further that we define the height of our mass at this instant to be zero, and therefore it has no gravitational potential energy. The only type of potential energy our system has initially is spring or elastic potential energy. This is equal to one-half the spring constant 𝑘 times the displacement of our mass 𝑥 from equilibrium squared. So, the mechanical energy of our system at this initial instant is one-half 𝑘𝑥 squared.
Considering the system next when the spring has extended to its natural length and the mass is at rest, if we call the mechanical energy at this moment 𝐸 sub 𝑓, we once again find that the kinetic energy of the system at this instant is zero and that the system’s potential energy is not zero. However, the potential energy of this system has gone through an energy conversion. Initially, it was all elastic or spring potential energy, whereas now that the spring is at its natural length and the mass has been raised a distance 𝑥, spring potential energy is zero, but gravitational potential energy is not. That’s the energy conversion that has taken place. And let’s imagine that rather than being equal, one-half 𝑘𝑥 squared is greater than 𝑚 times 𝑔 times 𝑥.
In other words, our system started out with more mechanical energy than it ended up with. When this happens, we say that mechanical energy has been dissipated. In this instance, that might happen due to friction in the coils of the spring. In that case, energy that might have been converted to gravitational potential energy is instead dissipated as heat.
We can now make a statement about the conservation of mechanical energy. We can say that in a closed system, one where energy is neither added to nor taken away from the system, if that system starts and ends with only mechanical energy, then that mechanical energy is conserved. We’ve seen an example of such a system, our ball rolling down the incline, and also an example of the system where mechanical energy is not conserved. Knowing all this, let’s look now at an example exercise.
A car is initially at rest before it starts to roll along a downward-sloping road with its engine turned off. While rolling, the car’s velocity increased by 1.4 meters per second. What vertically downward distance does the car travel? Gravity is the only force that acts on the car.
Let’s say that this is our car at the initial moment when it’s at rest before it starts to roll downhill. After rolling for some time, we’re told that the car’s velocity is increased by 1.4 meters per second. We want to know what is the vertically downward distance, we’ll call it 𝑑, that the car has traveled for this change to take place. To solve for this distance, let’s recognize that this scenario involves the conservation of mechanical energy.
In general, a system’s mechanical energy equals the sum of its potential and kinetic energies. In our scenario, we’re working with a closed system, one where energy is neither added to the system nor taken away. Along with this, the initial and final energy of the car can be expressed purely in terms of mechanical energy. This means that the mechanical energy of our system consisting of the car and the road is conserved. We can write then that the initial mechanical energy of our system equals its final mechanical energy. We can expand this equation in terms of initial and final potential and kinetic energies.
We’re going to say that the initial moment in our system is when our car is positioned here at rest. The final moment we’ll say is when it has achieved a speed of 1.4 meters per second downhill. Considering our car at its initial position, we know that because it is at rest, it will have zero kinetic energy. Therefore, KE sub 𝑖 is zero. Likewise, if we choose to set the elevation of the car at its final moment at a height of zero, then at this final moment, the car’s gravitational potential energy will be zero. Since the car possesses no other kind of potential energy, for example, spring potential energy, we can say that PE sub 𝑓 is zero.
All of that then brings us to this expression, the initial potential energy of our system, specifically gravitational potential energy, equals its final kinetic energy. Clearing a bit of space on screen, we can recall that in general, an object’s kinetic energy equals one-half its mass multiplied by its speed squared. And along with this, an object’s gravitational potential energy equals its mass times the acceleration due to gravity multiplied by its height relative to some reference level. In our case, this height ℎ is the distance 𝑑 we’re trying to solve for.
So then, based on this equation, we can write that 𝑚 times 𝑔 times 𝑑, the car’s initial mechanical energy, equals one-half 𝑚 times its final speed squared, its final mechanical energy. Note that in this equation, mass is common to both sides and therefore can cancel out. If we then divide both sides by the acceleration due to gravity 𝑔, that value cancels on the left and we find that 𝑑 equals 𝑣 squared over two times 𝑔. 𝑣 is equal to the final speed of our car, 1.4 meters per second. And we remember that the acceleration due to gravity is 9.8 meters per second squared. Therefore, 𝑑 equals 1.4 meters per second all squared divided by two times 9.8 meters per second squared. This is exactly equal to 0.1 meters. That’s the vertically downward distance this car needed to travel for its speed to increase by 1.4 meters per second.
Let’s look now at a second example exercise.
Which of the graphs (a), (b), (c), and (d) correctly shows the changes in kinetic energy shown in red and the gravitational potential energy shown in blue for a ball being thrown vertically upward and falling back to earth? The time axis of the graph starts with the instant the ball leaves the thrower’s hand and the energy values cease to be plotted at the instant that the ball falls back to the height that it was released from. Air resistance is negligible.
Okay, so knowing that gravitational potential energy is plotted in blue and kinetic energy is plotted in red on these graphs, we want to identify which one correctly shows how these energies change over time for a ball that’s being thrown directly upward in the air and then falling back down to earth. Let’s begin by clearing some space to work.
And recalling that the kinetic energy of an object is equal to one-half that object’s mass times its speed squared, while gravitational potential energy is given as an object’s mass times the acceleration due to gravity multiplied by its height relative to some reference level. Because this height value could be positive or negative, that means gravitational potential energy in general can be positive and negative. But note that an object’s kinetic energy can never be negative; the lowest value it can have is zero. Looking at our four graphs, we see that on graph (b) the red curve representing kinetic energy goes into negative values of energy. We’ve seen, though, that based on the equation for an object’s kinetic energy, this isn’t possible. Therefore, we can cross off option (b).
To see which of the remaining three graphs is correct, let’s consider our ball at both the lowest point and the highest point of its trajectory. At the lowest point, we know the ball’s speed will be greatest and its height will be lowest. This means we would have maximum kinetic energy and minimum gravitational potential energy. Then, at the ball’s highest point, it will have a minimum of kinetic energy — indeed, its kinetic energy at this point will be zero — and a maximum of GPE. These two types of energy, kinetic energy and gravitational potential energy, are the only energy types possessed by this ball. Therefore, we can completely describe the ball’s energy as mechanical energy, the sum of its potential and kinetic energies.
More than this, in this system, the ball’s mechanical energy is conserved. We know that because air resistance is negligible, and there’s no other place for the ball’s energy to transfer to. This fact that mechanical energy is conserved is very important for our solution. It means that if we add together the ball’s maximum kinetic energy and its minimum gravitational potential energy, then that sum will be equal to its minimum kinetic energy and its maximum gravitational potential energy. And not only that, but this value of mechanical energy will equal the mechanical energy of the ball at any point in its flight. This means that the correct graph, either (a) or (c) or (d), will show a constant mechanical energy. That is a constant sum of gravitational potential and kinetic energy of the ball for any time value 𝑡.
Let’s consider what this means, for example, for graph (c). Say that we choose this time value where kinetic energy is at its minimum and gravitational potential energy is at its maximum. If we add these values together to solve for the mechanical energy of the ball at this instant, we will be adding zero, the kinetic energy of the ball at this time, to this value of gravitational potential energy. Zero plus whatever this value is will simply equal that value itself. At this instant in time then, this is the total mechanical energy of the ball.
But now let’s look at another instant. Here, gravitational potential energy and kinetic energy both have the same value. If we add them together to solve for the mechanical energy of the ball, we get a value around here. This is greater than the mechanical energy we solved for earlier. In other words, this graph shows a system in which mechanical energy is not conserved. That tells us that graph (c) is not our final answer.
Similarly, for graph (d), if we add together kinetic and gravitational potential energies at this instant, we get a sum that is a mechanical energy for the ball of this value. Doing the same thing, however, at this instant in time where both kinetic and gravitational potential energies are here gives us a sum total of approximately this value. Once again then, we’re seeing a system in which mechanical energy is not conserved, at least according to this graph. We know that in reality, though, the ball’s mechanical energy is conserved. And therefore, we cross off graph (d).
That leaves us with graph (a). And here, if we apply a similar test solving for the total mechanical energy of the ball at different time values, we find that no matter what time we choose, that total is the same. Therefore, graph (a) does show a system in which mechanical energy is conserved. We choose this graph as our answer.
Let’s finish our video now by summarizing a few key points. We saw in this lesson that a system’s mechanical energy is the sum of its potential and kinetic energies. We saw further that mechanical energy can be converted from one type of energy to another, say a ball’s gravitational potential energy changing into kinetic energy. Lastly, we saw that in a closed system that starts and ends with only mechanical energy, that mechanical energy is conserved.