Video: Elastic Potential Energy

In this video we learn how compressed or stretched springs possess elastic potential energy, and how this energy equals the work done on a spring.

11:13

Video Transcript

In this video, we’re going to learn about elastic potential energy. We’ll learn what this type of energy is, how it relates to the work done on an object, and we’ll see how to combine elastic potential energy with other energy types.

To start out, imagine that you have made a tree fort in your backyard. And as the fort is one of your favorite places to spend time, you find yourself climbing up and down the steps to get there frequently. After a while though, you get tired of all this climbing and you have an idea. You purchase what is essentially a human-sized elastic slingshot and string it up between two small trees near the tree fort. In order to figure out just how far back you’ll need to stand in the human slingshot so that it will shoot you up into the tree fort, we’ll want to know a little bit about elastic potential energy.

The first thing we can say about elastic potential energy is that whenever a spring is stretched or compressed, it possesses potential energy, that is the ability to do work. And when we say spring, that can include not just literal springs, but springy materials or substances that rebound after they’re deformed.

If we were to take a spring resting at its equilibrium length and then stretch that spring out at distance 𝑥 displaced from its natural length, we know that the stretched spring would exert a restoring force, we can call 𝐹 sub 𝑠, whose magnitude is equal to the spring constant 𝑘 times that displacement from equilibrium 𝑥.

We recall a basic equation for work done, which is that it’s equal to force times displacement. But this equation for work assumes a force which does not vary with displacement, while our spring force does. It depends on the displacement 𝑥. For the work done then by our spring force, that work will be equal to the integral from zero to the maximum displacement 𝑥 of that force times 𝑑𝑥.

Plugging in for the magnitude of that force, we now have an expression for the work done by a stretched or compressed spring. Integrating this expression from zero to 𝑥, we find a result of one-half 𝑘 times 𝑥 squared. This is equal to the work done by a spring and it’s also equal to the spring potential energy, or elastic potential energy.

One way to understand elastic potential energy is to think of it like gravitational potential energy. We know that if we take a mass and elevate it in a gravitational field some amount of height, then that mass now has potential energy and, therefore, the ability to do work. Likewise, if we put a mass on the end of a spring and then compress that spring, that mass likewise has potential energy. In this case, due to the elastic energy of the spring. And just like gravitational potential energy, elastic potential energy fits within energy conservation schemes. Let’s get some practice with elastic potential energy through a couple of examples.

The spring of a spring gun has a force constant 𝑘 equals 12 newtons per centimeter. When the gun is aimed vertically upward, a 15-gram projectile is shot to a height of 5.0 meters above the end of the expanded spring, as shown in the accompanying diagram. How much was the spring compressed initially?

In the diagram, we’ve called this distance 𝑑. And we’re also told that the spring constant of the spring is 12 newtons per centimeter and the mass of the shot being fired is 15 grams. Looking at the diagram of this scenario, we see that the initial and final conditions are connected through energy conservation.

If we call the state where the spring is compressed the initial state and the point where the ball is at its highest point in flight the final state, then we can write that the initial sum of kinetic plus potential energy is equal to the final amount of kinetic plus potential energy. Both at the start and at the finish, the ball isn’t in motion, so its kinetic energy in both those cases is zero.

When we consider the initial state of the ball, we know that its potential energy is due to the elastic energy compressed in the spring. We can call this PE sub 𝑠, for spring potential energy. On the other hand, the final potential energy of our system is entirely gravitational as the ball is motionless a known height above the expanded spring. So, we can write that spring potential energy equals gravitational potential energy. And we recall that spring potential energy is equal to one-half the spring constant times the displacement from equilibrium squared and that gravitational potential energy is equal to 𝑚 times 𝑔 times ℎ.

For our scenario, we write the spring potential energy is one-half 𝑘 times 𝑑 squared, where 𝑑 is the distance we want to solve for. And on the potential energy side, we let the height ℎ be equal to 5.0 meters, the height of the ball when it’s above the fully extended spring and not in motion.

Rearranging this expression to solve for 𝑑, we find it’s equal to the square root of two times 𝑚 times 𝑔 times ℎ all divided by 𝑘. 𝑔, the acceleration due to gravity, we’ll treat as exactly 9.8 meters per second squared. When we plug in for these values, we’re careful to convert our mass into a value in units of kilograms and our spring constant 𝑘 into a value in units of newtons per meter.

Calculating this result, we find 𝑑 is 3.5 centimeters. That’s the amount the spring must be compressed in order to fire the ball up to this given height ℎ. Now, let’s look at another elastic potential energy example.

A string has a force constant of 53.0 newtons per meter. An object, initially at rest, with a mass of 0.960 kilograms is suspended from it. The object descends, stretching the spring, oscillates, and then comes to rest. How much is the spring stretched when the object has come to rest after oscillating? Calculate the decrease in the gravitational potential energy of the object between its position at the point at which it is attached to the unextended spring and its position at the point at which it comes to rest after oscillating. Calculate the energy stored in the spring by its extension.

In this three-part exercise, we first wanna solve for the displacement of the spring after the mass has been hung on it and comes to rest. Next, we want to solve for the change in gravitational potential energy of the mass between the two points from which it is hung on the unextended spring and the point at which it comes to rest. And finally, we wanna calculate the energy stored in the spring by its extension. We’ll call this PE sub 𝑠.

In this scenario, we start off with a vertically oriented spring at its natural, or equilibrium, length. The spring, we’re told, has spring constant 𝑘 of 53.0 newtons per meter. Then, we take a mass, we’ve called it 𝑚, of value 0.960 kilograms and attach it to the free end of the spring. In response to this, the spring stretches, oscillates, and then comes to rest at a displacement we’ve called Δ𝑥 from its original position.

To solve for Δ𝑥, let’s start by considering the forces that are acting on the mass 𝑚. When the mass has come to rest after oscillating on the end of the spring, we know that the gravitational force acting on the mass is equal in magnitude to the spring force on the mass. So, we can write 𝑚 times 𝑔 is equal to the spring constant 𝑘 times Δ𝑥. Rearranging this expression, we see that Δ𝑥 is equal to 𝑚 times 𝑔 over 𝑘, where 𝑔 we’ll treat as exactly 9.8 meters per second squared.

Plugging in for these three values, when we calculate Δ𝑥, we find that, to three significant figures, it’s 17.8 centimeters. That’s the amount the spring is stretched because of the mass 𝑚. Now, let’s move on to solving for the change in gravitational potential energy of the mass 𝑚 over this displacement.

Recalling that, in general, gravitational potential energy is equal to an object’s mass times the acceleration due to gravity times its height relative to some reference level. Since we’re calculating the change in gravitational potential energy of the mass, we can write that it’s equal to 𝑚 times 𝑔 times Δ𝑥. No negative sign needed.

When we plug in for these three values, being careful to write our displacement Δ𝑥 in units of meters, we find that ΔPE sub 𝑔 is 1.67 joules. That’s the change in the mass’ gravitational potential energy. Finally, we want to solve for the potential energy stored up in the spring as result of this extension.

We recall that in general that potential energy is equal to half a spring‘s constant multiplied by its displacement from equilibrium squared. So, we can write that PE sub 𝑠 is equal to one-half 𝑘 times Δ𝑥 squared, where 𝑘 the spring constant is known and Δ𝑥 we solved for earlier. Substituting in for these values using a distance in meters for our displacement Δ𝑥, we find that PE sub 𝑠 is 0.835 joules. That’s how much potential energy the spring has gained by virtue of being extended.

We may wonder why the spring potential energy is not equal to the change in gravitational potential energy. Recall that, before coming to rest, our mass 𝑚 oscillated up and down on the spring. During that up-and-down motion, it lost energy through friction with the air as well as through the spring. That helps account for the difference between these two values.

Let’s summarize what we’ve learnt so far about elastic potential energy. We’ve seen that a stretched or compressed spring has potential energy. We’ve said that that potential energy, we could call it PE sub 𝑠, is equal to one-half the spring constant 𝑘 multiplied by its displacement from its equilibrium, or natural length, 𝑥 squared.

We’ve also seen that the potential energy stored in a spring, PE sub 𝑠, is equal to the work done on the spring. And we’ve seen that elastic, or spring, potential energy in an isolated system is part of the overall energy conservation picture, including kinetic energy and gravitational potential energy. In other words, it’s an energy term just like other energy terms.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.