Video: Elastic Potential Energy

In this lesson, we will learn how to calculate the elastic potential energy stored in springs that are not at their equilibrium length.

15:45

Video Transcript

In this video, we’re talking about elastic potential energy. This sort of potential energy gets my vote for being the most fun kind, because it’s behind things like trampolines, rubber bands, and bows and arrows. Of all the objects we associate with elastic potential energy though, the one that comes up over and over, probably more than any other, is the spring. In fact, springs show up when we talk about this kind of potential energy so often that sometimes elastic potential energy is called spring potential energy. So if we hear that phrase, it means the same thing as elastic potential energy. But we’ll keep the term elastic because it’s a bit more general.

Now to get a better understanding of this term, let’s consider it word by word, starting with this last word “energy.” We can think of energy as the ability of something to do work, that is, to exert a force on some object over some distance. So that’s energy. And one type of energy we know is potential energy. This word potential means stored up or held in reserve. And then finally, this word elastic. This term describes an object that if we change its shape, then the object returns quickly to its original shape afterward. So, for example, considering our spring here, which is an elastic object. If we were to push on the end of the spring and compress it in some distance. Then the moment we would release our hand, the spring would spring back to its original or natural length. In that way, it returns to its original shape after it’s been deformed. So this spring is an elastic object.

So when we talk about elastic potential energy, we’re talking about the ability to do work. Which is stored up in some object due to or because of some deformation of that object and its tendency to elastically return to its original shape. So in the case of the spring, we would need to either compress it or extend it beyond its natural length in order to give it elastic potential energy.

Now, in order to get a deeper understanding of this topic, it’s helpful to talk not about energy, but about force. Say that we do this, say we take our hand. And we start to push to the left on the spring. We exert a force on it to compress it. Let’s say that as we press on the spring, we compress it at distance 𝑥 from its natural length or its equilibrium length. Now, there’s a helpful law, which tells us just how much force we have to exert on the spring in order to do this. That law is known as Hooke’s law. And here’s what it tells us.

It says that the force we need to apply to an elastic object, like a spring, is directly proportional to the compression or extension of that spring. In other words, if we needed to apply a force 𝐹 to compress a spring in amount 𝑥, then if you wanted to double our compression. Say we wanted to compress it by two 𝑥 that distance, we would need to apply twice as much force. So the force, Hooke’s law says, varies with the compression or the extension. It could go the other way too. And in particular, that force is proportional to the change in the length of the spring from its natural length.

Now there’s another equivalent mathematical way of writing this expression. We could also write it like this. We could say that the force we apply is equal to some constant of proportionality, typically called 𝑘, multiplied by displacement. These two relationships are saying the same thing. Often though, when we see Hooke’s law written out, we see it written this way. 𝐹 is equal to 𝑘 times 𝑥. And when we’re working with springs such as we are over here. Then this constant of proportionality 𝑘 has a special name. It’s called the spring constant.

The spring constant 𝑘 of a spring is unique to that spring. If we look at the units of 𝑘, those units are newtons per meter. If we want to stretch or compress a given spring by a distance of one meter, then the spring constant 𝑘 tells us how much force in newtons we would need to do that. So this spring we have here as well as any spring has some spring constant. And that tells us just how much force would be necessary in order to extend or compress the spring.

Now, one last thing about Hooke’s law. This expression of it, 𝐹 is equal to 𝑘𝑥. We can see that the force that we need to apply to a spring in order to stretch or compress it depends on how much the spring has been stretched or compressed. For example, over here, if we wanted to compress our spring even farther than the distance we have so far. We would need to press that much harder. We would need to apply more force. This is a way of saying that the force in Hooke’s law depends on 𝑥, the displacement. And we can write that mathematically this way. We can say that 𝐹 is a function of 𝑥. That is, 𝐹 depends on 𝑥.

Okay, now that we’ve talked about force, let’s remember what we said about energy earlier. We said that energy is the capacity of something to do work. Work, which we usually symbolize using a capital 𝑊, is equal to the force applied to an object multiplied by that object’s displacement. Now, we bring this up because we’re about to do a little experiment on our spring, which will help us understand elastic potential energy. Say that we set up a two-dimensional plot, where on the horizontal axis, we plot out distance. That’ll be the distance that our spring travels from its equilibrium length. And then, on the vertical axis, we plot out the force being applied to the spring in order to displace it this given distance.

So here we go. Starting out, we put our hand to the end of the spring. And we start to push in. We start to compress it. After we apply some amount of force, the spring compresses in some distance. And let’s say we plot that force and distance point on our graph. Let’s say that point is right here. So, in other words, we’re applying this much force to the spring. And that’s causing it to compress by this distance. Then, let’s say we double the force we’re applying to the spring and therefore compress it even further, twice that original distance. If we plot that point in the graph, it will go right there.

At this point, if we were to hold steady and continue to apply the same force, the spring wouldn’t keep compressing. But rather we would just hold it in place. Remember that the force we need to apply to compress or extend a spring is proportional to that compression or extension. So if at any moment we keep the force applied the same, that will mean that the spring extension or compression stays the same too. In other words, if we want more compression, we need to push with greater force. So we do. We push harder and we get more compression. And that data point looks like this.

Say then, as part of this experiment that we keep going. That we keep pressing harder and harder and compressing the spring farther and farther. By this point, we’ve compressed our spring some overall distance from its equilibrium position. We could call this distance 𝑥 and mark it in on our graph so that there is 𝑥. And then, if we look back over at Hooke’s law, we see that the force needed in order to displace our spring at distance 𝑥 is equal to the spring constant 𝑘 times 𝑥. So we can mark that value in on our vertical axis. Now, what we have is a plot of force versus displacement of our spring from equilibrium. And according to our equation for work, we know that if we multiply these two values together, force times distance. We’ll get the work done on the spring. Note that each time we do, we’re using the specific force needed to stretch the spring a specific distance. That these values change as the spring extends. And we need to take that into account when we calculate work.

The question is, what does that look like graphically. To figure that out, let’s do this. Let’s draw a line of best fit through these five data points. Since these data points all lie along the same line, this line moves exactly through them. And it also passes through the origin. From a graphical perspective, if we were to multiply the value on the vertical axis, force, by the value on the horizontal axis, distance. That product would be equal to the area under this line of best fit that we just drew in. In other words, it would be equal to the area of this right triangle, an area we can call capital 𝐴. We’re interested in solving for this area because if we can, our equation for work tells us that we’ve solved for the work done on the spring.

So what is that area under that curve? What is the area of the triangle? We can recall that a triangle’s area, we’ll call it 𝐴 sub 𝑡, is equal to one-half its base multiplied by its height. So that means that 𝐴 is equal to one-half the base of this triangle, which we know is 𝑥, multiplied by its height, which we know is 𝑘 times 𝑥. If we multiply these terms together, we find that this area is equal to one-half the spring constant 𝑘 times 𝑥 squared. And as we said, this value is equal to the total work done by our hand on the spring to compress it. And here’s where things get interesting. Because the work done by our hand on the spring is also equal to the elastic potential energy we give the spring. That is, this spring’s elastic potential energy, now that we’ve compressed it, is equal to one-half the spring constant multiplied by its displacement from equilibrium squared.

Now, we developed this equation for the elastic potential energy based on a specific spring and a specific experiment that we did. But notice that the equation is quite general. We haven’t specified what the spring constant is. We’ve left that as 𝑘. And also this displacement from equilibrium 𝑥 in our case was a compression. But nothing about this equation constrains it to only being a compression. It could also be an extension of a spring or other elastic object. What we found then is a general equation for elastic potential energy. So long as we know the displacement of an object from equilibrium and we know its spring constant, we can solve for this energy.

In just a minute, we’ll get some practice with this equation. But before we do, it’s important to realize what this 𝑥 is and what it is not. We’ve been saying that this 𝑥 is a displacement from equilibrium. That means that if we have some spring, which is sitting at its natural length, say that natural length is 𝐿. Then 𝐿 is not equal to 𝑥 in this equation. That’s because, as it is, this spring is neither extended nor compressed. It’s at the length that it naturally wants to be at. But if we do change the length of the spring, say we extended some distance beyond that equilibrium length. And that extension is 𝑥. Then that’s the distance that we’ll use in this equation for elastic potential energy. That’s just something to watch out for as we move on using this equation. Alright. All that being said, let’s try out an example exercise involving elastic potential energy.

A spring has 50 joules of energy stored in it when it is extended by 2.5 meters. What is the spring’s constant?

Okay. In this example, we have a spring. Let’s say that this is our spring. And we can imagine that it starts out at its natural or its equilibrium length. In other words, in this position, the spring isn’t being stretched or compressed at all. Now, by the way, if a spring is completely unstretched or uncompressed, that means that no energy is stored in it. We can say there are zero joules of energy stored in the spring when it’s at its natural length. But then the spring is stretched out. And we’re told it’s extended a distance of 2.5 meters. And that, under these conditions, the spring then has 50 joules of energy stored in it. So as a result of this change, the spring has gained internal energy. And we know that it’s elastic potential energy because it resulted from a stretching of the spring. This stretching was resisted by the spring, according to a value known as the spring’s constant, symbolized lowercase 𝑘. It’s that value that we want to solve for. And we can do it by recalling an equation for elastic potential energy in terms of spring constant.

The elastic potential energy of an object, sometimes also called its spring potential energy, is equal to one-half 𝑘, the spring constant, multiplied by the displacement of that object from equilibrium squared. And it’s important to know that this distance 𝑥 is not equal to the natural length of a spring or other elastic object. Instead, it’s equal to the displacement of that spring from equilibrium, whether by compression or, in our case, extension. So understanding this equation, it’s not the elastic potential energy we want to solve for. But it’s the spring constant 𝑘. To solve for that, we can rearrange this equation. We can multiply both sides by two divided by 𝑥 squared. Now on the right-hand side, the factor of one-half cancels with the two. And 𝑥 squared divided by 𝑥 squared is equal to one. We find then that two times the elastic potential energy of an object divided by its displacement from equilibrium squared is equal to its spring constant.

In our case, to solve for this particular spring constant applying to our particular spring. We’ve been told that the elastic potential energy stored in the spring is 50 joules. And that this much energy is stored in the spring when it’s displaced from equilibrium by 2.5 meters. So to solve for 𝑘, we only need to calculate the right-hand side of this equation. When we compute it, we find a result of 16 newtons per meter. In other words, for this particular spring, if we wanted to compress or extend it by a distance of one meter, we would need to apply 16 newtons of force. That’s the spring constant of the spring. Let’s look now at another example involving elastic potential energy.

A spring with a constant of 80 newtons per meter is extended by 1.5 meters. How much energy is stored in the extended spring?

In this example, we start out with this spring. Let’s say this is it right here. And we’re told that our spring has a spring constant. We’ll symbolize it with a lowercase 𝑘 of 80 newtons per meter. This means that if we want to extend or compress the spring by a distance of one meter, we would need to apply a force of 80 newtons. But at the moment, our spring isn’t extended or compressed. We can say that it’s at its natural or its equilibrium length. But we’re told that it doesn’t stay that way. Instead, the spring is extended from this original length. And that extension, we can call it 𝑥, is given as 1.5 meters. Based on all this, we want to know how much energy is stored in the extended spring.

Now, the reason there’s energy at all is because the spring wants to return to its natural length. It has a native capacity to do that. In doing so, this currently extended spring is capable of doing work. And that amount of work it can do is equal to how much energy is stored in it while it’s extended. When we consider what type of energy this is then, we know that it’s potential energy. The energy isn’t manifest now. But it will be if the spring is released from its extended length. And we also know the energy is elastic potential energy. That’s because it has its source in the compression or, in our case, extension of an elastic body away from equilibrium.

To solve for this amount of energy then, we can recall a mathematical relationship describing elastic potential energy. An object’s elastic potential energy is equal to one-half its spring constant multiplied by its displacement from equilibrium squared. As we solve for the elastic potential energy stored in our spring. The great thing is that we’re given 𝑘, the spring constant, as well as the spring’s extension. So the elastic potential energy stored in this extended spring is equal to one-half the spring constant, 80 newtons per meter, multiplied by the displacement from equilibrium, 1.5 meters squared. When we calculate the right-hand side of this expression, we find a result of 90 newton meters. But at this point, we can recall that one newton, the base unit of force, multiplied by a meter, the base unit of distance, is equal to a joule, the base unit of energy. So our final answer is 90 joules. That’s how much energy is stored in this extended spring.

Let’s take a moment now to summarize what we’ve learned in this lesson about elastic potential energy. We saw, first of all, that elastic potential energy is stored-up ability to do work due to the deformation. That is, the extension or compression of a body that will return to its natural shape. That’s a mouthful. But it basically means the energy due to an elastic object, like a spring or a rubber band being either stretched out or compressed. We also learned this. Hooke’s law says that the force on an elastic object is equal to the displacement of that object from equilibrium multiplied by something called the spring constant. And that that spring constant, which is unique to any spring or other elastic object, is a measure of how much force is required to stretch or compress the object a certain distance.

It was analyzing Hooke’s law and combining it with the notion of work that led us to an equation for elastic potential energy. That it’s equal to one-half the spring constant of the object multiplied by the object’s displacement from equilibrium squared. And lastly, we noted that 𝑥, in that equation for elastic potential energy, is not the natural length of an elastic object. But rather its displacement from that length, whether through stretching or through compression.

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