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In this video, we’re talking about elastic potential energy.
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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.
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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.
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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.
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So if we hear that phrase, it means the same thing as elastic potential energy.
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But we’ll keep the term elastic because it’s a bit more general.
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Now to get a better understanding of this term, let’s consider it word by word, starting with this last word “energy.”
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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.
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So that’s energy.
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And one type of energy we know is potential energy.
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This word potential means stored up or held in reserve.
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And then finally, this word elastic.
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This term describes an object that if we change its shape, then the object returns quickly to its original shape afterward.
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So, for example, considering our spring here, which is an elastic object.
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If we were to push on the end of the spring and compress it in some distance.
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Then the moment we would release our hand, the spring would spring back to its original or natural length.
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In that way, it returns to its original shape after it’s been deformed.
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So this spring is an elastic object.
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So when we talk about elastic potential energy, we’re talking about the ability to do work.
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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.
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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.
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Now, in order to get a deeper understanding of this topic, it’s helpful to talk not about energy, but about force.
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Say that we do this, say we take our hand.
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And we start to push to the left on the spring.
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We exert a force on it to compress it.
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Let’s say that as we press on the spring, we compress it at distance 𝑥 from its natural length or its equilibrium length.
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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.
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That law is known as Hooke’s law.
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And here’s what it tells us.
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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.
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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.
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Say we wanted to compress it by two 𝑥 that distance, we would need to apply twice as much force.
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So the force, Hooke’s law says, varies with the compression or the extension.
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It could go the other way too.
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And in particular, that force is proportional to the change in the length of the spring from its natural length.
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Now there’s another equivalent mathematical way of writing this expression.
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We could also write it like this.
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We could say that the force we apply is equal to some constant of proportionality, typically called 𝑘, multiplied by displacement.
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These two relationships are saying the same thing.
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Often though, when we see Hooke’s law written out, we see it written this way.
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𝐹 is equal to 𝑘 times 𝑥.
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And when we’re working with springs such as we are over here.
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Then this constant of proportionality 𝑘 has a special name.
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It’s called the spring constant.
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The spring constant 𝑘 of a spring is unique to that spring.
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If we look at the units of 𝑘, those units are newtons per meter.
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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.
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So this spring we have here as well as any spring has some spring constant.
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And that tells us just how much force would be necessary in order to extend or compress the spring.
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Now, one last thing about Hooke’s law.
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This expression of it, 𝐹 is equal to 𝑘𝑥.
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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.
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For example, over here, if we wanted to compress our spring even farther than the distance we have so far.
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We would need to press that much harder.
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We would need to apply more force.
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This is a way of saying that the force in Hooke’s law depends on 𝑥, the displacement.
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And we can write that mathematically this way.
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We can say that 𝐹 is a function of 𝑥.
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That is, 𝐹 depends on 𝑥.
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Okay, now that we’ve talked about force, let’s remember what we said about energy earlier.
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We said that energy is the capacity of something to do work.
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Work, which we usually symbolize using a capital 𝑊, is equal to the force applied to an object multiplied by that object’s displacement.
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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.
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Say that we set up a two-dimensional plot, where on the horizontal axis, we plot out distance.
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That’ll be the distance that our spring travels from its equilibrium length.
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And then, on the vertical axis, we plot out the force being applied to the spring in order to displace it this given distance.
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So here we go.
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Starting out, we put our hand to the end of the spring.
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And we start to push in.
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We start to compress it.
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After we apply some amount of force, the spring compresses in some distance.
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And let’s say we plot that force and distance point on our graph.
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Let’s say that point is right here.
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So, in other words, we’re applying this much force to the spring.
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And that’s causing it to compress by this distance.
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Then, let’s say we double the force we’re applying to the spring and therefore compress it even further, twice that original distance.
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If we plot that point in the graph, it will go right there.
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At this point, if we were to hold steady and continue to apply the same force, the spring wouldn’t keep compressing.
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But rather we would just hold it in place.
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Remember that the force we need to apply to compress or extend a spring is proportional to that compression or extension.
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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.
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In other words, if we want more compression, we need to push with greater force.
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So we do.
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We push harder and we get more compression.
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And that data point looks like this.
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Say then, as part of this experiment that we keep going.
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That we keep pressing harder and harder and compressing the spring farther and farther.
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By this point, we’ve compressed our spring some overall distance from its equilibrium position.
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We could call this distance 𝑥 and mark it in on our graph so that there is 𝑥.
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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.
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So we can mark that value in on our vertical axis.
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Now, what we have is a plot of force versus displacement of our spring from equilibrium.
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And according to our equation for work, we know that if we multiply these two values together, force times distance.
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We’ll get the work done on the spring.
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Note that each time we do, we’re using the specific force needed to stretch the spring a specific distance.
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That these values change as the spring extends.
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And we need to take that into account when we calculate work.
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The question is, what does that look like graphically.
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To figure that out, let’s do this.
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Let’s draw a line of best fit through these five data points.
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Since these data points all lie along the same line, this line moves exactly through them.
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And it also passes through the origin.
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From a graphical perspective, if we were to multiply the value on the vertical axis, force, by the value on the horizontal axis, distance.
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That product would be equal to the area under this line of best fit that we just drew in.
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In other words, it would be equal to the area of this right triangle, an area we can call capital 𝐴.
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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.
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So what is that area under that curve?
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What is the area of the triangle?
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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.
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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.
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If we multiply these terms together, we find that this area is equal to one-half the spring constant 𝑘 times 𝑥 squared.
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And as we said, this value is equal to the total work done by our hand on the spring to compress it.
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And here’s where things get interesting.
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Because the work done by our hand on the spring is also equal to the elastic potential energy we give the spring.
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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.
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Now, we developed this equation for the elastic potential energy based on a specific spring and a specific experiment that we did.
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But notice that the equation is quite general.
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We haven’t specified what the spring constant is.
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We’ve left that as 𝑘.
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And also this displacement from equilibrium 𝑥 in our case was a compression.
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But nothing about this equation constrains it to only being a compression.
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It could also be an extension of a spring or other elastic object.
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What we found then is a general equation for elastic potential energy.
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So long as we know the displacement of an object from equilibrium and we know its spring constant, we can solve for this energy.
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In just a minute, we’ll get some practice with this equation.
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But before we do, it’s important to realize what this 𝑥 is and what it is not.
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We’ve been saying that this 𝑥 is a displacement from equilibrium.
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That means that if we have some spring, which is sitting at its natural length, say that natural length is 𝐿.
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Then 𝐿 is not equal to 𝑥 in this equation.
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That’s because, as it is, this spring is neither extended nor compressed.
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It’s at the length that it naturally wants to be at.
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But if we do change the length of the spring, say we extended some distance beyond that equilibrium length.
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And that extension is 𝑥.
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Then that’s the distance that we’ll use in this equation for elastic potential energy.
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That’s just something to watch out for as we move on using this equation.
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Alright.
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All that being said, let’s try out an example exercise involving elastic potential energy.
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A spring has 50 joules of energy stored in it when it is extended by 2.5 meters.
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What is the spring’s constant?
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Okay.
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In this example, we have a spring.
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Let’s say that this is our spring.
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And we can imagine that it starts out at its natural or its equilibrium length.
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In other words, in this position, the spring isn’t being stretched or compressed at all.
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Now, by the way, if a spring is completely unstretched or uncompressed, that means that no energy is stored in it.
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We can say there are zero joules of energy stored in the spring when it’s at its natural length.
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But then the spring is stretched out.
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And we’re told it’s extended a distance of 2.5 meters.
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And that, under these conditions, the spring then has 50 joules of energy stored in it.
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So as a result of this change, the spring has gained internal energy.
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And we know that it’s elastic potential energy because it resulted from a stretching of the spring.
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This stretching was resisted by the spring, according to a value known as the spring’s constant, symbolized lowercase 𝑘.
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It’s that value that we want to solve for.
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And we can do it by recalling an equation for elastic potential energy in terms of spring constant.
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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.
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And it’s important to know that this distance 𝑥 is not equal to the natural length of a spring or other elastic object.
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Instead, it’s equal to the displacement of that spring from equilibrium, whether by compression or, in our case, extension.
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So understanding this equation, it’s not the elastic potential energy we want to solve for.
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But it’s the spring constant 𝑘.
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To solve for that, we can rearrange this equation.
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We can multiply both sides by two divided by 𝑥 squared.
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Now on the right-hand side, the factor of one-half cancels with the two.
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And 𝑥 squared divided by 𝑥 squared is equal to one.
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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.
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In our case, to solve for this particular spring constant applying to our particular spring.
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We’ve been told that the elastic potential energy stored in the spring is 50 joules.
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And that this much energy is stored in the spring when it’s displaced from equilibrium by 2.5 meters.
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So to solve for 𝑘, we only need to calculate the right-hand side of this equation.
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When we compute it, we find a result of 16 newtons per meter.
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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.
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That’s the spring constant of the spring.
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Let’s look now at another example involving elastic potential energy.
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A spring with a constant of 80 newtons per meter is extended by 1.5 meters.
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How much energy is stored in the extended spring?
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In this example, we start out with this spring.
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Let’s say this is it right here.
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And we’re told that our spring has a spring constant.
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We’ll symbolize it with a lowercase 𝑘 of 80 newtons per meter.
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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.
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But at the moment, our spring isn’t extended or compressed.
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We can say that it’s at its natural or its equilibrium length.
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But we’re told that it doesn’t stay that way.
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Instead, the spring is extended from this original length.
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And that extension, we can call it 𝑥, is given as 1.5 meters.
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Based on all this, we want to know how much energy is stored in the extended spring.
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Now, the reason there’s energy at all is because the spring wants to return to its natural length.
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It has a native capacity to do that.
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In doing so, this currently extended spring is capable of doing work.
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And that amount of work it can do is equal to how much energy is stored in it while it’s extended.
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When we consider what type of energy this is then, we know that it’s potential energy.
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The energy isn’t manifest now.
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But it will be if the spring is released from its extended length.
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And we also know the energy is elastic potential energy.
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That’s because it has its source in the compression or, in our case, extension of an elastic body away from equilibrium.
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To solve for this amount of energy then, we can recall a mathematical relationship describing elastic potential energy.
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An object’s elastic potential energy is equal to one-half its spring constant multiplied by its displacement from equilibrium squared.
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As we solve for the elastic potential energy stored in our spring.
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The great thing is that we’re given 𝑘, the spring constant, as well as the spring’s extension.
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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.
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When we calculate the right-hand side of this expression, we find a result of 90 newton meters.
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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.
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So our final answer is 90 joules.
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That’s how much energy is stored in this extended spring.
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Let’s take a moment now to summarize what we’ve learned in this lesson about elastic potential energy.
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We saw, first of all, that elastic potential energy is stored-up ability to do work due to the deformation.
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That is, the extension or compression of a body that will return to its natural shape.
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That’s a mouthful.
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But it basically means the energy due to an elastic object, like a spring or a rubber band being either stretched out or compressed.
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We also learned this.
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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.
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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.
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It was analyzing Hooke’s law and combining it with the notion of work that led us to an equation for elastic potential energy.
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That it’s equal to one-half the spring constant of the object multiplied by the object’s displacement from equilibrium squared.
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And lastly, we noted that 𝑥, in that equation for elastic potential energy, is not the natural length of an elastic object.
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But rather its displacement from that length, whether through stretching or through compression.