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Potential energy stored in a spring

Learn about the force required to compress a spring, and the work done in the process, and how this relates to Hooke's Law, which defines the restorative force of a spring. Using a graph, see how force increases proportionally with displacement, and how one can use the area under the graph to calculate the work done to compress the spring. Finally, relate this work to the potential energy stored in the spring. Created by Sal Khan.

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  • blobby green style avatar for user Matt
    Spring constant k will vary from spring to spring, correct?
    (21 votes)
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    • duskpin sapling style avatar for user Alina Chen
      Yes, the word 'constant' might throw some people off at times. The k constant is only constant for that spring, so a k of -1/2 may only apply for one spring, but not others depending on the force needed to compress the spring a certain distance. Hope this helps!
      (23 votes)
  • leaf green style avatar for user pumpkin.chicken
    if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it). So where does the other half go?
    (18 votes)
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    • spunky sam blue style avatar for user Ethan Dlugie
      You're analysis is a bit off here. This is because the force with which you pull the spring is not 4N the entire time.
      What do I mean by this? Imagine you were only stretching the spring by small incremental amounts, say increments of 10 cm. The first 10 cm would correspond to an applied force of 0.2 N. This force will steadily increase until you get to your 2 m and 4 N.
      Because the applied force is not constant, you cannot use the simple W=F*d. You must instead use this idea of chopping the stretch up into little bits and add up the work done during each of these little bits. If you are familiar with calculus, you should recognize this as a simple definite integration problem. W = ∫(kx) dx from x=0 to d. With this method, you will find that all of your work goes into spring energy.
      (29 votes)
  • blobby green style avatar for user akibshahjahan
    why is work work area under the line? if work = f*d and if f= kx and d = x then shouldn't work=kx^2 why is it just the triangle and not the square?
    (12 votes)
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    • starky seedling style avatar for user deka
      the formula we've learnt here is assuming F_initial to the spring is 0, not the same as F_final which you may be given in the problem description

      if you want to use squre to get W, you need a constant F. thus F_initial = F_final = kx, which can't be 0 at starting of spring's motion
      (1 vote)
  • aqualine ultimate style avatar for user milind
    At sal says thw work is going to be the area below the slope ... a triangle.... but why not the whole rectangle ?
    (5 votes)
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    • blobby green style avatar for user Brandon Corrales
      We are looking for the area under the force curve. We only have a rectangle-like graph when the force is constant. Unfortunately, the force changes with a spring. Basically, we would only have a rectangle graph if our force was constant!

      Say you're pushing a box 5 meters. If you apply a CONSTANT force of 13N, then the graph of work (which would be a graph of force on the y axis and displacement on the x axis) would be a rectangle; the vertical component (the force) would be all the way up to 13N (again, on the y axis), while the x component would just be all the way up to 5m.

      Why is it not the whole rectangle in the case of a spring? In the case of a spring, the force that one must exert to compress a spring 1m is LESS than the force needed to compress it 2m or 3m, etc. The force needed CHANGES; this is why we are given an EQUATION for the force: F = kx, yes? If the F = a constant, we would, indeed, have a rectangle. When we are given an equation for the force, that means the force is changing and we cannot straight up do W=Fd because F is changing. Because F is changing, there will be different y values for different x's. To make it a little easier, we can revert to integration (which is a b word), because when we integrate a function (say, a function for a force), then we are able to account for the differing Y.

      I'm sorry if this was too long of a response, I have had so much trouble with this topic and I still do... Hope this helped! :D
      (13 votes)
  • leaf orange style avatar for user rose  watson
    why is the restorative force -kx, negative
    (4 votes)
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  • leafers ultimate style avatar for user Tejas Tuppera
    How would you calculate the equation if you were putting force on the spring from both directions?
    (2 votes)
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  • leafers tree style avatar for user Will Boonyoungratanakool
    So, if the work done is equal to the area under the graph, couldn't the equation just be force times extension divided by 2? Why use a more complex version of the equation, or is it used when the force value is not known?
    (6 votes)
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  • leaf green style avatar for user Shunethra Senthilkumar
    What happens to the potential energy of a bubble whenit rises up in water? Also explain y it is so
    (4 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Shunethra,

      I like this question but I'm going to turn it back to you.

      Assume you are in a swimming pool. Would it take work to "push" a balloon to the bottom?

      This is very hard to do. As a kid I remember trying to move a basketball to the bottom of the pool. I couldn't do it. The force on the ball is equal to the weight of the water it displaces. As you go deeper the weight of this water increases. Perhaps you have seen ho air bubbles in crease in size as they rise in the water column...


      (2 votes)
  • marcimus pink style avatar for user j.elizabeth
    Why is the height=Xo times k?
    (4 votes)
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  • blobby green style avatar for user kristiana thomai
    i dont understand how to find the force constant k of a spring. can you give me some tips on how to start a problem like that.
    (2 votes)
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    • orange juice squid orange style avatar for user hidden
      So you have F=kx, say you had a 2m spring
      F = the force applied, and say I'm pulling the string with a force of 2N
      and that stretched the spring 0.2m, so x would be 0.2m
      divide both sides by 0.2
      k = 10
      now you can apply that to find the force required to stretch the spring any distance, or how much force you need to stretch the string a certain distance
      (4 votes)

Video transcript

Welcome back. So we have this green spring here, and let's see, there's a wall here. This connected to the wall. And let's say that this is where the spring is naturally. So if I were not to push on the spring, it would stretch all the way out here. But in this situation, I pushed on the spring, so it has a displacement of x to the left. And we'll just worry about magnitude, so we won't worry too much about direction. So what I want to do is think a little bit-- well, first I want to graph how much force I've applied at different points as I compress this spring. And then I want to use that graph to maybe figure out how much work we did in compressing the spring. So let's look at-- I know I'm compressing to the left. Maybe I should compress to the right, so that you can-- well, we're just worrying about the magnitude of the x-axis. Let's draw a little graph here. That's my y-axis, x-axis. So this axis is how much I've compressed it, x, and then this axis, the y-axis, is how much force I have to apply. So when the spring was initially all the way out here, to compress it a little bit, how much force do I have to apply? Well, this was its natural state, right? And we know from-- well, Hooke's Law told us that the restorative force-- I'll write a little r down here-- is equal to negative K, where K is the spring constant, times the displacement, right? That's the restorative force, so that's the force that the spring applies to whoever's pushing on it. The force to compress it is just the same thing, but it's going in the same direction as the x. If I'm moving the spring, if I'm compressing the spring to the left, then the force I'm applying is also to the left. So I'll call that the force of compression. The force of compression is going to be equal to K times x. And when the spring is compressed and not accelerating in either direction, the force of compression is going to be equal to the restorative force. So what I want to do here is plot the force of compression with respect to x. And I should have drawn it the other way, but I think you understand that x is increasing to the left in my example, right? This is where x is equal to 0 right here. And say, this might be x is equal to 10 because we've compressed it by 10 meters. So let's see how much force we've applied. So when x is 0, which is right here, how much force do we need to apply to compress the spring? Well, if we give zero force, the spring won't move, but if we just give a little, little bit of force, if we just give infinitesimal, super-small amount of force, we'll compress the spring just a little bit, right? Because at that point, the force of compression is going to be pretty much zero. So when the spring is barely compressed, we're going to apply a little, little bit of force, so almost at zero. To displace the spring zero, we apply zero force. To displace the spring a little bit, we have to apply a little bit more force. To displace soon. the spring 1 meter, so if this is say, 1 meter, how much force will we have to apply to keep it there? So let's say if this is 1 meter, the force of compression is going to be K times 1, so it's just going to be K. And realize, you didn't apply zero and then apply K force. You keep applying a little bit more force. Every time you compress the spring a little bit, it takes a little bit more force to compress it a little bit more. So to compress it 1 meters, you need to apply K. And to get it there, you have to keep increasing the amount of force you apply. At 2 meters, you would've been up to 2K, et cetera. I think you see a line is forming. Let me draw that line. The line looks something like that. And so this is how much force you need to apply as a function of the displacement of the spring from its natural rest state, right? And here I have positive x going to the right, but in this case, positive x is to the left. I'm just measuring its actual displacement. I'm not worried too much about direction right now. So I just want you to think a little bit about what's happening here. You just have to slowly keep on-- you could apply a very large force initially. If you apply a very large force initially, the spring will actually accelerate much faster, because you're applying a much larger force than its restorative force, and so it might accelerate and then it'll spring back, and actually, we'll do a little example of that. But really, just to displace the spring a certain distance, you have to just gradually increase the force, just so that you offset the restorative force. Hopefully, that makes sense, and you understand that the force just increases proportionally as a function of the distance, and that's just because this is a linear equation. And what's the slope of this? Well, slope is rise over run, right? So if I run 1, this is 1, what's my rise? It's K. So the slope of this graph is K. So using this graph, let's figure out how much work we need to do to compress this spring. I don't know, let's say this is x0. So x is where it's the general variable. X0 is a particular value for x. That could be 10 or whatever. Let's see how much work we need. So what's the definition of work? Work is equal to the force in the direction of your displacement times the displacement, right? So let's see how much we've displaced. So when we go from zero to here, we've displaced this much. And what was the force of the displacement? Well, the force was gradually increasing the entire time, so the force is going to be be roughly about that big. I'm approximating. And I'll show you that you actually have to approximate. So the force is kind of that square right there. And then to displace the next little distance-- that's not bright enough-- my force is going to increase a little bit, right? So this is the force, this is the distance. So if you you see, the work I'm doing is actually going to be the area under the curve, each of these rectangles, right? Because the height of the rectangle is the force I'm applying and the width is the distance, right? So the work is just going to be the sum of all of these rectangles. And the rectangles I drew are just kind of approximations, because they don't get right under the line. You have to keep making the rectangle smaller, smaller, smaller, and smaller, and just sum up more and more and more rectangles, right? And actually I'm touching on integral calculus right now. But if you don't know integral calculus, don't worry about it. But the bottom line is the work we're doing-- hopefully I showed you-- is just going to be the area under this line. So the work I'm doing to displace the spring x meters is the area from here to here. And what's that area? Well, this is a triangle, so we just need to know the base, the height, and multiply it times 1/2, right? That's just the area of a triangle. So what's the base? So this is just x0. What's the height? Well, we know the slope is K, so this height is going to be x0 times K. So this point right here is the point x0, and then x0 times K. And so what's the area under the curve, which is the total work I did to compress the spring x0 meters? Well, it's the base, x0, times the height, x0, times K. And then, of course, multiply by 1/2, because we're dealing with a triangle, right? So that equals 1/2K x0 squared. And for those of you who know calculus, that, of course, is the same thing as the integral of Kx dx. And that should make sense. Each of these are little dx's. But I don't want to go too much into calculus now. It'll confuse people. So that's the total work necessary to compress the spring by distance of x0. Or if we set a distance of x, you can just get rid of this 0 here. And why is that useful? Because the work necessary to compress the spring that much is also how much potential energy there is stored in the spring. So if I told you that I had a spring and its spring constant is 10, and I compressed it 5 meters, so x is equal to 5 meters, at the time that it's compressed, how much potential energy is in that spring? We can just say the potential energy is equal to 1/2K times x squared equals 1/2. K is 10 times 25, and that equals 125. And, of course, work and potential energy are measured in joules. So this is really what you just have to memorize. Or hopefully you don't memorize it. Hopefully, you understand where I got it, and that's why I spent 10 minutes doing it. But this is how much work is necessary to compress the spring to that point and how much potential energy is stored once it is compressed to that point, or actually stretched that much. We've been compressing, but you can also stretch the spring. If you know that, then we can start doing some problems with potential energy in springs, which I will do in the next video. See