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### Course: Physics archive>Unit 5

Lesson 2: Springs and Hooke's law

# 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.

## Want to join the conversation?

• Spring constant k will vary from spring to spring, correct?
• 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!
• 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?
• 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.
• 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?
• 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)
• At sal says thw work is going to be the area below the slope ... a triangle.... but why not the whole rectangle ?
• 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
• why is the restorative force -kx, negative
• Because it is in the opposite direction of the displacement, x.
• How would you calculate the equation if you were putting force on the spring from both directions?
• You are always putting force on the spring from both directions. If you weren't, it would move away from you as you tried to push on it.
• 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?
• Because you can't measure the force always. It's not so easy to measure force. Whereas to measure distance you just slap a ruler next to the spring. This is also true for theoretical problems where only k and x are known. But yeah your form is valid. U = kx^2 / 2 = Fx/2 = F^2 / 2k
(1 vote)
• What happens to the potential energy of a bubble whenit rises up in water? Also explain y it is so
• 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...

Regards,

APD
• Why is the height=Xo times k?
• I think its bc the eqn for a straight line is y = mx + b, which becomes height = kx + 0
so height = kx
• 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.