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So the UK exam board specifications (AQA GCSE) clearly state "...the work done on the spring and the elastic potential energy stored are equal" Here's my problem,

So work done = Force x displacement

Force = Spring constant x extension

Elastic potential energy = 0.5 x spring constant x extension squared

Extension = displacement for a strecthed spring

However if I take some sample values and calculate the work done on the spring, and then the elastic potential energy stored. The Elastic potential is always exactly half the work done. This contradicts the statement in the specifications.

I have looked all over the place but can't find a satisfactory answer to this question. What am I missing? I get that elastic potential energy is equal to area under the Fx graph but why does that not equal the work done?

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  • $\begingroup$ Note that if you hang a weight on a spring, it will overshoot and oscillate around its equilibrium point. Ultimately, one-half of the potential energy of the weight will go into the spring strain energy and one-half will be lost as dissipative heat. See here, here, and here. (The same issue arises when charging a capacitor to a voltage $V$.) $\endgroup$ Sep 16, 2020 at 17:53

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Work is not "force times displacement". Work is an integral $$W=\int\mathbf F\cdot \text d\mathbf x$$ which becomes $W=Fx$ under certain conditions.

The work done by a conservative force is always equal to the negative change in potential energy associated with that force:

$$W_\text{cons}=\int\mathbf F\cdot\text d\mathbf x=\int-\nabla U\cdot\text d\mathbf x=-\Delta U$$

The area under the Fx graph should be the work done by the spring, which is the negative change in potential energy.

Your mistake is most likely thinking that $W=Fx$ holds here, but it does not because the spring force varies with displacement. You are probably getting a discrepancy of a factor of $2$.

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  • $\begingroup$ Thank you. I understand most of that, I think. However if the spring has an applied force from a constant weight hanging from it, does this mean that the force applied to spring is not constant as the spring stretches? $\endgroup$
    – Alex
    Sep 16, 2020 at 12:12
  • $\begingroup$ Based on your knowledge on this topic I would highly recommend you to study the chapter work power and energy $\endgroup$ Sep 16, 2020 at 12:17
  • $\begingroup$ What "chapter"? I thought this was like a question forum. $\endgroup$
    – Alex
    Sep 16, 2020 at 12:28
  • $\begingroup$ @Alex The force applied to the spring would be constant. The force the spring exerts would not be constant if the spring was stretching/compressing. $\endgroup$ Sep 16, 2020 at 12:47
  • $\begingroup$ That's what I was thinking. Ok, I've got it now. Thank you $\endgroup$
    – Alex
    Sep 16, 2020 at 14:18
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The work is $Fx$ if the force is constant during the displacement from 0 to $x$.

In the case of an elastic spring, the force is a function of the displacement: $F = kx$

So the work for a small displacement when the spring is streched in a given $x$ position: $\Delta W = kx\Delta x$

Integrating from 0 to x: $W = \frac{1}{2}kx^2$, which is the stored elastic energy.

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