# Proof 37: Elastic potential energy

In a spring that is elongated/compressed the distance $\Delta L=L_\text{end}-L_\text{relaxed}$, an elastic force (due to Hooke’s law) $F_\text{elastic}=k\Delta L$ is pushing/pulling on the surroundings in order to return the spring to its relaxed size. That force can thus supply energy to, or do work on, the surroundings; this energy was stored in it as elastic potential energy $U_\text{elastic}$. So, if we can find the work that can be done via the work formula, $W=Fd$, then we have the stored elastic potential energy.

Since the relaxed length is the situation of zero energy stored, let’s place our coordinate system at that point. Then $L_\text{relaxed}=0$, and the elongation/compression is $\Delta L=L_\text{end}-\cancel{L_\text{relaxed}}=L_\text{end}$.

The work formula $W=F_\text{elastic}\underbrace{\,\Delta L}_d$ has to be calculated during a very, very tiny moment, because the elastic force changes as we go. And then all those small portions of work must be added together. Such a tiny moment only gives a tiny displacement, so instead of $\Delta L$, let’s symbolise it $\mathrm dL$: $W=F_\text{elastic}\mathrm d L$. And we’ll sum them all up with an integral:

$$W=\int F_\text{elastic}\mathrm d L=\int kL\,\mathrm d L=\frac12 kL^2$$

The final $L$ value equals the final $L_\text{end}$, which was equal to $\Delta L$. So, we’ll swap $\Delta L$ back in instead of $L$. And finally, we’ll remember that the work that was found here is stored as elastic potential energy $U_\text{elastic}$ before being released. So:

$$U_\text{elastic}=\frac12 k\,\Delta L^2\qquad\blacksquare$$

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