Elastic Potential Energy in a Stretched Wire

Elastic Potential Energy

When a wire is stretched, the work done by the deforming force is stored inside the material as elastic potential energy. This is the energy that gets released when the force is removed and the wire snaps back to its original shape. The key question is: how much energy is stored for a given stretch?

Consider a wire of length $L$ and cross-sectional area $A$. You apply a gradually increasing force $F$ along its length. At any stage, the wire is stretched by an amount $\Delta L$. The force $F$ at that instant is related to the extension by Hooke's law:

where $Y$ is Young's modulus. Notice that $F$ is not constant — it grows linearly with $\Delta L$, starting from zero when $\Delta L = 0$ and reaching its final value when the extension is $\Delta L$.

The work done by the applied force in stretching the wire by a small additional amount $d(\Delta L)$ is:

To find the total work done in stretching from zero to a final extension $\Delta L$, integrate:

This work is stored as elastic potential energy $U$ in the wire. So:

The last step follows because the final force $F = \frac{YA}{L} \Delta L$, so $\frac{1}{2} F \Delta L = \frac{1}{2} \frac{YA}{L} (\Delta L)^2$.

Energy Density — A More Useful Form

The energy stored depends on the size of the wire (its length and area). A more fundamental quantity is the elastic potential energy per unit volume, or energy density $u$. Since volume $V = AL$:

But $\frac{\Delta L}{L}$ is the longitudinal strain $\varepsilon$. So:

Using the relation stress $\sigma = Y \varepsilon$, we can also write:

Physical Interpretation …