Determine the elastic potential energy stored in stretched wire.

When a wire is stretched, work is done against the internal restoring forces acting between particles of the wire. This work done appears as elastic potential energy in the wire.

Consider a wire of length $\mathrm{L}$ and area of cross section $A$. Let deforming force F be the applied on the wire and $l$ be the increase in length of the wire.

$\mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}}$ here $\mathrm{Y}$ is the Young's modulus of wire.

$\Delta l=l$

Let work $d \mathrm{~W}$ is needed to increase in small length $d l$.

$\therefore d \mathrm{~W}=\mathrm{F} \times d l$ OR $\mathrm{YA} l \frac{d l}{\mathrm{~L}}$

Work $\mathrm{W}$ is done for increase the length of wire from $\mathrm{L}$ to $\mathrm{L}+l$. It is work for $l=0$ to $l=l$

$\mathrm{W}=\int_{0}^{l} \frac{\mathrm{YA} l}{\mathrm{~L}} d l=\frac{\mathrm{YA}}{2} \frac{l^{2}}{\mathrm{~L}}$

$\mathrm{W} \frac{1}{2} \times \mathrm{Y} \times\left(\frac{l}{\mathrm{~L}}\right)^{2} \times \mathrm{AL}$

$=\frac{1}{2} \times \text { Young modulus } \times(\text { Strain })^{2} \times \text { Volume of wire }$

$\mathrm{W}=\frac{1}{2} \times \text { Stress } \times \text { Strain } \times \text { Volume of wire }$

Hence, work stored in wire is the elastic potential energy. So the stored elastic potential energy per unit is obtained from $u=\frac{1}{2} \sigma \varepsilon$

$SI$ unit of elastic potential density is $\mathrm{N} \mathrm{m}^{-2}$ or $\mathrm{P} a$ and dimensional formula $\left[\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\right]$.