# Energy stored in a stretched wire

​When a body is deformed, work is done by the deforming forces against the intermolecular force of attraction between the molecules of the body. This work done is stored in the body in the form of elastic potential energy.
​Consider a body of length L and of cross section area A. Suppose on applying force F, elongation x is produced in the body.
​Then,  the young's modulus of elasticity  of the material of the body is given by,
$Y=\frac{FL}{Ax}$
$\therefore F=\frac{YA}{L}$

Then, small work done to produce an infinitesimally small extension dx is given by,
$dW=Fdx$
Integrating within limits from 0 to x,
$W={\int }_{0}^{x}Fdx$
$W={\int }_{0}^{x}\frac{YA}{L}dx$
$W=\frac{YA}{L}{\int }_{0}^{x}xdx$
$W=\frac{YA}{L}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{x}$
$W=\frac{1}{2}\cdot \frac{YA}{L}\cdot x²$
$W=\frac{1}{2}\cdot \frac{YA}{L}x\cdot x$
$W=\frac{1}{2}Fx$
$W=\frac{1}{2}×force×extension$
Since, work done = energy stored
$\therefore \text{Energy stored}=\frac{1}{2}×force×extension$

# Energy Density

Energy density is defined as the elastic potential energy per unit volume of the deformed body.
​Thus, we have,
$\text{Energy Density (U)}=\frac{1}{2}\cdot \frac{Fx}{V}$
here, V = Volume of the deformed body and
$V=\text{Area of cross section}×length$
$V=A×L$
$\therefore \text{Energy Density (U)}=\frac{1}{2}\frac{Fx}{AL}$
Here,
$\frac{F}{A}=\text{normal stress}$

and,
$\frac{x}{L}=\text{Longitudinal strain}$