Work Done In Stretching A Wire - Practice Questions & MCQ

When a body is in its natural shape, its potential energy corresponding to the molecular forces is minimum. When deformed, internal forces appear and work has to be done against these forces. Thus, the potential energy of the body is increased. This is called the elastic potential energy.

Suppose a wire having natural length L and cross-sectional area A is fixed at one end and is stretched by an external force applied at the other end. When the extension is x,the wire is under a longitudinal stress F/A. The strain is x/L.

If Young's modulus is Y, then

$
\frac{F / A}{x / L}=Y \Rightarrow F=\frac{A Y}{L} x
$

So, the work done for an additional small increase dx in length will be:-

$
d W=F d x \Rightarrow d W=\frac{A Y}{L} x d x
$

The total work done by the extermal force in increasing the length from 0 to $\Delta L$ will be:-

$
W=\int_0^{\Delta L} \frac{A Y}{L} x d x=\frac{1}{2} \frac{Y A}{L}(\Delta L)^2
$

This work done is stored into the wire as its elastic potential energy. So, the elastic potential energy of the stretched wire is:

$
U=\frac{1}{2} \frac{Y A}{L}(\Delta L)^2
$

We can also write,

$
\begin{aligned}
& W=\frac{1}{2}\left[\frac{Y A \Delta L}{L}\right] \Delta L=\frac{1}{2}(\text { maximum stretching force }) \times \text { extension } \\
& W=\frac{1}{2} Y(A L)\left[\frac{\Delta L}{L}\right]^2=\frac{1}{2} \times Y \times \text { Volume } \times(\text { strain })^2
\end{aligned}
$

Also, Potential energy per unit volume $=\frac{1}{2} \times$ strain $\times$ stress