Question

A wire $2m$ in length suspended vertically stretches by $10mm$ when mass of $10kg$ is attached to the lower end. The elastic potential energy gain by the wire is (take $g=10m/s^2)$

• A. 0.5 J
• B. 5 J
• C. 50 J
• D. 500 J

Answer 1

Answer: (A)

Potential energy per unit volume $=\frac{1}{2}\times$ stress $\times$ strain
$=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta L}{L}$
So, Potential energy $=$ potential energy per unit volume $\times$ volume
$=\frac{1}{2}\times \frac{F\cdot \Delta L}{A\cdot L}\times A\cdot L$
{ Volume $=$ Length $\times$ cross-sectional area }
$\Delta U=\frac{1}{2}\cdot F\cdot \Delta L$
$F=10\times 10N$
$\Delta L=10mm=10\times 10^{-3}m$
Substituting values
$\Delta U=\frac{1}{2}\times 100\times \frac{10}{1000}$
$\Delta U=0.5J$