Question

A massless wire $2\text{m}$ in length suspended vertically stretches by $10\text{mm}$ when mass of $10\text{kg}$ is attached to the lower end. The elastic potential energy gained by the wire is (Take $g=10{\text{m/s}}^2$)

• A. $0.5\text{J}$
• B. $5\text{J}$
• C. $50\text{J}$
• D. $500\text{J}$

Answer 1

The correct option is A $0.5\text{J}$

Change in elastic potential energy is given by:
$\Delta U=\frac{1}{2}\times F\times \Delta L...\left(i\right)$
Change in length of wire is,
$\Delta L=10\text{mm}=10\times {10}^{-3}\text{m}$
Force/Load acting on wire is its weight:
$F=mg=10\times 10=100\text{N}$
From Eq. $\left(i\right)$
$\Delta U=\frac{1}{2}\times 100\times 10\times {10}^{-3}$
$\therefore \Delta U=0.5\text{J}$