Question

When a block of mass $2\text{kg}$ is suspended by a long wire of length $1\text{m}$, the length of wire becomes $1.02\text{m}$. The elastic potential energy stored in wire is
(Take $g=10m/s^2$)

• A. $0.1\text{J}$
• B. $0.2\text{J}$
• C. $0.4\text{J}$
• D. $0.8\text{J}$

Answer 1

The correct option is B $0.2\text{J}$

Elastic potential energy of stretched wire,
$\Delta U=\frac{1}{2}\times \text{Stretching force}\times \text{extension in wire}$
Here body is stretched due to its weight $mg$
$\Rightarrow \Delta U=\frac{1}{2}\times F\times \Delta L$
$\Delta U=\frac{1}{2}\times mg\times (1.02-1)$
$\Delta U=\frac{1}{2}\times 2\times 10\times 0.02$
$\therefore \Delta U=0.2\text{J}$