Question

A steel wire of length $5\text{m}$ is suspended vertically, stretches by $20\text{mm}$ when mass of $60\text{kg}$ is attached to the lower end. Find the elastic potential energy gained by the wire. (take $g=10{\text{m/s}}^2$)

• A. $0.6\text{J}$
• B. $6\text{J}$
• C. $60\text{J}$
• D. $600\text{J}$

Answer 1

The correct option is B $6\text{J}$

Given, $L=5\text{m}$
$\Delta L=20\times {10}^{-3}\text{m}$ (extension in the wire)


Stretching force $\left(F\right)=mg=60\times 10$
$\Rightarrow F=600\text{N}$
We know, elastic potential energy of stretched wire,
$U=\frac{1}{2}\times \text{Stretching force}\times \text{Extension in wire}$
$=\frac{1}{2}\times F\times \Delta L$
$=\frac{1}{2}\times 600\times 20\times {10}^{-3}$
$\Rightarrow U=6\text{J}$
$\therefore$ Elastic potential energy gained by the wire is $6\text{J}$.