Question

A steel wire of length $2.0m$ is stretched through $2.0mm$. The cross-sectional area of the wire is $4.0m{m}^2$. Calculate the elastic potential energy (in $J$) stored in the wire in the stretched condition. Young's modulus of steel $=2.0\times {10}^{11}N{m}^{-2}$.

Answer 1

Formula used:

Stress$=Y\times \left(\text{Strain}\right)$

Given,

$l=2.0m$

$\Delta l=2.0mm$

$A=4m{m}^2$

Young's modulus of steel $=2.0\times {10}^{11}N{m}^{-2}$

As we know,

Strain in the wire,

$\frac{\Delta l}{l}=\frac{2.0\times {10}^{-3}}{2.0}={10}^{-3}$

Stress in the wire $=Y\times \text{Strain}$

$=2.0\times {10}^{11}\times {10}^{-3}$

$=2.0\times {10}^{8}N{m}^{-2}$

Volume of the wire,

$=(4\times {10}^{-6}{m}^{-2})\times \left(2.0m\right)$

$=8.0\times {10}^{-6}{m}^3$

Elastic potential energy stored,

$=\frac{1}{2}\times \text{Stress}\times \text{Strain}\times \text{Volume}$

$=\frac{1}{2}\times 2.0\times {10}^8\times {10}^{-3}\times 8.0\times {10}^{-6}$

$=0.8J$
Final Answer: (0.8)