Question

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

Answer 1

Here, $l=4.0m;\Delta l=2\times {10}^{-3}m;a=2.0\times {10}^{-6}{m}^2,Y=2.0\times {10}^{11}N/m^2$
(i) The energy density of stretched wire
$u=\frac{1}{2}\times$ stress $\times$ strain
$=\frac{1}{2}\times Y\times (strain{)}^2$
$=\frac{1}{2}\times 2.0\times {10}^{11}\times (2\times {10}^{-3})/4{)}^2$
$=0.25\times {10}^5=2.5\times {10}^{4}J/m^3$.
(ii) Elastic potential energy $=$ energy density $\times$ volume
$=2.5\times {10}^4\times (2.0\times {10}^{-6})\times 4.0J=20\times {10}^{-2}=0.20$J.