Question

A steel wire of length $2$m is hanging from a rigid horizontal support. How much energy is stored in it when a load of $5$ kg is suspended on it? If the load is increased to $10$ kg then by how much the energy stored will increase? ($Y=2\times {10}^{11}N/m^2$, area of cross-section $={10}^{-6}{m}^2$).

Answer 1

We know the energy stored per unit length

$U=\frac{1}{2}\times stress\times strain$

$5Kg$ load

$U_1=\frac{1}{2}\times \frac{5g}{{10}^{-6}}\times \frac{5g/{10}^{-6}}{2\times {10}^{11}}$ $[y=\frac{stress}{strain}\Rightarrow strain=\frac{stress}{y}]$

$U_1=\frac{1}{2}\times \frac{5g}{{10}^{-6}}\times \frac{5g}{{10}^{-6}}\times \frac{1}{4\times {10}^{-1}}=\frac{250g}{4}$

Energy stored in wire $=U_1\times 2=\frac{250g}{4}\times 2=125g$

$10kg$ load

$U_2=\frac{1}{2}\times \frac{10g/{10}^{-6}}{2\times {10}^{11}}=\frac{100g}{4\times {10}^{-1}}=\frac{1000g}{4}$

Energy stored in wire $=\frac{U}{2}\times 2=\frac{1000g}{4}\times 2=500g$

Thus, the difference in energy i.e, amount of energy increased is

$500g-125g=375g=375\times 9.8=3675J$