Question

Calculate the work done in stretching steel wire of length 2m and of cross sectional area 0.0225 $m{m}^2$, when a load of 100 N is applied slowly to its free end. (Young's modulus of steel = 20 x ${10}^{10}N/m^2$)

Answer 1

Given : $L=2m$ $A=0.0225m{m}^2=0.0225\times {10}^{-6}{m}^3$ $F=100N$

Stress $=\frac{F}{A}=\frac{100}{0.0225\times {10}^{-6}}=4.4\times {10}^{9}N/m^2$

Energy stored $U=\frac{1}{2}\times \frac{(stress{)}^2}{Y}\times AL$

Or $U=\frac{1}{2}\times \frac{(4.4\times {10}^9{)}^2}{20\times {10}^{10}}\times (0.0225\times {10}^{-6})\left(2\right)=2.222J$