Calculate the work done in stretching a steel wire $100\;cm$ in length and of cross-sectional area $0.03\;cm^2$ when a load of $100\;N$ is slowly applied before the elastic limit is reached.

Given,
Length of the steel $(l) = 100\;cm = 1 \; m$
Cross-sectional area $(A) = 0.03\,cm^2 = 0.03 * 10^{-4}\;m^2$
Load applied $(F) = 100\;N$
Young's modulus of elasticity $(Y) = 2 * 10^{11}\,N/m^2$
Work done $(W) =$

Now,

$Y = \frac{F/A}{e/l} = \frac{F\,*\,l}{e\,*\,A}$

⇒ $e =$ $\frac{F\,*\,l}{A \,*\,Y} = \frac{100\,*\,1}{0.03\,*\,10^{-4}\,*\,2\,*\,10^{11}}$ $= 1.677\,*\,10^{-4}\;m$

Now, Work done to stretch wire is,

$W =$ $\frac{1}{2}\,F\,.\,e = \frac{1}{2}\;*\;100\;*\;1.677 \; * \; 10^{-4} = 8.33\;*\; 10^{-3}\;J$

Return to Main Menu