Question

If Young’s modulus for steel is 20.0 x 10^11 Nm^-2, 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 without the elastic limit being reached.

Answer 1

As per formula:

$E=\frac{F\times l}{A\times \delta l}$

where E - Young's module, F - force, l - length of wire, A - cross-sectional area of wire, $\delta l$ - change of length of wire over load.

Findind change of length:

$\delta l = \frac{F\times l}{A\times E} = \frac{100 \times 1}{3\times 10^{-7} \times20\times 10^{11}}=1.7\times10^{-4}metres$

Finding work:

$A=F\times \delta l = 100\times1.7\times 10^{-4}= 0.017J$