Find the work done in stretching a wire of cross-sectional area $10^{-2}\;cm^2$ and $2\;m$ long through $0.1\;mm$, if $Y$ for the material of wires is $2\;*\;10^{11}\;N/m^2$.

Given,
Cross-sectional area $(A) = 10^{-2}\;cm^2 = 10^{-6}\;m^2$
Length of wire $(l) = 2\;m$
Elongation of wire $(e) = 0.1\;mm = 0.1 * 10^{-3}\;m$
Young's modulus $(Y) = 2 * 10^{11}\;N/m^2$
Workdone $(W) = ?$

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

or,

$F = \frac{Y.e.A}{l}$

Then,

Workdone $(W) = \frac{1}{2}F.e = \frac{1}{2}\frac{Y.e^2.A}{l}$ 

$\;\;\;\; \;\;\;\; \;\;\;\;  = \frac{1}{2} * \frac{2 \;*\; 10^{11}\;N/m^2 \;*\; (0.1 \;*\; 10^{-3})^2 \;*\; 10^{-6}}{2}$ $= 5 * 10^{-4}\;J$

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