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Q.
A wire of length $L$ and cross-sectional area $A$ is made of a material of Youngs modulus $Y$. The work done in stretching the wire by an amount $x$ is given by
AMUAMU 2009
Solution:
In stretching a wire work is done against internal restoring forces.
This work is stored in the wire as elastic potential energy or strain energy.
If a force $F$ acts along the length $L$ of the wire of cross-section $A$ and stretches it by $x$, then
$Y=\frac{\text { stress }}{\text { strain }}=\frac{F / A}{x / L}=\frac{F L}{A x}$
$\Rightarrow F=\frac{Y A x}{L}$
So, the work done for an additional small increase $dx$ in length,
$d W=F d x=\frac{Y A}{L} x d x$
Hence, the total work done in increasing the length by $l$,
$W=\int\limits_{0}^{x} d W=\int\limits_{0}^{x} F d x=\int\limits_{0}^{x} \frac{Y A}{L} x d x=\frac{1}{2} \frac{Y A}{L} x^{2}$
This work done is stored in the wire.
$\therefore $ Energy stored in wire
$U=\frac{1}{2} \frac{Y A x^{2}}{L}$