JEE Main & Advanced Physics Elasticity Work Done in Stretching a Wire

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

\[Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{x/L}=\frac{FL}{Ax}\]\[\Rightarrow \] \[F=\frac{YA}{L}x\]

So the work done for an additional small increase dx in length, \[dW=Fdx=\frac{YA}{L}x\,.\,dx\]

Hence the total work done in increasing the length by l,       \[W=\int_{0}^{l}{dW}=\int_{0}^{l}{Fdx}=\int_{0}^{l}{\frac{YA}{L}.x\,dx}=\frac{1}{2}\frac{YA}{L}{{l}^{2}}\]

This work done is stored in the wire.

\[\therefore \] Energy stored in wire \[U=\frac{1}{2}\frac{YA{{l}^{2}}}{L}=\frac{1}{2}Fl\]  \[\left[ \text{As }F=\frac{YAl}{L} \right]\]

Dividing both sides by volume of the wire we get energy stored in unit volume of wire.  

\[{{U}_{V}}=\frac{1}{2}\times \frac{F}{A}\times \frac{l}{L}\]\[=\frac{1}{2}\times \text{stress}\times \text{strain}=\frac{\text{1}}{\text{2}}\times Y\times {{(\text{strain})}^{2}}\] \[=\frac{1}{2Y}{{(\text{stress})}^{2}}\]      [As AL = volume of wire]

Note :

• If the force on the wire is increased from F1 to F2 and the elongation in wire is l then energy stored in the wire \[U=\frac{1}{2}\frac{({{F}_{1}}+{{F}_{2}})}{2}l\]
• Thermal energy density = Thermal energy per unit volume =\[\frac{1}{2}\] \[\times \] Thermal stress \[\times \] strain

= \[\frac{1}{2}\frac{F}{A}\frac{l}{L}\] = \[\frac{1}{2}(Y\alpha \Delta \theta )(\alpha \Delta \theta )\] = \[\frac{1}{2}Y{{\alpha }^{2}}{{(\Delta \theta )}^{2}}\]