Work Done in Stretching a Wire
Category : JEE Main & Advanced
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]
Total energy stored in wire (U) | Energy stored in per unit volume of wire (UV) |
\[\frac{1}{2}Fl\] | \[\frac{1}{2}\frac{Fl}{\text{volume}}\] |
\[\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}\] | \[\frac{1}{2}\times \text{stress}\times \text{strain}\] |
\[\frac{1}{2}\times Y\times {{(\text{strain})}^{2}}\times \text{volume}\] | \[\frac{1}{2}\times Y\times {{(\text{strain})}^{2}}\] |
\[\frac{1}{2Y}\times {{(\text{stress)}}^{\text{2}}}\times \text{volume}\] | \[\frac{1}{2Y}\times {{(\text{stress})}^{2}}\] |
Note :
= \[\frac{1}{2}\frac{F}{A}\frac{l}{L}\] = \[\frac{1}{2}(Y\alpha \Delta \theta )(\alpha \Delta \theta )\] = \[\frac{1}{2}Y{{\alpha }^{2}}{{(\Delta \theta )}^{2}}\]
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