Question
What is meant by elastic potential energy? Derive an expression for the elastic potential energy of a stretched wire. Prove that its elastic energy density is equal to $\frac{1}{2}\text{stress}\times\text{strain}.$
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Answer
When a wire is stretched, some work is done against the internal restoring forces acting between particles of the wire. This work done appears as elastic potential energy in the wire. Consider a wire of length l and area of cross-section a. Let F be the stretching force applied on the wire and $\Delta\text{l}$ be the increase in length of the wire. Initially, the internal restoring force was zero but when the length is increased by $\Delta\text{l},$ the internal force increases from 0 to F(applied force). Thus, average initial force on an increase in length $(\Delta\text{l},)$ of the wire, $=\frac{0+\text{F}}{2}=\frac{\text{F}}{2}.$ Hence, work done on the wire, W = Average force × Increase in length, $=\frac{\text{F}}{2}\times\Delta\text{l}$ This is stored as elastic potential energy U in the wire, $\therefore\text{U}=\frac{1}{2}\text{F}\times\Delta\text{l}=\frac{1}{2}\frac{\text{F}}{\text{a}}\times\frac{\Delta\text{l}}{\text{l}}\times\text{al}$ $=\frac{1}{2}(\text{Stress})\times(\text{Strain)}\times\text{Volume of the wire}$ $\therefore$ Elastic potential energy per unit volume of the wire is given by, $\text{u}=\frac{\text{U}}{\text{al}}=\frac{1}{2}\times\text{Stress}\times\text{Strain}$
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