Elastic Potential Energy

Stretching an elastic body stores energy because the applied force does work against internal restoring forces, and within the elastic region stress rises linearly with strain. Consider a small extension $dx$ of a rod of length $L$ and area $A$ under an instantaneous force $F=(YA/L)x$, where $x$ is the current extension. The total work done is $W={\int }_0^{\Delta L}\frac{YA}{L}xdx=\frac{1}{2}\frac{YA}{L}(\Delta L{)}^2$. Dividing by the volume $AL$ gives energy density $u=\frac{1}{2}Y{\left(\frac{\Delta L}{L}\right)}^2=\frac{1}{2}Y{\epsilon }^2$. Since stress $\sigma =Y\epsilon$, this becomes $u=\frac{1}{2}\sigma \epsilon$, the area under the stress–strain line. The option $\sigma \epsilon$ forgets the factor one-half from integrating a linear force. The option $2\sigma \epsilon$ overcounts the work fourfold. The option $\frac{1}{2}{\sigma }^2\epsilon$ is dimensionally inconsistent for energy density. This matches the NCERT result that elastic energy equals the triangular area under the stress–strain graph, and the same factor of one-half appears in the spring energy $\frac{1}{2}k{x}^2$ because the restoring force likewise builds up linearly from zero. The stored energy is fully recoverable only while the material stays within its elastic limit; beyond that point part of the work goes into permanent plastic deformation and is dissipated as heat rather than returned. Dimensionally, $\sigma \epsilon$ has units of ${\text{N/m}}^2={\text{J/m}}^3$, confirming it is an energy density rather than a total energy.