Elastic Potential Energy in a Stretched Wire and a Spring

Stretching a wire, or extending a spring, requires doing work against the body's internal restoring force. Because the deformation is elastic, this work is not dissipated (lost as heat, as it would be for a plastic or frictional deformation) -- it is stored inside the body as elastic potential energy, and is fully recovered (converted back into kinetic energy, or into work done on something else) when the deforming force is removed and the body springs back.

Elastic PE in a stretched wire. Consider a wire of natural length $L$ and cross-sectional area $A$, being stretched by a gradually increasing force, from an extension of $0$ up to a final extension $\Delta L$, entirely within the elastic (Hooke's-law) region. At any intermediate extension $x$ (where $0 \le x \le \Delta L$), Hooke's law gives the restoring force in the wire as directly proportional to $x$: writing $F(x) = \left(\dfrac{YA}{L}\right) x$ (since, by Young's modulus's own definition, $Y = \dfrac{F/A}{x/L}$, so $F = YAx/L$), the force-versus-extension graph is a straight line through the origin, of slope $YA/L$, up to the point $(\Delta L,\ F)$ where $F = YA\,\Delta L/L$ is the final force. The work done in stretching the wire from $0$ to $\Delta L$ -- and hence the elastic PE stored in it -- equals the area under this straight-line graph, which is a right triangle of base $\Delta L$ and height $F$:

Substituting $F = (\text{stress}) \times A$ and $\Delta L = (\text{strain}) \times L$, and noting that $A \times L$ is simply the original volume of the wire,

Dividing through by the volume gives the elastic potential energy stored per unit volume, often the more directly useful quantity since it does not depend on the particular size of wire being considered:

and, using $\text{stress} = Y \times \text{strain}$ (Hooke's law once again), this can equally be written purely in terms of the strain and Young's modulus alone:

Elastic PE in a stretched (or compressed) spring. A spring obeying Hooke's law has restoring force $F = -kx$ (or, ignoring the sign convention and just considering magnitudes, the applied stretching force needed to hold the spring at extension $x$ is $F = kx$), where $k$ is the spring's force constant (SI unit $\text{N/m}$) and $x$ is the extension (or compression) from the spring's natural length. Exactly the same "area under a straight-line force-extension graph" argument used for the wire applies here too -- the force-versus-extension graph is a straight line of slope $k$ through the origin, so the work done (and elastic PE stored) in stretching the spring from $0$ to a final extension $x$ is the area of the resulting right triangle:

$U = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(x)(kx) = \frac{1}{2} k x^2$ …