Q.Using the steel rod of Example 1 (length $1.5\ \text{m}$, area $2.5\times10^{-4}\ \text{m}^2$, stretched by $75{,}000\ \text{N}$, elongation $2.25\ \text{mm}$), calculate

When a wire is stretched, work is done against the internal restoring force developed in it, and this work is stored in the wire as ELASTIC POTENTIAL ENERGY. Consider a wire of un-stretched length L and cross-sectional area A, stretched (within the elastic limit, with no energy lost) by a force that produces a final extension l. The small increment of work done in stretching the wire a further dl is $dW=F\,dl$, so the TOTAL work done in stretching it from zero extension to l is $W=\int_0^l F\,dl$. Since Young's modulus gives $Y=\dfrac{F/A}{l/L}$, the force at any intermediate extension $l'$ is $F=\dfrac{YAl'}{L}$; substituting and integrating, $W=\int_0^l\dfrac{YAl'}{L}dl'=\dfrac{YAl^2}{2L}=\dfrac12\left(\dfrac{YAl}{L}\right)l=\dfrac12 Fl$. So the ELASTIC POTENTIAL ENERGY stored in a stretched wire is $W=\tfrac12 Fl$ -- exactly half the product of the final force and the final extension (because the force grew linearly from zero up to F as the wire stretched, by Hooke's law). Dividing by the wire's volume $AL$ gives the ENERGY DENSITY (elastic energy stored per unit volume): $u=\dfrac{W}{AL}=\dfrac12\left(\dfrac{F}{A}\right)\left(\dfrac{l}{L}\right)=\dfrac12\times\text{stress}\times\text{strain}$; since stress equals Young's modulus times strain (Hooke's law), this can equally be written $u=\dfrac12 Y(\text{strain})^2$ -- a form that is especially convenient when only the strain (not the raw force) is known or asked for. This elastic energy is what is relea …