Physics - Elasticity Question with Solution | TestHub

Young's modulus $Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}$. From this, the force $F=\frac{YA}{L}\Delta L$. This is analogous to $F=kx$ for a spring, where $k=\frac{YA}{L}$. The work done in stretching the wire (elastic potential energy stored) is $$W=\frac{1}{2}F\Delta L=\frac{1}{2}\left(\frac{YA}{L}\Delta L\right)\Delta L=\frac{1}{2}\frac{YA}{L}(\Delta L{)}^2$$