The elastic potential energy stored in a steel wire of length 20m...

The formula to calculate the Young's modulus of the material is given by

$\begin{cases}Y & = & \dfrac{stress}{strain} \\ & = & \dfrac{stress}{\dfrac{∆ l}{L}} . . . \left(1\right)\end{cases}$

The formula to calculate the potential energy \left(U\right)$stored into the wire is given by</p><p>$\begin{cases}U & = & \dfrac{1}{2} \times stress \times strain \times volume \\ & = & \dfrac{1}{2} \times stress \times \dfrac{∆ l}{L} \times A \times L \\ & = & \dfrac{1}{2} \times stress \times A \times ∆ l . . . \left(2\right)\end{cases}$</p><p>From equation (1) and equation (2), it can be written that</p><p>$U = \dfrac{1}{2} \times Y \times \dfrac{{\left(∆ l\right)}^{2}}{L} \times A . . . \left(3\right)$</p><p>Substitute the values of the known parameters into equation (3) and solve to calculate the required cross-sectional area of the wire.</p><p>$\begin{cases}80J & = & \dfrac{1}{2} \times 2 . 0 \times {10}^{11}{Nm}^{- 2} \times \dfrac{{\left(0 . 02m\right)}^{2}}{20m} \times A \\ & ⇒ & A = \dfrac{2 \times 80J \times 20}{2 . 0 \times {10}^{11} \times 4 \times {10}^{- 4}{Nm}^{- 1}} \\ & = & 4 \times {10}^{- 5}{m}^{2} \times \dfrac{{10}^{6}{mm}^{2}}{1{m}^{2}} \\ & = & 40{mm}^{2}\end{cases}