Question

Calculate the energy stored in the stretched wire. With an extension of 0.30mm a steel wire with a length of 4.0m and a cross sectional area of 2.0x 10^6m*

Answer 1

Given data,

Change in length,

$\Delta L=0.\text{30 mm}$

Length of the steel wire,

$L=4.0\text{ m}$

Area,

$A=2.0\times10^{-6}m^2$

Young modulus,

$\text{young modulus=2}.1\times10^{11}\text{ pa}$

Calculate the strain in the wire,

$\begin{gathered} \text{Strain}=\frac{\Delta L}{L} \\ Strain=\frac{0.30\times10^{-3}^{}\text{ m}}{\text{4.}0\text{ m}} \\ Strain=0.075\times10^{-3}\text{ m} \end{gathered}$

Calculate the stress in the wire,

$\begin{gathered} \text{Stress}=young\text{ modulus}\times strain \\ \text{Stress}=2.1\times10^{11}\times0.075\times10^{-3} \\ \text{Stress}=0.1575\times10^8Nm^{-2} \end{gathered}$

Calculate the volume of the wire,

$\begin{gathered} V=2\times10^{-6}\times4 \\ V=8\times10^{-6}m^3 \end{gathered}$

Calculate the elastic potential energy stored.

$\begin{gathered} U=\frac{1}{2}\times stress\times strain\times\text{volume} \\ U=\frac{1}{2}\times0.1575\times10^{8\times}0.075\times10^{-3}\times\text{8}\times10^{-6} \\ U=0.004725\text{ J} \end{gathered}$