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*10^(-6)m^2`

Young modulus,

`\text{young modulus=2}.1*10^(11)\text{ pa}`

Calculate the strain in the wire,

`\begin{gathered} \text{Strain}=(\Delta L)/(L) \\ Strain=\frac{0.30*10^(-3)^{}\text{ m}}{\text{4.}0\text{ m}} \\ Strain=0.075*10^(-3)\text{ m} \end{gathered}`

Calculate the stress in the wire,

`\begin{gathered} \text{Stress}=young\text{ modulus}* strain \\ \text{Stress}=2.1*10^(11)*0.075*10^(-3) \\ \text{Stress}=0.1575*10^8Nm^(-2) \end{gathered}`

Calculate the volume of the wire,

`\begin{gathered} V=2*10^(-6)*4 \\ V=8*10^(-6)m^3 \end{gathered}`

Calculate the elastic potential energy stored.

`\begin{gathered} U=(1)/(2)* stress* strain*\text{volume} \\ U=(1)/(2)*0.1575*10^(8*)0.075*10^(-3)*\text{8}*10^(-6) \\ U=0.004725\text{ J} \end{gathered}`