# A steel rod of 0.5 cm diameter and 10 m length is stretched 3 cm. Young?s modulus for this steel...

## Question:

A steel rod of 0.5 cm diameter and 10 m length is stretched 3 cm. Young?s modulus for this steel is 21 kN/cm2 . How much work, in kJ, is required to stretch this rod?

## Work Done In Stretching A Wire:

In order to change the dimension of the rod, we need to apply some amount of force at the cross-section of the rod. Then, the work done by the stretching forces for the change in length of the rod is given by the following formula:

{eq}W = \dfrac{EAL}{2} \left( \dfrac{\Delta L}{L} \right)^2\\ W = \dfrac{1}{2}* \text{Stress} * \text{Strain}* \text{Volume} {/eq}

Where,

• E is Young's modulus.
• A is the cross-sectional area.
• L is the length of the rod.

We're given the following information in the problem:

• Diameter of the rod is, {eq}D = 0.5\ cm {/eq}
• Length of the rod is, {eq}L = 10\ m = 1000\ cm {/eq}
• Young's modulus of the steel rod is, {eq}E = 21\ kN/cm^2 = 21000\ N/cm^2 {/eq}
• Change in the length of the rod is, {eq}\Delta L = 3\ cm {/eq}

The crossectional area of the rod is,

{eq}A = \dfrac{\pi}{4}D^2\\ A = \dfrac{\pi}{4}(0.5\ cm)^2\\ A = 0.196\ cm^2 {/eq}

The amount of work required to stretch the rod is,

{eq}W = \dfrac{EAL}{2} \left( \dfrac{\Delta L}{L} \right)^2\\ W = \dfrac{(21000\ N/cm^2)(0.196\ cm^2)(1000\ cm)}{2} \left( \dfrac{3\ cm}{1000\ cm} \right)^2\\ W = 18.522\ N-cm\\ W = 18.522* 10^{-5}\ kJ {/eq}