Question

A mass m is hanging from a wire of cross sectional area A and length L. Y is Young’s modulus of wire. An external force F is applied on the wire which is then slowly further pulled down by $\Delta$x from its equilibrium position. Find the work done by the force F that the wire exerts on the mass.

• A. $mg\frac{\Delta x}{2}+\frac{yA{\left(\Delta x\right)}^2}{L}$
• B. $mg\frac{\Delta x}{2}+\frac{yA{\left(\Delta x\right)}^2}{2L}$
• C. $mg\Delta x+\frac{yA{\left(\Delta x\right)}^2}{L}$
• D. $mg\Delta x+\frac{yA}{2L}{\left(\Delta x\right)}^2$

Answer 1

The correct option is D $mg\Delta x+\frac{yA}{2L}{\left(\Delta x\right)}^2$

Work done by applied force and gravity
$w=\underset{0}{\overset{\Delta x}{\int }}K(x+\Delta l)d\alpha =k{[\frac{{\alpha }^2}{2}+x\cdot \Delta l]}_0^{\Delta x}=\frac{yA}{L}[\frac{{\left(\Delta x\right)}^2}{2}+(\Delta x\left)\right(\Delta l\left)\right]$

It will be the work done by the force that the wire exerts on the mass
$=w=\frac{yA}{L}[\frac{{\left(\Delta x\right)}^2}{2}+\Delta x\cdot \Delta l]$

But $\Delta l=\frac{mgL}{Ay}$

So, $w=\frac{yA}{L}\Delta x[\frac{\Delta x}{2}+\frac{mg{L}^2}{Ay}]$

$=\frac{yA}{L}\frac{{\left(\Delta x\right)}^2}{2}+mg\cdot \Delta x$