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 $F$ the wire exerts on the mass.

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

Answer 1

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

Draw a labelled diagram


Find the work done by force F

Formula used: $Y=\frac{FL}{Al}$
Work done by applied force and gravity will be,
$W=\int F\cdot dx$
$W={\int }_0^{\Delta x}K(x+\Delta l)dx$
$=K{[\frac{x^2}{2}+x\Delta l]}_0^{\Delta x}$
$=\frac{YA}{L}[\frac{{\left(\Delta x\right)}^2}{2}+\left(\Delta x\right)\left(\Delta l\right)]$
It will be work done by the force the wire exerts on the mass.
$=|-W|=W=\frac{YA}{L}[\frac{{\left(\Delta x\right)}^2}{2}+\Delta x\Delta l]$
Elongation in steel wire, $Y=\frac{FL}{Al}$
But $\Delta l=\frac{mgl}{AY}$
So, $W=\frac{YA}{L}\Delta x[\frac{\Delta x}{2}+\frac{mgl}{AY}]$
$=\frac{YA}{L}\frac{{\left(\Delta x\right)}^2}{2}+mg\Delta x$
Final Answer: (a)