Question

A mass m is hanging from a wire of cross sectional are A and length L. Y is young's modulus of wire. An external force F is applied on he wire which is then slowly further pulled down by $△x$ from its equilibrium position. Find the work done by the force F that the wire exerts on the mass:

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

Answer 1

The correct option is A $mg△x+\frac{YA}{2L}(△x{)}^2$

When external force is applied, work done is given by

Work done $W=mg\Delta x$

And

$W_2=\frac{1}{2}\times stress\times strain\times volume$

$\frac{1}{2}\times Y\times {\left(strain\right)}^2\times V$

$\frac{1}{2}\times Y\times {\left(\frac{\Delta x}{L}\right)}^2\times AL$

$\frac{Y\Delta x^{2}A}{2L}$

So, total work done is

$W=mg\Delta x+\frac{Y\Delta x^{2}A}{2L}$