Question

A uniform chain of length $L$ and mass $M$ overhangsa horizontal table with its two third part on the table. The friction coefficient between the table and the chain is $\mu$. Find the work done by the friction during the period the chain slips off the table.

• A. $\frac{2\mu MgL}{9}$
• B. $-\frac{2\mu MgL}{9}$
• C. $\frac{-\mu MgL}{9}$
• D. None of these

Answer 1

The correct option is A $\frac{2\mu MgL}{9}$

Force of friction is against the movement (sliding) of the chain on the table.

$d{F}_f$ on element $dx$ of mass $dm$ due to friction $=\mu k\times dm\times g$

$=\mu g\times \left(\frac{m}{L}\right)dx$

Element $dx$ gets displaced by $x$. When the chain slides fo $f$.

So, work done on $dx$ by friction $d{F}_f\times x==\mu g\times \left(\frac{m}{L}\right)dx\times x$

Total work $\underset{0}{\overset{\frac{2L}{3}}{\int }}d{F}_f\times x$

$\begin{array}{c}=\frac{\mu gm}{L}{\int }_0^{\frac{2L}{3}}xdx=\frac{\mu gm}{L}{\left[\frac{x^2}{2}\right]}_0^{\frac{2L}{3}}\\ =\frac{2}{9}\mu gL\end{array}$