Question

A chain of mass $M$ and length $l$ rests on a rough table with part overhanging. The chain starts sliding down by itself if overhanging part is $\frac{l}{3}$. What will be the work performed by the friction force acting on the chain at the moment it slides completely off the table?

• A. $\frac{Mgl}{3}$
• B. $\frac{2Mgl}{3}$
• C. $\frac{2Mgl}{9}$
• D. $\frac{Mgl}{9}$

Answer 1

Answer: (D)

$\frac{Mgl}{9}$

$\mu \frac{2}{3}mg=\frac{m}{3}g$ or $\mu =\frac{1}{2}$
Assume at any instant the length of the chain on the table is x then force of friction $=\mu N=\frac{\mu M}{l}xg$
Work done against friction ${\int }_0^{2l/3}\mu \frac{M}{l}xg\cdot dx$
$=\mu \frac{Mg}{l}\frac{x^2}{2}{|}_0^{2l/3}=\frac{1}{2}\frac{M}{l}\frac{g}{2}\left(\frac{4l^2}{9}\right)=\frac{Mgl}{9}$