Question

A uniform chain of length $L$ and mass $M$ overhangs with two third of its length on the table. The coefficient of friction between table and chain is $\mu$. The work done by the friction during the time, the chain slips off the table is

• A. $-\frac{2\mu MgL}{9}$
• B. $-\frac{\mu MgL}{9}$
• C. $-\frac{\mu MgL}{18}$
• D. $-\mu MgL$

Answer 1

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


Let $x$ be the length of the chain on the table at a certain instant.

$\therefore$ work done by friction for a small displacement $dx$ is

$dW=fdx=\left(\mu N\right)dx=\mu \left(\frac{M}{L}xg\right)dx$

$dW=\frac{\mu Mg}{L}xdx$

Thus, the total work done by friction during the given interval is

$W=\int dW=\underset{2l/3}{\overset{0}{\int }}\frac{\mu Mg}{L}xdx$

$W=\frac{\mu Mg}{L}{\left[\frac{x^2}{2}\right]}_{2l/3}^0=\frac{\mu Mg}{2L}(0-\frac{4L^2}{9})$

$\therefore W=-\frac{2\mu MgL}{9}$

Therefore, option $\left(A\right)$ is the correct choice.