Question

A uniform chain of mass \(M\) and length \(L\) rests on a rough table such that one of its ends hangs over the edge. The chain starts sliding off the table by itself when overhanging part is equal to \(\dfrac{L}{4}\). The total work done by frictional force, till the moment the other end of chain slips-off the table, is

\( \Leftrightarrow M _{2} L \)

• A. $\frac{-MgL}{}$
• B. $\frac{-3MgL}{32}$
• C. $\frac{-MgL}{8}$
• D. $\frac{-3MgL}{40}$

Answer 1

The correct option is B $\frac{-3MgL}{32}$

$\mu \times \left(\frac{3Mg}{4}\right)=\frac{Mg}{4}$
$\Rightarrow \mu =\frac{1}{3}$
$\therefore W_f={\int }_0^{\frac{3L}{4}}\mu \frac{M}{L}\times x\times g\times dx=\frac{-3MgL}{32}$