Question

A uniform chain of length L and mass M overhangs a 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 friction during the period the chain slips off the table.

Answer 1

Let x length of the chain be on the table at a particular instant.
Consider a small element of length 'dx' and mass 'dm' on the table.
dm = $\frac{M}{L}dx$
Work done by the friction on this element is
$dW=\mu \mathit{}Rx=\mu \left(\frac{M}{L}\times gx\right)dx$

Total work done by friction on two third part of the chain,
$$\begin{gathered} W={\int }_{2L/3}^0\mu \frac{\mathrm{M}}{L}gxdx \\ \therefore \mathrm{W}=\mu \frac{M}{L}g{\left[\frac{x^2}{2}\right]}_0^{2L/3} \\ =-\mu \frac{M}{L}g\left[\frac{4L^2}{18}\right] \\ =-\frac{2\mu MgL}{9} \end{gathered}$$
The total work done by friction during the period the chain slips off the table is $-\frac{2\mu MgL}{9}$.