Question

A uniform chain of mass m and length l overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table.

Answer 1

Let 'dx' be the length of an element at distance x from the table.
Mass of the element, 'dm'$=\left(\frac{m}{l}\right)dx$
Work done to putting back this mass element on the table is

$dW=\left(\frac{m}{l}\right)\times x\times g\times dx$

So, total work done to put $\frac{1}{3}$ part back on the table

$$\begin{gathered} W={\int }_0^{1/3}\left(\frac{m}{l}\right)gxdx \\ \Rightarrow W=\left(\frac{m}{l}\right)g{\left[\frac{x^2}{2}\right]}^{1/3} \\ =\frac{mgl}{18l}=\frac{mgl}{18} \end{gathered}$$

The work to be done by a person to put the hanging part back on the table is $\frac{mgl}{18}$.