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 of mass 'dx' length

$=\left(\frac{m}{l}\right)$dx work done to put back on the table.

$w=\left(\frac{m}{l}\right)dxg\left(x\right)$

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

$W={\int }_0^{1/3}\left(\frac{m}{l}\right)gxdx$

$\Rightarrow w=\left(\frac{m}{l}\right)g{\left[\frac{x^2}{2}\right]}_0^{1/3}$

$=\frac{mg{l}^2}{18l}=\frac{mgl}{18}$