Question

A uniform chain of length $l$ and mass $m$ lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. How much work need be done to pull the hanging part back to the table ?

Answer 1

Suppose length of an element at distance $x$ from the table $=dX$

Let us consider that the mass of the hanging part is connected at the centre of gravity of hanging part$.$

Work done require to pull the hanging part on the table$.$

$W=\frac{M}{L}dxg\left(x\right)$

The total work done to pull $1/3$ part on the table$,$

$W={\int }_0^{\frac{1}{3}}\frac{M}{L}g\left(x\right)dx$

$W=\frac{M}{L}g{\int }_0^{\frac{1}{3}}xdx=\frac{MgL}{18}$

We get, $W=\frac{MgL}{18}.$