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

Explanation

Mass of hanging part of chain `= m//3`
Position of centre of gravity below table `= L//6`
`:.` Work done = Change in potential energy
or Work done `= (m/3) g(L/6) = (mgL)/(18)`
or Work required to be done = `(mgL)/(18)`