A uniform chain of length `L` and mass `M` is tying on a smoth table and one third of its length is banging vertically down table the edge of the table if g is acceleration the to gravity , the work required to pull the hanging part on the table is

Explanation

`U_1=-(1)/(2)((M)/(L)xx(L)/(3))g(L)/(3)=-(MgL)/(18)`
`U_2=0`
`W=-W_(gr)=U_2-U_1=(MgL)/(18)`