A uniform chain of length L and mass M overhangs a horizontal table with its two third part n the table. The friction coefficient between the table and the chain is `mu`. Find the work done by the friction during the period the chain slips off the table.

Explanation

Let at some instant the hanging length is `x`. Friction acting on the chain
`f=mu(m)/(L)(L-x)g`
Small work done by friction during displacement `dx`
`dW=-fdx`
Total work done from `x=(L)/(3)` to `x=L`
`W=-int_((L)/(3))^(L)(mumg)/(L)(L-x)dx`
`=-(mumg)/(L)|Lx-(x^2)/(2)|_((L)/(3))^(L)`
`=-(mumg)/(L)[{LxxL-(L^2)/(2)}-{Lxx(L)/(3)-(1)/(2)((L)/(3))^2}]`
`=-(mumg)/(L)[((L^2)/(2))-((5L^2)/(18))]`
`=-(mumg)/(L).(4L^2)/(18)=-(2mumgL)/(9)`