A unifrom chain of mass M and length L lies on a table with nth part of it hanging off the table. Find the work required to slowly pull the hanging part up to the table.

Explanation

Here work is done against the gravity.
`therefore W=DeltaU=U_(f)-U_("in")`
Taking table as reference level, i.e. `U_(f)=0`
`therefore W=-U_("in")=-[-mgh_(C.M.)]`
`=Mgh_(C.M.)`
here mass of hanging part is m = `(M)/(n)`
and its centre of mass is `h_(C.M.)` hight below the table
where `h_(C.M.)=((L//n))/(2)=(L)/(2n)`
`therefore W=((M)/(n))g((L)/(2n))=(MgL)/(2n^(2))`