A uniform chain of length 'L' and mass 'm' is on a smooth horizontal table, with `(1)/(n)` th part of its length hanging from the edge of the table. Find the kinetic energy of the chain as it completely slips off the table

Explanation

With respect to the top of the table, the initial potential energy of the chain
`U_(1)`= PF of the chain lying on the table + PE of the hanging part of the chain
`= L (1- (1)/(n)) mg (0) (m)/(n)g (-(L)/(2n))= - (mgL)/(2n^(2))`
P.E of the chain, when it just slips off the table.
`U_(2)= mg ((-L)/(2)) = - (mgL)/(2)`
From law of conservation of energy
`Delta K= - Delta U K_(r)- K_(i) = - (U_(f)- U_(i))`
`because K_(i)= 0, K_(f)= - [-(mgL)/(2)- (-(mgL)/(2n^(2)))]`

If .V. is the velocity of the chain, then, `(1)/(2) mv^(2)= (mgL)/(2) [1-(1)/(n^(2))]`