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

A uniform chain of length 'L' and mass 'm' is on a smooth horizontal t

1.	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
Answer» SOLUTION :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))]`