A uniform chain of mass m and length l is on a smooth horizontal table with (1/n)^th part of its leght is hanging from one end of the table. The velocity of the chain, when it completely slips off the table is

A uniform chain of mass m and length l is on a smooth horizontal table

1.	A uniform chain of mass m and length l is on a smooth horizontal table with (1/n)^th part of its leght is hanging from one end of the table. The velocity of the chain, when it completely slips off the table is
Answer» Solution :With respect to the TOP of the table, the initial potential energy of the chain, `U_(1)`= PE 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 jaw of conservation of energy `Delta K = - Delta U K_(f) - K_(i) = - (U_(f) - U _(i))` `thereforeK_(i)=0,K_(f)=-[-(mgL)/(2)-(-(mgL)/(2n^(2)))]` `K_(f)= (mgL)/(2)[1-(1)/(n^(2))]` It .V. is the velocity of the chain, then, `(1)/(2)mv^(2)=(mgL)/(2)[1-(1)/(n^(2))]` `thereforev=sqrt(gL[1-(1)/(n^(2))])`