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)= 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 law os conservation of energy `DELTA K =- Delta U "" K_(f)-K_(i)=-(U_(f)-U_(i))` `because K_(i)=0 , K_(f)=-[-(mgL)/(2)-(-(mgL)/(2n^(2)))]` `K_(f)=(mgL)/(2)[1-(1)/(n^(2))]` If .V. is the velocity of the chain, then, `(1)/(2)mv^(2)=(mgL)/(2)[1-(1)/(n^(2))] THEREFORE v=sqrt(gL[1-(1)/(n^(2))]`