A chain of length L and mass M is held on a frictionless table with `(1//n)th` of its length hanging over the edge. When the chain is released, find the velocity of chain while leaving the table.

Explanation

Taking surface of table as a reference level (zero potential energy)

Potential energy of chain when `1//nth` length hanging from the edge `U_(i nitial)=(-MgL)/(2n^2)`
Potential energy of chain when it leaves the table,
`U_(i nitial)=-(MgL)/(2)`
Using conservation of mechanical energy
`DeltaK+DeltaU=0`
or Kinetic energy of chain = Loss in potential energy
`implies 1/2Mv^2=(MgL)/(2)-(MgL)/(2n^2)`
`=(MgL)/(2)[1-(1)/(n^2)]`
Therefore, velocity of chain `v=sqrt(gL(1-(1)/(n^2)))`