A uniform chain of length l and mass m overhangs a smooth table with its two third part lying on the table. Find the kinetic energy of the chain as it completely slips off the table.

Explanation

Let us take the zero of potential energy at the table. Consider as part dx of the chain at a depth x below the surface of the table. The mass of this part is `dm/m/l dx` and hence its potentil energy is `-(m/ldx)gx`.

The potential energy of the l/3 of the chain that overhangs is `U_1=int^_(1/3)/_0-m/l gx dx`
`=-[m/lg(x^2/2)]^(1/3)_0=-1/18mgl`.
this is also the potential energy of the full chain in the initial position because the part lyng on the table has zero potential energy. The potential energy of the chain when it completely sklips off the table is
`U_2=int^2_0-m/lgx dx=-1/2mgl`
The loss in potential energy `=(-1/18 mgl)-(-1/2mgl)`
`=4/9 mgl`
Ths shoud be equal to the gain in the kinetic energy. But the initial kinetic energy is zero. Hence the kinetic energy of the chain as it completely slips off the table is `4/9` mgl.