Question

A uniform chain of length $2$ m and mass $0.1$ kg overchangs 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.

Answer 1

$W=\frac{mgl}{2n^2}$

Here $\frac{2}{3}th$ part is lying on the table it means $\frac{1}{3}th$is hanging

Remember n should always be greater than 1

Here n is 3

So $W=\frac{mgl}{18}$

Now W=kinetic energy

$W=\frac{mgl}{18}=\frac{0.1\times 9.81\times 2}{18}=0.10$