Question

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.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let us consider the zero of PE at the table.Consider a part dx of the chai x below the table.Its mass dm = $\frac{m}{l}$dx and hence its potential energy is $-\left(\frac{m}{l}dx\right)gx$.

Initial PE is only due to the overhanging part.

Therefore,

$U_1={\int }_0^{\frac{1}{3}}-\frac{m}{l}gxdx=-\frac{1}{18}mgl$

Final PE when it completely slips off,$U_2={\int }_0^l-\frac{m}{l}gxdx=-\frac{1}{2}mgl$

KE = $U_1-U_2=\frac{4}{9}mgl$