Question

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

• A. $\frac{4}{9}mgl$
• B. $\frac{13}{18}mgl$
• C. $Zero$
• D. $\frac{11}{18}mgl$

Answer 1

The correct option is A $\frac{4}{9}mgl$


Let us take the zero of potential energy at the table. Consider a part dx of the chain at a depth x below the surface of the table. The mass of this part is $dm=\frac{m}{l}dx$ and hence its potential energy is $-\left(\frac{m}{l}dx\right)gx$
The potential energy of the $\frac{l}{3}$ of the chain that overhangs is
$U_1={\int }_0^{l/3}-\frac{m}{l}gxdx=-{\left[\frac{m}{l}g\left(\frac{x^2}{2}\right)\right]}_0^{l/3}=-\frac{1}{18}mgl$.
This is also the potential energy of the full chain in the initial position because the part lying on the table has zero potential energy. The potential energy of the chain when it completely slips off the table is
$U_2={\int }_0^l-\frac{m}{l}gxdx=-\frac{1}{2}mgl$
The loss in potential energy = $(-\frac{1}{18}mgl)-(-\frac{1}{2}mgl)=\frac{4}{9}mgl$.
This should 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 $\frac{4}{9}mgl$.