Question

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

• A. $\frac{5}{9}mgl$
• B. $\frac{6}{9}mgl$
• C. $\frac{4}{9}mgl$
• D. $\frac{8}{9}mgl$

Answer 1

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


The potential energy of the $l/3$ part of the chain that overhangs is $U_x={\int }_0^{\frac{l}{3}}-\frac{m}{l}gxdx=-{{\left(\frac{m}{l}g\left(\frac{x^2}{2}\right)\right]}_0}^{\frac{l}{3}}=-\frac{1}{18}mgl$
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 $=\left(-\frac{1}{18}mgl\right)-\left(-\frac{1}{2}mgl\right)=\frac{4}{9}mgl$
This should be equal to the gain in kinetic energy. But the initial kinetic energy is zero. Hence, the kinetic energy of the chain as it completely slips off the table is $\frac{4}{9}mgl$