Q. A uniform chain of length $L$ and mass $M$ overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is $\mu$. The work done by the friction during the period the chain slips off the table is
A) $- \frac{1}{4}\ \mu MgL$
B) $- \frac{2}{9}\ \mu MgL$
C) $- \frac{4}{9}\ \mu MgL$
D) $- \frac{6}{7}\ \mu MgL$
Explanation
(b) $dW = - \mu\left\lbrack \frac{M}{L} \right\rbrack gl\ dl$
$W = \int_{0}^{\frac{2L}{3}}{- \frac{\mu Mg}{L}l\ dl}$
or $W = - \frac{\mu Mg}{L}\left| \frac{l^{2}}{2} \right|_{0}^{\frac{2L}{3}}$
or $W = - \frac{\mu Mg}{L}\left| \frac{4L^{2}}{9} - 0 \right|$
or $W = - \frac{2}{9}\mu MgL$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.