Q. A cord is used to lower vertically a block of mass $M$ by a distance $d$ with constant downward acceleration $\frac{g}{4}$. Work done by the cord on the block is
A) $Mg\frac{d}{4}$
B) $3Mg\frac{d}{4}$
C) $- 3\ Mg\frac{d}{4}$
D) $Mgd$
Explanation
(c) When the block moves vertically downward with acceleration $\frac{g}{4}$ then tension in the cord
$T = M\left( g - \frac{g}{4} \right) = \frac{3}{4}Mg$
Work done by the cord $\overrightarrow{F}.\overrightarrow{S} = FS\cos\theta$
$= Td\cos{180{^\circ}}$
$= \left( - \frac{3}{4}\ Mg \right) \times d = - 3Mg\frac{d}{4}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.