• # question_answer 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                        [CPMT 1972] A)             $Mg\frac{d}{4}$        B)             $3Mg\frac{d}{4}$ C)             $-3Mg\frac{d}{4}$ D)               Mgd

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 )$$=-\left( \frac{3Mg}{4} \right)\times d$$=-\frac{3}{4}Mgd$