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

Correct Answer: C

Solution :

                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 = \[\overrightarrow{F}.\overrightarrow{s}=Fs\cos \theta \]             = \[Td\cos (180{}^\circ )\]\[=-\left( \frac{3Mg}{4} \right)\times d\]\[=-\frac{3}{4}Mgd\]