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