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Q.

A block of mass 'M' is lowered with the help of a rope of negligible mass through a distance 'd' with an acceleration of $\frac{g}{3}$ . Work done by the rope on the block is

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a

- \frac{2Mgd}{3}

b

- \frac{Mgd}{3}

c

\frac{2Mgd}{3}

d

\frac{Mgd}{3}

answer is B.

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Detailed Solution

W = \bar{F} . \bar{S} - Td (θ = 180 °)

$here T = M (g - a)$

$W = - M (g - a) d$

$= - M (g - \frac{g}{3}) d$

$= - \frac{2Mgd}{3}$

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