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

\frac{2 M g d}{3}

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- \frac{2 M g d}{3}

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\frac{M g d}{3}

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- \frac{M g d}{3}

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Solution

The correct option is B $- \frac{2 M g d}{3}$
Tension and acceleration due to gravity acts on the block. Such that

$T - m g = m a$

$T = m (g - a)$

$T = m (\frac{g}{3} - g)$

$= \frac{- 2 m g}{3}$

Work done = $f o r c e \times d i s p l a c e m e n t$

$= \frac{- 2 m g}{3} . d$

$= - \frac{2 m g d}{3}$

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