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A Cord is used to lower vertically a block of mass m a distance d at a constant downward acceleration of $\frac{g}{4}$ , then the work done by the Cord on the block is

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$m g \frac{d}{4}$

$3 m g \frac{d}{4}$

$- 3 m g \frac{d}{4}$

$m g d$

answer is C.

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Detailed Solution

The block is coming down with acceleration

$\frac{g}{4}$ thus $M g - F = M a = \frac{M g}{4}; F = \frac{3}{4} M g$

the workdone by the card on the block is
$w = F d \cos θ = (\frac{3}{4} M g) \times d \cos 180^{0}$

$w = \frac{- 3}{4} M g d$

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