A rope is used to lower a block of mass M ver | vedclass

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Question

(C) Let $T$ be the tension in the rope acting upwards and $Mg$ be the weight of the block acting downwards.The block is moving downwards with an accel

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$-\frac{1}{2}Mgx$

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(C) Let $T$ be the tension in the rope acting upwards and $Mg$ be the weight of the block acting downwards.The block is moving downwards with an acceleration $a = g/2$.Applying Newton's second law of motion: $Mg - T = Ma$Substituting $a = g/2$,we get: $Mg - T = M(g/2)$$T = Mg - Mg/2 = Mg/2$The displacement $x$ is downwards,while the tension $T$ acts upwards.Therefore,the angle between the force $T$ and the displacement $x$ is $180^{\circ}$.Work done by the rope $W = T \cdot x \cdot \cos(180^{\circ}) = (Mg/2) \cdot x \cdot (-1) = -\frac{1}{2}Mgx$.

$A$ rope is used to lower a block of mass $M$ vertically downwards by a distance $x$ with a constant downward acceleration of $g/2$. What is the work done by the rope on the block?

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