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

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(C) Let $T$ be the tension in the rope acting upwards on the block.Applying Newton's second law for the downward motion of the block:$Mg - T = M a$Giv

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

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(C) Let $T$ be the tension in the rope acting upwards on the block.Applying Newton's second law for the downward motion of the block:$Mg - T = M a$Given that the acceleration $a = \frac{g}{2}$,we have:$Mg - T = M \left(\frac{g}{2}\right)$$T = Mg - \frac{Mg}{2} = \frac{Mg}{2}$The tension $T$ acts upwards while the displacement $x$ is downwards. Therefore,the angle between the force and displacement is $180^\circ$.The work done by the rope is:$W = T \cdot x \cdot \cos(180^\circ)$$W = \left(\frac{Mg}{2}\right) \cdot x \cdot (-1)$$W = -\frac{1}{2} Mgx$

$A$ rope is used to lower a block of mass $M$ vertically by a distance $x$ with a constant downward acceleration of $\frac{g}{2}$. The work done by the rope on the block is:

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