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(C) Let $T$ be the tension in the string.According to Newton's second law of motion for the downward motion of the body:$mg - T = ma$Given that the ac

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$-\frac{5 m g x}{6}$

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(C) Let $T$ be the tension in the string.According to Newton's second law of motion for the downward motion of the body:$mg - T = ma$Given that the acceleration $a = \frac{g}{6}$,we substitute this into the equation:$mg - T = m(\frac{g}{6})$$T = mg - \frac{mg}{6} = \frac{5mg}{6}$The tension $T$ acts upwards while the displacement $x$ is downwards. Therefore,the angle between the force (tension) and the displacement is $180^{\circ}$.The work done by the string is given by:$W = T \cdot x \cdot \cos(180^{\circ})$$W = (\frac{5mg}{6}) \cdot x \cdot (-1)$$W = -\frac{5mgx}{6}$

$A$ body of mass $m$ is allowed to fall with the help of a string with downward acceleration $\frac{g}{6}$ to a distance $x$. The work done by the string is .............

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