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A cable in the form of a spiral roll as shown in the figure has a linear density 0.25 kg/m. It is uncoiled at a uniform speed of 2 m/s. If the total length of the cable is 15 m, the work done in uncoiling is (neglect friction)

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15 J

7.5 J

30 J

3.75 J

answer is B.

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Detailed Solution

$W = ΔKE = \frac{1}{2} {Mv}^{2} = \frac{1}{2} \times 3 .75 \times 2^{2} = 7 .5 J oule$

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