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A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m as shown in fig. (7).
Its speed at the bottom is 2 m/s. The work done by the frictional force is

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-8J

-4J

answer is B.

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Detailed Solution

$W_{f} = Δ U + Δ K = (0 - m g h) + (\frac{1}{2} m v^{2} - 0) = - (1) (10) (1) + \frac{1}{2} (1) (2)^{2} = - 10 + 2 = - 8 J$

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