A block of mass 2 kg slides down a curved track that is one quarter of a circle of radius 1 m [see fig. (1)]. Its speed at the bottom is 4 m,/s. Work done by frictional force is (take g = 10 m / s2 )
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a
-4J
b
8J
c
-8J
d
20J
answer is A.
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Detailed Solution
In absence of friction, the velocity at the bottom of quarter circleV=2gR K.E. =12mv2=mgR=2×10×1=20J Actual K.E. =12×2×(4)2=16J Work done by frictional force = decrease in K E. =16J−20J=−4J