The block of mass 1kg slides down a curved track which forms one quadrant of a circle of radius 1m as shown in the figure. The speed of block at the bottom os the track is v=2msā1. The work done by the force of friction is:
A
+4J
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B
−4J
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C
−8J
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D
+8J
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Solution
The correct option is C−8J Work done by the friction is the change in energy of the block. Work done by friction=K.E−P.E=12mv2−mgh ⇒ work done=(0.5)(1)(22)−(1)(10)(1)=−8J