A block of mass 1 kg slides down a curved track which forms one quadrant of a circle of radius 1m as shown in figure. The speed of block at the bottom of the track is v= 2 ms-1. The work done by the force of friction is (g=10m/s2)
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a
+4J
b
-4J
c
-8J
d
+8J
answer is C.
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Detailed Solution
Work done by friction = Efinal-Einitial=KEfinal-PEinitial KE=12×m×(v)2=12×1×(2)2 PE=mgh=1×10×1 W= 12×1×(2)2-1×10×1 Work done by friction=-8J