A solid sphere is rolling on a frictionless horizontal surface, with a translational velocity v m/s. If it is to climb a rough inclined surface of height h, when v should be?
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Suppose the body just about climbs the inclined surface and stops and rolls back. Then the Kinetic energy in the body at the bottom of the incline is equal to the potential energy at the top.
1/2 m v² = m g h
=> v = √[2 g h ]
1/2 m v² = m g h
=> v = √[2 g h ]
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15
Thats the correct answer..verified answer is wrong
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