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PhysicsW.E.P.Vertical circular motionEasy2 minPYQ_2013
PhysicsEasysingle choice

Paragraph: A small block of mass is released from rest at the top of a rough track. The track is a circular arc of radius . The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point , as shown in the figure below, is . (Take the acceleration due to gravity, ).

Question: The magnitude of the normal reaction that acts on the block at the point is

Options:

Answer:
A
Solution:

From FBD
N-mgsin30°=mv2RN=1×10240+1×102N=7.5 N

Stream:JEE_ADVSubject:PhysicsTopic:W.E.P.Subtopic:Vertical circular motion
2mℹ️ Source: PYQ_2013

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