JEE MainPhysicsWork, Power and Energy
A block of mass m is pushed slowly up a smooth vertical quarter-circular track of radius R , from its lowest point to the horizontal level of its center. The pushing force applied on the block is strictly horizontal at all times. The total work done by this horizontal external force is:
Options
- AmgR ( 2 )
- BmgR 2
- C0
- DmgR
Correct answer
D. mgR
Step-by-step solution
Since the block is pushed slowly, its kinetic energy remains constant, so the change in kinetic energy is zero ( K = 0 ). According to the work-energy theorem, the net work done by all forces acting on the block is zero: W_ net = W_ ext + W_ gravity + W_ normal = K = 0 The normal force is always perpendicular to the displacement of the block, so the work done by the normal force is W_ normal = 0 . The work done by gravity as the block moves up by a vertical height R is W_ gravity = -mgR . Substituting these into th