Question

A block of mass M slides down the smooth surface of a bowl of radius R from its rim to the bottom. The kinetic energy of the block at the bottom is

Answer 1

By conserving the mechanical energy of the block of mass 'M'

Initial potential energy + Initial Kinetic energy = Final potential energy+ Final Kinetic energy
(considering potential energy equal to zero at the bottom of the bowl)
So,
MgR + 0 = 0 + Final Kinetic energy
Therefore,
Final Kinetic energy = MgR