A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.50 m. Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls.
If the normal force exerted on the rock as it slides through its lowest point (point B) is twice the weight of the rock, how much work did friction do on the rock as it moved from A to B?

0(zero)

Explanation:

Given that =0.20kg, radius of the bowl, and that the mass slides from point A to B.

-Since the normal force is perpendicular to the displacement at every point of the motion,then work  done by the normal force is zero.

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

v=ir = 3x15 =45 v