Question

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 the rock slides rather then rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J.

(A) Between point A and B, how much work is done on the rock by the normal force and gravity?
(B) What is the speed of the rock as it reaches point B.
(C) of the three forces acting on the rock as it slides down the bowl, which is constant and which are not? Explain.
(D) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl.

Answer 1

a)

Since the normal force will always be perpendicular to the motion .Therefore work done by normal force is

W_N= 0

work done by gravity

W= mgR=0.2*9.81*0.5

W=0.981 J

b)

From Work energy theorem

0.981 -0.22= (1/2)mV²

0.761= 0.5*0.2*V²

V=2.76 m/s

c)

SInce gravitational force will remain constant ,as mass remains as constant

Normal force Will not remain constant ,since normal will remain changing as the particle moves towards bottom, due to centripetal force acting on the partical & because of normal force , friction will also remain changing due to its dependency on normal force

d)

normal at the bottom

F= mg + mV²/r

F=0.2*9.81+0.2*2.76²/0.5

F=5.006 N