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# A Small Rock with Mass 0.20 kg is Released

## Question

A small rock with a mass of 0.20 kg is released from rest at point A which is at the top edge of a large, hemispherical bowl with a radius R = 0.5 m. Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle. And that the rock slides rather than 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 a magnitude of 0.22 J.

a) Between points A to point B, how much work is done on the rock by;

i). The normal force and

ii). The 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, if any, are 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?

### Solution a

i) The work done by the normal force is zero. This is because a normal force is a perpendicular force when two forces came in contact with each other.

Therefore, the rock is moving in the horizontal direction, the normal force is perpendicular to the direction of the dispalcement.

ii) To find the gravity, we consider the point “A small rock with a mass of 0.20 kg is released from rest at point A”

Now, apply the formula

W = F x s

and if s = distance = h = radius

where F = mg

we will now have

W = mgh

and

m = mass = 0.2 kg

g = gravitational acceleration = 10 ms-2

h = radius = height = 0.5 m

W = work done by the gravity = ?

Thus, w = mgh = 0.2 x 10 x 0.5 = 1 Joule = 1 J

Therefore, the workdone by the gravity is one joule.

### Solution b

To calculate the speed of the rock as it reaches B, we need to first calculate the kinetic energy of the rock

Since M.E = K.E + P.E

K.E = 0.98 – 0.22 = 0.76 Joules

Now apply the formula for kinetic energy which says

K.E = (1/2)mv2

to find the speed, we need to make v subject of the formula

(2 x K.E)/m = v2

v = √(2 x K.E)/m

we insert our data from the question to get

v = √[(2 x 0.76)/0.2] = √(1.52/0.2) = √7.6 = 2.757

Therefore,

v = 2.8 ms-1

### Solution c

Of the three forces acting on the rock as it slides down the bowl, the gravitational force is constant. While the frictional force and the normal force between the surface of the rock and the bowl is not constant.

d) The normal force N is

N = mg + mv2/R= 0.2 x 10 + (0.2 x 2.82)/0.5= 2 + (0.2 x 7.84)/0.5 = 2 + (1.568/0.5)

Therefore,

Normal force, N = 2 +3.136 = 5.136 N

Thus, the normal force is 5.136 newtons.

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