Posted by **Frankie** on Tuesday, October 23, 2012 at 3:24pm.

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 = .5m. 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.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 0.22 J.

A. Between points A and B, how much work is done on the rock by (i) the normal force and (ii) 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 due to the normal force on it due to the bottom of the bowl?

- Physics -
**abdulla nabeel **, Saturday, November 7, 2015 at 12:23pm
Between points A and B, how much work is done on the rock by the normal force?

-There is no work done by the normal force because all of the work is being done by gravity.

W=0J

Between points A and B, how much work is done on the rock by gravity?

-Work done by gravity is just the gravitational potential energy lost by moving in the vertical direction, so you use the equation Wg=mgh. In this case, h is just the radius so I'll sub that instead

Wg=mgR

Wg=(.26)*(9.8)*(.50)

Wg=1.274 J

What is the speed of the rock as it reaches point B?

-Find the speed by using the equation W=(.5)mv^2. For this part, you have to take into account the friction, Wt=Wg-Wf. So, solving for v:

v=sqrt((Wg-Wf)/(.5*m))

v=sqrt((1.274-.22)/(.5*.26))

v=2.85

Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

-When any object reaches the bottom of a loop or circle, the normal force is equal to the weight(mg) added to the mass times the radial acceleration(Arad). The equation looks like this: n=mg+m(Arad). First, you need to find the radial acceleration which is equal to v^2/R.

Arad=v^2/R

Arad=(2.85^2)/.5

Arad= 16.25

Now, you just plug in everything you know:

n=mg+m(Arad)

n=(.26*9.8)+(.26*16.25)

n=6.77N

hope that help ;)

## Answer This Question

## Related Questions

- Physics - A small block of mass m = 194 g is released from rest at point A , ...
- PHYSICS - A 2.5 g ice flake is released from the edge of a hemispherical bowl ...
- physics - A 224 g particle is released from rest at point A along the diameter ...
- physics - ik I just asked one, but I have more A 314 g particle is released from...
- physics - A small block of mass m = 225 g is released from rest at point circled...
- physics - A 244 g particle is released from rest at point A inside a smooth ...
- Physics-please help!!! - A 190 g particle is released from rest at point A ...
- Physics - A hemispherical bowl of radius R is rotated about its axis of symmetry...
- Physics - A 215 g particle is released from rest at point A (the middle of one ...
- physics - A person standing at the top of a hemispherical rock of radius R = 5....

More Related Questions