# Finding velocity with energy equations

1. Jun 11, 2017

### BlueQuark

1. The problem statement, all variables and given/known data
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. 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.

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

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

2. Relevant equations
$U_i = K_f$ (I believe this is correct)
$W_{other} + W_{grav} = K_2 + K_1$
$.5mv_1^2 + mgy_1 + W_{other} = .5mv_2^2 + mgy_2$
$U_{grav} = mgh$
$K = .5mv^2$

3. The attempt at a solution

b. The potential energy at point A should be equal to the kinetic energy at point B.
The potential energy at point A is $U_{grav,1} = (.20)(9.81)(.5) = .98J$
So $.98J = .5(.20)v^2 = .1v^2$

Solving for V, I end up with approximately 3.1m/s. The problem is that in the back of the book, the answer key says 2.8m/s. I'm not sure if I did something wrong or the book is wrong (It's been wrong before).

d. At the bottom, the normal force should be equal to its weight.
$W = mg = (.20)(9.81) = 1.96 N$
So the normal force is equal to 1.96 Newtons. The back of the book says 5 Newtons, so I'm wrong here as well.

Any help is appreciated, thanks!

2. Jun 11, 2017

### PeroK

for part b) the book is correct. What happened to the friction?

3. Jun 11, 2017

### PeroK

For part d) you need to think about the shape of the motion of the rock.