# Finding a new speed after adding a friction force? Help

1. Oct 17, 2012

### monikraw

1. The problem statement, all variables and given/known data
A 8.00kg block of ice, released from rest at the top of a 1.04m--long frictionless ramp, slides downhill, reaching a speed of 2.76m/s at the bottom.

What is the angle between the ramp and the horizontal?
This I calculated as 21.9 degrees

What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.9N parallel to the surface of the ramp?

2. Relevant equations
F = ma
w = mg
vx = v0 + axt

i might be missing several
3. The attempt at a solution

so I drew a diagram which i probably can't move to here, but--
I aligned the x and y axis to match the normal force and the friction force

I calculated the force of the block of ice going down the ramp by using "mgsin(theta)" (where theta is 21.9) and got 29.24 N. I subtracted 10.9N from it to get the net force of 18.34N. I used F = ma to find acceleration and I got 2.293m/s^2.

I tried putting in 2.293m/s^2 and it says close but not quite.
How do I find t so I can use the vx = v0 + axt???

I got stuck up to there and some(or all of it) may be wrong since physics is my weakest subject haha.

Last edited: Oct 17, 2012
2. Oct 17, 2012

### CAF123

Correct

Why not use $v^2 = v_{o}^2 + 2as ?$ Could also use energy methods.

3. Oct 17, 2012

### monikraw

I got the answer after dividing by 1.04. I don't know why that happened to be the answer nor do I know why that was a necessary step. Someone please explain?

4. Oct 17, 2012

### CAF123

That would not make any sense. Dividing an acceleration by a length does not give a speed. (Check the dimensions)
Use what I suggested in my previous post.

5. Oct 17, 2012

### monikraw

where s is seconds, I'm assuming?
so v^2 = 0 + 2*(2.293m/s^2)(0.377m/s?)
I got time by dividing 2.76 by 1.04 since d = speed * time
square root each side and I get v = 1.314556188

the answer was 2.19 :x...probably did something wrong.

6. Oct 17, 2012

### CAF123

The $s$ in this eqn is the displacement of the body. The value of $s$ here is the 1.04m, since we are looking for the speed at the bottom of the slope and the body moved 1.04m from where it was released.
Check this again, given what I said above.
From the moment the body is released, it will gain speed. The eqn d=vt only applies if the body is not accelerating.