# Magnitude of Force of Friction

1. Feb 28, 2009

### septum

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

Suppose a slide similar to Der Stuka is 36.0 meters high, but is a straight slope, inclined at 45° with respect to the horizontal.
(a) Find the speed of a 51.1 kg thrill seeker at the bottom of the slide, assuming no friction.
26.5 m/s

(b) If the thrill seeker has a speed of 22.6 m/s at the bottom, find the change in mechanical energy due to friction.
-4978 J

(c) Find the magnitude of the force of friction, assumed constant.
____ N

2. Relevant equations/attempts
Magnitude of F = W/change in X, but no X is given, just Y.
I know F= mu/N and I found normal force to be [51.1 kg*9.8 m/s^2 cos(45)]=354 N

I am stuck there and cannot think of a way to calculate mu.

2. Feb 28, 2009

### LowlyPion

You know how fast he was going frictionless. Now they tell you how fast he was going under less than ideal conditions. i.e with friction.

The difference in speed then must have been from the friction. So what was the retarding force, given by first determining the difference from ideal acceleration?

3. Mar 2, 2009

### septum

I still don't get this at all.

4. Mar 2, 2009

### Delphi51

You aren't given the coefficient of friction, so you can't work it out with F = uN.
Work with the energy lost due to friction, which you do have from part (b).
This is the work done against the force of friction, W = F*d.

5. Mar 2, 2009

### septum

I... finally came to the answer after about 8 pieces of paper. I think far too into this. Thank you for your help

6. Mar 2, 2009

### Delphi51

Speed will come with experience! The lesson on this one is that if you can't work forwards, try working backwards.