# Coefficient of friction for surface

1. Dec 22, 2008

### Mini_Mashi

a) A block weighing 9.1N requires a force of 2.8N to push it along at constant velocity. What is the coefficient of friction for the surface?
b) A weight "W" is now place on the the block and 8N is needed to push them both at constant velocity. What is weight "W" on the block?

This is what I got:
Since it is constant velocity, there is no acceleration, therefore Fnet is 0, therefore Ff is 2.8N.
Normal force is (9.1)(9.8) =89.18N
Formula: Ff=uN

2.8=u(89.18)
2.8/89.18=u
Therefore u= 0.031
, but that's not the right answer... Why? what did I do wrong?

I take it I need to know a before I can do b?

2. Dec 22, 2008

### jamesrc

Why did you multiply the weight of 9.1 N by 9.8 to get the normal force?

3. Dec 22, 2008

### Mini_Mashi

Hm? Don't you need to?
Fg=mg so Fg=(9.1)(9.8), and normal force is going the other direction and Fnet=0, so normal force is also 89.18.

Isn't that right?

4. Dec 22, 2008

### jamesrc

The weight would equal the mass times g, but you weren't given the mass, you were given the weight which is equal to the normal force. Look at your units: N is a unit of force, if you were given mass in kg and multiplied that by g which has units [m/s^2], you would end up with Newtons (again, a force). But what is 1N * 1m/s^2?

5. Dec 22, 2008

### Mini_Mashi

ohhh. I see what I did wrong, thanks!! You're great help!
I can't believe my mistake!