# Coefficient of kinetic friction on car

1. Nov 2, 2004

### unknownfrost

Ok here is my situation. There is a problem that states:

Find the friction force if the acceleration of a 500 kg car being towed on a flat surface is 5 m/s2 and a tow truck is pulling with a force of 3550 N.

I do not where to go from here. Furthermore, the next question asks:

What would the coefficient of kinetic friction be?

If someone could help me out with this I would be extremely thankful!

This has got me going and

2. Nov 2, 2004

### Galileo

First, notice that, according to Newtons 2nd law, if the net force on the truck were to be 3550 N, the acceleration of the truck would be:
$$a=\frac{F}{m}=\frac{3550}{500}=7.1 m/s^2$$
So the frictional force is of that magnitude to bring the acceleration down to $5 m/s^2$.
Can you find the force needed to do this?
(There's a slightly quicker way. But I find this more instructive)

3. Nov 3, 2004

### unknownfrost

Well...
I know that f = (mk)(n)
Would I use the 3500 from earlier and take mk = f/n to get the coeff. of kinetic?

4. Nov 3, 2004

### Phymath

its a simple force diagram

ma = (Force tow truck) - Friction

(500kg)(5m/s^2) = 3500 N - uN
2500 N - 3500 N = -1000 = -umg

1000 N/((500kg)(9.81 m/s^2)) = u = 0.220 < friction coffecient

5. Nov 3, 2004

### unknownfrost

ok i'm with you on the 3500 part, but where did you get the 2500 from?

I understand now how to plug it in. i just don't know how you got it!

I could just take the awnser but I am totally lost on the awnser!
Physics is tough man!

Last edited: Nov 3, 2004
6. Nov 4, 2004

### Galileo

This net force is equal to its mass times its acceleration: $F=ma=500 \cdot 5 = 2500 N$.