Coefficient of kinetic friction

[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 :eek: and :cry:

[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)

[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?

[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

[unknownfrost]

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



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!

[Galileo]

Think about it like this.
You are given that a 500 kg object has an acceleration of 5 m/s^2.
Then Newton's law says: "Aha, it accelerates. So there must be a net force acting on it."
This net force is equal to its mass times its acceleration: F=ma=500 \cdot 5 = 2500 N.
So the net force on the car is 2500 N. Since the tow truck is pulling with a force of 3550 N, something must be exerting a force in the other direction (which is the force of friction). Can you take it from here?