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Coefficient of kinetic friction on car

  1. Nov 2, 2004 #1
    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:
  2. jcsd
  3. Nov 2, 2004 #2


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    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:
    [tex]a=\frac{F}{m}=\frac{3550}{500}=7.1 m/s^2[/tex]
    So the frictional force is of that magnitude to bring the acceleration down to [itex]5 m/s^2[/itex].
    Can you find the force needed to do this?
    (There's a slightly quicker way. But I find this more instructive)
  4. Nov 3, 2004 #3
    I know that f = (mk)(n)
    Would I use the 3500 from earlier and take mk = f/n to get the coeff. of kinetic?
  5. Nov 3, 2004 #4
    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
  6. Nov 3, 2004 #5

    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
  7. Nov 4, 2004 #6


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    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: [itex]F=ma=500 \cdot 5 = 2500 N[/itex].
    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?
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