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Finding the coefficient of frictional force

  1. Apr 16, 2009 #1
    You are driving 2500 Kg car at a constant speed of 14 m/s along an icy, but stright and lvel road. While approaching a traffic light, it turns red. You slam on the brakes. You wheels lock, the tires begin skidding and the car slides to a halt in a distance of 25 m. What is the coefficient of sliding friction between the tires and the icy road?

    -------------------------------------
    Calculations:

    m = 2500 Kg
    Vi = 14 m/s
    Vf = 0 m/s
    deltaD = 25 m
    a = (vf^2 - vi^2)/(2 x deltaD)
    = -14 m/s / 50
    = 2.8 m/s^2

    W = (2500) (9.81)
    = 24525 N

    Fnet = (2500)(-0.28)
    = -700 N

    I`m not sure what to do after this.
     
  2. jcsd
  3. Apr 17, 2009 #2

    Nabeshin

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    Looks good so far.

    You know that friction is the force producing this acceleration (and thus net force), so what is the equation for the force of friction?
     
  4. Apr 17, 2009 #3
    F = uFn

    700 = u(24525)

    Is that what I am supposed to be doing?
     
  5. Apr 17, 2009 #4

    Nabeshin

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    Yup. Have a little confidence :)
     
  6. Apr 17, 2009 #5
    I have an answer sheet and it says the answer is supposed to be 0.4 so I'm a little confused.
     
  7. Apr 17, 2009 #6

    Nabeshin

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    Sorry I wasn't checking your arithmetic, the approach was correct though.

    The acceleration should be 3.92m/s^2:
    [tex]14^2/(50)=3.92[/tex]

    Also, for some reason in your f=ma you used your a value divided by ten instead of the a you derived...

    But yeah, see if you can get the .4 now.
     
  8. Apr 17, 2009 #7
    Thank you so much :)
     
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