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Coefficient of kinetic friction between a car and the road

  1. Jan 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Just found this site ,looks really helpful , so i am doing some revision using past exam papers and have a few head scratches you guys maybe able to help with .

    A car of mass 1000 kg is travelling at a speed of 25 m/s when it applies its brakes suddenly.
    The car immediately skids. If the coefficient of kinetic friction between the car and the
    road is 0.58, calculate:

    i. the magnitude of the force of friction acting on the car during the skid

    ii. the time taken from when the brakes are applied for the car to come to a complete
    stop.

    (Use g = 9.81 m/s2. You may neglect air resistance and assume level ground)

    2. Relevant equations
    (i) F = ma

    Fk = μk Fn
    (ii)

    Dynamic equations


    3. The attempt at a solution

    so first i worked out that the force of the car is its mass (1000kg) x its acceleration (25 m/s) and that this is x the coefficient of 0.58 = 14500n

    but when starting to work on the 2nd i became confused on what information i actually have . If i wanted to use one of the dynamic equations for solve the time taken to stop do i have enough to do so?

    a= ?
    v = 0 m/s
    u =25m/s
    t = ?
    s=?

    this led me to think that 25 m/s is not actually the acceleration like I have used it in the first part and now im very confused .
     
  2. jcsd
  3. Jan 1, 2015 #2

    Doc Al

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    Staff: Mentor

    That's right. 25 m/s is the velocity of the car, not its acceleration. (Acceleration would be in units of ##m/s^2##.)

    First step is to calculate the friction force on the car. (You'll then use that to figure out the acceleration.)
     
  4. Jan 1, 2015 #3
    thanks for the reply ,

    see this is what confused me , i know that i need force and the coefficient to work out the friction force but i cant work out the force with out the acceleration ?
     
  5. Jan 1, 2015 #4

    Doc Al

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    Staff: Mentor

    Sure you can. Use your second equation, the one that describes Fk.
     
  6. Jan 1, 2015 #5
    Ohh iv just noticed its the normal force I need to work out the friction .

    so the normal force is 1000 x 9.81 and then this x 0.58 ( μk) = 5689.8 N

    But how to i work out the acc from this?
     
  7. Jan 1, 2015 #6

    Doc Al

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    Staff: Mentor

    Good. Now it's time to apply Newton's 2nd law. (Your first equation.)
     
  8. Jan 1, 2015 #7
    AHH i see , so i can use my new fk as my force in terms of a = f/m ?

    a = 5689.8/1000 = 5.6898 m/s2?

    and was i right with my velocities ?

    thanks again for your help .
     
  9. Jan 1, 2015 #8

    Doc Al

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    Staff: Mentor

    Yes.

    Good.

    You haven't used the velocity yet. You'll need that to figure out the time.
     
  10. Jan 1, 2015 #9
    well using v = u + at

    a=5.6898
    v = 0 m/s
    u =25m/s

    I got 20.60 s
     
  11. Jan 1, 2015 #10

    Doc Al

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    Staff: Mentor

    The acceleration will be negative, since it's opposite to the velocity.

    Show how you got that.
     
  12. Jan 1, 2015 #11
    Ah ok I left that minus out

    0 = 25 + (-5.698)t
    -25 = -5.698t

    t = 4.393 .
     
  13. Jan 1, 2015 #12

    Doc Al

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    Staff: Mentor

    Good!
     
  14. Jan 1, 2015 #13
    Great thanks for your help :-)
     
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