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More friction

  1. Nov 3, 2003 #1
    A police man investigating an accident measures the skid marks left by a car. He determines that the distance between the point that the driver slammed on the brakes and the point where the car came to a stop was 25.2 m. from a reference manual, he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.330. How fast was the car going when the driver applied the brakes? (This car was not equipped with anti-lock brakes)
     
  2. jcsd
  3. Nov 3, 2003 #2

    chroot

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    The definition of kinetic friction allows you to find the force experienced by the vehicle (you must assume it is independent of velocity).

    Once you know the force, you can use Newton's second law to determine the acceleration -- it is constant.

    Since the acceleration is constant, you can use the equation:

    v(t) = v0 + a t

    to calculate the time it took the car to stop (a is negative, of course), and the equation

    s(t) = v0 t + 1/2 a t2

    to solve for v0.

    If these equations are new to you, let me know and I'll guide you through their derivation.

    - Warren
     
    Last edited: Nov 3, 2003
  4. Nov 3, 2003 #3
    ?

    but all i have to start with is the 0,.330..how do i get the force from that?
     
  5. Nov 3, 2003 #4

    chroot

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    FLANKER,

    There's a simple equation relating two quantities: the friction force, and the normal force. The coefficient of (static or kinetic) friction is simply the proportionality constant between them. The equation is this one:

    F = [mu]k (or s) N
    = [mu]k (or s) m g

    Can you calculate F now?

    - Warren
     
  6. Nov 3, 2003 #5

    chroot

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    FLANKER,

    Sorry, I should be more specific. You don't need to actually compute a NUMBER for the force. You can leave it in symbolic form, like this:

    F = [mu]k m g
    F = 0.33 m g

    And just use it like that.

    The next step is to calculate the acceleration experienced by the vehicle, which is given by Newton's second law:

    F = m a

    Plugging in your value for F yields this:

    0.33 m g = m a

    The m's cancel:

    0.33 g = a

    And you have the acceleration.

    With me so far?

    - Warren
     
  7. Nov 3, 2003 #6
    ?

    yes, i understand that part..and then?
     
  8. Nov 3, 2003 #7

    chroot

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    FLANKER,

    Now that you have the acceleration, you can plug it into this equation:

    v(t) = v0 + a t

    to find the time taken for the car to stop.

    In other words,

    0 = v0 - 0.33 g t

    t = v0 / (0.33 g)

    Now you can plug t into the equation

    s(t) = v0 t + 1/2 a t2

    And solve for v0.

    - Warren
     
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