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Kinetic Friction Problem (Truck Braking)

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    "A bread truck is traveling at 18 m/s on a horizontal road. The brakes are applied and the truck skids to a stop in 3.6s. The coefficient of kinetic friction between the tires and the road is:"

    2. Relevant equations

    Kinetic Friction = coefficient of kinetic friction x normal force

    3. The attempt at a solution

    I'm not sure how you can manipulate velocity and time to get kinetic friction..
     
  2. jcsd
  3. Sep 18, 2008 #2

    Doc Al

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    Hint: What's the truck's acceleration?
     
  4. Sep 18, 2008 #3
    Well if brakes are being applied, the truck is decelerating until velocity is 0
     
  5. Sep 18, 2008 #4

    Doc Al

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    Find the deceleration.
     
  6. Sep 18, 2008 #5
    Well acceleration is in units of m/s^2, so 18 m/s / 3.6s = 5 m/s^2 if I'm not mistaken
     
  7. Sep 18, 2008 #6

    Doc Al

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    Good! Now make use of it. Hint: What does Newton's 2nd law tell you?
     
  8. Sep 18, 2008 #7
    Well F = MA, but I can't solve for force or mass.. unless of course there's something big that I'm missing..
     
  9. Sep 18, 2008 #8

    Doc Al

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    Make use of what you had in your first post. The "F" is the kinetic friction force.
     
  10. Sep 18, 2008 #9
    Alright so far I've established that the acceleration is 5 m/s^2, the velocity was 18 m/s, and that the time to stop was 3.6s.

    Since Kinetic Friction Force = coefficient of kinetic friction x normal force

    I can break the normal force into mass and acceleration, giving the equation
    Kinetic Friction Force = coefficient of kinetic friction x mass x acceleration (5 m/s^2)

    But I still don't see what unknowns I can solve for..
     
  11. Sep 18, 2008 #10

    Doc Al

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    The normal force does not equal "mass X acceleration". To find the normal force add up the vertical components of all forces and set it equal to zero.
     
  12. Sep 18, 2008 #11
    Oh man I feel like beating myself over the head with a stick right about now.. quite a eureka moment.

    Alright so: kinetic frictional force = coefficient of k x normal force
    coefficient of k = kinetic frictional force / normal force

    Based off that, mass is irrelevant, so it is ultimately acceleration / gravity

    So the coefficient is 0.51.. man.. thanks for spurring the ole noggin!
     
  13. Sep 18, 2008 #12

    Doc Al

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    Sweet. :wink:
     
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