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Frictional forces

  1. Oct 29, 2006 #1
    The minimal stopping distance for a car moving at an initial speed of module 100km/h is 60m on a flat surface. What is the stopping distance when the car is moving a) down at 10 degrees b) up at 10 degrees? We assume that the initial velocity and surface don't change.

    I was thinking I could find the coefficient of friction of the flat surface (Uk = V^2/(2ag) which would give me Uk = 0.657. Then I would find the acceleration of the car by adding all the vectors together (mgsin(theta) - f = mA). The mass cancels out therefore giving me a = gsin(theta) - Ukgcos(theta) which in terms gives me a = -4.64. I then substitute the values in Vf^2 = Vi^2 + 2ad which gives me d = 83.3m. The answer to b would be mostly the same work switching up the sin/cos. Does that work?
     
  2. jcsd
  3. Oct 29, 2006 #2

    OlderDan

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    Your idea is correct. Your equation for Uk is dimensionally inconsistent. If you find the correct coefficient of friction, then you can find the stopping distance up and down the plane.
     
  4. Oct 29, 2006 #3
    Oops, I meant Uk = V^2 /(2gs) not ag. Sorry. This, I think, fixes the inconsistency problem. Was the rest of my work ok?
     
  5. Oct 30, 2006 #4

    OlderDan

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    Looks good for down the plane. Up the plane is not switching sin and cos. It is just getting the forces actin in the right direction.
     
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