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Check my solution - Does it make sense?

  1. May 26, 2007 #1
    Find μs of the car.
    - A Car takes 10 seconds to accelerate to 80km before the driver shifts the car into neutral.
    - The cars mass = 1810kg
    - The car coats 0.9km before coming to a complete stop.
    - The total time = 86 seconds.

    Here is the work I have done.

    μs = ma/mg
    to find a
    V2 = V1 +at
    22.2 = 0 +a10
    a = 2.22m/s2

    μs = 1810 * 2.22 / 1810 * 9.8
    μs = 0.22

    Does this make sense. This is the μs for the road right? How can I find it for the car? I figure I need to use the total distance it travel for something but I'm not sure how. Any help is welcome

  2. jcsd
  3. May 26, 2007 #2


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    I'm not sure I understand this. You're saying that the coefficient of friction multiplied by the normal force (which is just equal to the weight of the car in this case) is equal to some force, i.e.:

    [tex] \mu_s mg = F [/tex]

    But why should this "F" be the net force on the car, ma?

    This looks reasonable, assuming constant acceleration.

    Umm...:uhh:...a frictional force is a force that opposes the motion of two surfaces in contact that are trying to slide relative to each other. So it only makes sense to speak of a coeffcient of friction between two surfaces. In this case, the only [itex]\mu[/itex] you could calculate would be the coefficient of friction between the tires and the road (i.e. the coefficient of friction for rubber on asphalt). It doesn't make sense to speak of mu only for the car, or only for the road. Are you sure this is the exact wording of the problem?

    EDIT: Ohhhh!! I think I understand. In this case, the frictional force is the forward force of the road on the tires that causes the car to move forward instead of its tires just spinning in place, right? So that explains why you were setting the frictional force F equal to ma! It also explains why you were using the coefficient of static friction rather than kinetic friction (the two surfaces never actually slide relative to each other...unless the tires lose traction). Okay, in that case, I think you are on the right track, but you need to address the coasting phase of the trip too...not just the initial phase in which the engine is powering the car. During this phase, the situation is slightly different, right?
    Last edited: May 26, 2007
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