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Kinetic Friction Pbm

  1. Jan 9, 2006 #1
    An Olympic skier moving at 28 m/s down
    a 24 degrees slope encounters a region of snowless
    ground of coefficient of kinetic friction 0.67.
    The acceleration of gravity is 9.8 m/s^2
    How far down the slope does she travel
    before coming to a halt? Answer in units of

    I came up with an equation of:

    mgsin(theta) - Fk*d = ma

    I would solve for "d" but i'm not sure how to get the "a".

    help appreciated.

  2. jcsd
  3. Jan 9, 2006 #2
    Try this,

    The sum of the forces parallel to the slope is mgcos62(gravity) - mgUk(friction)

    mg(Uk-cos62) = - 0.21 mg = -2.06m/s^2 * m

    so your acceleration, a, is 2.06 m/s^2 going against the skier.

    From there you use t = dv/a = 28/2.06 so t = 13.6 s

    then L traveled is L = 1/2 a t^2 = 44.8 m

    Or something like that.
  4. Jan 9, 2006 #3
    Thanks a lot. Your math is off in the last step but you had the right concepts.

    I did mgsin(theta)-Ukmgcos(theta) = ma

    a = 2.01...

    t = 28/2.01...

    L = .5at^2 = 195.25 (not precise but within 1% margin of error)

    Thanks again. :)
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