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Another friction question

  1. Oct 4, 2006 #1
    A 25kg skier on a "bunny" hill that is inclined at 5 degrees has an intial velocity of 3.5 m/s. The coefficient of kinetic friction is 0.20

    a) calculate the time taken for the skier to come to a stop
    b) the distance travelled down the hill..

    what i have so far is..

    ive calculated is
    Fg =mg = 25*(9.8) = 245N
    Fn = mgcos 5 = 25*(9.8) * cos 5 = 244.07N
    Friction = (mu) Fn = 0.20* 244.07 = 48.81N
    F=ma thus 48.81 = 25a thus a= 1.95 m/s^2

    a = Vf - Vi / delta T ...... delta T = 1.79s

    am i right so far?
  2. jcsd
  3. Oct 4, 2006 #2
    Check your net force equation. According to yours, friction is the only force acting on the skier. This would mean that he is being pushed by friction and friction only.
  4. Oct 4, 2006 #3
    ok so i have force normal, force of gravity and fictional force acting on the skier, so how do i add up those three?
  5. Oct 4, 2006 #4
    You add up the ones parallel to the direction of acceleration. The normal force is not relevant, but the other component that makes up mg is relevant.
  6. Oct 5, 2006 #5
    so what i have now is Fg + Fk - mg sin 5 = net force
    is what i have now correct?
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