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Forces - distance of a skier on an incline

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A skier on a slope inclined at 4.7 degrees to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7 m/s. The coefficient of kinetic friction between skis and snow is 0.11. Determine how far the skier will slide before coming to rest.

    2. Relevant equations

    uK= 0.11

    v initial= 2.7 m/s
    v final= 0.0 m/s

    Fk = uK*Fnormal


    (v final)^2 = (v initial)^2 +2*a*d

    3. The attempt at a solution

    (Refer to system diagram for notations and stuff!)

    So from what I can tell, the applied force is the horizontal component of gravity, Fgx, and the normal force is the vertical component of gravity, Fgy.

    The horizontal component of gravity is:
    Fgx= (sin4.7)(9.8)(m)

    The force of kinetic friction would come out to be:

    The net force would be:
    Fnet = Fapplied - Fkinetic
    Fnet = Fgx - Fk
    Fgx - Fk = m*a

    So, solving, I get:

    (sin4.7)(9.8)(m) - (0.11)(cos4.7)(9.8)(m) = m*a
    a = 8.67 m/s^2

    Now I can use acceleration to find the distance traveled by the skier:

    (v final)^2 = (v initial)^2 +2*a*d
    0 = (2.7)^2 +(2)(8.67)(d)
    d= 0.42 m

    The actual answer is 13m and I'm sure I've done something wrong :redface:.
    Can someone guide me through this, please?

    Attached Files:

  2. jcsd
  3. Jul 11, 2010 #2

  4. Jul 11, 2010 #3

    You have some calculation error here. Be sure to do you calculations using degrees not radians.
  5. Jul 11, 2010 #4
    Oooh, whoops. So my method was right all along?

    Thank you !
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