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Speed skier travelling up a hill

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A speed skier is travelling horizontally at a constant speed of 5.80 m/s when she approaches a snow-covered hill that has a slope of 23.0° above the horizontal. The coefficient of kinetic friction between the skis and the hill is 0.350 and the combined mass of her and her skis is 77.0 kg. If she decided to glide up the hill, how far would she make it before she comes to a complete stop?

    2. Relevant equations
    Fnet = m*a
    Fk = μ Fn
    Fg = m(-g)
    Fgx = m(-g)sinθ
    Fgy = m(-g)cosθ

    3. The attempt at a solution
    So first I solved for the y component of the force due to gravity:
    Fgx = (77.0kg)(-9.8m/s2)*cos23 = -694.61296

    Then I solved for the kinetic friction with the the y component of the gravity force equaling the normal force:
    Fk = (0.350)(-694.61296) = -243.11454

    Then I went on to solve for acceleration using the Fnet formula with the x component of gravity subtracting the kinetic friction force:

    m*a = Fgx - Fk
    (77.0kg)*a = [(77.0kg)(-9.8m/s2)*sin23] - 243.11454
    a = -6.98650

    I was just wondering is this right so far?
     
    Last edited: Oct 13, 2016
  2. jcsd
  3. Oct 13, 2016 #2

    haruspex

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    Looks right. (With that coefficient of friction for skis on snow, the skiing industry would collapse.)
     
  4. Oct 13, 2016 #3
    So I am supposed to use negative 9.8 (or whatever decimal places I use), depending on whether I make up or down positive (I made down negative), on both x and y component?
    Also is that Fnet equation right Fgx subtract Fk? I'm having a lot of doubts and confusions in my calculations from which value should be positive or negative.
     
  5. Oct 13, 2016 #4

    haruspex

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    Clearly both the friction and the downslope component of gravity will contribute to the slowing of the skier, so the important thing is that they should have the same sign. The other main opportunity for errors is in confusion between sine and cosine, but I confirm you have that right too.
     
  6. Oct 13, 2016 #5
    I plugged in that acceleration into Vf2 = Vi2 +2*a*d and I ended up getting 4.81500 m and that answer was wrong apparently did I do something wrong?
    Final velocity I'm almost certain is 0.
    Initial velocity was given at 5.80 m/s
    acceleration I solved to be -6.98650

    Edit: Oops I forgot to multiply the acceleration by 2 before I divided the initial velocity by that
     
  7. Oct 13, 2016 #6

    haruspex

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    I get about half that. Did you forget the 2 in 2ad?
     
  8. Oct 13, 2016 #7
    Yeah I did ooops lol thanks for quick reply though.
     
  9. Oct 13, 2016 #8

    haruspex

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    Ok. Fwiw, a typical coefficient for ski on snow would be less than 0.1.
     
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