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Dynamics Skier on a Slope Problem

  1. Mar 29, 2016 #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.

    Answer on the book: 13m

    2. Relevant equations
    Fnetx = (-Ff) + Fa + (-Fgx)
    Fnety = Fn + (-Fgy)


    v2^2 = v1^2 + 2ad

    3. The attempt at a solution
    I went by creating two equations:

    Fnetx = (-Ff) + Fa + (-Fgx)
    ma = -Fn(u) + Fa + mg(sin4.7)
    m was canceled out on both sides.
    Fn(u) = Fa + g(sin4.7) - a
    (1) Fn = Fa + g(sin4.7) - a / u

    Fnety = Fn + (-Fgy)
    ma = Fn + (m)(-g)(cos4.7)
    again canceled out m on both sides
    (2) Fn = a + gcos4.7

    Correction: Equation 2 is Fn = gcos4.7

    And so I did substitution and subbed in both equations to solve for acceleration (a) but the problem is that Fa is also unknown:

    (9.8)cos4.7 = Fa + (9.8)(sin4.7) - a / 0.11

    So I'm wondering if there is a better way to do this problem, because I feel like I'm doing to wrong.

    I included a fbd diagram:
    pPdlBr4.jpg
     
    Last edited: Mar 29, 2016
  2. jcsd
  3. Mar 29, 2016 #2

    haruspex

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    In your equation (2), what acceleration does a represent? Acceleration into the snow?
     
  4. Mar 29, 2016 #3
    Looking back at equation 2, I feel like the net force on the y-axis is 0. Because from my understanding, the skier is only moving on the x-axis being on the incline, where the forces on the y-axis are cancelling each other out.
    Correction of equation 2: Fn = gcos4.7
    But even then that still leaves two unknowns, both Fa and a.
     
  5. Mar 29, 2016 #4

    haruspex

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    And what applied force does Fa represent?
     
  6. Mar 29, 2016 #5
    Fa is the applied force that the skier is pushing on the ski poles thus moving up the incline.
     
  7. Mar 29, 2016 #6

    haruspex

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    The skier is moving down the incline and only uses the poles to get going. Having reached 2.7m/s, the poles are no longer used, and the skier slides to a stop.
     
  8. Mar 29, 2016 #7
    I have redone my equations, with:
    Equation 1 : Fn = a + gsin4.7 / u
    Equation 2 : Fn = mgcos4.7

    But again same situation, there are still two unknowns being m and a, since I can't cancel out m on the y-axis as shown below:
    Fnety = Fn + (-Fgy)
    0 = Fn + (m)(-g)(cos47)
    Fn = mgcos4.7
     
  9. Mar 29, 2016 #8
    Algebra Question
    In the equation as shown below:
    (m)(a) = (u)(g)(m)(cos4.7) + (m)(-g)(sin4.7)
    Can I cancel out all the m's on both sides and result into this?:
    a = (u)(g)(cos4.7)+(-g)(sin4.7)
    Or is that not possible?
     
  10. Mar 29, 2016 #9

    haruspex

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    Of course you can cancel the m's. All you need to assume is that m is not zero, and you can safely do that.
     
  11. Mar 29, 2016 #10
    Oh okay, now it makes sense. Thank you.
     
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