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Draw free body diagramm of all forces acting on skier

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A skier of mass m is skiing down a frictionless slope. the skier starts form rest at t = 0 and is subject to a velocity dependant drag force due to air resistance of the form F = -bv, where b is a constant.

    a)draw free body diagramm of all forces acting on skier

    b) wire a differential equation that can be used to solve for the velocity of the skier as a function of time

    c) find expression for Terminal velocity

    d) solve b and ditermine the velocity of the skier as a function of time

    2. Relevant equations



    3. The attempt at a solution

    a) note the attached diagramm. does the the wind come paralel to the hill of horizontally

    b)

    Fx = ma = mgsin[tex]\theta[/tex] - bv

    v = (mgsin[tex]\theta[/tex] - ma)/b

    i belive i need to find an equation for a which involvs t, do i just use one of the kinematic equations

    the answer is v(t) = (mg sinθ / b) (1 – e-bt/m) where did the who e part come from
     
  2. jcsd
  3. Oct 14, 2009 #2

    Redbelly98

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    Re: Skiing

    Since the air resistance force is -bv, it is directly opposite to the direction of the skier's velocity. Is the skier's velocity parallel to the hill or horizontal?
    The attachment seems to be missing.
    It cannot come from the kinematic equations, which only work when acceleration is constant.

    It comes from the equation you wrote in part (b),

    ma = mg sinθ - bv​

    Hint: what equation defines a in terms of v?
     
  4. Oct 14, 2009 #3
    Re: Skiing

    Sorry, heres the attachment

    the velocity is parallel to the ground.

    What do you mean by it came from the equation from part b, i cant plug the inquation into itself
     

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  5. Oct 14, 2009 #4

    Redbelly98

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    Re: Skiing

    Your free body diagram looks good.

    Correct.

    True, lol. No, you need to substitute another expression in for a first. Refer to my hint:
     
  6. Oct 15, 2009 #5
    Re: Skiing

    a = dv/dt
     
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