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Skier going down a hill force and motion question

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A sight seen on many "bunny" hills across Ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest/ Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give herself an initial velocity of 3.5m/s^2. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20, calculate

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


    2. Relevant equations


    V final ^2 = V initial ^ 2 + 2 a t


    3. The attempt at a solution

    a)

    Given:
    m = 25kg
    V initial = 3.5 m/s^2
    Angle of hill = 5 degrees
    coefficient kinetic = 0.20

    I need to find acceleration to use that formula to find time, acceleration down the hill with no initial velocity and just gravity is Fg parallel to the hill - Fk kinetic friction acting in the opposite direction.

    Given:
    m = 25kg
    V initial = 3.5 m/s^2
    Angle of hill = 5 degrees
    coefficient kinetic = 0.20

    Fg parallel to hill = mg sin 5
    Fk parallel to hill = mg cos 5 0.20

    Fg = (25kg)(9.8m/s^2)(sin5) = 21.35kgm / s^2 < Why are my units weird here?
    Fk = (25kg)(9.8m/s^2)(cos5)(0.20) = 48.81kgm / s^2

    21.35 - 48.81 = -27.46

    The kinetic friction is greater than gravity parallel to hill, does that mean the skier can't move down with just gravity alone?

    Or maybe the value I found for "Fk" is not kinetic friction because I think it might just be the Fn acting perpendicular to the hill. So then how do I find Fk?

    Fk = Fn 0.20 ?

    But I've already applied the friction coefficient to find Fn.

    So then what is -27.46? Is it the acceleration parallel to hill due to gravity?



    Assuming I have found the correct value for acceleration,
    V final ^2 = V initial ^2 + 2at
    V final is 0
    V initial is 3.5m/s

    t = -V initial ^ 2 / 2 a
    t = (-3.5m/s)^2 / 2(-27.46m/s^2)
    t = 0.22s


    b)

    Assuming I have found the correct value for t

    d = v average t
    Do I use this formula to find d?
    If V final = 0 is V average = 3.5m/s + 0 / 2 = 1.75m/s?
    So,
    d = (1.75m/s)(0.22s)
    d = 0.39m
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 28, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    makes sense. So
    ma = mg sin 5 - mg cos 5 0.20 downhill
    a = g(sin5 - cos5)
    Negative means it is decelerating.
     
  4. Feb 28, 2009 #3
    I don't really understand how you arrived at this. You mean that total force going down parallel to hill = Fg - Fk, this is Fa, and Fa = ma, so ma = mg sin 5 - mg cos 5 0.20.
    And then you factored out the mg so it becomes mg (sin 5 - cos 5 0.20) = ma, and then you cancel out the m, so should it be a = g(sin 5 - cos 5 0.20) ? How come there is no mu kinetic (0.20) in your equation?

    If this is so, then

    a = (9.8m/s ^ 2)(-909038955)(0.20) = -1.7m/s ^ 2
    Which is not what I get if I don't do the factoring and canceling of m. Does this mean I always have to factor and cancel out terms to get the right answer?


    So assuming now I have the correct value for a = -1.7m/s^2, I can find the time with this formula?

    t = -V initial ^ 2 / 2 a
    t = (-3.5m/s)^2 / 2(-1.7m/s^2)
    t = 3.6s

    Is this correct?
     
  5. Feb 28, 2009 #4

    Delphi51

    User Avatar
    Homework Helper

    Quite right - the .2 slipped away on me - good catch!
    The idea is to write that the sum of the forces along the ramp equals ma, then find a.
    You don't need to cancel the m's, but it saves calculations.
    I'm getting a = -1.10. I don't know if you or I have the error.

    There is something the matter with this. The basic formula is Vf = Vi + a*t
    so t = (Vf - Vi)/a
     
  6. Feb 28, 2009 #5
    Ok, so this is what I'm doing
    a = g[(sin 5) - (cos 5)](0.20)
    a = (9.8m/s^2)[(0.087155742) - (0.996194698)](0.20)
    a = (9.8m/s^2)(-0.909038956)(0.20)
    a = (9.8m/s^2)(-0.181807791)
    a = -1.781716344 m/s^2

    And yes I see I have copied the formula for t wrong, thank you.
     
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