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Kinetic Friction of skier

  1. Oct 16, 2009 #1
    1. A skier slides horizontally along the snow for a distance of 29 m before coming to rest. The coefficient of kinetic friction between the skier and the snow is µk = 0.010. Initially, how fast was the skier going?


    2. f kinetic = Coeff. of friction * Normal Force
    Work Energy Theorem = (1/2 mass * Final Velocity^2)-(1/2 mass * Initial Velocity^2)



    3. In the WET, the masses cancel out, I get an equation of Initial Velocity^2= 2Final Velocity ^2.

    Only problem is the final velocity is zero so I get an initial velocity of zero as well, but that's is impossible...
     
  2. jcsd
  3. Oct 16, 2009 #2

    cepheid

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    First of all, "Work Energy Theorem" is the name of a theorem, not some physical quantity, so it makes no sense to write, work energy theorem = 'blah.' The work energy theorem is a theorem which *states* that:

    work done = change in kinetic energy.

    And no, you don't get an equation which states that initial velocity^2 = final velocity^2. You should be more careful with your algebra.

    If

    W = 1/2m(v_f)^2 - 1/2m(v_i)^2)

    then you get the result that:

    m* [ (v_f)^2 - (v_i)^2 ] = 2W

    and since v_f = 0,

    m*(v_i)^2 = 2W

    You know that in this case, the work (W) was being done by friction, which allows you to calculate it.
     
  4. Oct 16, 2009 #3
    So how would I calculate it without the mass? We know how far it travels and the coefficient of Kinetic Friction, but not the mass.
     
  5. Oct 16, 2009 #4

    cepheid

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    How would you calculate W, the work done by friction? That will make the answer to your question clear.
     
  6. Oct 16, 2009 #5

    rl.bhat

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    Two accelerations are acting on the skier. They are in the opposite direction. One due to gravitation, and another due to friction. Find the net acceleration. Using kinematic equation, find the initial velocity.
     
  7. Oct 16, 2009 #6
    Work is equal to force times displacement, and the force of friction is equal to the coefficient of kinetic friction times the normal force (which is equal to mass times gravity). But I'm still confused how I would get it without having the mass.

    And thank you for helping me out, I'm really struggling with this stuff and my algebra isn't that great...
     
  8. Oct 16, 2009 #7

    cepheid

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    No, the skier is not on an incline. Try not to confuse the OP. The skier is merely being decelerated to rest from some initial velocity due to a frictional force. To find the initial velocity, all the OP has to do is calculate the work done by friction and equate it to the initial KE. That's it.
     
  9. Oct 16, 2009 #8

    cepheid

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    Exactly right. The mass appears in the expression for the work done as well. Which is why it cancels from both sides of the equation.
     
  10. Oct 16, 2009 #9

    rl.bhat

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    Sorry.
     
  11. Oct 16, 2009 #10
    Work is equal to force times displacement, and the force of friction is equal to the coefficient of kinetic friction times the normal force (which is equal to mass times gravity).

    So the equation would look like m*(v_i)^2 = 2 (m * g) but then since the masses cancel, the equation would then be Vi^2 = 2 g?
     
  12. Oct 16, 2009 #11

    cepheid

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    Don't forget the coefficient of kinetic friction (mu) on the right hand side, as well as the displacement. You definitely have the right idea though.
     
  13. Oct 16, 2009 #12
    So Vi = Square root (2*g*Coeff. of friction*displacement)?

    Which I get to be Vi = Square root (5.684) which equals 2.384 m/s?
     
  14. Nov 24, 2009 #13
    How did you get the equation - Vi= Square root (2*g*Coeff. of friction*displacement?
     
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