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Homework Help: Problem With Maximum Acceleration up Hill

  1. Mar 16, 2010 #1
    Problem With Maximum Acceleration up Hill!!

    1. The problem statement, all variables and given/known data

    cross-country skier is going up a slope at angle 5º to the horizontal. She is skating so only her skis provide the propulsion (i.e. she does not push with her ski poles). The static and kinetic friction coefficients for this situation are μs = 0.12, μk = 0.07 respectively.

    Find the magnitude of her maximum possible uphill acceleration?


    2. Relevant equations

    Fnet=MA

    3. The attempt at a solution

    So Far:

    Fnet(x)= F(static) - F(kinetic) - Fg = ma

    Fnet(x) = (0.12 x m x 9.8cos85) - (0.07 x m x 9.8cos85) - (9.8cos85 x m) = ma

    Divide by m you get:

    Fnet(x) = (0.12 x 9.8cos85) - (0.07 x 9.8cos85) - 9.8cos85 = a

    Therefore a= - 0.81m/s^2

    But this gives me a negative accelleration, Where did I go wrong? as it asks for the Max acceleration up the hill and that acceleration would point down hill.

    HELP PLEASE!!!
     
  2. jcsd
  3. Mar 16, 2010 #2

    ideasrule

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    Re: Problem With Maximum Acceleration up Hill!!

    How can she possibly accelerate uphill?
     
  4. Mar 16, 2010 #3
    Re: Problem With Maximum Acceleration up Hill!!

    She uses her skis to push against the ground which propells her forward through static friction.
     
  5. Mar 16, 2010 #4

    ideasrule

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    Re: Problem With Maximum Acceleration up Hill!!

    I thought the question said "i.e. she does not push with her ski poles".
     
  6. Mar 16, 2010 #5
    Re: Problem With Maximum Acceleration up Hill!!

    she DOES NOT use her poles, but she DOES use her skis - the same way people ice skate
     
  7. Mar 17, 2010 #6

    PhanthomJay

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    Re: Problem With Maximum Acceleration up Hill!!

    Please do not double post.

    https://www.physicsforums.com/showthread.php?t=386703

    You should first correct your geometry/trig/normal force calculation errors. Friction is a function of the normal force. The normal force is not mgcos85; what should it be?
    Then, once you make your correction for the normal force equation, you must make certain assumptions about the distribution of the skiers weight ....that all of it is one one ski as she pushes back against the snow, causing static friction to propel her forward, while the gravity component downhill (which you have calculated correctly) is at the same time working against her. But what about the sliding friction on the other ski..what's the normal force on the other ski with all her weight initially shifted to the 'propelling' ski??
     
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