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Work nergy theorem question

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    At the base of a frictionless icy hill that rises at 25 degree above horizontal, a toboggan has a speed of 12m/s toward the hill. How high vertically above the base will it go before stopping?


    2. Relevant equations
    WORK ENERGY THEOREM


    3. The attempt at a solution
    I know the answer but the steps were quite confusing.
    W= Delta K. Where W should be the NET work. Weight does a work at 180 from the displacement. So Wmg = -mgh.. Fine! Now, the book equated this with the kinetic energy. Isn't there a normal force acting too. Why didn't the book consider the work done by the normal force. I feel i should add W done by N= (mgcos theta) x cos theta x h ... Why can't i do that... I need good replies. Thanks to whoever contributes with good answers. :)
     
  2. jcsd
  3. Nov 19, 2012 #2

    TSny

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    Careful. The displacement vector of the taboggan is in the direction of the slope of the hill; that is, at 25 degrees above the horizontal. But the force of gravity is vertically downward. So, the angle between the weight and the displacement is not 180 degrees.
     
  4. Nov 19, 2012 #3
    But now, he is asking me about the vertical height. So shouldn't we think of the y axis as a reference now?
     
  5. Nov 19, 2012 #4

    TSny

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    I think it's best to stick with the definition of work as W = Fdcosθ

    If you can find the distance d along the hill, then you can use that to find the height h.

    So, what would be the value of θ for the weight force and what would be the value of θ for the normal force?
     
  6. Nov 19, 2012 #5
    Ok... I get your point and let me solve it with your convention. W done by gravity along s (which is the hypotanious) is equal to = -mgcos theta X s = - .5 mv^2 .... s would be h/sin theta... The answers turns out to be wrong because over all we equate and have - mg h/tan theta = .5 m v^2 ... Kindly Clarify!
     
  7. Nov 19, 2012 #6

    TSny

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    Please specify what you used for the value of θ in the expression mgscosθ for the work done by the weight force.
     
    Last edited: Nov 19, 2012
  8. Nov 19, 2012 #7
    Oh my godness... I knew where my mistake is... It is W=F.s ... So we take mg as a whole , s as a whole and the angle between them is sin theta... We then mltiply it by h/sin theta .. so sin theta cancels... WOW! I owe you bro. Cs i didn't wan't to solve it with potential energy rather than with kinetic energy.. Thanks!
     
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