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Question regarding distance traveled up an incline?

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    After a bobsled race, the sled and riders have been partially slowed down up in an icy incline, but they need to be brought to a stop. This happens on a portion of track inclined at 30 degrees that has a coefficient of kinetic friction of 0.60. The sled enters the incline at 25.0 m/s. How far does it travel along the incline before stopping?


    2. Relevant equations
    (1/2)mv(final)^2 - (1/2)mv(initial)^2 = Ffriction


    3. The attempt at a solution
    Im not sure how to incorporate the angle of the incline, but as an attempt, i did
    1/2vi^2=FG(cos 30) d
    and i got 88.3 m. I'm not sure if that's right...
     
  2. jcsd
  3. Nov 1, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks like you're using an energy approach. Good. But don't forget gravitational PE, which increases as the sled goes up the incline.
     
  4. Nov 1, 2008 #3
    How would you incorporate gravitational PE? and also, would the formula
    d = vi^2 / [2g(sin30 + ucos30)] work?
     
  5. Nov 1, 2008 #4

    Doc Al

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    Staff: Mentor

    What's the definition of gravitional PE? How do you calculate it?
    I would advise against looking for a formula to plug into unless you fully understand what the formula means. Better to stick with the basics and derive your own formula, tailor-made to the specific problem. (That formula is close, but not quite right, anyway.)

    The basic idea that you want to use is energy conservation:
    Energy(initial) + Work done by friction(which is negative) = Energy(final)

    Here "Energy" means total mechanical energy: KE + PE.
     
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