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Work of a car's engine as it moves up a slope

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A car of mass 900 kg accelerates up a slope. The velocity of the car at the bottom of the slope is 5 m/s. By the time the car reaches the top of the slope, its velocity is 15 m/s. The slope is 8m high from the ground and has a length (radius - angled length - not base) of 500m. The average frictional force experienced by the car between A and B is 50N.

    Calculate: The total work done by the car's engine between the bottom and top of the slope.


    2. Relevant equations
    F = ma
    m = mass in kilograms
    a = acceleration in m/s
    F = force in newtons
    W = F x s
    s = distance
    W = work
    3. The attempt at a solution

    I am quite at a loss with this question but this is my attempt.
    Fbottom = m x a = 900 x 5 = 4500N

    Work (bottom) = f x s
    = 4500 x 500 (slope is 500m)
    = 2250000J

    Ftop = F x ma
    = 900 x 15
    = 13500N
    Work (top) = f x s
    = 135000 x 500
    = 6750000J

    W (top) - W (bottom) = 450000J

    I am sure that my answer is incorrect but I am unsure as to how to solve it and the memo is missing the answer, any help would be appreciated. Sorry for the n00b question! :)
     
    Last edited: Aug 21, 2012
  2. jcsd
  3. Aug 21, 2012 #2

    CAF123

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    There is a typo here somewhere, you mean its velocity is 15m/s?
     
  4. Aug 21, 2012 #3
    Yeah, sorry about that!
     
  5. Aug 21, 2012 #4

    CAF123

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    It is difficult to follow your solution because you have inputted velocities in place of accelerations.
    I think this is meant to be solved using the work-energy theorem, but also incorporating the effects of an external force, ie friction, so; [tex] W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 + F_f d = ΔK.E + F_f d, [/tex] where [itex] F_f [/itex] is the force of friction.
     
  6. Aug 22, 2012 #5

    CWatters

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    Don't forget the gain in PE as well as KE?
     
  7. Aug 22, 2012 #6

    CAF123

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    In taking into account the change in potential energy, i believe we add on, to the eqn above;
    [tex] W = \vec{F}.\vec{d} = mgdcos(90 +θ) = -mgdsinθ [/tex]

    Apologies for forgetting that earlier, I must have done the problem ignoring the incline!
     
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