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Resolving power on an inclined slope

  1. Sep 21, 2007 #1
    This one the last questions on my mechanics paper, its hard to visualize without drawing a diagram i found, but i still cant solve it
    1.
    A tractor of 6000kg approaches a slope at arcsin(0.05) to the horizontal. The non-gravitational resistance to motion on this slope is 2000N. The tractor accellorates uniformly from 3m/s to 3.25m/s over a distance of 100m while climbing the slope

    1. Calculate the time taken to travel this distance of 100m, and the average power required over this time period.


    2. Relevant equations

    p = Fv
    p= F/t
    f = ma
    Work Done = 0.5mv^2

    3. The attempt at a solution

    i got as far as using s = (u+v)/2 x t, which gives the time as 32 seconds. i'm just not sure how to find the average power because i dont know where to put the 2000N resistance in the equation, plus the angle of the slope confuses the issue somewhat
     
    Last edited: Sep 21, 2007
  2. jcsd
  3. Sep 21, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Find the work done in traversing that distance against (1) gravity, and (2) the resistive force of 2000 N. Add them up and use that total to find the average power.
     
  4. Sep 21, 2007 #3
    cheers i never thought of using GPE. height is 5m so

    WD is 0.5 x 6000 x (3.25^2 - 3^2)

    + 6000 x 9.8 x 100 x 0.05

    +2000 * 100 = 498687.5 W


    i presume that thats the total power from that total?, and the average power will be 498687.5 / 32 which is 15.6 KW?
     
  5. Sep 21, 2007 #4

    Doc Al

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    Staff: Mentor

    Sounds good to me.
     
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