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Power/work questions (with inclines)

  1. Jan 18, 2006 #1

    dnt

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    the question states at 120 kg tractor goes up a 21 degree incline of 12.0 m in 2.5 seconds. what is the power?

    ok i know power is W/T which is basically Fd/T

    my question is how do you use the incline. work is Fdcos0 but i just dont see why its cosine in this question. the work is going UP so shouldnt you want to use the vertical component? meaning use sine insteand of cosine? doesnt make sense to me.



    a similar question is a 185 fridge goes up 10.0 m at an angle of 11 degrees, what is the work done?

    again, its going up so should we use cosine or sine? i think it should be sine but the equation says use cosine.

    dont quite get it. please help. thanks.
     
  2. jcsd
  3. Jan 18, 2006 #2

    andrevdh

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    The work under discussion here is that done by the force pushing the object up the incline. This force has to cancel the weight component of the object, parallel to the incline, out. Is this component given by the sine or the cosine of the inclination?
     
  4. Jan 18, 2006 #3

    dnt

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    i believe thats the sine. is that correct?
     
  5. Jan 19, 2006 #4

    andrevdh

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    Correct. So the you want to calculate the work done by a force equal to [itex]W\sin(\theta)[/itex], but in the opposite direction, that pushes the object up the incline. The cosine in the work equation is applicable when the angle [itex]\theta[/itex] is given as the angle between the force and the displacement vectors. The reason why there is a [itex]\cos(\theta)[/itex] in the work equation is to get the component of the force that is parallel to the displacement. It is only this component that does work as the object is displaced. The other component, perpendicular to the displacement, does no work as the object is displaced.
     
    Last edited: Jan 19, 2006
  6. Jan 19, 2006 #5
    Just use potential energy. But I'll show you why you use cosine anyway.
    [tex]W = \int_{s_i}^{s_f} F_s ds = \int_{s_i}^{s_f} F cos(\theta)ds[/tex]
    F and [tex]\theta[/tex] are constant and can be moved out of the integral. F_s is the component of F(vector) in the direction of motion, and F makes an angle [tex]\theta[/tex] with respect to a displacement [tex]\Delta r[/tex], so the component of the force vector is the direction of motion is [tex]F_s = Fcos\theta[/tex].
    [tex]W = Fcos(\theta)\int_{s_i}^{s_f}ds = F cos\theta (s_f - s_i) = F cos\theta (\Delta s) = F(\Delta r) cos\theta[/tex]
     
    Last edited: Jan 19, 2006
  7. Jan 21, 2006 #6

    dnt

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    so isnt that a little misleading? why do they include cosine in the work equation when sometimes you use sine?
     
  8. Jan 22, 2006 #7

    andrevdh

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    When we want to calculate the work done by a force applied to an object we need to get the component of the force that is in line with the displacement - therefore the [itex]\cos(\theta)[/itex] in the work equation. Where [itex]\theta[/itex] is the angle between the direction of the force and displacement vectors. In this case weve got the force parallel to the incline, so there is no need for the cosine. The sine comes from gettting the component of the weight parallel to the incline, which involves the angle of the incline, which is a completely different angle than the one in the work equation.
     
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