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Mmmm Good ol' work and power

  1. Oct 22, 2006 #1
    Mmmm....Good ol' work and power

    A 1200N force is applied parallel to a horizontal surface. It pushes an 80N box 10m across the surface. What work is done? What power is dissipated in 3 minutes?

    What i did for this one is subtract the 80N force from the 1200N force and multiply it by 10m because W= FxD
    1200-80=1120x10m=11200J, problem is im not sure if im supposed to subtract the 80N force of the box. For power, i did 11200/180s=62.22 watts, but im not sure if thats just power or if its power dissipated
    any help would be greatly appreciated, thanks
     
  2. jcsd
  3. Oct 22, 2006 #2
    You should not be subtracting the 80N from 1200N. The 80N does have some significance, but in this situation the force vector of the 80N (its weight) is perpendicular to the displacement vector so it will come out to zero work. What I'm trying to say is that since it is not on a sloped plane, the work done by gravity is 0. Therefore, the only work done is by the 1200N force.

    Also remember that the equation for work is [tex]w=\vec{F}\cdot\vec{D}[/tex] which means [tex]w=|F||D|cos\theta[/tex] theta being the angle between the force and the displacement vectors.
     
    Last edited: Oct 22, 2006
  4. Oct 22, 2006 #3
    So ill just find power by multiplying 1200 by 10m to get 12000J, but what about the power dissipated? i found power to be 66.67 by dividing work by time, but im not sure if thats the dissipation or not
     
  5. Oct 22, 2006 #4
    You get the work by multiplying the two. I don't think the word "dissipated" has any significant meaning other than how much power was exerted/used/etc, I believe all they are asking for is the standard work/time for power.
     
  6. Oct 22, 2006 #5
    Cool, thanks a lot man :)
     
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