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Work of pushing box horizontally - including force of friction

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data

    A man slides a 100.0 kg box along the floor for a distance of 4.0m. If the coefficient of kinetic friction is 0.250, and the man does the job in 3.6s, what is his power output in watts?

    2. Relevant equations


    3. The attempt at a solution

    I dont know how to calculate the force used to push the box. Fn should equal Fg so Ff = 0.25 x 980 = 245. But from what force do I subtract this
  2. jcsd
  3. Jul 18, 2007 #2


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    Homework Helper

    Well, the force needs to be at least equal to or larger than the frictional force does'nt it?
  4. Jul 18, 2007 #3
    thats the force you are looking for:tongue2:
  5. Jul 18, 2007 #4
    heheh, *(that drum thing)*...well, its obvious as to what to do next, so this was just a random a post.
  6. Jul 18, 2007 #5
    Well, you have been given the kinetic friction force. This is the friction as long as the box is moving !

  7. Jul 18, 2007 #6
    The answer is 2.7 x 10^2 W.

    So if my Ff = 245, I need to subtract that from the horizontal force to get Fnet which I can then use to get W= Fd and to me it is obvious from there. But how do I find the horizontal force of moving the box?

  8. Jul 18, 2007 #7
    Oh I think i understand. The net force must equal the force of friction as it is moving so we use the Ff as the force. Thanks :D
  9. Jul 18, 2007 #8
    bingo, now u get it
  10. Jul 18, 2007 #9
    Just walk yourself through the math, okay? It's nothing bad.

    [tex] \overline{P} = \frac{ \Delta E}{ \Delta t}[/tex]

    [tex]\Delta E = F \Delta x[/tex]

    You've calculated force, right? So, what's the box's position at t=1? 0, right? So, time 1 = 0, right? It's your starting out point. Go from there.
    Last edited: Jul 18, 2007
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