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Homework Help: 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

    W=Fd
    P=W/t

    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

    andrevdh

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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 !

    marlon
     
  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?

    Thanks
     
  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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