Work of pushing box horizontally - including force of friction

AI Thread Summary
A man pushes a 100.0 kg box over 4.0 meters with a coefficient of kinetic friction of 0.250, taking 3.6 seconds. The frictional force is calculated as 245 N, which must be overcome by the applied force to maintain motion. The net force is equal to the frictional force while the box is moving. To find power output, the work done (force times distance) is divided by time. The final power output is determined to be approximately 270 watts.
avsj
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Homework Statement



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?


Homework Equations



W=Fd
P=W/t

The Attempt at a Solution



I don't 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
 
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Well, the force needs to be at least equal to or larger than the frictional force doesn't it?
 
avsj said:
. Fn should equal Fg so Ff = 0.25 x 980 = 245.

thats the force you are looking for:-p
 
heheh, *(that drum thing)*...well, its obvious as to what to do next, so this was just a random a post.
 
avsj said:
I don't 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
Well, you have been given the kinetic friction force. This is the friction as long as the box is moving !

marlon
 
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
 
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
 
bingo, now u get it
 
avsj said:

Homework Statement



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?

I don't 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

Just walk yourself through the math, okay? It's nothing bad.

\overline{P} = \frac{ \Delta E}{ \Delta t}

\Delta E = F \Delta x

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.
 
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