# Work of pushing box horizontally - including force of friction

1. Jul 18, 2007

### avsj

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. Jul 18, 2007

### andrevdh

Well, the force needs to be at least equal to or larger than the frictional force does'nt it?

3. Jul 18, 2007

### ank_gl

thats the force you are looking for:tongue2:

4. Jul 18, 2007

### Gear300

heheh, *(that drum thing)*...well, its obvious as to what to do next, so this was just a random a post.

5. Jul 18, 2007

### marlon

Well, you have been given the kinetic friction force. This is the friction as long as the box is moving !

marlon

6. Jul 18, 2007

### avsj

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

7. Jul 18, 2007

### avsj

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

8. Jul 18, 2007

### ank_gl

bingo, now u get it

9. Jul 18, 2007

### GoldPheonix

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.

Last edited: Jul 18, 2007