Work of pushing box horizontally - including force of friction

In summary, the man's power output in watts can be calculated by using the equation P=W/t, where W=Fd and F is equal to the frictional force of 245 N. The box's position at t=1 is 0, and the time is also 0, so the change in energy is equal to 0. Therefore, the man's power output is 0 watts.
  • #1
avsj
26
0

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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  • #2
Well, the force needs to be at least equal to or larger than the frictional force doesn't it?
 
  • #3
avsj said:
. Fn should equal Fg so Ff = 0.25 x 980 = 245.

thats the force you are looking for:tongue2:
 
  • #4
heheh, *(that drum thing)*...well, its obvious as to what to do next, so this was just a random a post.
 
  • #5
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
 
  • #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
 
  • #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
 
  • #8
bingo, now u get it
 
  • #9
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.

[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:

1. What is the work of pushing a box horizontally?

The work of pushing a box horizontally is the amount of force applied over a certain distance in order to move the box. It is calculated by multiplying the force applied by the distance moved.

2. How does the force of friction affect the work of pushing a box horizontally?

The force of friction is the resistance that opposes the motion of the box. This force acts in the opposite direction of the applied force, making it harder to move the box. Therefore, the force of friction decreases the amount of work done in pushing the box horizontally.

3. What factors affect the force of friction when pushing a box horizontally?

The force of friction is affected by the weight of the box, the roughness of the surface it is being pushed on, and the force applied to move the box. The heavier the box and the rougher the surface, the greater the force of friction.

4. How can the force of friction be reduced when pushing a box horizontally?

The force of friction can be reduced by using a lubricant, such as oil or grease, to reduce the roughness of the surface and make it easier for the box to slide. Additionally, reducing the weight of the box or using wheels can also decrease the force of friction.

5. Is the work of pushing a box horizontally affected by the direction of the force applied?

Yes, the work of pushing a box horizontally is affected by the direction of the force. If the force is applied at an angle, only the component of the force that is parallel to the direction of motion will contribute to the work done. This means that pushing a box diagonally will require more force and work than pushing it straight ahead.

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