Work of pushing box horizontally - including force of friction

Click For Summary

Homework Help Overview

The problem involves calculating the power output of a man pushing a 100.0 kg box over a distance of 4.0 m, considering the effects of kinetic friction with a coefficient of 0.250. The task is to determine the necessary force and subsequent power output without providing a complete solution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the force needed to overcome friction and how to apply this to find the net force and power output. Questions arise about the relationship between normal force, gravitational force, and frictional force.

Discussion Status

Some participants have offered insights into the relationship between the forces acting on the box, particularly regarding the frictional force. There is an ongoing exploration of how to calculate the horizontal force required to move the box and its implications for power output.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for calculations. The discussion reflects uncertainty in the application of physics concepts to derive the necessary values for the solution.

avsj
Messages
26
Reaction score
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
 
Physics news on Phys.org
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.

[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:

Similar threads

  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
2K