Work of pushing box horizontally - including force of friction

Click For Summary
SUMMARY

The discussion revolves around calculating the power output of a man sliding a 100.0 kg box over a distance of 4.0 meters with a coefficient of kinetic friction of 0.250 in 3.6 seconds. The frictional force (Ff) is calculated as 245 N using the formula Ff = μ * Fn, where Fn equals the gravitational force (Fg = 980 N). The net force required to move the box must equal the frictional force, leading to the conclusion that the power output is 270 watts, calculated using the formula P = W/t.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of force, work, and power
  • Knowledge of kinetic friction and its calculation
  • Ability to perform basic algebraic manipulations
NEXT STEPS
  • Study the principles of Newton's second law of motion
  • Learn about work-energy principles in physics
  • Explore the calculations of power in mechanical systems
  • Investigate different coefficients of friction and their applications
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to force, work, and power calculations.

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.

\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:

Similar threads

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