Calculating Work Done for a Box on Horizontal Floor

Click For Summary

Homework Help Overview

The problem involves calculating the work done on a box being pulled along a horizontal floor, given its weight, the distance it is moved, and the coefficient of kinetic friction. The context is within the subject area of mechanics, specifically focusing on forces and work.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the applied force and the frictional force, noting that the box moves at a constant speed, which implies zero acceleration. There is an exploration of how to calculate the applied force based on the frictional force.

Discussion Status

Some participants have provided calculations for the frictional force and have noted that the applied force must equal the frictional force in magnitude. There is an ongoing exploration of the relationship between these forces, with some participants confirming the calculations and others reiterating the principles involved.

Contextual Notes

Participants are working within the constraints of the problem statement, which specifies the weight of the box, the distance moved, and the coefficient of friction, but does not provide explicit instructions on how to approach the calculation of work done.

chawki
Messages
504
Reaction score
0

Homework Statement


A box, whose weight is 182 kg, is drawn with constant speed 21 m along a horizontal concrete floor. The kinetic friction between the box and the floor is 0.58.
(acceleration due to gravity, g = 9,81 m/s2).

Homework Equations


Calculate the work done when the box is drawn with a horizontal force.

The Attempt at a Solution


W = F*d
i don't know how to find F
i could only find Ff which is the frictional force
Ff= 0.58*182*9.81
Ff= 1035.54 N
 
Physics news on Phys.org
hi chawki! :smile:
chawki said:
W = F*d
i don't know how to find F
i could only find Ff which is the frictional force
Ff= 0.58*182*9.81
Ff= 1035.54 N

the speed is constant, so the acceleration is zero,

so Ftotal = 0,

so F + Fr = 0 :wink:
 
applied force is equal to force of friction in magnitude. so job is done!
 
yesyes,
by applying Newton second law and by projection on xx, we get:
F-Ff=0
F=Ff
F=0.58*182*9.81
F=1035.5436 N

W=F*d
W=1035.5436*21
W=21746.4156 J
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K