Calculate Work Friction in Figure 8-16 | Walker2 8.P.002

  • Thread starter Thread starter tony873004
  • Start date Start date
  • Tags Tags
    Friction Work
Click For Summary

Homework Help Overview

The problem involves calculating the work done by friction as a box slides along a floor from point A to point B, considering different paths. The subject area pertains to mechanics, specifically the concepts of work, friction, and forces acting on an object in motion.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the work done by friction using the coefficient of kinetic friction and normal force, but expresses uncertainty about the correctness of their answer. Some participants question whether vertical movements should be considered in the calculation of work done by friction, leading to a discussion about the role of normal force in those directions.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of how to account for vertical movements in the context of work done by friction. There is no explicit consensus, but participants are engaging with the problem and questioning assumptions related to the forces involved.

Contextual Notes

Participants note that the problem specifies the work done by friction, which raises questions about the relevance of vertical components of movement and the normal force in those directions. Previous problems involving work done by gravity are referenced for comparison.

tony873004
Science Advisor
Gold Member
Messages
1,753
Reaction score
143
Figure 8-16: http://www.webassign.net/walker/08-16.gif

12. [Walker2 8.P.002.] Calculate the work done by friction as a 2.1 kg box is slid along a floor from point A to point B in Figure 8-16 along paths 1, 2, and 3. Assume that the coefficient of kinetic friction between the box and the floor is 0.21.

Since work can have a negative number, I'm assuming that all 3 answers will be the same, since the displacement is 3 meters. Path 1 travels 4 - 1 = 3 meters, Path 1 travels 2 + 1 = 3 meters, and Path 3 travels 3 meters. The vertical travels will not be affected by friction.

W = f * d
d = 3
f = friction
friction = mu * N (force Normal)
N = ma
N = 2.1 * 9.81
N = 20.601
friction = 0.21 * 20.601 = 4.32621
W = 4.32621 * 3
W = 12.98 j

but this is wrong, and so is -12.98.

Thanks in advance for your input :biggrin:
 
Physics news on Phys.org
Work is done in the direction of an applied force. In this case you can't neglect the vertical lines. Work is done on them too. You should add them up.
 
CartoonKid said:
... In this case you can't neglect the vertical lines. Work is done on them too...

But is work done on them by friction? All it asks for is the work done by friction. And friction is mu * N, but in the vertical direction doesn't N = 0? meaning that work done by friction would also be 0? I can see why there would be work done by gravity, but we have to ignore that. I think :confused:

In the previous problem, we had to figure out work done by gravity, but not friction and I got it right:
figure: http://www.webassign.net/walker/08-15.gif

[Walker2 8.P.001.] Calculate the work done by gravity as a 1.7 kg object is moved from point A to point B in Figure 8-15 along paths 1, 2, and 3.
The answers are -33j for all 3 parts.
 
Last edited:
This question is different, it is actually a plan. You look the movement of the object from above. The object is always on the floor.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
21
Views
3K
Replies
8
Views
9K
  • · Replies 2 ·
Replies
2
Views
18K