Inclined plane, Calculate friction force

Click For Summary

Homework Help Overview

The discussion revolves around calculating the friction force on an inclined plane using the work-energy theorem. Participants are exploring the relationship between net work, gravitational work, and frictional work in the context of a physics problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss their calculations and seek confirmation on the correctness of their answers. There is a focus on the application of the work-energy theorem and the components of work involved.

Discussion Status

Some participants have provided their working calculations and are questioning the accuracy of their results. There is an indication of productive dialogue, with one participant acknowledging a mistake in their earlier calculation and presenting revised work for consideration.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can provide or the methods they can use. There is an emphasis on verifying calculations and understanding the underlying physics principles.

TheRedDevil18
Messages
406
Reaction score
2

Homework Statement



work.jpg


Homework Equations



Wnet = Ek

The Attempt at a Solution



1) That was fine.
2) I just want to confirm if this answer is right?, I got 12.28N. Used work energy theorem.

Sorry for double post, edited the title
 
Physics news on Phys.org
TheRedDevil18 said:

Homework Statement



work.jpg


Homework Equations



Wnet = Ek

The Attempt at a Solution



1) That was fine.
2) I just want to confirm if this answer is right?, I got 12.28N. Used work energy theorem.

Sorry for double post, edited the title

Seems a bit high. Pls post your working.
 
haruspex said:
Seems a bit high. Pls post your working.

Sorry my mistake, forgot something in my calculation. Anyway here's my working

Wnet = Ek
Wg + Wf + Wn = 1/2*m(vf^2 - vi^2)
(5*9.8*sin30*20) - 20Ff + 0 = 1/2*5(12^2 - 6^2)...That 20 is the length of the incline
Ff = 11N...Now correct?
 
TheRedDevil18 said:
Sorry my mistake, forgot something in my calculation. Anyway here's my working

Wnet = Ek
Wg + Wf + Wn = 1/2*m(vf^2 - vi^2)
(5*9.8*sin30*20) - 20Ff + 0 = 1/2*5(12^2 - 6^2)...That 20 is the length of the incline
Ff = 11N...Now correct?

Yes, that looks right.
 
  • Like
Likes   Reactions: 1 person

Similar threads

Replies
2
Views
891
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
Replies
8
Views
3K
Replies
5
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 19 ·
Replies
19
Views
3K