Physics Graph question -- Pushing a block on a frictionless surface

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a block moving on a frictionless surface, influenced by a varying force along the x-axis. The original poster presents two questions: calculating the work done by the force as the block moves from the origin to 8.0 m, and determining the block's velocity at that position given its initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the work done by finding the area under the force vs. position graph, detailing their calculations for different segments. Some participants question the original poster's understanding of work-energy relationships and suggest considering the work-energy theorem.

Discussion Status

The discussion is ongoing, with participants exploring the original poster's calculations and questioning their familiarity with relevant concepts. Guidance has been offered regarding the work-energy theorem, but no consensus or resolution has been reached yet.

Contextual Notes

The original poster expresses uncertainty about their skills related to the work-energy relationships, indicating a potential gap in understanding that is being addressed in the discussion.

Bdb1331
Messages
2
Reaction score
0
The diagram shows how a force along the x-axis varies with location. A block of mass 7.20 kg moves along the x-axis on a horizontal frictionless surface, influenced by this force. (Graph is attached)
A) How much work is done by the force as the block moves from the origin to 8.0 m?

B) If the block has a velocity of 3.80 m/s at the origin, what is its velocity at 8.0 m?

For part A, I worked out the work to be 33 J by finding the area under the curve...
in the first two seconds

W = F(avg)d
W = [(2 + 8) / 2](2 - 0)
W = 10 J

from 2 to 4 seconds

W = [(8 + 4)/2] (4 - 2)
W = 12 J

from 4 to 6 s

W = 4(6 - 4)
W = 8 J

from 6 to 8 s

W = [(4 + -1)/2](8 - 6)
W = 3 J

Total work

W = 10 + 12 + 8 + 3
W = 33 J

But I am sort of lost when it comes to B)
 

Attachments

  • Screen Shot 2014-10-27 at 12.55.37 PM.png
    Screen Shot 2014-10-27 at 12.55.37 PM.png
    7.8 KB · Views: 557
Physics news on Phys.org
Are you familiar with work-energy relationships?
 
PhanthomJay said:
Are you familiar with work-energy relationships?
Sorta, but not exactly confident in my skills
 
Bdb1331 said:
Sorta, but not exactly confident in my skills
Well, just try using the work - energy theorem. What does that theorem say?
 

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
7
Views
2K
Replies
7
Views
2K
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K