What is the final position of the object?

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Homework Help Overview

The discussion revolves around determining the final position of an object given its initial position and the work done on it. The context involves concepts from mechanics, particularly work and force.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between initial position, work done, and force direction. Questions arise regarding the strength and direction of the force acting on the object and how it affects the work done.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about force direction and the implications of the work done. Some have proposed potential final positions based on their calculations, while others are seeking clarification on the reasoning behind these values.

Contextual Notes

There is a discussion about the direction of movement required to achieve the specified work done, with some participants noting the need to consider changes in force along the path. The calculations presented involve assumptions about the work being negative when moving backwards.

Angela_vaal
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Homework Statement


What is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to -0.19 J?
Walker4e.ch07.Pr036.jpg


Homework Equations



work(J)= Fx ΔX

The Attempt at a Solution


I do not know where to begin? please give me a hint where to begin.
 
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It starts at 0.4m. What strength of force acts on it there, and in which direction?
If it were then moved by the force to 0.5m, how much work has the force done?
 
haruspex said:
It starts at 0.4m. What strength of force acts on it there, and in which direction?
If it were then moved by the force to 0.5m, how much work has the force done?
work would be .08 J in positive direction
 
Angela_vaal said:
work would be .08 J in positive direction
Right. Is that going in the right direction to reach a work done of -0.19 J?
 
haruspex said:
Right. Is that going in the right direction to reach a work done of -0.19 J?

No it would not. You would have to go to the left side.
 
Angela_vaal said:
No it would not. You would have to go to the left side.
Right, so go to the left, but stop when you get to a change in the force and see how the total work is going.
 
haruspex said:
Right, so go to the left, but stop when you get to a change in the force and see how the total work is going.

I believe it is either between .12m or .11 m? Am i right?
 
Angela_vaal said:
I believe it is either between .12m or .11 m? Am i right?
Do you mean between x=.11m and x=.12m, or between .11m and .12m left of the 0.25m mark?
Either way, please post your working.
 
haruspex said:
Do you mean between x=.11m and x=.12m, or between .11m and .12m left of the 0.25m mark?
Either way, please post your working.

from x=.40m to x=.25m work is= .12J
from x=.25m to x=.13m work is = .072 j

total work =.192 J. since I am working backwards I am assuming the answer is negative.

Therefore, my final position would be .13 m?
 
  • #10
Angela_vaal said:
from x=.40m to x=.25m work is= .12J
from x=.25m to x=.13m work is = .072 j

total work =.192 J. since I am working backwards I am assuming the answer is negative.

Therefore, my final position would be .13 m?
Yes.
 
  • #11
haruspex said:
Yes.

Thank you for the help! :)
 

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