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What is the final position of the object?

1. 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


2. Homework Equations

work(J)= Fx ΔX
3. The Attempt at a Solution
I do not know where to begin??? please give me a hint where to begin.
 

haruspex

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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?
 
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
 
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.
 

haruspex

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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.
 
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?
 

haruspex

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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.
 
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?
 

haruspex

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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.
 

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