What is the final position of the object?

In summary, the final position of the object would be 0.13 m if its initial position is 0.40 m and the work done on it is -0.19 J. This is determined by calculating the total work done in the reverse direction, which is found to be -0.192 J, and then finding the position at which this amount of work is reached, which is 0.13 m.
  • #1
Angela_vaal
59
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


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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  • #2
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?
 
  • #3
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
 
  • #4
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?
 
  • #5
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.
 
  • #6
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.
 
  • #7
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?
 
  • #8
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.
 
  • #9
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! :)
 

1. What is the difference between final position and displacement?

Final position refers to the exact location of an object at the end of its motion, while displacement is the straight-line distance between the initial and final positions of an object. Displacement takes into account the direction of motion, while final position does not.

2. How is final position calculated?

Final position is calculated by adding the initial position of the object to its displacement. The displacement can be determined using distance-time graphs, velocity-time graphs, or equations of motion.

3. Can the final position of an object be negative?

Yes, the final position of an object can be negative if the object has moved in the opposite direction of the initial position. This is known as a negative displacement.

4. Is final position affected by the path taken by an object?

No, final position is not affected by the path taken by an object. It only depends on the initial position and the displacement of the object.

5. How does the final position of an object change with time?

The final position of an object can change with time if the object's velocity changes. The object's position will increase if it is moving in the same direction as its velocity and decrease if it is moving in the opposite direction. The rate of change in position is determined by the object's velocity.

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