Solving for final position with a negative work value

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SUMMARY

The discussion focuses on calculating the final position of an object subjected to a negative work value of -0.29 J, starting from an initial position of 0.50 m. The user initially attempted to solve the problem using a modified equation that assumed negative force values, leading to an incorrect final position of 1.7 m. Key insights reveal that displacement can indeed be negative, and the force acting on the object is positive, indicating that other forces may be influencing the object's movement in the negative direction.

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Selayna
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Homework Statement



An object is acted on by the force shown in the picture shown below:

07-16.gif

What is the final position of the object if its initial position is x = 0.50 m and the work done on it is equal to -0.29 J?

Initial position = Xi = 0.5 m
Work = W = -0.29 J
Force = F
displacement = d
Final position = Xf = ? m

Homework Equations



W = F(d - Xi) + F(d-d) + F(Xf - d)
W = Fd

The Attempt at a Solution


[/B]
I worked on the first part of this problem, where work was equal to 0.12 J and the initial position was 0.5 m, using this equation:
0.12 = 0.8(0.5 - 0.5) + 0.4(0.75-0.5) + 0.2(Xf - 0.75)

I assumed, since W = Fd, and displacement can't have a negative value, Force would have to have a negative value to get a negative value for Work. Using this logic, I modified the original equation to use a negative Force:
W = -F(d - Xi) - F(d-d) - F(Xf - d)

Using this equation, I went through these steps:
1. Plug in values
-0.29 = -0.8(0.5 - 0.5) - 0.4(0.75-0.5) - 0.2(Xf - 0.75)
2. Reduce
-0.29 = 0 - 0.1 + 0.15 - 0.2Xf
3. Subtract 0.15 and add 0.1 to both sides
-0.34 = -0.2Xf
4. Divide both sides by 0.2
1.7 = Xf

However, the answer I got for the final position (1.7), was incorrect. I'm assuming my initial logic using negative values for force was incorrect, but I'm not sure how else I would solve the problem with a negative work value.

Any help is appreciated. Thanks!
 
Last edited:
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Hello, and welcome to PF!

The force values are all positive as indicated in the graph. This means that the force is always directed toward the positive x direction.

You mentioned that you thought that displacement cannot be negative. But this is not true. You may assume that the object can move in either direction along the x axis. For example, the object might have been given an initial velocity in the negative x direction so that it will move toward negative x even though there is a force acting toward positive x. (Similar to throwing a ball upward. The ball travels upward while the force of gravity acts downward.)

If a force acts in the positive x direction while an object moves toward the negative x direction (i.e., negative displacement), what is the sign of the work?

Think about how the object can start at x = 0.50 m and move such that the work done on the object is negative.
 
Last edited:
Selayna said:
displacement can't have a negative value
It starts at 0.5m. If at some later time its position is < 0.5m then that's a negative displacement.
The only way I can make sense of the question is that there are other forces acting on the object, and these can overcome the force F to result in the object's displacement being opposite to F.
 

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