Solving for final position with a negative work value

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The discussion revolves around calculating the final position of an object given an initial position of 0.50 m and negative work of -0.29 J. The user initially assumed that negative work required a negative force, but was corrected to understand that displacement can be negative even with a positive force acting in the opposite direction. It was clarified that the object could move in the negative x direction, resulting in negative displacement despite the force being positive. The conversation emphasized the importance of considering all forces acting on the object to accurately determine the final position. Understanding these dynamics is crucial for solving the problem correctly.
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


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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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