Solving for final position with a negative work value

In summary: So the work done by F can be negative.In summary, the object's final position cannot be determined solely based on the given information as other forces and factors can affect the displacement and work done on the object. The equation W = Fd can be used to solve for the final position, but it must be taken into account that displacement can be negative and the sign of the work done by F can be negative if the displacement is in the opposite direction of the force.
  • #1
Selayna
1
0

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:
Physics news on Phys.org
  • #2
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:
  • #3
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.
 

1. How can a negative work value affect the final position of an object?

A negative work value indicates that the force applied to an object is in the opposite direction of its motion. This means that the work done on the object is negative, resulting in a decrease in the object's final position.

2. Is it possible for an object to have a negative final position due to negative work?

Yes, if the work done on an object is negative and there are no other external forces acting on the object, it will have a negative final position. This means that the object will move in the opposite direction of the applied force.

3. How does the magnitude of the negative work value affect the final position of an object?

The magnitude of the negative work value indicates the amount of work done on the object in the opposite direction of its motion. The larger the negative work value, the greater the decrease in the object's final position will be.

4. Can a negative work value be canceled out by a positive work value to determine the final position of an object?

Yes, if there are both positive and negative work values acting on an object, they can cancel each other out. The final position of the object will then depend on the net work done, which is the difference between the positive and negative work values.

5. How can we use the equation for work to solve for the final position of an object with a negative work value?

The equation for work (W = F * d * cosθ) can be rearranged to solve for the final position (d = W / F * cosθ). By plugging in the known values for work and force, we can determine the final position of an object with a negative work value. The angle (θ) between the force and displacement vectors will also affect the final position, as indicated by the cosine function.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
216
  • Introductory Physics Homework Help
Replies
3
Views
962
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
779
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top