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Work question 2

  1. Aug 2, 2009 #1
    Hello, I got another inquiry about work. Supposing an external agent gives potential energy to an object by pulling it up for instance, he's doing work equal to:

    [tex]W=mgh_{f} - mgh_{i}[/tex]

    Now, at the same time the earth is doing the same work, only negative. Only, when the object is let go, it still possesses the same energy obtained by the external agent, without subtracting the work done by the earth. Why is that? How does the work done by the earth at that time manifest?
    Thanks for the stellar help.
     
  2. jcsd
  3. Aug 2, 2009 #2

    Doc Al

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    Staff: Mentor

    Here's how I would put it. If an agent lifts an object with a force equal to the object's weight, it does work on the object equal to mghf - mghi. By doing this work against gravity, the gravitational PE is increased. If you look at gravity as another force, then the work done by gravity is negative of the work done by the agent: Thus the change in KE is zero, no net work was done.

    Either consider the work done by gravity or consider the change in gravitational PE. But not both, since they are just two ways of saying the same thing.
     
  4. Aug 2, 2009 #3
    OK, so the work done by gravity is actually manifesting in the fact that no KE is gained by the object?
    If no net work is done though, how can it be that there's energy gained by the object?
     
  5. Aug 2, 2009 #4

    Doc Al

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    A conservative force, like gravity, can be represented by a potential energy. But in accounting for energy you either consider the work done by gravity or the change in gravitational PE. Not both!

    Using gravitational PE: The agent does work on the system, increasing its energy. That energy is gravitational PE.

    Using work done by gravity: The agent does work; gravity does negative work. There's no change in KE, since the net work is zero. (No concept of gravitational PE here.)
     
  6. Aug 2, 2009 #5
    OK, I think I got it then. So, when considering an agent increasing the potential energy you disregard the work done by the conservative force, and when considering the work done by the conservative force, you take a look at how the kinetic energy changed.
    Thanks for clarifying!
     
  7. Aug 2, 2009 #6

    Doc Al

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    You are, of course, free to look at the change of kinetic energy whether or not you use gravitational PE. (In this example, it's zero no matter how you look at it.)
     
  8. Aug 5, 2009 #7
    Its something like your income(work done by the applied external force) and bank deposit(gravity). When the object stays at rest before and after being lifted through a height h, its like your entire income had been invested in the bank; which you could later collect in the absence of income(increase in KE with only gravity pulling the object down). Recession effect! :smile:
     
  9. Aug 5, 2009 #8
    Are you asking about the reaction of the Earth as the object is lifted? In the inertial frame of the object and the Earth, the center of mass of the object and the Earth is motionless. So m dx = - M dX.
     
  10. Aug 5, 2009 #9

    rcgldr

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    Imagine a very large massive plate creating a constant gravitational field, and a very large charged plate creating a constant electrical field. The intensity of the fields are equivalent, such that an electron will move at constant velocity between the plates. Say the electron is moving at constant velocity away from the gravitational plate towards the charged plate. How much work is being peformed per unit time?
     
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