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Negative work

  1. May 29, 2013 #1
    I have just read a thread here:https://www.physicsforums.com/showthread.php?t=382031
    Here is part from the thread.
    Please see the bold sentence in the quote. In this case, does it mean that my force(push) is larger than the push force from fixed charge. If so, how can I calculate the work that I used to push a charge from infinity to a specific point?
     
  2. jcsd
  3. May 29, 2013 #2
    In the context of the post you quoted from, the charge has some finite velocity in the direction you're pushing it (let's assume its movement is 1-dimensional in spacial coordinates):
    (1) If your applied force was greater in magnitude than the Coulomb force on the charge, it would accelerate in the direction of your applied force.
    (2) If your applied force and the Coulomb force were equal in magnitude, the charge would remain in motion at a constant velocity.
    (3) If your applied force was less in magnitude than the Coulomb force on the charge, it would accelerate in the opposite direction of your applied force.

    (1) and (2) would let you displace the charge from its initial position to a point arbitrarily close to the charge you're moving towards.

    In general, you'd calculate it using a path integral. Do you have a good book that treats subjects like electric fields and potential?
     
  4. May 29, 2013 #3
    Thanks, I read some books calculating the work. In all case, they say that the force that I push the charge and Coulomb force are equal as your #2 case above. Then they use path integral from A to infinity of Coulomb force.
    But what I on't understand is that as in your #2 case, two forces are equal and the charge move with constant velocity. How the charge get the velocity first? Because the charge starting at rest, when it changes to a constant velocity, the two forces are no longer equal during this time and the work done calculating using the condition two forces are equal is not right.
     
  5. May 29, 2013 #4
    In the post above, the postulate is that the charge starts with some finite velocity, not necessarily at rest. In an arbitrary Newtonian inertial reference frame, where Newton's laws are valid, every object with 0 net force on them has some constant velocity, which may be 0 if you choose to work in that frame. In case #2, since there is 0 net force on the charge, the charge remains at its initial velocity, which may be "at rest", if that is the frame you choose to work in.
     
  6. May 29, 2013 #5
    Ah ok, I think I know the examples you're talking about. I suspect they're all purposed for deriving an expression for the potential energy of a system that has some conservative field associated with it.

    In your examples, they assume that the object being displaced starts out with some velocity and doesn't accelerate, which means the applied force and the conservative force acting against it are always equal in magnitude. That way, the system has no change in kinetic energy, so the work done on the system must equal its change in potential energy (assuming no nonconservative forces act within it etc.). It's then straightforward, since they have an expression for the conservative force, to derive an expression for the potential energy of the system.

    Now, that's just a way to derive an expression. It doesn't mean the applied force and the conservative force always have to be equal in magnitude and opposite in direction.

    It doesn't matter that they're not equal in magnitude at some point in time. You can still calculate the work done just fine. An example:

    I have a book lying on the ground - everyone likes an example with a book and gravity. I apply a constant force of 1 million newtons in an upright direction. At an altitude of 1 km (I can fly btw.) I have done work equal to 1*106*1000 J. That's a lot more than the magnitude of the work done by gravity, but I'm pretty sure the book is also approaching lightspeed.
     
  7. May 29, 2013 #6
    Thanks for helpful answer.:smile:
    I feel relieved knowing that. And I have just found this (which was confusing me a lot):
     
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