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Work = Energy?

  1. Aug 21, 2006 #1
    Hi,
    Now, we know that the work done on a system equals the change of energy of that system, amd if there is no work, then the energy change must be zero.
    Now, suppose if I hold a stone by a string and let it drop while still holidng
    it so that it falls at a constant speed to the ground.
    My questions are:
    1. Where did the potentail energy go here? it is not converted to kinetic energy as the speed is constant. To what other form of energy did it get converted to?
    2. The net work done on the object is zero because the net force is zero, so there can't be any net work. The work of gravity equals my work, the net is zero work. If the work is zero, how could the energy of the stone decrease in the time no net work is applied on it?

    If the gravity does positive work and the potential energy decreases, then what about my negative work?
     
  2. jcsd
  3. Aug 21, 2006 #2

    Integral

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    Did you really do this experiment? If so what did you feel as the string slides between your fingers? Did it not get warm? There is the missing work. Your fingers are doing work on the string. If the velocity is constant you are exerting a force on the string exactly equal and opposite in direction to the force of gravity, thus no net force and no acceleration.
     
  4. Aug 21, 2006 #3
    ok...that seems a reasonable answer :)
    but what about the stone. It does not experience a net work, but its energy state is reduced.
    The decrease of an energy must be brought about by an energy crossing or passing/leaving the boundary of the system by means of an energy transfer mechanism, like the work.

    thanks.
     
  5. Aug 21, 2006 #4

    Integral

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    The stone, the string and your fingers are the system, so I am not sure just what your question means.
     
  6. Aug 21, 2006 #5
    ah..I see..the problem is in assigning the system.
    So, if all of these compose the system, and the energy lost as potential energy is gained as heat, then there is an "internal transfer" of energy between 2 components of the system.
    I was trying to tackle the problem by assuming the system as the stone only, and fingers and hands and gravity are outside acting agents on the system. In that case, the net work on the system(stone)is zero, but the system still losing its potential energy without net acting force taking this energy out.

    I think however you assign the system, the physics should be right on the system you choose, whether you include some objects or not.
    I suppose my assignation was not quite right because I considered the stone ONLY in my system (having ititial potential energy), and regarding gravity as outside the system. I turns out this wrong because when I assume the system has a potential energy, that potential energy is not separable from the earth(gravity). Therefore, I can't say there is a potential energy inside the system and gravity outside it.

    Therefore, the situation becomes; the system is the stone and the gravity(earth) having an initial potential energy, and the work done by my hand converted the potential energy into heat...THERE is a net work here.

    sorry for writing alot and repeating myself.
     
  7. Aug 21, 2006 #6
    The way you define your system only affects your bookkeeping, not the outcome. Proper system boundaries allow you to get the answers to the questions your asking, in other words gives you information that is useful.

    As for you title: work = energy
    Energy is the ability to do work. Doing work on a system converts one type of energy into another one, the total energy being conserved. The rate of change of a type of energy being converted is the work.
     
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