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I Question about the physics term "work"

  1. Dec 23, 2017 #1
    I have a question about work. If Newtons 3rd law of motion states there is an equal and opposite reaction for any force. Now if for example I apply 500 newtons of force on my dresser and move it 5 meters I used 2500 joules of energy. Now because the dresser is equally putting that much force in return for the same distance does this also mean the dresser used 2500 joules of energy resisting my force? So I used 5000 joules of energy?
     
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  3. Dec 23, 2017 #2

    QuantumQuest

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    Initially the dresser is at rest. Now, when you exert a force on it, it won't get into motion till the force you exert overcomes the resultant reactive force. From this point on the dresser is in motion - i.e. the dresser does not exert an equal force anymore, in the direction of the force you exert and the work you consume becomes kinetic energy minus the work that is consumed by the resultant reactive force (i.e. friction). It is always a good idea to draw a two-axes diagram in order to see what are the forces in both axes and see it in more detail.
     
  4. Dec 23, 2017 #3

    russ_watters

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    Kind of. It applied that force to you in the opposite direction you did. But it also applied a force in the same direction, against the ground (friction).
    Huh? You just said you used 2500N (correctly)!
     
  5. Dec 23, 2017 #4
    You did 2500 J work on the dresser, and the dresser did -2500 J of work on you. So??
     
  6. Dec 23, 2017 #5

    sophiecentaur

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    It’s better to include the Momentum change. Pushing something with your feet firmly on the floor involves a near zero movement of your feet / Earth. So Force X Distance is Zero.
    Momentum is the only conservation law that you can lean on in this kind of argument.
     
  7. Dec 23, 2017 #6
    Are you saying that what I said is incorrect, to wit: the work that the dresser does on you is equal in magnitude and opposite in sign to the work you do on the dresser?
     
  8. Dec 23, 2017 #7

    sophiecentaur

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    If you consider where the Energy ends up then there is no reason to suppose that the Energy from the muscles should be shared equally.
    You will say it’s all a matter of reference frames, no doubt. But Newton 3 only refers to force and not work.
     
  9. Dec 23, 2017 #8
    I don't follow. Isn't the work I do on the the dresser equal to force I apply times the distance over which I apply it? And isn't the work that the dresser does on me equal to the force that it applies on me times the distance over which the force is applied (with a minus sign because the force and displacements are in opposite directions)? We seem to be talking about two different things.
     
  10. Dec 23, 2017 #9

    sophiecentaur

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    Yes, I think we are. The strict definition of Work, which is Force times Distance would suggest that you are right but the share of the energy is not equal. Perhaps it's just a bad idea to try to resolve this question. The change in energy of two objects, mutually repelling is not necessarily equal so saying that the Work Done is equal is not really relevant. It raises a question without answering it satisfactorily. The very fact that we (and so many people in history) have this conversation from time to time is because there is a perceived paradox and it makes people uneasy.
    The earlier posts in which the Energy is discussed show the problem of going down this particular road because it leads to implications about Energy Conservation that are not true.
     
  11. Dec 23, 2017 #10

    Mister T

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    The problem occurs when considerations of work done are generalized to considerations of energy. There is nothing wrong, for example, with saying that 2500 joules of work is done by the boy on the dresser, which is of course equivalent to saying that -2500 joules of work is done by the dresser on the boy. This is nothing more than saying that when 2500 joules of work is done on the dresser, -2500 joules of work is done by the dresser.

    Problems occur when we make statements about energy transfer, like saying that 2500 joules of energy was transferred to the dresser by the boy. Such a statement is valid only if the dresser can be modeled as a particle, unable to absorb internal energy and undergo a temperature change. For example, if the dresser moves at a steady speed the 2500 joules of work done by the boy on the dresser may not constitute a transfer of 2500 joules of energy to the dresser, some of that 2500 joules may be transferred to the dresser, increasing its temperature, but the rest may be transferred to the floor raising its temperature. And statements such as the work done "goes into heat" are also incorrect, as in this case there is no transfer of heat, the process is adiabatic.
     
  12. Dec 23, 2017 #11

    jbriggs444

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    The work done by each of two objects on the other when interacting with a contact force (e.g. a normal force) where the contacting surfaces are not in relative motion will always be equal and opposite.

    Kinetic friction and non-contact forces present opportunities for the two energy changes to differ.
     
  13. Dec 24, 2017 #12

    sophiecentaur

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    We sort of dealt with that. Whilst it is numerically true, it does not imply anything about the Energy.
    Conserving Momentum is sacrosanct and both the contact times and distances are the same. Work Done is not always an important quantity and it isn't particularly relevant in many circumstances. Friction and "non contact Forces" do not affect the general principle.
    Work done is fine for ' one sided' processes like a ballista or a gun but I wouldn't know where to start to do an impact calculation by replacing Impulse with Work. I wouldn't even like to do the sums on an artillery shell by choosing the shell's reference frame, despite the fact that Work can be calculated in both directions.
     
  14. Dec 24, 2017 #13

    jbriggs444

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    It does not imply anything. It states it flat out. Work is a transfer of energy.
     
  15. Dec 24, 2017 #14

    sophiecentaur

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    Yes, you are stating something that is more like a mathematical identity which needs some context to have physical meaning. "Transfer of Energy" from what to what? Starting with a coiled spring between the Earth and a stone, where does the energy end up? Your answer must satisfy momentum conservation.
    We are still arguing in quadrature. :smile: The fact that the question is asked so often, serves to demonstrate that the Work Done idea is not always very helpful and needs to be helped along.
     
  16. Dec 24, 2017 #15

    jbriggs444

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    One can examine the energy transferred by work at each interface.

    Pick a reference frame. The frame in which the Earth is initially at rest is convenient.

    The interface between spring and Earth is [at least approximately] immobile. No energy is transferred between the two. The interface between spring and stone moves in the direction of the force on the stone. Energy is transferred to the stone. The stone gains energy from the interaction. Energy is transferred from the spring. The spring loses energy from the interaction.

    If you prefer to use a different frame, that can be done. It turns out that the net energy loss in the spring is a classical invariant. The energy changes in Earth and stone are not invariant, but will sum to a figure that is invariant and is equal and opposite to the energy change in the spring.
     
  17. Dec 24, 2017 #16

    sophiecentaur

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    That's the one I would choose.
    Absolutely. Always.
    The point I am making is that there is confusion about this. If, like you, people follow the Maths scrupulously, then the conclusion is what it is. The bare facts of Work being done both ways is just not helpful for anyone who's mind strays from the 'literal'. Rather than insisting that the definition is what it is, why not help the situation along if it doesn't actually involve introducing an error? What I am introducing is no more than stating the sort of problem that rocket engines encounter at low speeds. All the energy goes into the ejecta and virtually none goes into the rocket at the start.
     
  18. Dec 24, 2017 #17

    A.T.

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    Look at the exact vector definition of work, via the dot product: Since force and displacement are opposite, the work done by the dresser on you is negative.
     
  19. Dec 24, 2017 #18

    A.T.

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    Not really what the OP asks about.
     
  20. Dec 24, 2017 #19

    jbriggs444

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    I think I understand your point better now. At the risk of oversimplifying: If the concept of work is confusing, there are two possible approaches. One is to clarify the confusion. The other is to use a different concept.
     
  21. Dec 25, 2017 #20

    Mister T

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    This is a perfectly valid dynamical relation.

    Suppose a person on roller skates launches himself into motion across the floor by pushing on a wall. This would seem to qualify as an example of what you mention above, but in what way are energy changes equal?
     
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