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Question about calculating work done

  1. Feb 6, 2012 #1
    We all know that work done corresponds to energy used and calculated as force*distance
    So if we are pushing against a wall, it does not count as work done, as there is no motion
    However, aren't we still wasting energy when we are pushing against a wall? Since a certain amount of energy is being expended shouldn't it count as work being done?
    For example, if we pass current through a resistor, there is no motion, still energy is used up as heat for the resistor. Shouldn't this energy expended be counted as some equivalent work being done?

  2. jcsd
  3. Feb 6, 2012 #2


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    Homework Helper

    SHort answer - NO.

    Work is Force*displacement [dot product of the two vectors] The dot product means that if the two vectors [force and displacement] at perpendicular, then no Work is done.

    Note that if you swing a 10kg mass in a circle, the force you apply [centripetal Force] is at all times perpendicular to the motion - especially if you use a string - so no Work is done there either. You will certainly get tired, but the circling mass will not gain Kinetic Energy: is travels at constant speed in a circle.

    Many people go to work every morning, then come home every evening, but, depending on their field of employment, may not do any Work at all - in the sense that we define it in Physics.
    Notice that I called it Work, not work [capital letter].
  4. Feb 6, 2012 #3
    so we should say that work requires energy, but use of energy does not always imply work
  5. Feb 6, 2012 #4
    Mechanical work is not the only way to expend energy, but your muscles are probably doing work. They are still contracting (so the cells are moving) and they are creating heat which is moving particles.
    I think that in modern physics the idea of energy trumps the idea of work. Usually "mechanical work" is what the macroscopic [itex]\int F\cdot\vec{dx}[/itex] work is called. At a molecular and atomic level that definition becomes meaningless.
    If you define Work as macroscopic mechanical work (which is how my profs use it) then you are correct. I think the use of work to define energy is a historical artifact as well as a pedagogical tool. But I may be wrong.
  6. Feb 6, 2012 #5


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    Homework Helper

    I note you have not referred to Work in your statement, merely work - which has far reaching "general-population" implications.
    Remember; Work is ours, and has a very specific interpretation.
  7. Feb 7, 2012 #6
    Work is spelled with a lower case "w", not Work; as this is a physics forum, I would assume most of the posters here know that when talking about work, the physics definition is implied.
  8. Feb 7, 2012 #7
    I believe Work represents the energy that is transferred through distance,not the energy is used.
    In the case of pushing a wall although you use energy...you don't transfer energy to the wall hence you don't do Work.
    The Work is differentiated from energy expenditure by the factor of efficiency(e=W/E).In wall pushing you have zero efficiency.
  9. Feb 7, 2012 #8
    Thanks to everyone for the informative replys
  10. Feb 7, 2012 #9
    Work is being done but not on the wall.The work is internal and due to the actions required of the muscles in order to do the pushing.
  11. Feb 7, 2012 #10


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    Staff: Mentor

    Correct. Energy in doesn't usually equal work out due to losses: it is always less. You've just discovered the concept of efficiency.
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