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Work and weight

  1. Apr 1, 2015 #1
    Everybody experiments fatigue holding a weight, and almost everybody knows that points of applications of the involved forces don't move.

    We also know that we cannot use the standard equation of the conservation of energy ( ΔK + ΔU = Wext ) because the system (Body+weight) is composed by objects that have an internal structure.
    (Here K stays for Kinetic Energy, and U is potential energy of the inner forces, supposing they're conservatives)

    According to this observation we write, more generally : ΔK + ΔU + ΔEint = Wext

    Let's see what happens to the system (Body+Weight)

    ΔK = 0, obviously.
    Wext = 0 too, according to the fact that the the external forces (weights and reactions of the soil) don't move their point of a.
    ΔEbody<0 ; we burn calories to avoid fatigue, decreasing our internal energy.
    ΔU=0 because the system does not change configuration.

    I'm ok with the microscopic behaviour (sarcomeres contract and strecht continuously to achieve the 'rigid' stand, so they actually do work) but how it can be explained with the conservation of energy?
     
  2. jcsd
  3. Apr 1, 2015 #2

    russ_watters

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    Welcome to PF!
    Eout/Ein = efficiency
    [Edit: had it upside down]

    So, you might consider the human body, in this case, to be a 0% efficient machine. Consider the similar concept of a car using its engine and clutch to "hover" on a steep hill. The motor burns chemical energy and generates heat, but the car doesn't go anywhere.
     
    Last edited: Apr 2, 2015
  4. Apr 2, 2015 #3
    There's still something that I don't understand:

    If you consider that situation as a 0% efficient machine the whole calories burnt should convert to heat.
    But if you imagine that you lift the weight doing a certain amount of motion against gravity, you should conclude that efficiency is not equal to 0 any more. So you'r 'loss' should be lower and system itself should be colder.

    I think that sarcomeres actually do an amount of work burning calories and part of the energy is lost because of friction between fibers but the other part of the energy goes 'successfully' but I don't know how it can be quantified.

    Think about a charge particle in an homogeoneus electrostatic field. You lose energy to keep that charge at a 'fixed' potential (imagine a finger which goes on the top of the paticle); the whole energy required doesn't go into heat according to me.

    -
    On Italian wikipedia I red:
    work done by every single force in a system (partial work) is not just equal to the scalar product between force and displacementent but between force and the displacementent that body should do if only that force acts. (If that's true, you should be able to calculate the amount of energy involved to keep the weight lifted for a certain amount of time t , calculating some sort of 'virtual displacement' )

    I'm a bit confused.
     
  5. Apr 2, 2015 #4

    CWatters

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    Not necessarily. Body temperature is regulated....

    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatreg.html

    When you eat the energy you consume isn't all used to do work or make heat. Some will be excreted, some converted to mass (fat, hair, finger nails etc). It's well known that if you do more exercise then there is less spare energy available to be stored in the form of body fat.
     
  6. Apr 2, 2015 #5

    russ_watters

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    Cwatter's caveat taken into account, I suspect it is well over 99%, so yes, close enough at least.
    Yes.
    Only if the calories burned is the same.

    Off the top of my head, my exercise bike says that I burn something like 700cal/hr and generate 120 watts of mechanical power. If you do the unit conversion and divide, that's about 15% efficiency.

    If, leaning against a wall, I burn 100 cal/hr at 0% efficiency, an input of 7x more energy results in 5x more heat dissipation and the remaining one part is the mechanical output.
    It may be a language issue, but there is no such thing as "virtual displacement". Partial work is partial force times actual displacement.

    The idea of calculating a "virtual displacement" by applying F=MA to an object in space doesn't work because you don't necessarily even have an "m" to apply it to. As far as your muscles know, pushing a car with its parking brake on is the same as pushing the Empire State Building.
     
    Last edited: Apr 2, 2015
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