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Does a person burn calories when lowering a weight?

  1. Apr 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A person exercises daily by lifting weights.
    When he lifts a weight, he is exerting a force on it to displace it. This means he is doing work and therefore burning calories. But when he lowers the same weight, does he do any work and burn calories?

    2. Relevant equations
    W = F x d
    W is -ve when F and d are in opposite direction
    W is +ve when F and d are in same direction

    3. The attempt at a solution

    When weight is lowered

    When weight is lowered, he still applies some force on the weight and therefore does work. But not sure if he burns calories.
    The person exerts an upward force on the weight while it's displacement is in a downward direction, which means that work done by this force is negative.
    I cannot understand how negative would mean he had to supply energy.

    When weight is raised
    When he raised the weight then force applied by him was in same direction as displacement and therefore work done on weight by him was positive which meant he had to supply the energy i.e. calories to do this work.
     
    Last edited: Apr 10, 2017
  2. jcsd
  3. Apr 10, 2017 #2

    haruspex

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    Muscles are not simple machines. Even to hold a weight steady takes work, all ending up as heat. I've not been able to find anything on the web that goes into this that's not behind a paywall.
    When it comes to lowering a weight, it clearly depends on how it is done. Lowering it slowly will cost effort, while dropping it suddenly does not.
     
  4. Apr 10, 2017 #3
    So, it doesn't matter even if work done is negative when lowering it slowly. Negative work could still mean work done by the person? Negative work when lowering indicates potential energy is being converted to work, so work should not come from energy of the person. I am thinking that Energy⇒Work or Work⇒Energy or Energy⇒Energy, but not sure if this is correct thinking. While lowering PE⇒Work, so there is no need of Person's Energy⇒ Work. (⇒ stands for converted to)

    I thought for energy to be spent by a person, he must do positive work on the weight and then that work would come from somewhere i.e. from the person lifting it.
     
    Last edited: Apr 10, 2017
  5. Apr 10, 2017 #4

    PeroK

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    There's not really such a thing as "negative" work in this context. If the weight is being raised and lowered at constant speed, then the situation is very similar: in both cases the muscles are applying a force equal to gravity to maintain the weight's constant speed.

    In raising the weight, the muscles must apply an initial extra force to get the weight moving, but can ease off at the top, as gravity slows the weight. Lowering the weight, gravity gets the weight going, but the muscles must apply the extra force at the end to stop the weight.

    The work-energy theorem is a red herring in this problem.
     
  6. Apr 10, 2017 #5

    haruspex

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    Yes, but while raising it the force and displacement are in the same direction, so clearly work is being done. But while lowering it the displacement and force are opposite, so how much work is the muscle doing now?
     
  7. Apr 10, 2017 #6

    PeroK

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    Yes, that's a good point. The "work" in this case must be internal in order to transform the PE/KE of the weight into heat of the muscles. It's a good question, therefore, how many calories are required to do this. So, we're back to biomechanics again!
     
  8. Apr 10, 2017 #7

    haruspex

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    Even if the person does work while lowering the weight, it is much less than distance x weight.
    Whatever work the person does, plus the work done by the lowered weight, goes to heat.
     
  9. Apr 10, 2017 #8
    So, PE==>Work==>Heat, when weight is being lowered?
    It seems true in classical mechanics that if we find mechanical energy i.e. KE or PE being converted to work, then that work must eventually get converted to another form of energy.
     
    Last edited: Apr 10, 2017
  10. Apr 10, 2017 #9

    PeroK

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    I think the difficulty with this problem is how to associate work being done with calories being burned. This, the more you think about it, is almost entirely biomechanical.

    If we first look at a mechanical system using springs instead of arms. Energy is needed to get the weight into the starting position, after which the weight bounces up and down on the springs. Note that no additonal energy input is needed, to move the weight up or down. In this sense, positive and negative work is not relevant. The only energy input is to replace energy dissipated. The mechanical system is not burning calories once started.

    Whatever is true of a human lifting a weight must make assumptions about biomechanics. It's an obvious assumption that calories must be burned to lift a weight. But, for repeated lifts, that assumes that little or no energy can be stored usefully in the muscles. An obvious assumption, perhaps, but still an assumption. Is it easier, for example, if you let the weight "bounce" on compressed muscle at the bottom?

    In terms of lowering a weight, it's possible that few calories are burned. It's also possible that a lot of calories are burned. But, there is no way to determine this from the external work-energy theorem. For example: holding a weight at the top of a lift requires no energy (and is not very tiring). However, holding a weight with bent arms is very tiring (although no external work is being done). Are we burning calories to do this or are the muscles tiring not a sympton of energy usage but of something else?

    Experimentally, it appears that you "work" your muscles lowering a weight. But, how this translates to calories being burned is almost entirely biomechanical.
     
  11. Apr 10, 2017 #10

    CWatters

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    Humm....

    Presumably you get hotter on the way down than on the way up because on the way down the body also has to dissipate the PE of the weight. (Eg the body doesn't have regenerative braking).
     
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