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An athlete pumping iron

  1. Jul 24, 2013 #1
    An athlete needs to lose weight and decides to do it by "pumping iron". How many times must an 80 kg weight be lifted a distance of 1 m in order to burn off .5 kg of fat, assuming that that much fat is equivalent to 3500 Cal?

    [itex]Q = n\left(m\cdot g \cdot h\right)[/itex]
    [itex]3500\,10^3\,4.18=n\left(80\,9.8\,1\right)[/itex]
    [itex]n = 18660.7 \text{ times}[/itex]

    That was my book answer (as I imagined), but wouldn't it be more correct to assume that the athlete would use energy (the same energy) to lift iron up and then to support it down? Thus he would only need to lift it [itex]N = \frac{n}{2} = 9330.36[/itex]

    I don't mean to be pedantic or anything like that, I'm genuinelly curious to know if it's physically correct to assume that.
     
  2. jcsd
  3. Jul 24, 2013 #2
    There is a difference between physical work done and actual calories burned. The question is clearly asking you to ignore all efficiency effects Additional calories are burned for stabilization, returning the weight to the start position and even rebuilding muscles and recovery processes later.

    Also, the question is worded to ask only how many times must the weight be lifted. It says nothing about requiring the athlete to lower the weight himself (or herself). This is a natural assumption you made because you know that's what typically is done.
     
  4. Jul 25, 2013 #3
    Right. I agree and understood. But considering a real-life situation, would it be plausible? Or is the fact that it's a constant deceleration (decelerating the iron until initial point, I mean) somehow resultant in smaller work done?
     
  5. Jul 25, 2013 #4
    It's hard for me to say. I think we need someone knowledgeable about physiology of the muscles to answer that. From a physics point of view, gravity is doing work when we lower the weight, and the person is just restricting the rate of work done (power). But, obviously we know we burn calories in this process, so that's not a good view to take.

    My guess is that the (internal) work (or calories burned) is smaller, but then again, this type of "negative resistance training" is often thought to build muscle even more effectively than the positive work direction of the exercise. So, I can't say for sure. I would say that the longer you take on the down stroke, the more calories are burned.

    I know from personal experience that actual calories burned from weight training far exceeds the estimate given by a physics calculation like this. Seriously, who does tens of thousands of repetitions with weight training? :smile: I find weight training more effective than cardiovascular exercises for losing weight. So, something very significant happens with physiology that goes way beyond the physics calculation we are talking about.
     
    Last edited: Jul 25, 2013
  6. Jul 25, 2013 #5
    From a physiology point of view, calories are burned just holding the weight in a constant heigh. Clearly there's no work being done to the weight in this situation (you're not adding gravitational potential energy to it) but there's some muscle exertion. There's some work being done, but it's at the cellular level and has to do with how muscle fibres work. I think this is what you're is getting at.

    That said, it's not correct to equate the work done to raise the weight (adding gravitational potential energy to it) and the "work" (chemical processes in the muscle) done when lowering the weight. For example, you can imagine a situation where he raises the weight and lowers it over a very long time. The energy expended lowering it can be a large amount - maybe he takes an hour, or a day. The energy expended lowering the weight is variable. Really, the energy expended in this second way (lowering the weight) is outside the scope of the question. This is an annoying thing about these questions: you can't think too hard!

    I think question is asking you to imagine this scenario: the lifter raises the weight with perfect efficiency and lets it fall to the floor.
     
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