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    How do you think you could calculate the number of calories that you have burnt after you have gone for a run?
    How would u use physics knowledge to answer this?
    Thanks
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    (Original post by runny4)
    How do you think you could calculate the number of calories that you have burnt after you have gone for a run?
    How would u use physics knowledge to answer this?
    Thanks
    Hello,

    To answer this question, we unfortunately require a little bit more data. If we knew the average drag experienced when running at a certain, we could state the force that is required to maintain this velocity and therefore be able to calculate the work done over a given distance (as \Delta E = F \Delta x). Another method may be to use the average power output during exercise. I would estimate between 50W and 100W for running (source: https://en.wikipedia.org/wiki/Human_...ered_transport). This can then be multiplied by the time over which the person is exercising (as  \Delta E = P \Delta t ).

    Of course, both of these solutions return the answer in J, assuming SI units are used. This should then be converted to calories. I hope that this has been helpful and if I have not been clear, please do not hesitate to ask.
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    What I think they do IRL is make you run in a face mask and measure the amount of CO2 you exhale - the amount of CO2 corresponds accurately to an amount of energy.

    energy used during exercise will be greater than work done because the body isn't 100% efficient - and there's a amount of energy used doing nothing - just staying alive.
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    (Original post by Joinedup)
    What I think they do IRL is make you run in a face mask and measure the amount of CO2 you exhale - the amount of CO2 corresponds accurately to an amount of energy.

    energy used during exercise will be greater than work done because the body isn't 100% efficient - and there's a amount of energy used doing nothing - just staying alive.
    Hello,

    Yes, what you say is true. Like I have said, we need a little bit more information. The first method that you described is probably amongst the most effective anyway, however I thought that we were to be seeking a more mechanical method. You are correct in saying that the energy will in fact be greater due to a lack of efficiency, however it is still perhaps useful to obtain a lower bound. To conclude, I think that we should stick to the traditional methods; however there was nothing wrong with trying something different .
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    (Original post by Smithenator5000)
    Hello,

    To answer this question, we unfortunately require a little bit more data. If we knew the average drag experienced when running at a certain, we could state the force that is required to maintain this velocity and therefore be able to calculate the work done over a given distance (as \Delta E = F \Delta x). Another method may be to use the average power output during exercise. I would estimate between 50W and 100W for running (source: https://en.wikipedia.org/wiki/Human_...ered_transport). This can then be multiplied by the time over which the person is exercising (as  \Delta E = P \Delta t ).

    Of course, both of these solutions return the answer in J, assuming SI units are used. This should then be converted to calories. I hope that this has been helpful and if I have not been clear, please do not hesitate to ask.
    how would you estimate the power output during exercise if u didn't have the internet to look it up?
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    (Original post by runny4)
    how would you estimate the power output during exercise if u didn't have the internet to look it up?
    Hello,

    Of course, if I was feeling somewhat snide, I would say 'use a book'. I do however understand your complaint. As I have said in my reply to Joindedup, it is probably best to estimate energy output by measuring CO2 output. If we are looking for the output from the exercise alone, we could of course use some sort of exercise machine to show the mechanical output of a particular person. This does however have its own associated problems. I'm sorry that my responses haven't been all that helpful.
 
 
 
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