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B Total work/effort/energy spent to climb a hill

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  1. Mar 10, 2016 #1
    https://scontent.fath3-2.fna.fbcdn.net/hphotos-xtf1/v/t1.0-0/s526x296/12804814_10208727068471849_3639991170291956856_n.jpg?oh=a60b7b49070602e57e99d4dfdf3545dd&oe=57640241
    Yesterday my bf asked me to calculate the calories he burns to go up a hill.
    So of course the first thing I thought was to take the difference of the potential energy to estimate that work and so I said that work equals W=mgh, so for a person of 60kgr, assuming that g=10 m/s^2 and h=200m I said that work is W=120.000 J. So then I thought that this calculation is only for vertical lifting of his weight but still is correct since I estimated it with the potential energy from level zero (Point A) to point B which has potential energy. But still is this calculation enough? Or should I add the work from point A to C and then add the vertical work?
    I want to know the total work needed to climb the hill and the energy consumption to move his weight from A to B.
     
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  3. Mar 10, 2016 #2
    in physics we deal with-
    work done by a force ,on a body, from initial to final pos
    mgh = work done by gravity , on bf, in climbing hill
    theres no way equations be framed for calories burnt
    suppose ur bf wears 10 sweaters on midday June
    or just a vest on midnight december ?
     
  4. Mar 10, 2016 #3
    i never mentioned clothes, weather or ground because I take them a negligible. I just want to know the work and effort to just move the weight from A to B. Please, consider metabolism activity like sweating negligible as well :P
     
  5. Mar 10, 2016 #4
    if u want to reduce problem to -
    work done by minimum force required to push a block (60 kg) up an inclined plane (vertical height 200m , hori length 3000m)
    then work by that force is (mg.sinx + umg.cosx) S
    s is slant height u coeff of friction & x is incline angle
     
  6. Mar 10, 2016 #5

    Drakkith

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    You're right in that just calculating the work required to lift his mass up the vertical distance is insufficient. The human body is an inefficient machine, and a lot of energy is lost in a variety of ways during movement. Using a random calories burned calculator I found on google, I get that he would burn somewhere around 200 calories (actually kcals) just by running 3 km horizontally. 200 kcals = 840 kJ.
     
  7. Mar 10, 2016 #6
    but in this case it takes forever for block to reach top
     
  8. Mar 10, 2016 #7

    Drakkith

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    Unfortunately people aren't blocks so you can't really use this method.
     
  9. Mar 10, 2016 #8
    loss in chemical energy is not negligible as compared to loss in total energy
    and negligible loss in chemical energy(sweating etc) would mean 0 calories burnt
     
  10. Mar 10, 2016 #9
    You did calculate the work done against gravity correctly. The horizontal motion does not contribute to work done by (or against) gravity.
    But, as other posts mentioned already, this work will give just a minimum amount of energy spent, assuming 100% efficiency of the body "machine".
    We burn calories even when doing nothing, sleeping in the bed, not mentioning when we walk on a horizontal road.
    So the actual number of calorie burnt may be several times higher than the minimum you get from just mechanical considerations.
     
  11. Mar 10, 2016 #10
    even this loss in potential energy can occur without any need of burning calories for instance in elevator
     
    Last edited: Mar 10, 2016
  12. Mar 10, 2016 #11

    russ_watters

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    There are websites that calculate calories burned when hiking that you can find via google.
     
  13. Mar 10, 2016 #12
    And do you think this is relevant for the OP?
     
  14. Mar 10, 2016 #13
    probably op must consult a heath forum not a physics forum
     
  15. Mar 11, 2016 #14
    Yeah that's right about the work against gravity. So I guess the horizontal motion only contains effort and of course energy loss. I wanted to know about work vs calories and if there is a difference. But now I see that work is not the same as effort. Because work depends only from the difference of point A and B whereas effort counts all the way (like if I go around circles and then go to point B). That's why I assumed metabolism etc negligible.
    I was just curious if I needed to know if the horizontal motion will contribute to the work from W=F dx and then the total work would be W=Fdx+mgh or if W=mgh is enough.
     
  16. Mar 11, 2016 #15

    CWatters

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    What would F be ?

    Presumably F is some notional or equivalent "force required to walk horizontally". Thats mentioned in here but I haven't read it all...

    http://www.sigmadewe.com/fileadmin/user_upload/pdf-Dateien/Bergaufgehen_engl.pdf
     
  17. Mar 11, 2016 #16
    I guess F is created by our bodies to move. So F is the effort we put to walk and is linked with the energy we consume :)
     
  18. Mar 11, 2016 #17

    PeroK

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    What about the energy consumed walking downhill?
     
  19. Mar 11, 2016 #18
    I think it would be a force of the same kind, against grtavity to hold your body from falling while in the previous case was the force to lift your weight against gravity. In a theoretical model it would be the same force but in different direction.
    Now, as far as calories are concerned, i dont know which would cause more calories consumption but let's stick to the physics problem and not the "sports" aspect :P
     
  20. Mar 12, 2016 #19

    CWatters

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    That's also mentioned in the article I posted above.
     
  21. Mar 12, 2016 #20

    Merlin3189

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    But it still depends on what you think is the physics problem.
    If ur bf were a particle being pushed up a smooth incline, the physics work is just the mgh as you first calculated and there is nothing more to consider.
    You have already raised the question of sideways movement, so you are not content with mgh. Once you start down this path, you are getting into the "sports" aspect, even for old codgers like me tottering to the shops at 2km/hr. Russ Watters' link mentions the work done, when walking or running on the flat, by raising and lowering your CG with each step. In any normal gait you must lift your leg & foot from the ground and lower it again to make a pace.
    IMO this is the crux of biomechanical work. There is no mechanism for reclaiming energy that has been expended by a muscle (though some energy can be stored elastically), so every necessary movement does work and requires energy. Even relaxing a muscle causes some work to be done by the antagonist muscle. The question is, which movements you ignore.
    By avoiding the "sports" aspects, presumably you ignore work done by muscles to operate heart, lungs and balance. Probably also work done by muscles purely in reconfiguring the body during different phases of gait. Ultimately you're back to mgh, but you have ignored the majority of work done by muscles in the body.

    So I would say you need to consider these "sports" aspects, because that is where most of the energy goes and he could not move his mass from A to B without them. But they are the most difficult to calculate - and won't even be the same for him when he does it on different occasions. AFAIK the way this is usually calculated is by inference from the measured CO2 expired during and after exercise compared with resting levels.

    Incidentally, in your equation W=Fdx+mgh if Fdx is the work required to travel horizintally for a distance dx, then I would say it should be W=Fds+mgh where ds is the distance travelled along the route chosen.
    Even if you climbed a ladder, you would have Fds for the height of the ladder, ds = h, and F = (the vertical force applied) - mg . The vertical force here must be greater than mg since you have to lift your leg on each step. Generally F is part of the "sports" aspect involved in bodily reconfiguration and applies throughout the distance actually travelled and of course also changes in magnitude depending on the direction of travel.
     
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