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An energy problem

  1. Nov 14, 2006 #1
    Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kcal = 4186 J. Metabolizing 1 g of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 80 steps, each 0.150 m high, in 65 s. For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is 20.0%. Therefore when your body converts 100 J from metabolizing fat, 20 J goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume that the student's mass is 50.0 kg. How many times must she run the flight of stairs to lose 1 lb of fat

    I got an answer for this question, but the number is too large, so I thought I may have gotten something wrong in my calculation:

    okie, so if 1 g of fat can release 9.00 kcal, then 1lb of fat can release 17124545 J.

    0.5mv^2 + W = mgh

    However, to get the initial velocity (v):
    Vf^2 = Vi^2 + 2ax
    0 = Vi^2 + 2 * -9.8 * (0.15 * 80)
    Vi = 15.3m/s

    Plug it back in:
    0.5 * 50 * 15.3^2 + W = 50 * 9.8 * (0.15 * 80)
    W = 27.8 J

    However, since human is 20% efficiency...
    27.8 J * .2 = 5.6J

    So running 80 set of stairs will let you lose 5.6 J of fat. So to lose 1lb of fat:
    17124545J/5.6J = 3057954

    To lose 1 lb of fat, you need to run the 80 stairs 3057954 times! For some reason, I think this value is too large. Anyone caught a mistake on my calculation?
  2. jcsd
  3. Nov 14, 2006 #2


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    Kinetic energy should not be a factor in this problem. The student has to do a certain amount of work to climb the fight of stairs, and when the 20% efficiency is included, the energy expended to do that work will be 5 times the work accomplished. Your calculation of the amount of work to climb one flight of stairs is far less than the work output needed to accomplish that task, and on top of that you have inverted the effect of efficiency in the calculation. It takes more energy than the work accomplished, not less.
  4. Nov 14, 2006 #3
    So are you saying the student does not move initially... then my equation will be W = mgh???
  5. Nov 14, 2006 #4


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    The starting and stopping of the student is of little consequence. It is in fact a smaller effect than descending the stairs, which is being ignored in the problem. So yes, W = mgh for each climb of the stairs.
  6. Nov 15, 2006 #5
    Thank you very much. I thought that as you are climbing the stairs, you need initial velocity, so I'm having the wrong concept here. I was thinking more of motions. So here we go again:
    W = mgh
    = 50 * 9.8 * (0.15 * 80)
    = 5880 J
    Therefore, 5880 * 100%/20% = 29400J
    Then the final answer should be: 17124545 / 29400 = 583

    So the student must climb 583 flight of stairs to lose 1 lb of fat. Sounds realistic! Thanks. Can you please check if I did anything wrong here?
  7. Nov 15, 2006 #6


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    That looks OK. There will be a bit more work needed the first few steps to get up to speed, but that is offset by having to do a bit less work at the top couple of steps because of the momentum that is being carried. Because of the inefficiency factor, you could include startup work to build kinetic energy the first few steps as some extra energy being burned, but assuming the stairs are at a 45° angle the kinetic energy is only about 35 J. (Except for the 45° assumption, this is what you were calculatiing in your first attempt.) As you can see, this is very small compared to the mgh work, but it would save a few trips up those stairs.
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