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The efficiency of converting fat to mechanical

  1. Jul 23, 2003 #1

    sb

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    One kg of fat is equivalent to about 30 MJ of energy. The efficiency of converting fat to mechanical energu is about 20%.
    a. Suppose you lift a mass of 12kg 2.0m vertically, 500 times how much work do you do? (Assume that the work done by mass on you is disepated as heat to the surroundings).
    b. If asll the enery used to do the work comes from "burning" fat, how much fat is used up by the expercise

    This is what I did:
    1kg = 30MJ
    percentage efficinecy = 20%
    m = 12kg
    delta d= 2.0m
    w=?

    1kg = 1,000,000J
    12kg = 12,000,000J
    E = 12,000,000J
    IMA = 12,000,000J

    potencial efficiency = (AMA/IMA) * 100%
    20% = (AMA/12,000,000) * 100
    AMA = (20/100) * 12,0000
    AMA = 24000J

    FBD Diagam

    Eg = mgh
    Eg = (12kg)(9.8m/s^2 [D])(2.0)
    Eg = 235.2N

    Answers for this problem given at the end of the book are
    a. 1.2x10^2 kJ
    b. 20g
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jul 24, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    First rule: read the problem carefully!


    (a) asks only for the work you do- it says nothing about where the energy required comes from, nothing about "fat", and so the energy in a kg. of fat and the efficiency of your muscles is irrelevant to part a.

    Lift a 12 kg mass 2 meters increases its potential energy by "mgh" or (12)(9.8)(2)= 235.2 Joules. Doing that 500 times means you have done 500(235.2)= 117600 Joules of work or, to 2 significant figures
    1.2 x 10^5 J= 1.2x 10^2 kilo-Joules.

    (b) NOW the problem asks how much fat is burned to do that work (assuming all the work comes from burning fat which isn't generally true.)
    1.2x10^2 kilo- Joules = 1.2 x10^(-1) Mega-Joules (1000000 is 1000x1000 so there are 1000 kilo-joules in a Mega-Joule). Since one kg of fat is "equivalent" to 30 MJoules, if you had 100% efficienecy you would have to burn 1.2x10^(-1)/30 = 4.0x10^(-3) kgs of fat.

    Since conversion of fat to work is 20% efficient (the other 80% becomes waste heat), you will actually have to burn
    4.0 x 10^(-3)/0.2= 2.0 x 10^-2 kg of fat. Since there are 1000 g in a kilogram, this is, of course, 20 g of fat.

    (Now, how many kilograms do you have LEFT to lose!)
     
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