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Using mechanical energy?

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Converting the energy of one food calorie into mechanical energy, starting from rest, to what speed could you accelerate a 133-lbs person?

    2. Relevant equations
    1 calorie= 4.184J
    1 lbs= 0.453592kg
    K=1/2mv^2

    3. The attempt at a solution
    here what I did, I converted the lbs to kg which was 60.33kg. I was under the impression to use kinetic energy.
    4.184J=1/2(60.33kg)v^2
    4.184J/30.165kg=v^2
    sqrt 4.184J/30.165kg= 0.372m/s

    This is how I solved my problem, but I got 1 out of 10. Can someone explain to me what I did wrong?
     
  2. jcsd
  3. Oct 20, 2013 #2

    Simon Bridge

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    Well, 1cal = 4.184J, but 1 "food calorie" = 1k.cal
    I don't see you dropping 9 points just for a botched conversion.

    The model you used assumes that all the food energy goes to translation of the center of mass.
    There are other mechanical things - motion of arms and legs for eg, that the energy could go to.
    Were you supposed to account for the food energy lost to non-mechanical sources?

    Check your notes for the mechanical energy of a human being JIC you were expected to do something different.
    i.e. Should you have used only a percentage of the food energy to speed?

    http://en.wikipedia.org/wiki/Food_energy#Energy_usage_in_the_human_body

    Do you have a model answer?
     
  4. Oct 20, 2013 #3

    Dick

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    You didn't do much wrong at all. Except that a "food calorie" is actually 1000 of the calories that you are using.
     
  5. Oct 20, 2013 #4
    ok it seems that I made a mistake in my conversion with the calories. So with that 1000 calories do I convert that into joules? if so how would I do it?
     
  6. Oct 20, 2013 #5

    Dick

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    Multiply 4.184J by 1000 to get the number of Joules in a "food calorie". What else?
     
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