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Energy Problem

  • Thread starter raman911
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  • #1
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One kilogram of fat is equivalent to 30 MJ of energy. The efficiency of converting fat to mechanical energy is about 20%.

a) Suppose a 75 kg person runs up the stairs of tall building with a vertical height of the 450 m, how much work is done by the person?

b) If all the energy used to do the work comes from "burning" fat, how much fat is used up by the exercise?

c) If the person took 45 min. to run up the stairs, what was their power output.

My proof:

A. m= 75kg
d= 450m
w=75kg(9.8N/kg)*450m
w=3.3*10^5J

PLZ HELP ME IN B AND C AND TELL ME A IS RIGHT OR WRONG.
 

Answers and Replies

  • #2
Dick
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A is fine. So 1kg of fat can provide .20*30MJ of mechanical work. What fraction is w of this? That's the fraction of a kg that you will burn. For C, power is w/time. If you express time in seconds then what units is the power in?
 
  • #3
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A is fine. So 1kg of fat can provide .20*30MJ of mechanical work. What fraction is w of this? That's the fraction of a kg that you will burn. For C, power is w/time. If you express time in seconds then what units is the power in?
HOW CAN I FIND B?
So 75kg of fat can provide (.20*30MJ)75 KG
WHAT'S NEXT?
 
  • #4
Dick
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You don't have to shout. You have (.20*30MJ)/kg. So (.20*30MJ/kg)*x kg=3.03*10^5J. Solve for x. What's the problem? MJ=10^6J if that helps.
 
  • #5
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Anyone Explain Me
 
  • #6
Dick
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Think proportions. 3.03*10^5J is to (.20)*30MJ as x kg is to 1kg. Where x kg is the amount of fat you burn. Solve for x. If you don't get that, I give up.
 
  • #7
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Think proportions. 3.03*10^5J is to (.20)*30MJ as x kg is to 1kg. Where x kg is the amount of fat you burn. Solve for x. If you don't get that, I give up.
can u give me answer b and c?
 
  • #8
Dick
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No. Think.
 
  • #9
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Anyone Explain Me
 
  • #10
Dick
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You mean "Anyone Give Me The Answer". And I don't think anyone will. But you might get lucky... It's really easy.
 
  • #11
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You mean "Anyone Give Me The Answer". And I don't think anyone will. But you might get lucky... It's really easy.
it's not easy for me
 
  • #12
Dick
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it's not easy for me
Point taken. But A was actually harder. How do you solve a problem like 2/3=x/4. That is the problem you are facing. Just with odder numbers and units attached.
 
  • #13
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Point taken. But A was actually harder. How do you solve a problem like 2/3=x/4. That is the problem you are facing. Just with odder numbers and units attached.
thxxxxxxxx
i got u
MJ=10^6J
(.20*30MJ/kg)*x kg=3.3*10^5J
so 330000/6000000
Answer=55g
 
  • #14
Dick
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Absolutely right.
 

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