Calculating Energy and Fat Loss in Stair Climbing

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A 75 kg person climbing 450 m of stairs does approximately 330,000 J of work. To calculate the fat burned, the energy provided by 1 kg of fat is 6,000,000 J, and the fraction of fat used is determined by the ratio of work done to energy from fat. This results in the person burning about 55 grams of fat during the exercise. The power output, calculated as work divided by time, yields a specific value when time is expressed in seconds. Overall, the calculations demonstrate the relationship between energy expenditure, fat utilization, and power output in stair climbing.
raman911
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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.
 
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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?
 
Dick said:
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?
 
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.
 
Anyone Explain Me
 
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.
 
Dick said:
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?
 
No. Think.
 
Anyone Explain Me
 
  • #10
You mean "Anyone Give Me The Answer". And I don't think anyone will. But you might get lucky... It's really easy.
 
  • #11
Dick said:
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
raman911 said:
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
Dick said:
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
Absolutely right.
 
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