Calculating Energy and Fat Loss in Stair Climbing

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Homework Help Overview

The discussion revolves around calculating energy expenditure and fat loss associated with stair climbing, specifically focusing on a scenario involving a 75 kg person ascending a height of 450 m. The problem encompasses concepts from physics, particularly work, energy conversion, and power output.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of work done in climbing stairs and the subsequent energy derived from fat. Questions arise regarding the conversion of energy from fat to mechanical work and the relationship between work and power output.

Discussion Status

Some participants confirm the correctness of the initial calculation of work done. Others provide guidance on how to approach the calculation of fat burned and power output, emphasizing the use of proportions and energy conversion factors. There is an ongoing exploration of the relationships between the variables involved.

Contextual Notes

Participants express varying levels of understanding and seek clarification on specific calculations. The discussion reflects a mix of attempts to derive answers while adhering to the forum's learning-focused approach.

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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