Energy from food to Mechanical Energy & distance traveled

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Homework Help Overview

The discussion revolves around calculating the distance a person can walk using energy derived from carbohydrates in an energy bar. The subject area includes concepts of energy conversion, efficiency, and kinematics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conversion of velocity from km/h to m/s and the implications of efficiency on energy calculations. There are questions about the definitions of energy units and the correct interpretation of kinetic energy in this context.

Discussion Status

Some participants have offered clarifications on unit conversions and the interpretation of energy values. There is an ongoing exploration of the calculations involved, with no explicit consensus on the correctness of the final displacement value.

Contextual Notes

Participants are navigating potential misunderstandings regarding energy units and efficiency, as well as the proper setup of the problem. There is an acknowledgment of the need to consider efficiency in the energy calculations.

KAC
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Homework Statement


An "energy bar" contains 22 g of carbohydrates. If the energy bar was his only fuel, how far could a 68 kg person walk at 5 km/h?

Mass(person) = 68 kg; v = 5km/h; 22g carb = 4 Cal/1 carb = 88 Cal; P(power of man walking) = 380W (this is derived from my textbook).

Homework Equations


How long is the time interval that the man walks at; what is the KE of the man due to his efficiency.

The Attempt at a Solution


velocity needs to be converted to m/s rather than km/h, so 5km/h*1000m/1km*1h/60s=83m/s. (This seems incorrect to me, but I don't understand how the answer would be different).

22g carb* 4Cal/1 Carb= 88 Cal; 88 Cal*4190J/1 Cal = 368720J; 368720 = Ecarb (I am also confused in this portion of the problem; does 1 carb = 1 Cal or 1 cal? Am I multiplying by 4.19J for 1 cal or 4190 for 1 Cal?)

efficiency = output/input; e = KE/Ecarb; KE = Ecarb*e; KE = 368720* .25 (.25 for the efficiency of a person, derived from book/lecture); KE = 92180 J.

P= KE/time interval; time interval = KE/P; t = 92180J/380W; t= 242.57s.

displacement=velocity*time interval; d = 83m/s*242.57s; d=20133.31m.

I know this final answer for displacement is completely wrong, but I cannot figure out where I made my mistakes in my calculations; I believe I am just setting the problem up incorrectly. Can somebody help me out?

Thanks!
 
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Hello KAC. Welcome to PF!

Are there 60 s in one hour?

I believe 1 carb = 4 Cal (not 4 cal)

I'm not sure why you are using the symbol KE here. But your basic procedure looks correct to me.
 
TSny said:
Hello KAC. Welcome to PF!

Are there 60 s in one hour?

I believe 1 carb = 4 Cal (not 4 cal)

I'm not sure why you are using the symbol KE here. But your basic procedure looks correct to me.
TSny,

Thank you for your welcome! And thank you for pointing out my white mistake. LOL. So let me try this out again:

v = 5km/h*1000m/1km*1h/3600s= 1.39 m/s.
e = output/input; efficiency*Ecarb = KE; .25*368720J; KE = 92180J.
P = K/time interval; t = K/P; t= 92180J/380W; t = 242.58 s.
displacement = v*t; d = 1.39m/s * 242.58s = 337.2 m

But this displacement was also incorrect... Have I made any other mistakes in my calculations or used the wrong formulas?
 
Hope I'm not overlooking something. But your method looks correct to me. I wonder if you are meant to take into account the 25% efficiency. I don't know.

The 92180 J is not the kinetic energy (KE) of the person. It's the total energy used to walk the distance.