Energy from food to Mechanical Energy & distance traveled

AI Thread Summary
The discussion focuses on calculating how far a 68 kg person can walk at 5 km/h using energy from a 22 g carbohydrate energy bar. The initial calculations included incorrect unit conversions and misunderstandings about energy units, specifically between calories and joules. After correcting the velocity to 1.39 m/s and applying the efficiency of 25%, the total energy used was calculated to be 92180 J. The final displacement calculation yielded 337.2 m, but there was confusion regarding the interpretation of energy and efficiency. Clarifications were provided on the use of terms and the importance of efficiency in determining the actual distance traveled.
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.
 
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