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Energy from food to Mechanical Energy & distance traveled

  1. Jul 18, 2015 #1

    KAC

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    1. The problem statement, all variables and given/known data
    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).

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

    3. 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!
     
  2. jcsd
  3. Jul 18, 2015 #2

    TSny

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    Homework Helper
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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.
     
  4. Jul 18, 2015 #3

    KAC

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    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?
     
  5. Jul 18, 2015 #4

    TSny

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    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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