Energy is conventionally measured in Calories as well as in joules. One Calorie

Question

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is 1 kilocalorie, which we define as

1 kcal = 4,186 J.

Metabolizing 1 gram of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. The student plans to run up and down the stairs in a football stadium as fast as possible and as many times as necessary.

To evaluate the program, suppose the student runs up a flight of 85 steps, each 0.150 m high, in 61.0 s. For simplicity, ignore the energy used in coming down (which is small). Assume that a typical efficiency for human muscles is 20.0%. This means that when your body converts 100 J from metabolizing fat, 20 J goes into doing mechanical work (here, climbing stairs). The remainder goes into internal energy. Assume the student's mass is 107.0 kg.

(a) How many times must the student run the flight of stairs to lose 1 pound of fat?

b) is the average power output, in watts and in horsepower, as the student is running up the stairs?

Explanation / Answer

Given t=61.0s, m=107.0kg, k=0.20, g=9.81m/s²
h = 85(0.150) = 12.75m
E = (1 lb-fat)(37.8 kJ/g-fat)(454 g/lb) = 17.1 MJ
The energy needed to climb stairs once is given by:
E = mgh/k
So #n times student must run stairs to lose 1 lb fat is:
nE = 17.1MJ
n = 17.1MJ*k/mgh = 225.54

Average power (energy per time) output is:
P = E/t = mgh/kt = 1097 W

incase of any doub, please comment.

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