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One Calorie in nutrition is 1 kilocalorie, which we define in Chapter 11 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. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary.

To evaluate the program, suppose she runs up a flight of 65 steps, each 0.150 m high, in 72.0 s. For simplicity, ignore the energy she uses 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 65.0 kg.

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

(b) What is her average power output, in watts and in horsepower, as she is running up the stairs?

Answer 1

Step 1: Determine the amount of energy that must be burned to lose 1 lb of fat.
From: Metabolizing 1 gram of fat can release 9.00 kcal.
1 lb = 453.6 g, so metabolizing 1 lb of fat releases 4082.4 kcal.
4082.4 kcal * 4.186 kJ/kcal = 17089 kJ. Thus, she must use 17089 kJ to burn 1 lb of fat.

Step 2: Determine the amount of mechanical work done per flight:
W = Fd
d = 65 * .15 m = 9.75m
F = mg = 65 kg * 9.8 = 637 N
W = 637 * 9.75 = 6210.75 J = 6.211 kJ

Step 3: Determine the amount of energy burned for work done.
6.211 kJ / 20% = 31.055 kJ metabolized per flight of stairs.

Step 4: Divide the total energy that must be metabolized by the energy burned per flight, to find the number of flights:
17089 kJ / 31.055 kJ/flight = 550.28 flights, or 551 flights (round up to a full flight).

so, a.) is 551 flights of stairs

b.) Power = W/t = 31.055 kJ/flight / 72.0 s/flight = 0.431 watts.
1 watt = 0.00134102209 hp
.431 watt = 0.000578 hp