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Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kcal = 4 186 J. Metabolizing 1 g of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. He plans to run up and down the stairs in a football stadium as fast as he can and as many times as necessary. To evaluate the program, suppose he runs up a flight of 85 steps, each 0.150 m high, in 50.0 s. For simplicity, ignore the energy he uses in coming down (which is small). Assume that a typical efficiency for human muscles is 20.0%. Therefore when your body converts 100 J from metabolizing fat, 20 J goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume that the student's mass is 85.0 kg.
(a) How many times must the student run the flight of stairs to lose 1.00 kg of fat?
(b) What is his average power output, in watts and horsepower, as he runs up the stairs?
(c) Is this activity in itself a practical way to lose weight?

Respuesta :

gennerandrescarrillo

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

[tex]W= F*d= m*g*d[/tex]

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

[tex]W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J][/tex]

Converting from Joules to Kcals:

[tex]\frac{10620.75}{4186} =2.537 Kcal[/tex]

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

[tex]x=\frac{2.537*100}{20} =12.685[/tex]

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

[tex]\frac{9000Kcals}{12.685Kcals} =709.5 times[/tex]

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

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For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

[tex]Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\[/tex]

[tex]P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp][/tex]

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