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 95 steps, each 0.150 m high, in 63.0 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 74.0 kg.(a) how many times must the student run the flight of stairs to lose 1.00 kg of fat? times(b) what is his average power output, in watts and horsepower, as he runs up the stairs? whp(c) is this activity in itself a practical way to lose weight?
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]
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.
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
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 )
Converting from Joules to Kcals:
Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%
So the student is consuming 12.685 KCals each time he runs up the stairs.
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:
He must run up the stairs 709.5 times, to burn 1 Kg of fat.
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)
The answers to the questions are;
a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.
b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.
c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.
To solve the question, we write out the known variables as follows
1 g of fat = 9.00kcal
Number of steps the student climbs = 95 steps
Height of each step = 0.150 m
Time it takes for the student to reach the top of the stairs = 57.5 s.
Efficiency of human muscles = 20 %
Mass of student, m = 65 kg
a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.
The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..
That is work done, W, = P. E. = m·g·h
h = The total height climbed by the student
g = Acceleration due to gravity = 9.81 m/s²
h = Height of each step × Number of steps the student climbs =
= 0.150 m/(step) × 95 steps = 14.25 m
Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²
= 9086.5125 J
We remember that the efficiency of the muscle is 20 %
The formula for efficiency is
The work produced by the muscle = Energy Output = 9086.5125 J
Energy input is given by
= 9086.5125 J/ (0.2) = 45432.5625 J
= 45.432 kJ
From the question, 1 g of fat = 9.00 kcal and
1 kcal = 4186 J
Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J
Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ
To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs times to make up the 37674 kJ energy contained in 1 kg of fat
That is = 829.23 times
b. Power is the rate of doing work
That is Power output = = = 158.026 watts
c. No as the activity student will have to spend a total time of
829.23 × 57.5 s = 47680.67 s climbing up the stairs alone and
47680.67 s = 13.24 Hours climbing up of which if the time to climb down is the same s climbing up, then we ave total time = 2× 13.24 Hours
= 26.49 hrs = 1.104 days exercising which is not humanly possible.