A spring scale hung from the ceiling stretches by 5.6 cm when a 2.0 kg mass is hung from it. the 2.0 kg mass is removed and repl
1 answer:
When the mass of 2.0 kg is attached to the spring, the force applied to the spring is the weight of the mass, so
This force stretches the spring by
, so we can use Hook's law to find the spring constant:
Then the mass of 2.0 kg is removed and replaced with another mass m2=2.7 kg. The force produced by this mass is equal to its weight:
And so, we can use again Hook's law to calculate the new stretch of the spring:
This force stretches the spring by
Then the mass of 2.0 kg is removed and replaced with another mass m2=2.7 kg. The force produced by this mass is equal to its weight:
And so, we can use again Hook's law to calculate the new stretch of the spring:
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Answer:
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.
Explanation:
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
Where:
h = The total height climbed by the student
g = Acceleration due to gravity = 9.81 m/s²
Therefore;
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
Efficiency =
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.