- #1
sb
One kg of fat is equivalent to about 30 MJ of energy. The efficiency of converting fat to mechanical energu is about 20%.
a. Suppose you lift a mass of 12kg 2.0m vertically, 500 times how much work do you do? (Assume that the work done by mass on you is disepated as heat to the surroundings).
b. If asll the energy used to do the work comes from "burning" fat, how much fat is used up by the expercise
This is what I did:
1kg = 30MJ
percentage efficinecy = 20%
m = 12kg
delta d= 2.0m
w=?
1kg = 1,000,000J
12kg = 12,000,000J
E = 12,000,000J
IMA = 12,000,000J
potencial efficiency = (AMA/IMA) * 100%
20% = (AMA/12,000,000) * 100
AMA = (20/100) * 12,0000
AMA = 24000J
FBD Diagam
Eg = mgh
Eg = (12kg)(9.8m/s^2 [D])(2.0)
Eg = 235.2N
Answers for this problem given at the end of the book are
a. 1.2x10^2 kJ
b. 20g
a. Suppose you lift a mass of 12kg 2.0m vertically, 500 times how much work do you do? (Assume that the work done by mass on you is disepated as heat to the surroundings).
b. If asll the energy used to do the work comes from "burning" fat, how much fat is used up by the expercise
This is what I did:
1kg = 30MJ
percentage efficinecy = 20%
m = 12kg
delta d= 2.0m
w=?
1kg = 1,000,000J
12kg = 12,000,000J
E = 12,000,000J
IMA = 12,000,000J
potencial efficiency = (AMA/IMA) * 100%
20% = (AMA/12,000,000) * 100
AMA = (20/100) * 12,0000
AMA = 24000J
FBD Diagam
Eg = mgh
Eg = (12kg)(9.8m/s^2 [D])(2.0)
Eg = 235.2N
Answers for this problem given at the end of the book are
a. 1.2x10^2 kJ
b. 20g
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