A man trying to lose weight lifts a 10 kg mass, one thousand times to a height of 0.5m each time. Assume that the potential energy lost each time, he lowers the, mass is dissipated. What is the work does he do against the gratational force? Fat supplies 3.8 x 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the man use up?
Answers
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Answer:
Explanation:
Given: A man trying to lose weight lifts a 10 kg mass, one thousand times to a height of 0.5 m each time. Assume that the potential energy lost each time, he lowers the, mass is dissipated. What is the work does he do against the gravitational force? Fat supplies 3.8 x 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the man use up?
Given mass of weight m = 10 kg
The man lifts the weight , so h = 0.5 m
So weight is lifted number of times, so n = 1000
Now work done against gravitational force
= n x mgh
= 1000 x 10 x 9.8 x 0.5
= 49 kg
Now for 1 kg of fat the energy equivalent is
3.8 x 10^7 J
So rate of efficiency is 20%
Energy supplied will be
20/100 x 3.8 x 10^7 J
Now mass of fat lost by man will be
1/1/5 x 3.8 x 10^7 x 49 x 10^3
49 x 5 = 245 / 3.8 x 10^4
= 64.47 x 10^- 4 kg
= 6.44 x 10^-3 kg