The efficiency of converting fat to mechanical

[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

 

[HallsofIvy]

First rule: read the problem carefully!


(a) asks only for the work you do- it says nothing about where the energy required comes from, nothing about "fat", and so the energy in a kg. of fat and the efficiency of your muscles is irrelevant to part a.

Lift a 12 kg mass 2 meters increases its potential energy by "mgh" or (12)(9.8)(2)= 235.2 Joules. Doing that 500 times means you have done 500(235.2)= 117600 Joules of work or, to 2 significant figures
1.2 x 10^5 J= 1.2x 10^2 kilo-Joules.

(b) NOW the problem asks how much fat is burned to do that work (assuming all the work comes from burning fat which isn't generally true.)
1.2x10^2 kilo- Joules = 1.2 x10^(-1) Mega-Joules (1000000 is 1000x1000 so there are 1000 kilo-joules in a Mega-Joule). Since one kg of fat is "equivalent" to 30 MJoules, if you had 100% efficienecy you would have to burn 1.2x10^(-1)/30 = 4.0x10^(-3) kgs of fat.

Since conversion of fat to work is 20% efficient (the other 80% becomes waste heat), you will actually have to burn
4.0 x 10^(-3)/0.2= 2.0 x 10^-2 kg of fat. Since there are 1000 g in a kilogram, this is, of course, 20 g of fat.

(Now, how many kilograms do you have LEFT to lose!)

 

[M98Ranger]

of fat

a. The work done in this scenario would be 1.2x10^2 kJ (120 kJ). This can be calculated by multiplying the force (235.2N) by the distance (2.0m) and the number of repetitions (500).
Work = Force x Distance x Number of repetitions
Work = (235.2N)(2.0m)(500)
Work = 1.2x10^2 kJ

b. If all the energy used to do the work comes from burning fat, then 20g of fat would be used up. This can be calculated by converting the work done (120 kJ) to MJ and then dividing by the efficiency (20%).
Work in MJ = 1.2x10^2 kJ/1000 kJ/MJ = 0.12 MJ
Fat used = Work in MJ/efficiency
Fat used = 0.12 MJ/20%
Fat used = 0.6 MJ
1kg of fat = 30 MJ
0.6 MJ = (0.6/30)kg of fat
0.6 MJ = 0.02kg of fat
0.02kg = 20g of fat

Therefore, 20g of fat would be used up in this scenario. This shows that converting fat to mechanical energy is not a very efficient process, as a large amount of fat is needed to produce a relatively small amount of work.