- #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

Last edited by a moderator: