Question

A person trying to lose weight (dieter) 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 she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force ?
(b) Fat supplies 3.8 × 10$^7$J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Answer 1

Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5 m
Number of times the weight is lifted, n = 1000
∴Work done against gravitational force :
= n (mgh) = 1000 × 10 × 9.8 × 0.5
= 49 × 10 $^3$ J
= 49 KJ
Energy equivalent of 1 kg of fat = 3.8 × 10$^7$ J
Efficiency rate = 20%
Mechanical energy supplied by the person’s body :
= $\frac{20}{100}$ × 3.8 × 10$^7$ J

= $\frac{1}{5}$× 3.8 × 10 $^7$J
Equivalent mass of fat lost by the dieter :
= $\frac {1}{\frac 15 \times 3.8 \times 10^7} \times 49 \times 10^3$

= $\frac{245}{3.8}$ × 10 $^{-4}$

= 6.45 × 10 $^{-3}$ kg