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