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 × 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
(a) Given,
Mass of the weight, m = 10 kg
Height upto which the weight is lifted, h = 0.5 m
Number of times the weight is lifted, n = 1000
Work done against the gravitational force,
W = nmgh
⇒ W = 1000 × 10 kg × 9.8 m s-2 × 0.5
⇒ W = 49000 J = 49 kJ
(b) 1 kg of fat is equivalent to 3.8 × 107 J.
Efficiency of conversion, η = 20%
Mechanical energy supplied by body of the person,
E = (20/100)×3.8×107 J
⇒ E = 7.6 × 106 J
Equivalent mass of fat lost by dieter,
M = 
⇒ M = 6.45×10-3 kg = 6.45 g