A person trying to lose weight (dieter) lifts a 10 kg mass 1,000 times to a height of 0.5 m each time. Assume that the potential energy lost each time she lowest the mass is dissipated. a. How much work does she do against the gravitational force?b. Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficient rate. How much fat will the dieter use up?
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(a) Given,
Mass of the weight, m = 10 kg
weight lifted by dieter upto height, 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/s² × 0.5
W = 49000 J = 49 kJ
(b) 1 kg of fat is equivalent to 3.8 × 10^7 J.
Efficiency of conversion, η = 20%
Mechanical energy supplied by body of the person, E = 20 % of 3.8 × 10^7 J
E = 20/100 × 3.8 × 10^7 J
E = 7.6 × 10^6 J
Equivalent mass of fat lost by dieter = work done against gravitational force /mechanical energy supplied by body of the person
= 49000/7.6 × 10^6
= 6.45 × 10^-3 Kg
Mass of the weight, m = 10 kg
weight lifted by dieter upto height, 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/s² × 0.5
W = 49000 J = 49 kJ
(b) 1 kg of fat is equivalent to 3.8 × 10^7 J.
Efficiency of conversion, η = 20%
Mechanical energy supplied by body of the person, E = 20 % of 3.8 × 10^7 J
E = 20/100 × 3.8 × 10^7 J
E = 7.6 × 10^6 J
Equivalent mass of fat lost by dieter = work done against gravitational force /mechanical energy supplied by body of the person
= 49000/7.6 × 10^6
= 6.45 × 10^-3 Kg
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