6.22 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?

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Payal Gupta | Contributor-Level 10

9 months ago

6.22 Mass lifted, m = 10 kg

Height to which the mass lifted, h = 0.5 m

No of repetitions, n = 1000

(a) Work done against gravitational force,

W = nmgh = 1000 $* 10 * 9.81 * 0.5 =$ 49050 J

(b) Mechanical energy supplied by 1 kg fat, with 20% efficiency rate = 0.2 $*$ 3.8 $* 10^{7}$ = 0.76 $* 10^{7}$ J/kg

Fat used by dieter = 49050 / (0.76 $* 10^{7})$ kg =

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(b) conserving energy between “O” ans ”A”

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