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?

0 5 Views | Posted 5 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    5 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 *107 = 0.76 *107 J/kg

    Fat used by dieter = 49050 / (0.76 *107) kg = 6.45*10-3 kg

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