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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.
- How much work does she do against the gravitational force?
- Fat supplies 3.8 x 107J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution
Here, m = 10 kg, h = 0.5 m, n = 1000
(a) work done against gravitational force.
W = n(mgh)
= 1000 x (10 x 9.8 x 0.5) = 49000J.
(b) Mechanical energy supplied by 1 kg of fat = `3.8 xx 10^7 xx20/100`
= 0.76 x107 J/kg
∴ Fat used up by the dieter =`(1kg)/(0.76 xx 10^7) xx 49000`
= 6.45 x 10-3 kg
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