Ncert Exercises & Solutions

Question 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 × 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Given,

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 = $\frac{3.8 \times 10^{7} \times 20}{100}$ = 0.76 x10⁷ J/kg

.Thus, fat used up by the dieter = $\frac{1}{(0.76 \times 10^{7}) \times 49000} = 6.45 \times 10^{- 3} k g$

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