# Q22 (b)  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. Fat supplies $3.8 \times 10 ^ 2 J$ of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

D Devendra Khairwa

Efficiency is given to be 20 per cent.

Thus energy supplied by the person :

$=\ \frac{20}{100}\times 3.8\times 10^7$

Thus the amount of fat lost is :

$=\ \frac{49\times 10^3}{\frac{20}{100}\times 3.8\times 10^7}$

or                                                 $=\ 6.45\times 10^{-3}\ Kg$

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