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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?

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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                                                

Posted by

Devendra Khairwa

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