A person trying to lose weight by burning fat lifts a mass of $10$ kg upto a height of $1$ m $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7}$ J of energy per kg which is converted to mechanical energy with a $20 %$ efficiency rate. Take $g = 9.8 m s^{- 2}$ .

2.45 \times 10^{- 3}

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6.45 \times 10^{- 3}

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9.89 \times 10^{- 3}

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12.89 \times 10^{- 3}

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Solution

The correct option is D $12.89 \times 10^{- 3}$ kg
The net work done by the man will be $1000$ times the work done in lifting $10 k g$ to a height of $1 m$
Net work done $= 1000 \times 10 \times 9.8 \times 1$ $J$
Let's assume $x k g$ of fat is burnt in doing this work. Energy balance will give the following equation,
$x \times \frac{20}{100} \times 3.8 \times 10^{7} =$ Net work done
$x = 12.8947 \times 10^{- 3}$ $k g$

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