Work Power and Energy 5 Question 3
5. A person trying to loose 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, $g=9.8 ms^{-2}$ )
(2016 Main)
(a) $2.45 \times 10^{-3} kg$
(b) $6.45 \times 10^{-3} kg$
(c) $9.89 \times 10^{-3} kg$
(d) $12.89 \times 10^{-3} kg$
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Answer:
Correct Answer: 5. (d)
Solution:
- Work done in lifting mass $=(10 \times 9.8 \times 1) \times 1000$
If $m$ is mass of fat burnt, then energy
$$ =m \times 3.8 \times 10^{7} \times \frac{20}{100} $$
Equating the two, we get
$$ \therefore \quad m=\frac{49}{3.8} \approx 12.89 \times 10^{-3} kg $$
