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Physics Question on work, energy and power

A person trying to lose weight by burning at lifts a mass of 10 kg10\, kg upto a height of 1 m1\, m 10001000 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 3.8×107 J3.8 \times 10^7 \, J of energy per kg which is converted to mechanical energy with a 20%20\% efficiency rate. Take g=9.8 ms−2g = 9.8 \, ms^{-2} :

A

2.45×10−3 kg2.45 \times 10^{-3} \, kg

B

6.45×10−3 kg6.45 \times 10^{-3} \, kg

C

9.89×10−3 kg9.89 \times 10^{-3} \, kg

D

12.89×10−3 kg12.89 \times 10^{-3} \, kg

Answer

12.89×10−3 kg12.89 \times 10^{-3} \, kg

Explanation

Solution

0.2×3.8×107×m=10×g×1×10000.2 \times3.8 \times 10^{7} \times m = 10\times g \times1\times 1000
m=10×9.8×10000.2×3.8×107=1.289×10−2 kg=12.89×10−3 kgm = \frac{10 \times 9.8 \times 1000}{0.2\times 3.8\times 10^{7}} = 1.289 \times 10^{-2} \,kg = 12.89 \times 10^{-3} \,kg