Work Energy And Power Question 147
Question: A person trying to lose weight by burning fat lifts a mass of 10 kg up to 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=9.8m{s^{-2}} $ .
Options:
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 $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The change in potential energy during exercise $ \Delta PE=1000\times mgh $
$ =1000\times 10\times 9.8\times 1=98000J $
Mechanical energy per unit mass $ =\frac{2}{100}\times 3.8\times 10^{7} $
$ =7.6\times 10^{5}J/Kg $
$ \therefore $ Fat burn $ =\frac{9800}{7.6\times 10^{5}}=12.89\times {10^{-3}}kg $
