Discussion Forum



1)

A person trying to lose weight by burning fat lifts a mass of 10kg 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 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate (Take, g=9.8 ms-2)

 


A) $2.45 \times 10^{-3}kg$

B) $6.45 \times 10^{-3}kg$

C) $9.89 \times 10^{-3}kg$

D) $12.89 \times 10^{-3}kg$

Answer:

Option D

Explanation:

 Work done in lifting the mass

                         =(10 × 9.8 × 1)× 1000

if m is mass of far burnt, then energy

                   = $m\times 3.8\times 10^{7} \times\frac{20}{100}$

Equating the two , we get

      $\therefore$   $m= \frac{49}{3.8}=12.89 \times 10^{-3}$ kg