A person trying to loose weight by burning fat lifts a mass of 10kg up to a height of 1m 1000 times.   Assume that the P.E. lost each time he lowers the mass is dissipated.  How much mass of fat will be use up considering the work done only when the weight is converted to mechanical energy with a 20% efficiency rate.   Take  $g=9.8m/s^2$; Energy of fat per unit mass = $3.8\times {10}^{7}J/kg$

Here  $mass m =10kg;height h=1m;\text{\hspace{0.17em}acceleration due to \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}gravity g=9.8m/s^2$
   n   =    1000 times
   $\therefore$      Energy of fat    =$3.8\times {10}^{7}J/kg$        
   Efficiency=  n    =        $20\%=\frac{20}{100}=\frac{1}{5}$
   $\therefore$  Net work done  by the man in lifting the mass
   =     n  $\times$ gain in P.E. of the mass

   W=     n mgh   =    1000 x 10 x 9.8 x 1   =     98000 J
     $$\begin{gathered} \therefore n=\text{\hspace{0.17em}efficiency=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{Networkdonebyman}{Energyinthefat} \\ \text{let mass of the fat burnt =m} \\ Energy in the fat = m x 3.8\times {10}^{7}J \end{gathered}$$
     $n=\frac{98000}{m\times 3.8\times {10}^7}\Rightarrow \frac{1}{5}=\frac{98000}{m\times 3.8\times {10}^7}$
   $\therefore m=\frac{98000\times 5}{3.8\times {10}^7}$

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