$A$ person trying to lose weight (dieter) lifts a $10\; kg$ mass,one thousand times,to a height of $0.5\; m$ each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
$(a)$ How much work does she do against the gravitational force?
$(b)$ Fat supplies $3.8 \times 10^{7} \;J$ of energy per kilogram which is converted to mechanical energy with a $20\%$ efficiency rate. How much fat will the dieter use up?

(N/A) Given:
Mass $m = 10\; kg$
Height $h = 0.5\; m$
Number of times $n = 1000$
Acceleration due to gravity $g = 9.8\; m/s^2$
$(a)$ Work done against gravitational force:
$W = n \times m \times g \times h$
$W = 1000 \times 10 \times 9.8 \times 0.5$
$W = 49000\; J = 49\; kJ$
$(b)$ Energy supplied by $1\; kg$ of fat $= 3.8 \times 10^{7}\; J$
Efficiency $= 20\% = 0.2$
Mechanical energy available from $1\; kg$ of fat $= 0.2 \times 3.8 \times 10^{7}\; J = 7.6 \times 10^{6}\; J$
Mass of fat used $= \frac{\text{Total Work}}{\text{Mechanical energy per kg}}$
Mass of fat used $= \frac{49000}{7.6 \times 10^{6}}$
Mass of fat used $\approx 6.45 \times 10^{-3}\; kg$