Question

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

Answer 1

Here, $m=10kg,h=0.5m,n=1000$
1. Work done against gravitational force,
$W=n\left(mgh\right)=100\ast 10\ast 9.8\ast 0.5=49000J$
2. Mechanical energy supplied by $1kg$ of fat
$=3.8\ast {10}^7\ast \frac{20}{100}=0.7\ast {10}^{7}J/kg$
$\therefore$ Fat used up by the dieter $=\frac{1kg}{0.7\ast {10}^7}\ast 49000=6.45\ast {10}^{-3}kg$.