A person trying to lose weight (dieter) lifts a $10\; kg$ mass, one thousand times, to a hetght 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?
A person trying to lose weight (dieter) lifts a $10\; kg$ mass, one thousand times, to a hetght 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?
Mass of the weight, $m=10 kg$
Height to which the person lifts the weight, $h=0.5 m$
Number of times the weight is lifted, $n=1000 \therefore$
Work done against gravitational force:
$=n(m g h)$
$=1000 \times 10 \times 9.8 \times 0.5$
$=49 \times 10^{3} J =49 kJ$
Energy equivalent of $1 kg$ of fat $=3.8 \times 10^{7} J$
Efficiency rate $=20 \%$
Mechanical energy supplied by the person's body:
$=\frac{20}{100} \times 3.8 \times 10^{7} J$
$=\frac{1}{5} \times 3.8 \times 10^{7} J$
Equivalent mass of fat lost by the dieter
$=\frac{1}{\frac{1}{5} \times 3.8 \times 10^{7}} \times 49 \times 10^{3}$
$=\frac{245}{3.8} \times 10^{-4}$
$=6.45 \times 10^{-3} kg$
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