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?

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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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