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A person trying to lose weight lifts a $10 \mathrm{~kg}$ mass, to a height of $0.5 \mathrm{~m}$ each time for 1000 times. Assume that PE lost each time she lowers the mass is dissipated
(a) How much work does she do against the gravitational force?
(b) Fat supplies $3.8 \times 10^7 \mathrm{~J}$ of energy/kg which is converted to mechanical energy with a $20 \%$ efficiency rate. How much fat will the dieter use up?
(a) How much work does she do against the gravitational force?
(b) Fat supplies $3.8 \times 10^7 \mathrm{~J}$ of energy/kg which is converted to mechanical energy with a $20 \%$ efficiency rate. How much fat will the dieter use up?
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(a) $m=10 \mathrm{~kg}, \mathrm{~h}=0.5 \mathrm{~m}, n=1000$
Work done against the gravitational force $=n(\mathrm{mg} h)$ $=1000 \times 10 \times 9.8 \times 0.5=49000 \mathrm{~J}$.
(b) Mechanical energy supplied by $1 \mathrm{~kg}$ of fat with $20 \%$
$$
\begin{aligned}
&\text { efficiency }=3.8 \times 10^7 \times \frac{20}{100} \\
&=0.76 \times 10^7 \mathrm{~J}
\end{aligned}
$$
or $0.76 \times 10^7 \mathrm{~J}$ energy is supplied by $=1 \mathrm{~kg} \mathrm{fat}$.
$\therefore$ Fat used up by dieter in the exercise $=\frac{1}{0.76 \times 10^7} \times 49000=6.45 \times 10^{-3} \mathrm{~kg}$
Work done against the gravitational force $=n(\mathrm{mg} h)$ $=1000 \times 10 \times 9.8 \times 0.5=49000 \mathrm{~J}$.
(b) Mechanical energy supplied by $1 \mathrm{~kg}$ of fat with $20 \%$
$$
\begin{aligned}
&\text { efficiency }=3.8 \times 10^7 \times \frac{20}{100} \\
&=0.76 \times 10^7 \mathrm{~J}
\end{aligned}
$$
or $0.76 \times 10^7 \mathrm{~J}$ energy is supplied by $=1 \mathrm{~kg} \mathrm{fat}$.
$\therefore$ Fat used up by dieter in the exercise $=\frac{1}{0.76 \times 10^7} \times 49000=6.45 \times 10^{-3} \mathrm{~kg}$
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