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A person of mass $60 \mathrm{~kg}$ wants to lose $5 \mathrm{~kg}$ by going up and down a $10 \mathrm{~m}$ high stairs. Assume he burns twice as much fat while going up than coming down. If $1 \mathrm{~kg}$ of fat is burnt on expending $7000 \mathrm{k}$ cal, how many times must he go up and down to reduce his weight by $5 \mathrm{~kg}$ ?
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As given that, Energy produced by burning $1 \mathrm{~kg}$ of fat $=7000 \mathrm{kcal}$
So, energy produced by burning $5 \mathrm{~kg}$ of fat $=5 \times 7000=35000 \mathrm{kcal}=35 \times 10^6 \mathrm{cal}$
Height of the stairs $(h)=10 \mathrm{~m}$
As we know that,
Energy consumed to go up one time $=m g h$ Energy consumed to come down one time $=\frac{1}{2} m g h$ (given)
So, energy consumed to going up and down one time $=m g h+\frac{1}{2} m g h=\frac{3}{2} m g h=\frac{3}{2} \times 60 \times 10 \times 10$ $\therefore$ Energy consume one time $=9000 \mathrm{~J}=\frac{9000}{4.2} \mathrm{cal}$ Let he go up and down $n$ time to consumed energy $=35 \times 10^6 \mathrm{cal}$
$$
\begin{aligned}
&n \times \frac{9000}{4.2}=35 \times 10^6 \\
&n=\frac{35 \times 10^6}{(9000 / 4.2)}=\frac{35 \times 4.2 \times 10^6}{9000} \\
&n=\frac{4900}{3}=16.3 \times 10^3 \text { times }
\end{aligned}
$$
So, energy produced by burning $5 \mathrm{~kg}$ of fat $=5 \times 7000=35000 \mathrm{kcal}=35 \times 10^6 \mathrm{cal}$
Height of the stairs $(h)=10 \mathrm{~m}$
As we know that,
Energy consumed to go up one time $=m g h$ Energy consumed to come down one time $=\frac{1}{2} m g h$ (given)
So, energy consumed to going up and down one time $=m g h+\frac{1}{2} m g h=\frac{3}{2} m g h=\frac{3}{2} \times 60 \times 10 \times 10$ $\therefore$ Energy consume one time $=9000 \mathrm{~J}=\frac{9000}{4.2} \mathrm{cal}$ Let he go up and down $n$ time to consumed energy $=35 \times 10^6 \mathrm{cal}$
$$
\begin{aligned}
&n \times \frac{9000}{4.2}=35 \times 10^6 \\
&n=\frac{35 \times 10^6}{(9000 / 4.2)}=\frac{35 \times 4.2 \times 10^6}{9000} \\
&n=\frac{4900}{3}=16.3 \times 10^3 \text { times }
\end{aligned}
$$
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