Question

A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is bunt on expanding 7000 kcal, how many times must he go up and down to reduce his weight by 5 kg?

Answer 1

Height (h) = 10 m

Energy produced by burning 1 kg of fat

= 7000 kcal

∴ Energy produced by burning 5 kg fat

= 5 × 7000 = 35000 kcal

= 35 × 10⁶ cal

Energy utilized in going up and down one time

= mgh+\(\frac{1}{2}\)mgh=\(\frac{3}{2}\)mgh

= \(\frac{3}{2}\)× 60 × 10 × 10 J

= 9000 J

\(=\frac{90000}{4.2}\)cal

\(=\frac{3000}{1.4}\)cal

∴ Number of times, the person has to go up and down the stairs :

\(=\frac{35\times10^6}{(\frac{3000}{1.4})}\)

\(=\frac{3.5\times1.4\times10^6}{3000}\) 

= 16.3 × 10³ times.