Question

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

Answer 1

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

•Given :

m = 60kg

Height ,h= 10m and

g =10m/s²

Formulas and values :

1) work done my gravity = mgh

2) 1 cal = 4.2 J

Solution :

Energy produced by burning 1kg of fat = 7000 kilo calories

⇒Energy produced by burning 5kg of fat = 7000 ×5 = 35000 kilo calories

1 cal = 4.2J

⇒35000 k cal = 4.2×35000 kJ =$147×10{}^{6}$

Energy utilised in going up = mgh

Energy utilised in going down =$\frac{mgh}{2}$

Total energy utilised

$= mgh + \frac{mgh}{2}$

$= \frac{3mgh}{2}$

$= \frac{3}{2} \times 60 \times 10 \times 10$

$= 9000 \: j$

No. of time's person has to go up and down

$= \frac{147 \times 10 {}^{7} }{9000}$

$= \frac{147}{9} \times 10 {}^{4}$

$= 16.333 \times 10 {}^{4}$

$= 16334 \: (approx)$

Therefore , He must go up and down 16334 times to up and down to reduces his weight by 5 kg.

Answer 2

Given

$\bold{ Height \ of \ the \ stairs} = 10 \ m$

$\bold{ Weight \ of \ the \ man} = 60 \ kg$

Work done to burn 5 kg of fat $= 5 * 7000 * 10^{3} * 4.2$

                                                    $= 147 * 10^{6} J$

Work done to burn fat in one trip = $mgh + 1/2mgh = 3/2 mgh$

                                                          $= 3/2 * 60 * 10 * 10$

                                                          $= 9 * 10^{3} J$

No : of trips required

$\dfrac{147 * 10^{6} }{9 * 10^{3} }$

$= \boxed{\boxed{\bold{16300 \ times}}}}$      (appx)