Question

Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do extra exercise to avoid gaining weight. He was doing jumps: he leaves the ground with a speed of 3.3m/s at every jump. Assuming that his body turns energy to mechanical work with a 25 % efficiency, how many jumps he will have to make g

Answer 1

Step-by-step explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work is 25 % .

Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.

4000 Cals = 4.2 x 4000 = 16800 J .

Work done in one jump = kinetic energy while jumping

= 1/2 m v²

= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

= 44 .