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

2721

Explanation:

First, we find the change in calories

E = E(taken) - E(daily)

E = 3500 - 2500

E = 1000 cal

Next, we use the efficiency to find the available metabolic energy

E' = n.E, where n = efficiency

E' = 0.25 * 1000

E' = 250 cal

Converting this cal to J, we have

1 cal = 4.18 J

250 cal = 250 * 4.18 J

250 cal = 1045 kJ

Given that the final velocity is 0, we use equation of motion to find the height, h attained

u² = 2gh

h = u²/2g

h = 3.3² / 2 * 9.81

h = 10.89 / 19.62

h = 0.56 m

Potential energy, E = mgh

E = 70 * 9.8 * 0.56

E = 384 J

Total number of jumps, n =

n = 1045 / 384

n = 2721 jumps

12000j is work done

6/12=1/2

step-by-step explanation: