Question

11 HW Problem 11.14 - Enhanced with Feedback 9 of 25 Constants | Periodic Table Part A Each time he does one pushup, Jose, who has a mass of 89kg, raises his center of mass by 25 cm. He completes an impressive set of 150 pushups in 5 minutes, exercising at a steady rate If we assume that lowering his body has no energetic cost, what is his metabolic power during this workout? Express your answer with the appropriate units. alue Units Submit Request Answer Part B In fact, it costs Jose a certain amount of energy to lower his body-about half of what it costs to raise it. If you include this in your calculation, what is his metabolic power? Express your answer with the appropriate units. alue Units Submit Request Answer Provide Feedback Next >

Each time he does one pushup, Jose, who has a mass of 89kg, raises his center of mass by 25 cm.He completes an impressive set of 150 pushups in 5 minutes, exercising at a steady rate.

***I've tried solving for Part A and got 109.14 W. But it keeps saying I am wrong because, "It appears that you have not considered efficiency of the human body in your calculation. Recall the value of this efficiency." The efficiency is 25%...any help would be great!

Answer 1

part - A Mass m = 89 kg height h = 25 cm = 0 25 cm = 0 25 m acceleration due to gravity g = 9 81 m/s^2 Total no of push up = 150 In each push up work done = mgh so total work done = 150 mgh total time t = 5 mins = 5 times 60 s = 300 s S power required, p_R = 150 mgh/300 wat = 150 times 89 times 0 25 times 9 81/300 watt Now human body is 25% efficient That means, human body can use 25% of total power produced lay metabolism. so it P_m in the metabolic power, then we should have, P_m times 25/100 = P_R \r\n Therefore P_m = P_R times 100/25 = 150 times 89 times 0 95 times 9 81 times 100/25 times 300 watt = 436 54 watt So if we assume that lowering his body has no energetic cost