In an exercise physiology lab a 84 kg male ran on a treadmil

In an exercise physiology lab, a 84 kg male ran on a treadmill with a 4% grade at a speed of 17.0 km/hr for 23 minutes. For a conversion factor you can use 16.7 m/min for every 1 km/hr. (Don\'t forget about units)

a) What was his power output during this exercise expressed in kcal/min? SHOW ALL WORK!!!!

b) Suppose his oxygen consumption during the exercise was 2.12 L/min, his resting oxygen consumption was 0.36 L/min, and his RER was 0.91. Calculate his mechanical efficiency using the power output above. SHOW ALL WORK.

PLEASE PLEASE SHOW THE WORK MATHEMATICALLY AND SPECIFICALLY represent what number represent Pout and Pin etc. THANK YOU VERY MUCH!!!

Solution

a). Power output = Work done / minutes =?

Given,

Weight =84 Kg

Grade = 4 %

Speed = 17.0 Km/Hr = 17 x 16.7 m/min = 283.9m/min

Time = 23 min

Therefore,
Vertical Distance travelled = 283.9 m/min x 23 min x 4/100 = 6,529.7 m x 0.04 =261.188 m

Work Done = Body Weight x Vertical Distance travelled

=> W = 84 Kg x 261.188 m = 21,939.792kg.m ~ 215.155 Kj

Power output = Work Done /min

P = 21,939.792kg.m / 20 min = 1096.9896 Kg.m/min

Therefore,

Power output = 1096.9896 Kg.m/min Or 1096.9896 Watts

b) Mechanical efficiency = Power output/ Power input

VO₂ = 1.8 (Power input/ Mass) + 7

Delta VO2= VO2 during exercise - VO2 at rest = 2.12-0.36 =1.76 L/min or 1760mL/min

1760= 1.8(power input/ (84/9.8)) +7

OR

Power input/ 8.57= (1760-7)/1.8

=> Power Input = -2900 x 8.57

=> Power Input = 24950 [Power can never be negative]

Therefore,

Mechanical efficiency = Power output/ Power input

=>M. Efficiency = 1096.9896/ 24950 = 0.044

Therefore, the mechanical efficieny = 4.4%