(a) First, we need to calculate the total energy consumed by the man during the climb.
Total energy consumed = 9.8 kcal/min * 60 min/hr * 3 hr
Total energy consumed = 1764 kcal
Converting this to joules:
Total energy consumed = 1764 kcal * 4186 J/kcal
Total energy consumed = 7387424 J
Now, we can calculate the power consumption in watts:
Power consumption = Total energy consumed / time
Power consumption = 7387424 J / (3 hr * 3600 s/hr)
Power consumption = 683.34 W
Therefore, the power consumption of the man during the climb is 683.34 watts.
(b) The useful work done by the man during the climb is equal to the change in potential energy:
Useful work = mgh = 75 kg * 9.8 m/s^2 * 1000 m
Useful work = 735000 J
The power output in useful work is then:
Power output = Useful work / time
Power output = 735000 J / (3 hr * 3600 s/hr)
Power output = 67.36 W
Therefore, the power output of the man in useful work is 67.36 watts.
(c) The efficiency of the man during the climb can be calculated as the ratio of the power output to the power consumption:
Efficiency = (Power output / Power consumption) * 100%
Efficiency = (67.36 W / 683.34 W) * 100%
Efficiency = 9.86%
Therefore, the efficiency of this man during the climb is 9.86%.