Question

A 71 kg hiker climbs to the top of a 4300 m -high mountain. The climb is made in 5. 5 h starting at an elevation of 2800 m. Calculate the work done by the hiker against gravity. Calculate the average power output in watts. Calculate the average power output in horsepower. Calculate assuming the body is 16% efficient, what rate of energy input was required

Answer 1

A. 1.0x10^6 J

B. 53 W

C. 7.1x10^-2

D. 330 W

Step-by-step explanation:

A) work done = mgh

so 71*9.8*1500=1.0x10^6 J

B) power = Work done/time

So P=1.0x10^6/19800 sec

P= 53 W

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C) 1 HP =735 W

503 W= 7.1x10^-2 HP

D) efficiency =(output/ input) *100

so 16= (53/input) *100

input= 7.1x10^-2