Question #116407
(10) A 65 kg hiker climbs to the top of a 3700m high mountain. The climb is made in 5.0 h
starting at an elevation of 2300m. Calculate the following:
a) The work done by the hiker against gravity
b) The average power output in watt and in horsepower
c) Assuming the body is 15% efficient, what rate of energy input was required?
1
Expert's answer
2020-05-25T10:55:01-0400

a) work done is equal to change in the potential energy

work done = mgΔh=65×10×(37002300)=650×1400=910kJmg\Delta h=65\times10\times(3700-2300)=650\times1400=910kJ


b) power = Work donetime=910×1035×60×60=50.55W\frac{Work\ done }{time}=\frac{910\times10^3}{5\times60\times60}=50.55 W

1 HP =735 W

50.55 W= 0.068 HP


c) efficiency = outputinput×100\frac{output }{input}\times100

15=50.55input×100input=337W15=\frac{50.55}{input}\times100\\input=337 W


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