Answer to Question #116407 in Quantum Mechanics for Foibe Kambala

Question

(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?

Answer 1

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

work done = $mg\Delta h=65\times10\times(3700-2300)=650\times1400=910kJ$

b) power = $\frac{Work\ done }{time}=\frac{910\times10^3}{5\times60\times60}=50.55 W$

1 HP =735 W

50.55 W= 0.068 HP

c) efficiency = $\frac{output }{input}\times100$

$15=\frac{50.55}{input}\times100\\input=337 W$