- #1
amanda.ka
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Homework Statement
A pump draws 500 kg of water/min from a 12 m deep well. The water is ejected with a speed of 20 m/s.
a) What is the power output of the pump?
b) How high is the fountain (ignoring air resistance)
Homework Equations
The Attempt at a Solution
For part a I found the potential energy and kinetic energy:
PE: m*g*delta h = (500)(9.81)(12) = 58860 J
KE: 1/2mv^2(final) - 1/2mv^2(initial) = 1/2(500)(20)^2 - 0 = 100, 000 J
Then I used Power = change in Work/change in time
= KE+PE/60 seconds
= 2647.7 W
Is this correct?
And I'm stuck on part b, here is what I did but I am not sure if it is correct:
[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]
-1/2mv^2(initial) = mgh(final)
-1/2*20^2/9.81 = h (final)
so height final = 20.4 m
Any help is appreciated :)