In an ornamental fountain, water is squirted vertically upwards through a nozzle by a pump. 5kg of water pass through the nozzle each second, and the water reaches a height of 5m after leaving the nozzle. What is the power of the pump (assuming 100% efficiency), and at what speed does the water leave the nozzle? (Take g to be 10N/kg)

Answer:

23) G

It is important to remember that power = energy per second.
Potential energy (PE) of water when it reaches a height of 5m = m x g x h
PE = 5 x 10 x 5 = 250J

This is the energy of 5kg of water which is released per second.
Hence, power = 250 J/s or 250 W

We know that all kinetic energy is converted to potential energy when it reaches a height.

Therefore PE = KE = 250 J
KE = ½mv²

250 = 0.5 x 5 x v²

100 = v²

v = 10 m/s

Hence the answer is G.