# Homework Help: Kinetic energy of water pouring

1. Mar 6, 2013

### kox

Hello. It's not a homework but it is kind of exercise. I was wondering what's the kinetic energy of water flow (for a certain amount) from a faucet. But I'm not sure if I got it right. So I counted how much volume of water is dropped in 30 seconds. It turned out it's about 5.6liters.
m=5.6kg // as 1 liter of water is almost 1kg
t=30s
l=1.256*10E-3m^2 // area of water dropping, cross-sectional area
Vol=5.6*10E-3m^3 // volume of water
Ek=mv^2/2
v=s/t
s=Vol/l
s=5.6*10E-3/1.256*10E-3 [m^3/m^2=m]
s=4.45m
v=4.45m/30s
v=0.148m/s
Ek=5.6kg*(0.148m/s)^2/2
Ek=5.6*0.0219kg*m^2/s^2
Ek=0.0613J
Ek=61.3mJ
So I got that pouring 5.6 liters of water realeses 61.3 milijoules. I thought it woul be a little more when I see water dropping at high pressure.
Did I get that right?
I don't know if I posted it in the right section. I'm ew here. It's not a homework but I'm afraid I would get my post deleted if I posted it in classic physics. Sorry I didn't use the template provided,but they didn't fit my post too good.

2. Mar 6, 2013

### Staff: Mentor

The kinetic energy of slowly moving masses is tiny. You get the same kinetic energy if you drop the water by a height of just ~1mm.

3. Mar 6, 2013

### kox

I don't understand how is it equal to dropping water (how much?) from 1mm height? The water was dropping from more height and was at pressure. Could you explain?

4. Mar 7, 2013

### Staff: Mentor

In that case, some energy went to something else.

Dropping 5.6kg of water by a height of 1mm gives releases potential energy of mgh=5.6kg*10m/s2*0.001m = 0.056J = 56mJ.