# Homework Help: Work done to lift a mass of water

1. Mar 11, 2015

### henry3369

1. The problem statement, all variables and given/known data
A pump is required to lift 800 kg of water (about 210 gallons) per minute from a well 14.0 m deep and eject it with a speed of 18 m/s
(a) How much work is done per minute in lifting the water?
(b) How much work is done in giving the water the kinetic energy it has when ejected?

2. Relevant equations
Work = change in energy

3. The attempt at a solution
I don't understand the different between part a and part b.
Wouldn't the work done to lift it also be the work done to give it the final kinetic energy?

W = -ΔU or ΔK

2. Mar 11, 2015

### Staff: Mentor

They are breaking the total work required into two pieces:
(a) How much work is required just to lift the water? (Without imparting any kinetic energy.)
(b) How much additional work is needed to give it some kinetic energy?

3. Mar 11, 2015

### henry3369

(a) W = mgh
(b) W = (1/2)mv^2

Ok so I got those two. The next question asks for the total power output of the pump.
It seems that the answer comes from adding (a) and (b) then dividing by 60 seconds.
I'm confused because:
P = Net Work / time
Net Work = -ΔU or ΔK, and the former gives (a) while the latter gives (b).

Additionally, I tried using conservation of energy with nonconservative forces:
Ki + Ui + Wnc = Kf + Uf
Wnc = Kf + Uf. Using this for the power equation gives the correct answer, but this is the work due to ONLY the nonconservative forces. If I want to find the total power output, wouldn't I need net work?

4. Mar 11, 2015

### henry3369

I found another post that explained that work is actually the change in total mechanical energy. So does that mean Wnc = Net Work?

5. Mar 12, 2015

### Staff: Mentor

Yes and yes.

I would put it this way: The net work done must equal the change in mechanical energy (ΔU + ΔKE) of the water. (We are ignoring details like friction and so on.)