Work and kinetic energy comprehension question

[Lina]

Homework Statement

A pump is required to lift 790 kg of water (about 210 gallons) per minute from a well 14.1 m deep and eject it with a speed of 17.5 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? (c) What must be the power output of the pump?

Relevant Equations

Kinetic Energy, Work, Power

Hello,

I’ll start by saying I have the answers and the steps to the solutions, but there’s a comprehension disconnect somewhere that I’m trying to figure out. There are two parts to my question but the second one may not apply depending on the answer to the first. I wasn’t sure from the forum rules if I could include the actual solutions to the problem I‘m asking about, so I tried to do without, but let me know if it’d be better to add them.

KE: Kinetic Energy
W: Work
F: Force
s: Distance

1) Right now my understanding is that W = F * s and W = KE2 - KE1 are identical for the same scenario, however, (a) and (b) have different answers. I don’t fully understand why. Working theory is that the first two questions are actually looking at different parts in the system, but I would appreciate it if someone could confirm and clarify it a bit more.

2) If it’s that the problem is looking at the system in parts then I understand why the answer to (c) uses the sum of (a) and (b). If this is not the case though then I remain confused.

Thank you for taking a look.

 

[Orodruin]

(a)and (b) are different contributions to the work that needs to be done. There is no reason to believe they would be equal. First you need to lift the water against gravity. Then you need to add enough kinetic energy to give it the required speed. The total work input needed is the work required to do both those things.

 

[Lnewqban]

Welcome, Lina!
Could you please post the given answers to this problem?
Thank you.

 

[Steve4Physics]

my understanding is that W = F * s and W = KE2 - KE1 are identical for the same scenario

The 2nd equation (W = KE2 - KE1)is true only when W is the work done by the resultant force on an object.

For example, a force, F, could be used to lift a heavy object extremely slowly so that negligible kinetic energy is gained (you can go as slow as you want to make the kinetic energy gain as close to zero as you want).

But F has done work (W= Fs). F has increased the system’s potential energy. But since F wasn’t the resultant force we can’t say Fs = KE2 – KE1.

(The resultant force in the example is in fact the vector-sum of F and the object’s weight, F – mg.)

In your particular question about the water, it’s easiest to think only about the energy that has to supplied to the water each second – potential energy and kinetic energy. The question is best answered without thinking about the forces on the water.

 

[Lina]

The 2nd equation (W = KE2 - KE1)is true only when W is the work done by the resultant force on an object.

For example, a force, F, could be used to lift a heavy object extremely slowly so that negligible kinetic energy is gained (you can go as slow as you want to make the kinetic energy gain as close to zero as you want).

But F has done work (W= Fs). F has increased the system’s potential energy. But since F wasn’t the resultant force we can’t say Fs = KE2 – KE1.

(The resultant force in the example is in fact the vector-sum of F and the object’s weight, F – mg.)

In your particular question about the water, it’s easiest to think only about the energy that has to supplied to the water each second – potential energy and kinetic energy. The question is best answered without thinking about the forces on the water.

Thank you! That was a really clear explanation and it helped a bunch.