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There is an engine pumping liquid, for example, water, through a pipe. One way to figure out the engine's power is to times the force with the velocity of the liquid. The other way is to calculate the kinetic energy in the liquid per unit of time. Which is correct?

There is a similar question Work power energy contradiction in hose pipe problem [closed]. But it was closed due to the check-my-work and homework policy. Some answers on the Internet support the first way: times the force with the velocity.

However, some answers on physics.stackexchange.com support the second way: the engine power is equal to the kinetic energy.

Which is correct?

I didn't give any example in this post because I didn't want this post to get the same fate as Work power energy contradiction in hose pipe problem [closed] although I think this post has educational value. If you want to find examples, click the link. However, welcome to give examples. Examples are always the best way to demonstrate the concept and also a method we learn things from the fundamentals. This is true not only in the physics field.

Response from knzhou

knzhou is the author of the picked answer of Power of water pump. knzhou claimed the power of the engine should equal to the kinetic energy in the liquid per unit of time.

The physically correct answer is 400 W. As for what was marked "correct" on that test question, I have no idea -- it depends on whether the test writer knew what they were doing. Actually, many JEE prep problems are incorrectly posed, and don't have any correct answers.

Backup for the similar question

Work power energy contradiction in hose pipe problem [closed] has been deleted. I failed to cache the page with archive.org. To let others see the example, I copy-pasted that question from the page cached by Google.

  • Author: Neel Narlawar

Question: An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine?

Approach 1:

Force = vdm/dt = 2(2×100) = 400

Power= Force × Velocity = 800W

800 is the answer given in most of the books out there.

Approach 2:

Change in kinetic energy of water = 1/2 * m * v^2

Rate = 1/2 * m/t * v^2 = 1/2 * 200 * 4 = 400W

I am unable to find an error in either of the approaches so can someone tell me what's wrong in the second approach? Both the concepts seem correct, the former being related to the usual force power relation while the second uses kinetic energy work relation. Is there a problem in my understanding of the concept? Thank you!

Note: I have tried looking at other sources for clarification but nothing seems to answer my question properly.

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  • $\begingroup$ This question is conceptual and cannot be construed as "homework-type". You need not worry about it being closed due to that policy. Also, is there a reason why you think it has to be one method and not the other, as opposed to both being valid methods? $\endgroup$
    – joseph h
    Commented Jan 4, 2023 at 22:50
  • $\begingroup$ If we define engine power in one way, there should be only one correct answer. It's not possible that we input the same energy but get different velocity of liquid if other settings are the same. $\endgroup$
    – IvanaGyro
    Commented Jan 5, 2023 at 2:13
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    $\begingroup$ Mechanical power is change in kinetic energy per unit time. We also know that mechanical power is the product of force and velocity. Both are valid. That someone tried to do a calculation using both these definitions and got two different results does not change this. It just means that the calculations were done wrong. $\endgroup$
    – joseph h
    Commented Jan 5, 2023 at 2:29
  • $\begingroup$ Sorry, I can't understand what you mentioned. Can you explain more clearlier? $\endgroup$
    – IvanaGyro
    Commented Jan 5, 2023 at 2:37
  • $\begingroup$ I mean you can use both approaches: kinetic energy/time or $F.v$ Both methods will give the same results IF the calculations are done correctly. In the first question you linked, you did the calculations incorrectly and that’s why you get two different answers. Also, you should merge all your accounts into one. Go to meta.physics.stackexchange.com to see how. Cheers $\endgroup$
    – joseph h
    Commented Jan 5, 2023 at 22:38

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