How Much Work Does a Water Pump Do to Raise Water from a Well?

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Homework Help Overview

The problem involves calculating the work done by a water pump to raise water from a well that is 14.0 meters deep, with the water being pumped at a rate of 800 kg/minute and reaching a speed of 18.0 m/s at the top. Participants are discussing the implications of the information provided and the calculations involved.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the calculations related to acceleration, force, and work done. Some question the clarity of the problem statement and whether it provides sufficient information to determine the work done or if it is more appropriate to calculate power instead.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the problem's wording and the adequacy of the information provided. There is a recognition that the question may not lead to a definitive answer regarding work done, and some suggest that it might be more relevant to focus on power calculations.

Contextual Notes

Some participants note that the problem lacks clarity, particularly regarding the time frame for the acceleration of the water, which affects the interpretation of the work done versus power. There is also mention of discrepancies between the calculated results and the textbook answer.

student34
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Homework Statement



Water is pumped from a well at 800kg/minute. The well is 14.0 meters deep. By the time the water reaches the top of the well, it is traveling at 18.0m/s. How much work is done on the water by the pump?

Homework Equations



ƩFy = F - m*g = m*a
W = F*y
a = vf/t

The Attempt at a Solution



a = vf/t = (18.0m/s)/60.0s = 0.300m/s^2

F = m*a + m*g = 8080 N

W = F*y = 8080N*14.0m = 113,120 J

But the answer in my textbook is 1.30*10^5 J.
 
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The question does not make sense. The pump does some definite amount of work in some definite period of time. As time goes on, the work done by the pump increases without any limit.
 
As voko says, the question makes no sense as it stands. Some missing information?
student34 said:
a = vf/t = (18.0m/s)/60.0s = 0.300m/s^2

That calculation doesn't make sense either. The info does not say that it takes 1 minute to accelerate the water to 18m/s.
 
What is the kinetic energy per unit volume of the water at the top of the well? What is the kinetic energy per unit volume of the water before it is acted upon by the pump? What is the potential energy per unit volume of the water at the top of the well, compared to at the bottom of the well? What is the volume rate of flow of the water? What is the rate at which work is done by the pump on the water?
 
Something is wrong with the question or the textbook answer. The question only has enough info to calculate the power (Watts) not the energy (Joules).
 
CWatters said:
Something is wrong with the question or the textbook answer. The question only has enough info to calculate the power (Watts) not the energy (Joules).

I think the implication of the question was to get the power, although the question certainly was worded poorly.
 
In that case the textbook answer is also wrong.
 
CWatters said:
In that case the textbook answer is also wrong.

Yes. I agree. I get about 4 kW. Does that agree with your result?
 
Yes that's what I got.

I couldn't see an obvious way to modify the question to get the answer 1.30*10^5 J unless they missed out..
"How much energy is consumed if it pumps the water for 30 seconds".
 

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