Work Energy Theorem Homework: Water Pump Power Output

AI Thread Summary
The discussion revolves around calculating the work done by a pump lifting 800 kg of water from a depth of 14.0 m and ejecting it at a speed of 8 m/s. It highlights the use of the work-energy theorem, emphasizing that work can be calculated either through integrating tension over displacement or by finding the difference in energy. The conversation points out that while the book suggests using the formula work done = mgh, it is important to consider the tension and force applied by the pump as well. The key takeaway is that these exercises aim to demonstrate alternative methods for calculating work, especially when acceleration is not zero. Understanding these concepts is crucial for solving similar physics problems effectively.
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Homework Statement



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 8ms-1 (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?

Homework Equations



Work Energy theorem

The Attempt at a Solution



The book attempt was that work done = mgh... Why don't we add tension as well... Or force applied by pump?
 
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hi ehabmozart! :smile:
ehabmozart said:
The book attempt was that work done = mgh... Why don't we add tension as well... Or force applied by pump?

you can calculate the work done in two ways …

by integrating the tension T times the displacement, total ∫ T dh

or (applying the work energy theorem) by simply finding the difference in energy! :-p

(if the acceleration is 0, so that T is constant, then T = mg, and ∫ T dh = mgh :wink:)

the whole point of these work-energy exercises is to show you how to avoid using the usual work done formula …

which would be very difficult if the acceleration wasn't 0!​
 
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