Calculating Pump Motor Output: Where Did I Go Wrong?

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To calculate the required output rating for a pump motor lifting 16.0 kg of water through a height of 3.38 m per minute, only the change in potential energy needs to be considered, not kinetic energy. The formula for potential energy is mgh, where m is mass, g is gravitational acceleration, and h is height. The total energy required to lift the water is calculated as 16 kg * 9.8 m/s² * 3.38 m. This energy must be divided by 60 seconds to find the power in watts, as power is defined as energy per unit time. The initial negative output indicates a miscalculation, emphasizing the importance of focusing solely on potential energy in this scenario.
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A pump is to lift 16.0 kg of water per minute through a height of 3.38 m. What output rating (watts) should the pump motor have?

Potential Energy Initial = O
Kenetic Energy Final = 0

So,
KE-PE = W
1/2mv^2 - mgh = Work
1/2(16.0)(3.38)^2 - (16.0)(9.8)(3.38) = -439 W

The fact that my answer is negative automatically tells me that I did something wrong as the pump wouldn't have a negative output. Does anyone see where I went wrong?
 
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There is no kinetic energy involved in this problem at all, so that's your first trouble spot. The only thing that matters is the change in potential energy of the water when lifted 3.38 meters.

The pump has to lift 16 kg 3.38 m every minute. It has to expend energy to do this. It requires m g y joules to lift a mass of any substance y meters:

http://www.google.com/search?hl=en&q=16+kg+*+9.8+m/s/s+*+3.38+m&btnG=Google+Search

This energy is expended over the course of a minute; the pump thus has to expend a sixtienth of this energy per second. Power, in watts, is equivalent to one joule per second:

http://www.google.com/search?hl=en&lr=&q=(16+kg+*+9.8+m/s/s+*+3.38+m)+/+60+seconds&btnG=Search

- Warren
 
Wow, I was way off. Thanks for your help, it helped me with another problem I was stuck on to.
 
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