Watts as opposed to Joules?

strugglin-physics

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?

chroot

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 [itex]m g y[/itex] joules to lift a mass of any substance y meters:

http://www.google.com/search?hl=en&q...=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&l...ds&btnG=Search

- Warren

strugglin-physics

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