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**[SOLVED] work and power**

**1. Homework Statement**

the Grand Coulee Dam on the Columbia River is 1270 m long and 170 m high. The electrical power output from the generators at its base is approximately 2000 MW, How many cubic meters of water much flow from the top of the dam per second to produce this amount of power if 92% of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000kg)

**2. Homework Equations**

W=kf-ki=1/2mv^2

W=F*deltax*costheta

P=W/deltaT

F=m*a

**3. The Attempt at a Solution**

I attempted solving this problem using unknowns since I did not know another way to approach it. so I said

F=m(9.8)

W=m(9.8)(170 m) (costheta)

then I plugged Work into the Power equation:

P=[(m(9.8)(170 m) (costheta))/deltaT](.92)

and I multiplied the whole equation by .92 for the 92% in the problem

Now it seems that the only known in the problem is Power. I could solve for mass using the unknowns of time and theta, but I feel like that is the wrong approach and am not sure how to solve otherwise.

Thanks!