## University Physics with Modern Physics (14th Edition)

The required volume of water is $1300~m^3$.
The power output is 2000 MW, which is $2.0\times 10^9~W$ 92% of the work done by gravity on the water is converted to electricity. $0.92~Work = (2.0\times 10^9~W)(1~s)$ $Work = 2.17\times 10^9~J$ The work done by gravity on the water is $mgh$. $mgh = 2.17\times 10^9~J$ $m = \frac{2.17\times 10^9~J}{gh}$ $m = \frac{2.17\times 10^9~J}{(9.80~m/s^2)(170~m)}$ $m = 1.3\times 10^6~kg$ We can find the volume $V$ of water in cubic meters. $V = \frac{1.3\times 10^6~kg}{1000~kg/m^3} = 1300~m^3$ The required volume of water is $1300~m^3$.