Solution:
The input power is \(P_{\text{in}} = \frac{dm}{dt} gh = 15 \times 10 \times 60 = 9000\text{ W} = 9\text{ kW}\). Frictional losses are 10%, meaning the output efficiency is 90%. Thus, generated power is \(0.90 \times 9\text{ kW} = 8.1\text{ kW}\).
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