To find the power generated by the turbine, we first calculate the total input power available from the falling water and then account for energy losses.

1. Input Power ($P_{\text{in}}$): Power is defined as the rate of change of energy. For falling water, the input power is the rate at which gravitational potential energy ($mgh$) is delivered. $P_{\text{in}} = \frac{dm}{dt} \cdot g \cdot h$ Given the mass flow rate $\frac{dm}{dt} = 15 \text{ kg/s}$, $g = 10 \text{ m/s}^2$, and $h = 60 \text{ m}$: $P_{\text{in}} = 15 \times 10 \times 60 = 9,000 \text{ W} = 9.0 \text{ kW}$

2. Accounting for Losses: The problem states that 10% of the energy is lost to friction. Therefore, the turbine operates with an efficiency ($\eta$) of 90% ($0.9$).

3. Generated Power ($P_{\text{out}}$): $P_{\text{out}} = P_{\text{in}} \times \eta$ $P_{\text{out}} = 9.0 \text{ kW} \times 0.9 = 8.1 \text{ kW}$

Thus, the power generated by the turbine is 8.1 kW.