An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms^⁻¹. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 m s^⁻²)

Correct option is C

​$$\textbf{Given:} \\\bullet\ \text{Mass } m = 2000\,\text{kg} \\\bullet\ \text{Speed } v = 1.5\,\text{m/s} \\\bullet\ \text{Frictional force } f = 3000\,\text{N} \\\bullet\ \text{Gravitational force } mg = 2000 \times 10 = 20000\,\text{N}\\[2em]\text{When moving at \textbf{constant speed}, the net work is done \textbf{against gravity and friction}.}\\[1em]\text{Total opposing force } = mg + f = 20000 + 3000 = 23000\,\text{N} \\\text{Power } = \text{Total force} \times \text{Velocity} \\P = 23000 \times 1.5 = 34500\,\text{Watts}$$​​