To solve this problem, we need to calculate the minimum power delivered by the motor to the lift.$\\$Given:$\\$• Maximum load (lift + passengers) $= 2000$ kg$\\$• Constant speed $= 1.5$ m/s$\\$• Frictional force opposing the motion $= 3000$ N$\\$• Gravitational acceleration $g = 10$ m/s$^2 \\$First, calculate the gravitational force acting on the lift:$\\ F_{gravity} = m \cdot g \\ F_{gravity} = 2000 \cdot 10 \\ F_{gravity} = 20000$ N$\\$The total force that the motor needs to overcome is the sum of the gravitational force and the frictional force:$\\ F_{total} = F_{gravity} + F_{friction} \\ F_{total} = 20000 + 3000 \\ F_{total} = 23000$ N$\\$The power delivered by the motor is given by the product of the total force and the velocity:$\\ P = F_{total} \cdot v \\ P = 23000 \cdot 1.5 \\ P = 34500$ W$\\$Therefore, the minimum power delivered by the motor to the lift is $34500$ watts.$\\$This corresponds to Option 3.