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 ms^–2)

34500

To solve this problem, we need to calculate the minimum power delivered by the motor to the electric lift with a maximum load of 2000 kg. The lift is moving upwards with a constant speed of 1.5 m/s against a frictional force of 3000 N. Given the acceleration due to gravity, g, is 10 m/s², let's determine the power step by step.

Concept: Power is the rate of doing work or the rate at which energy is transferred. When an object moves with a constant speed, the motor must provide enough force to counter both the gravitational force acting on the lift and the frictional force. The total force required is given by:

Total Force (F) = Weight of the lift + Frictional Force
= (m × g) + f
= (2000 × 10) + 3000
= 20000 + 3000
= 23000 N

Calculation:
The power delivered by the motor is given by the formula:
P = F × v
Substituting the values:
P = 23000 × 1.5
P = 34500 W