NEET2022PhysicsWork, Power and EnergyActual
An electric lift with a maximum load of 2000   kg (lift + passengers) is moving up with a constant speed of 1 . 5   m   s - 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
Options
- A20000
- B34500
- C23500
- D23000
Correct answer
B. 34500
Step-by-step solution
Since the speed is constant the motor has to give equal and opposite upward force. ∴       F = 3000 N + 2000 × 10 = 23000   N P = F V = 23000 × 1 . 5 = 34500   W