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 ) [2022]A: 20000 B: 34500 C: 23500 D: 23000
Question
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 ) [2022]A: 20000 B: 34500 C: 23500 D: 23000
Solution
The power (P) needed to move the lift can be calculated using the formula:
P = F * v
where F is the total force and v is the velocity.
The total force is the sum of the weight of the lift and the frictional force. The weight of the lift can be calculated using the formula:
F = m * g
where m is the mass of the lift (2000 kg) and g is the acceleration due to gravity (10 m/s²). So,
F = 2000 kg * 10 m/s² = 20000 N
The total force is then:
F_total = F_weight + F_friction = 20000 N + 3000 N = 23000 N
Substituting F_total and v (1.5 m/s) into the power formula gives:
P = 23000 N * 1.5 m/s = 34500 W
So, the minimum power delivered by the motor to the lift is 34500 W. Therefore, the correct answer is B: 34500.
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