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 3000N. The minimum power delivered by the motor to the lift in watts is: (g =10 ms^-2)

• A. 23000
• B. 20000
• C. 34500
• D. 23500

Answer 1

The Correct Option is C

To find the minimum power delivered by the motor to the lift, we need to consider both the force needed to overcome gravity and the frictional force. The power \( P \) required by the motor can be calculated using the formula:

\( P = F_{\text{total}} \times v \)

where \( F_{\text{total}} \) is the total force opposing the lift's movement, and \( v \) is the velocity of the lift.

First, calculate the gravitational force \( F_g \):

\( F_g = m \times g \)

Given: \( m = 2000 \, \text{kg} \) (mass of the lift + passengers), \( g = 10 \, \text{ms}^{-2} \) (acceleration due to gravity).
Therefore,

\( F_g = 2000 \times 10 = 20000 \, \text{N} \)

Next, add the gravitational force and the frictional force to get the total force:

\( F_{\text{total}} = F_g + F_{\text{friction}} = 20000 + 3000 = 23000 \, \text{N} \)

Now, use the formula for power:

\( P = 23000 \, \text{N} \times 1.5 \, \text{ms}^{-1} = 34500 \, \text{W} \)

Thus, the minimum power delivered by the motor is 34500 watts.

Answer 2

Total force needed to go at a steady speed, 
Force is equal to weight plus friction. 
F = 20,000 + 3,000 
F equals 23000. N 

Power is equal to F ⋅ V. 
= 23000 × 1.5 
P is 34500. W

The correct answer is (C) : 34500.