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

Step 1: Understanding the Concept:
Power is defined as the rate of doing work. In mechanical systems involving linear motion, power is the product of the applied force and the velocity of the object.
When the lift moves at a constant speed, the acceleration is zero. According to Newton's Second Law (\(F_{net} = ma\)), the net force acting on the lift must be zero.
Therefore, the upward force exerted by the motor must exactly balance all the downward forces.
Key Formula or Approach:
1. Total Downward Force (\(F_d\)) = Weight of the lift (\(mg\)) + Frictional force (\(f\)).
2. Force provided by motor (\(F_m\)) = \(F_d\) (to maintain constant speed).
3. Power delivered (\(P\)) = \(F_m \times v\).
Step 2: Detailed Explanation:
1. Calculate the Weight force:
Mass \(m = 2000 \text{ kg}\).
Acceleration due to gravity \(g = 10 \text{ m/s}^2\).
Weight \(W = mg = 2000 \times 10 = 20000 \text{ N}\).
2. Identify the Frictional force:
The lift is moving up, so the frictional force acts downwards to oppose the motion.
Friction \(f = 3000 \text{ N}\).
3. Calculate total upward force required:
Since speed is constant, the upward pull must cancel both gravity and friction.
\(F_{up} = W + f = 20000 \text{ N} + 3000 \text{ N} = 23000 \text{ N}\).
4. Calculate Power:
Velocity \(v = 1.5 \text{ m/s}\).
\(P = F_{up} \times v\)
\(P = 23000 \times 1.5\)
\(P = 34500 \text{ Watts}\).
Step 3: Final Answer:
The minimum power delivered by the motor is 34500 W.