Question

An elevator can carry a maximum load of 2000 kg (elevator + passengers) and is moving up with a constant speed of 9 km/h. If the frictional force opposing the motion is \( 5 \times 10^3 \) N, the minimum power delivered by the motor to the elevator is (acceleration due to gravity = 10 ms\(^2\))

• A. 26.25 kW
• B. 52.5 kW
• C. 31.25 kW
• D. 62.5 kW

Hint:

Power calculations in mechanical systems often involve considering both the force required to overcome resistance and the velocity at which the system operates.

Answer 1

The Correct Option is D

Given: - Maximum load, \( m = 2000 \, {kg} \) - Speed, \( v = 9 \, {km/h} \) - Frictional force, \( f = 5 \times 10^3 \, {N} \) - Acceleration due to gravity, \( g = 10 \, {ms}^{-2} \) Step 1: Convert speed to meters per second \[ v = 9 \, {km/h} = \frac{9 \times 1000}{3600} = 2.5 \, {ms}^{-1} \] Step 2: Calculate the gravitational force The gravitational force \( F_g \) acting on the elevator is: \[ F_g = m \times g = 2000 \times 10 = 2 \times 10^4 \, {N} \] Step 3: Determine the total force required The motor must overcome both the gravitational force and the frictional force to move the elevator upward at a constant speed. Therefore, the total force \( F \) required is: \[ F = F_g + f = 2 \times 10^4 + 5 \times 10^3 = 2.5 \times 10^4 \, {N} \] Step 4: Calculate the power delivered by the motor Power \( P \) is given by the product of force and velocity: \[ P = F \times v = 2.5 \times 10^4 \times 2.5 = 6.25 \times 10^4 \, {W} \] Convert watts to kilowatts: \[ P = 6.25 \times 10^4 \, {W} = 62.5 \, {kW} \] Final Answer:
62.5 kW