Question

A lift with a load of \(1000kg\) is moving up against the frictional force \(2000N\). If the power delivered to it by the operating motor is 36000 W, then the speed of the lift is \((g = 10 m s^{-2})\)

• A. \(2 ms^{-1}\)
• B. \(4 ms^{-1}\)
• C. \(3 ms^{-1}\)
• D. \(6 ms^{-1}\)
• E. \(10 ms^{-1}\)

Hint:

When an object moves at constant speed, the net force is zero. The applied force equals the sum of opposing forces.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The motor must overcome both the weight of the lift and the frictional force. Power is the product of force and velocity (\(P = Fv\)).

Step 2: Detailed Explanation:
Mass of lift with load, \(m = 1000 kg\). Gravitational force (weight), \(F_g = mg = 1000 \times 10 = 10000 N\) (downwards). Frictional force, \(F_f = 2000 N\) (downwards). Total downward force the motor must overcome, \(F_{total} = F_g + F_f = 10000 + 2000 = 12000 N\). Power delivered, \(P = 36000 W\). We know \(P = F_{total} \times v\), where \(v\) is the constant speed. \[ v = \frac{P}{F_{total}} = \frac{36000}{12000} = 3 \, m/s \]

Step 3: Final Answer:
The speed of the lift is \(3 ms^{-1}\).