If the maximum load carried by an elevator is 1400 kg (600 kg - Passengers + 800 kg- elevator), which is moving up with a uniform speed of 3 ms-1 and the frictional force acting on it is 2000 N, then the maximum power used by the motor is ____ kW. (g=10 m/s2)
Correct Answer: 48
Solution and Explanation
Given
Mass of the elevator system = 1400 kg
Velocity (\(V\)) = \(3 m/s^-1\)
Frictional force (\(f\)) = 2000 N
The net force on the elevator is zero, as it is moving with uniform speed. So, the upward force (tension \(T\)) must balance the downward forces, which are the gravitational force (\(Mg\)) and the frictional force. 1. Tension in the string: \[ T = Mg + f = 1400 \times 10 + 2000 = 14000 + 2000 = 16000 \, \text{N} \] 2. The maximum power used by the motor is given by: \[ \text{Maximum Power} = F \times V = T \times V = 16000 \times 3 = 48000 \, \text{W} = 48 \, \text{kW} \] Thus, the maximum power used by the motor is 48 kW.
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