JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 10)
An elevator in a building can carry a maximum of 10 persons, with the average mass of each
person being 68 kg, The mass of the elevator itself is 920 kg and it moves with a constant speed
of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its
full capacity, the power delivered by the motor to the elevator (g = 10 m/s2) must be at least :
48000 W
62360 W
56300 W
66000 W
Explanation
Net force on motor will be
Fm = [920 + 68(10)]g + 6000 = 22000 N
So, required power for motor
P = Fm.V = 22000$$ \times $$3 = 66000 W
Fm = [920 + 68(10)]g + 6000 = 22000 N
So, required power for motor
P = Fm.V = 22000$$ \times $$3 = 66000 W
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