An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 k g . The mass of the elevator itself is 920 k g and it moves with a constant speed of 3 m /

Question

An elevator in a building can carry a maximum of $10$ persons, with the average mass of each person being $68 k g$ . The mass of the elevator itself is $920 k g$ 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 $(g = 10 m / s^{2})$ must be at least:

Options:

Answer:

D

Solution:

Net force on motor will be
$F_{m} = [920 + 68 (10)] g + 6000$
$= 22000 N$
So, required power for motor
$P_{m} = {\vec{F}}_{m} . \vec{v}$
$= 22000 \times 3$
$= 66000 w a t t$

TestHub · Accepted Answer

Net force on motor will be F m = 920 + 68 10 g + 6000 = 22000 N So, required power for motor P m = F → m . v → = 22000 × 3 = 66000 w a t t

Physics - Kinematics Question with Solution | TestHub

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