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Q. 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 $\left(g=10\, m / s ^{2}\right)$ must be at least

Work, Energy and Power

Solution:

Mass of elevator, $M=920\, kg$
Mass of all ' $10$ ' passengers carried by elevator $=10 \times m$
$=10 \times 68=680\, kg$
Total weight of elevator and passengers
$=(M+10 m) g=(920+680) \times 10=16000\, N$
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Force of friction $=6000\, N$
Total force $(T)$ applied by the motor of elevator
$=16000+6000=22000\, N$
Power delivered by elevator's motor,
$P =F \cdot v=22000 \times 3$
$\left[\because v=3\, ms ^{-1}\right]$
$=66000\, W$