Q. An elevator of mass $500kg$ is to be lifted up at a constant velocity of $0.4ms^{- 1}$ . What should be the minimum horse power of the motor to be used? (Take $g=10ms^{- 2}$ and $1hp=750$ watts $\left.:-$

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A

$8/3hp$

B

$4/3hp$

C

$2000hp$

D

$75hp$

Solution:

$P=F\cdot v=mgv=500\times 10\times 0.4W$ $=2000W$
$=\frac{2000}{750}hp=\frac{8}{3}hp$

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