A 60 HP electric motor lifts an elevator having a maximum total load capacity of $2000 \mathrm{~kg}$. If the frictional force on the elevator is $4000 \mathrm{~N}$, the speed of the elevator at full load is
close to: $\left(1 \mathrm{HP}=746 \mathrm{~W}, \mathrm{~g}=10 \mathrm{~ms}^{-2}\right)$
Correct Option: , 2
(2) Total force required to lift maximum load capacity against frictional force $=400 \mathrm{~N}$
against frictional force $=400 \mathrm{~N}$
$F_{\text {total }}=M g+$ friction
$=2000 \times 10+4000$
$=20,000+4000=24000 \mathrm{~N}$
Using power, $P=F \times v$
$60 \times 746=24000 \times v$
$\Rightarrow v=1.86 \mathrm{~m} / \mathrm{s} \approx 1.9 \mathrm{~m} / \mathrm{s}$
Hence speed of the elevator at full load is close to
$1.9 \mathrm{~ms}^{-1}$
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