# A 60 HP electric motor lifts an elevator having a maximum total load capacity

Question:

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: $\left(1 \mathrm{HP}=746 \mathrm{~W}, \mathrm{~g}=10 \mathrm{~ms}^{-2}\right)$

1. (1) $1.7 \mathrm{~ms}^{-1}$

2. (2) $1.9 \mathrm{~ms}^{-1}$

3. (3) $1.5 \mathrm{~ms}^{-1}$

4. (4) $2.0 \mathrm{~ms}^{-1}$

Correct Option: , 2

Solution:

(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

$1.9 \mathrm{~ms}^{-1}$