A $60 H P$ electric motor lifts an elevator with a maximum total load capacity of $2000 k g$ . If the frictional force on the elevator is $4000 N$ , the speed of the elevator at full load is close to (Given $1 H P = 746 W, g = 10 m / s^{- 2}$ )

$1.5 m s^{- 1}$

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$1.7 m s^{- 1}$

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$2.0 m s^{- 1}$

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$1.9 m s^{- 1}$

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Solution

The correct option is D
$1.9 m s^{- 1}$

Step 1: Given data
Power of electric motor, $P = 60 H P$ [ $1 H P = 746 W$ ]
$P = 60 \times 746 W$
$P = 44760 W$
Maximum total load capacity, $m = 2000 k g$
Frictional force on the elevator, $F = 4000 N$
Step 2: Find the force of the elevator to lift the load at constant velocity, $F_{E}$
$F_{E} =$ Work done, $W +$ Frictional force $F$
$F_{E} = m g + 4000$ [ $W = m g$ and $g = 10 m s^{- 2}$ is acceleration due to gravity]
$F_{E} = 2000 \times 10 + 4000$
$F_{E} = 24000 N$
Step 3: Find the speed of the elevator at full load, $v$
To find the speed, firstly write the equation for the power
Applied power, $P =$ force $F_{E} \times$ speed, $v$
Applied power, $P = F_{E} \times v$
$v = \frac{P}{F_{E}}$
$v = \frac{44760}{24000}$
$v = 1.9 m s^{- 1}$
Hence, option $(D)$ is correct.

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A 60HP electric motor lifts an elevator with a maximum total load capacity of 2000kg. If the frictional force on the elevator is 4000N, the speed of the elevator at full load is close to (Given 1HP=746W,g=10m/s-2)

A $60 H P$ electric motor lifts an elevator with a maximum total load capacity of $2000 k g$ . If the frictional force on the elevator is $4000 N$ , the speed of the elevator at full load is close to (Given $1 H P = 746 W, g = 10 m / s^{- 2}$ )