An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of . The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is [2022]

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Solution

Q.
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of $1.5$ $m$ $s^{- 1}$ . The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is $(g = 10$ $m$ $s^{- 2})$ [2022]

1 23000

2 20000

3 34500

4 23500

Ans.
(3)

Given that,

Mass of lift + passengers, $m = 2000 kg$

Speed, $v = 1.5 m/s$

Frictional force, $f = 3000 N$

Power delivered, $P = Force \times velocity$ ...(i)

Force acting

$F = m g + f$

$F = 2000 \times 10 + 3000$

$F = 23000$ $N$

Using value of $' F'$ in equation (i),

$P = 23000 \times 1.5 = 34500$ $W$

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