An electric lift with a maximum load of $2000 k g$ (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: $(g = 10 {m s}^{- 2})$

34500

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23000

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20000

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23500

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Solution

The correct option is A $34500$
Total force required to move with constant speed,

Force $=$ weight $+$ friction

$F = 20000 + 3000$

$F = 23000 N$

Power, $P = F \cdot V$

$= 23000 \times 1.5$

$P = 34500 W$

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An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms−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 ms−2)

The correct option is A 34500Total force required to move with constant speed, Force = weight + friction F=20000+3000 F=23000 N Power, P=F⋅V =23000×1.5 P=34500 W

An electric lift with a maximum load of $2000 k g$ (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: $(g = 10 {m s}^{- 2})$

The correct option is A $34500$
Total force required to move with constant speed,

Force $=$ weight $+$ friction

$F = 20000 + 3000$

$F = 23000 N$

Power, $P = F \cdot V$

$= 23000 \times 1.5$

$P = 34500 W$