Kelvi

An electric lift with a maximum load of $2000$ kg (lift + passengers) is moving up with a constant speed of $1.5$ m/s. 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 $^2$ )

medium

Work, Energy and Power

2022

physics

23000

20000

34500

23500

Explanation

To solve this problem, we need to calculate the minimum power delivered by the motor to the lift. $\\$ Given: $\\$ • Maximum load (lift + passengers) $= 2000$ kg $\\$ • Constant speed $= 1.5$ m/s $\\$

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An electric lift with a maximum load of 2000 2000 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 1.5 1.5 m/s. The frictional force opposing the motion is 3000 3000 3000 N. The minimum power delivered by the motor to the lift in watts is: (g=10 g = 10 g=10 m/s2^2 2)

Explanation

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An electric lift with a maximum load of $2000$ kg (lift + passengers) is moving up with a constant speed of $1.5$ m/s. 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 $^2$ )