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…

Question

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)

MARKS Content Team · Accepted Answer

Correct answer: (B) 34500. Since the speed is constant the motor has to give equal and opposite upward force. ∴ F=3000N+2000×10 =23000 N P=FV=23000×1.5 =34500 W

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 : $(g = 10 {m s}^{- 2})$

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Relevant Questions

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

Solution

Relevant Questions

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 : $(g = 10 {m s}^{- 2})$