An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms -1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by motor to the elevator.

Question

An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms -1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by motor to the elevator. (A) 18000 W (B) 27000 W (C) 36000 W (D) 44000 W

Tardigrade Admin · Accepted Answer

The downward force on the elevator is F=m g+Ff=(1800 × 10)+4000=22000 N. The motor must supply enough power to balance this force. Hence, P= barF ⋅ barv=22000 × 2=44000 W

Q. An elevator can carry a maximum load of $1800 k g$ (elevator + passengers) is moving up with a constant speed of $2 m s^{- 1}$ . The frictional force opposing the motion is $4000 N$ . Determine the minimum power delivered by motor to the elevator.

18000 W

27000 W

36000 W

44000 W

The downward force on the elevator is $F = m g + F_{f} = (1800 \times 10) + 4000 = 22000 N$ . The motor
must supply enough power to balance this force. Hence, $P = \overset{ˉ}{F} \cdot \overset{v}{ˉ} = 22000 \times 2 = 44000 W$

Q. An elevator can carry a maximum load of 1800kg (elevator + passengers) is moving up with a constant speed of 2ms−1. The frictional force opposing the motion is 4000N. Determine the minimum power delivered by motor to the elevator.