First slide

Power

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. What is minimum power delivered by the motor to the elevator?

Easy

A

22 kW
B

44 kW
C

66 kW
D

88 kW

Solution

Here, m = 1800 kg
Frictional force, f = 4000 N
Uniform speed, v = 2 ${ms}^{- 1}$
Downward force on elevator is
$F = mg + f$
= $(1800 kg \times 10 {ms}^{- 2}) + 4000 N = 22000 N$
The motor must supply enough power to balance this force. Hence,
$P = Fv = (22000 N) (2 {ms}^{- 1})$
= $44000 W = 44 \times 10^{3} W = 44 kW$

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