Question

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

• A. 22 kW
• B. 44 kW
• C. 66 kW
• D. 88 kW

Answer 1

Answer: (B)

44 kW

Answer 2

Here, m = 1800 kg

Frictional force, f = 4000 N

Uniform speed, v = 2m s^-1

Downward force on elevator is

F = mg+f

=$(1800kg \times 10ms^{- 2}) + 4000N = 22000N$

The motor must supply enough power to balance this force. Hence,

$p = Fv = (22000N)(2ms^{- 1})$

= 44000 W = 44 ×10³ W = 44 kW