Question

An elevator which 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?

• 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, $v = 2\, ms^{-1}$ 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, $P = Fv = (22000 \,N)(2\,ms^{-1})$ $= 44000 \,W = 44 \times 10^3\, W = 44\, kW$