Question

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms^-2)

(1) 1.7 ms^-1 

(2) 2.0 ms^-1 

(3) 1.9 ms^-1

(4) 1.5 ms^-1

Answer 1

Answer is (3) 1.9 ms^-1

Let elevator is moving upward with constant speed V. Tension in cable

T = 2000 g + f_r = 2000 + 4000

T = 24000 N

Power P = TV 

60 × 746 = (24000) V

Answer 2

Answer is (3) 1.9 ms^-1

4000 × V + mg × V = P

((60 x 746)/(4000 + 20000) = V

V = 1.86 m/s = 1.9 m/s