A 60 HP electric motor lifts an elevator of maximum total load 2000 kg. If the frictional force is 4000 N, find the elevator speed at full load. (1 HP=746 W, g=10 m/s²)
Options

Question

A 60 HP electric motor lifts an elevator of maximum total load 2000 kg. If the frictional force is 4000 N, find the elevator speed at full load. (1 HP=746 W, g=10 m/s²)

G1 Physics · Accepted Answer

Motor power is P=60×746=44760 W. At constant full-load speed the motor balances mg+f=2000×10+4000=24000 N. Hence v=P/F=44760/24000≈1.87 m/s≈1.9 m/s, option C.