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 4000N, the speed of the elevator at full load is close to:  1HP=746W,g=10ms−2

Both mg and friction are acting downwards. If elevator is moving up with uniform velocity, then  FfrV  +FmgV= P ⇒4000×V+(2000×10)×V=(60×746)                     ⇒V =  60×7464000+20000            ⇒ V=1.865m/s≈1.9m/s