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 ) ` )

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms^(-2) ).

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms^(-2) ).

A 120 HP electric motor lifts an elevator having a maximum total load capacity of 3000 kg. If the frictional force on the elevator is 6000N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms^(-2) ).

An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 m//s^2 )

An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 m//s^2 )

An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 m//s^2 )

If the tension in the cable of 1000 kg elevator is 1000 kg weight, the elevator

An elevator which can carry a maximum load of 1800kg is moving with a constant speed of 2ms^(-1) . The frictional force opposing the motino is 5000N . Calculate the minimum power delivered by the motor to the elevator in watt and hp ? Take g=10m//s^(2)

An elevator that can carry a maximum load of 1500 kg (elvator+ passengers) is moving up with a constant speed of 2 ms^(-1). The frictional force opposing the motion is 3000 N. Find the minmum power delvered by the motor to the elevator in watts as well as in horse power (g=10ms^(-2))