An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?

An elevator can carry a maximum load of 1800 kg (celevator + passengers) is moving up with a constant speed of 2 ms^(-1) . The friction force oppposite the motion is 4000 N .What is minimum power delivered by the motor to the elevator?

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))

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)

Sand drops from a stationary hopper at the rate of 5kg//s on to a conveyor belt moving with a constant speed of 2m//s . What is the force required to keep the belt moving and what is the power delivered by the motor, moving the belt?

An object of mass 10kg moves at a constant speed of 10 ms^(-1) . A constant force, that acts for 4 sec on the object, gives it a speed of 2ms^(-1) in opposite direction. The force acting on the object is :

A truck of mass 500kg is moving with constant speed 10ms^(-1) . If sand is dropped into the truck at the constant rate 10kg/min ,the force required to mainatain the motion with constant velocity is .