Question

An elevator weighing 500 kg is to be lifted up at a constant velociyt of 0.20 m/s. What would be the minimum horsepower of the motor to be used?

Answer 1

As the elevator is going up with a uniform velociyt, the total work done on it is zero in any time interval. The work done by the motor is, therefore, equal to the work done by the force of gravity in that interval (in magnitude). The rate of doing work, i.e., the power delivered is
`P=Fv-mgv`
`=(500kg)(9.8m/s^2)(0.2m/s)=980W` Assuming no loss against friction etc., in the motor, the minimum horsepower of the motor is `
P=980W=980/746hp=1.3hp.`