Determine the power input for a motor necessary to lift 300

Determine the power input for a motor necessary to lift 300 lb at a constant rate of 5 ft/s. The efficiency of the motor is belongsto = 0.65.

Solution

We know that Output power=Force * velocity

=300(4.448)*5(0.3048) becoz 1 lb=4.448 Newtons & 1 Feet=0.3048 meters

   =2033.6 Newton -meter/ sec

given Efficiency=0.65

Now Efficency= output power / input power

0.65= 2033.6/input power

now input Required to the Motor=2033.6/0.65=3128.65 watts