Searches related to a motor lifts an elevator of mass m at constant speed. The power output of the motor is P and there are no frictional losses so that all the power output from the motor goes into lifting the elevator mass. What is the correct expression for the time required for the motor to lift the elevator to a height h?
P=mgh/time
solve for h.
an elevator weighing400kg is to be lifted up at a constant velocity of 3cms-1 what would be the minimum power of motor to be used
To determine the expression for the time required for the motor to lift the elevator to a height h, we can use the work-energy principle. The work done on the elevator is given by the product of the force applied and the distance over which it is applied. In this case, the force is equal to the weight of the elevator, which is given by the mass m multiplied by the acceleration due to gravity g. The distance is the height h.
Therefore, the work done on the elevator is W = mgh, where m is the mass of the elevator, g is the acceleration due to gravity, and h is the height.
The power output of the motor is given by P = W/t, where P is the power output, W is the work done on the elevator, and t is the time required.
Rearranging this equation, we get t = W/P. Substituting the expression for work W, we have t = mgh/P.
So, the correct expression for the time required for the motor to lift the elevator to a height h is t = mgh/P.