Elevator starts from rest with constant acceleration

[splac6996]

Homework Statement

650kg elevator starts from rest. It moves upward for 3.00s with constant acceleration until it reches its cruising speed of 1.75m/s. What is the average power of the elevator motor during this period?

Homework Equations

use either force or conservation of energy

The Attempt at a Solution

I first used my kinematic equations to find the height which the elevator travels where I am getting stuck is by finding the work to solve for power is it as simple as using w=force*distance

 

[Dick]

It is that simple. The force is constant during the acceleration phase. You just have to find it. You already have distance from your kinematics.

 

[splac6996]

if I wanted to find work by using energy equation what would that look like. The reason i ask is because I should be able to use that equation as well and get the same answer.

 

[Dick]

I think you do need to do some kinematics. There's lots of ways to go from 0 to 1.75m/sec, all with different power requirements. You need both final time and final distance and the relation between those is kinematics.

 

[splac6996]

well I know that my final time is 3.00s and the final distance I found to be 2.63m. But for my energy equation should equal work

 

[Dick]

Ok, then you are all set. You could add the final kinetic energy and final gravitational potential energy to get total work. Or you could work out force and multiply by the distance. Both should give you the same result.