Solving Power/Work Problem: Average Power of Elevator

[Aubiefan]

I have a problem about average power that I am stuck on:
A 686 kg elevator starts from rest and moves upward for 3.10 s with constant acceleration until it reaches its cruising speed, 1.70 m/s.
a) What is the average power of the elevator during this period?
b) What is the average power during an upward cruise with constant speed?

I used (delta x) = (V+Vo)/2 x t to solve for delta X, and got 2.635 m. I used V=Vo+at to solve for acceleration, and got 0.55 m/s^2 (I also used delta V over delta t and got the same answer).

I know that power is Work/time, which is F(delta x)/(delta t). I used F=mg to find a force of 994.18 N, and plugged it into work = F(delta X) to get 320.705 J.

I tried dividing this by my time, 3.1s, and got 320.705, but the Webassign page told me I was not only wrong but was off by orders of magnitude. Any tips on what I'm doing wrong?
thanks for your time and patience!

 

[Doc Al]

I know that power is Work/time, which is F(delta x)/(delta t). I used F=mg to find a force of 994.18 N, and plugged it into work = F(delta X) to get 320.705 J.

Careful: While the elevator accelerates, the force on it must be greater than its weight.

Consider: Work = change in energy (PE + KE)

 

[radou]

...I know that power is Work/time, which is F(delta x)/(delta t). I used F=mg to find a force of 994.18 N, and plugged it into work = F(delta X) to get 320.705 J...

How did you use F = mg to find the force? Which force is it? Remember, Newton's law of motion states that the resultant of all forces equals the product of mass and acceleration. You got the acceleration and displacement right. But which forces are acting on the elevator? Make a sketch.

http://physics.bu.edu/~duffy/semester1/c05_elevator.html"

 

[SGT]

The work done must be equal to the increase of energy. In this case you have two energies involved:
Kinectic energy due to the acceleration of the mass by 0.55m/s^2.
Potential energy, since the elevator is going up.