When an elevator starts going up from rest, does the motor apply a force larger than mg for a brief moment?

Question

I'm new to physics.

Suppose an elevator is at rest and then starts going up at a constant velocity. I think what happens is, at first, the motor pulls on the elevator with a force larger than mg on the elevator in order to accelerate it.

$$a = \frac{F_{net}}{m} = \frac{F_{motor}-mg}{m} = C > 0$$

Immediately afterward, this force is reduced so that it is equal to mg to keep the velocity constant.

$$a = \frac{F_{net}}{m} = \frac{F_{motor}-mg}{m} = 0$$

Am I correct?

Does the same happen when I pick up an object from the floor and lift it up at a constant velocity? Does my arm also apply a force larger than mg and immediately adjust it to be equal to mg?

Thank you.

Answer 1

As long as you are accelerating any massive object upward against gravity, you are applying a force to it in excess of m x g. If the object is thence traveling upward at constant velocity, the force being applied is equal to m x g.

This excludes friction effects. Your mileage will be lower in California.

Answer 2

Yes, you are correct. Another way to think is by Newton's $1^{st}$ Law. An object will stays at rest, until and unless an external force is applied. So, if lift was at rest initially and starts going up with constant velocity, there must be a time duration where external force was applied. To answer your further, question yes your arm will also act as the same way. Though there are a lot more going on, but if you assume point particle analysis and Newtonian mechanics, your argument holds true.