For linear acceleration of a mass, the power required will depend on how quickly you want to accelerate it.
This is because the Kinetic Energy formula is $E_k = \frac{1}{2} m v^2$. Note the v squared. So going from 0 to 1 m/s will cost less energy than going from 1 to 2 m/s, even if it accelerates at the same rate both times.
So I'll just make up more details of your example to show you how to do the math. You have a mass of 50 kg and a final velocity of 0.5 m/s, and of course it starts at zero velocity.
I'll choose an a of 0.5 m/s$^2$. That means it takes 1 second to accelerate to the final velocity.
The final KE of this is $0.5*50*0.5^2$ = 6.25 J. Therefore you had to do 6.25 joules of work over a time of 1 second. The unit watt is just 1 joule per second. Therefore you need 6.25 watts to do this acceleration...
However, this is only taking into account the linear force assuming no resistance of any kind besides inertia. Your problem said it must move up, in other words, against gravity. So we must take into account the change in potential energy.
$E_p = mgh$. First we have to find h. There is a kinematic formula to determine distance traveled during constant acceleration, $x_f = x_i + v_i * t + \frac{1}{2} * a * t^2$. Starting position and starting velocity are zero, so it simplifies and distance is 0.25 m. So EP = 122.625 J.
So over a time of 1 second, you changed the EP from 0 to 122.625 joules, and therefore requires 122.625 watts.
The final power is the sum: 128.875 watts. Use whatever efficiency and safety factors on top of this that you think is best.
It is interesting here that this is less power to move it normally with no acceleration. Maybe it is too coincidental, but I can do one more example to show you how different things are when choosing a different a.
Let's make a ten times bigger. a = 5.0 m/s$^2$. Final velocity is still 0.5 m/s, mass is still 50 kg, but now time to reach that final velocity is 0.1 s.
Distance moved upwards during the acceleration phase is $0.5 * 5.0 * 0.1^2$ = 0.025 m, so potential energy at that height is 12.2625 J, but this time that had to be done over a time of 0.1 seconds. So the power is 122.625 W. This is the same as before, for now.
Now consider the change in KE. The final velocity is the same, so KE still is 6.25 joules. But it was done over 0.1 seconds, so the power is actually 62.5 W.
The sum is 185.125 watts. Still less than your requirement for the linear phase.
Just to be thorough, I want to calculate the power for the linear phase, in a different way than you did. We can use the change in potential energy again, $E_p = mgh$. And note that is linear this time, not squared, so if it's constant velocity then it doesn't matter where you start and end.
A 50 kg mass moving up at 0.5 m/s, so ever second it changes its PE by $50 * 9.81 * 0.5$ = 245.25 joules. Inputting that much energy per second is equivalent to 245.25 watts. You already calculated this with force times velocity, which works here, but I prefer changes in PE when going against gravity.
Anyway, since 245 watts is bigger than the power needed for a quick 0.1 second acceleration to final velocity, I think you have nothing to worry about here. You will probably see the motor jump to steady velocity faster than your eye can detect.
