Vertical ideal spring, streched by weight, different results working with forces and energy [duplicate]

Imagine we have a vertical spring with constant $$k$$ and load on it (like a weighing machine) a mass $$m$$.

Initially, the energy of the system is $$0$$, since the spring it's on it's original position and there's no speed (so kinetic energy equals $$0$$) and we choose potential energy to be null at that point. Once released, the mass reaches a new equilibrium position, where the spring is streched a distance $$-\Delta y$$. At this new point, the mass has no speed (kinetic energy equals $$0$$) and since the spring is streched there's a potential energy $$E_p=(1/2)k(-\Delta y)^2=(1/2)k\Delta y^2$$ and also a gravitational energy $$E_p=mg(-\Delta y) = -mg\Delta y$$, so since energy is conserved $$\Delta y=\frac{2mg}{k}$$

If we apply forces at the new rest point, the object has no acceleration, then no net force and the unique forces acting on it are weight (downwards, $$mg$$) and the elastic force ($$k\Delta y$$, upwards) and this leads us to $$\Delta y = \frac{mg}{k}$$

Both resualts are different so, what I'm missing when working with energies? I know that with forces it's Ok.

I guess that is something related about no conservative or external forces (the force done by my hand for dropping the object over the plate) but I have no confirmation

I know there's some related question like If string is stretched just by weight, where does the gravitational potential energy goes if only half is converted to elastic potential energy? or Inconsistencies in Vertical Spring however I think that both cases are a little different with my case

• How do you think your question differs from the ones you linked? Both of them have the correct answer in them.
– noah
Oct 4 '21 at 15:37
• I don't get why I have to take into account velocity in my case. There's no initial speed and at the end is static. In a spring HSM there's no quasi-statically movement and energies are conserved Oct 4 '21 at 15:39
• – noah
Oct 4 '21 at 16:30
• ”Once released, the mass reaches a new equilibrium position...At this new point, the mass has no speed” If you try this experiment, you’ll find that the mass has a nonzero speed. The only other explanations are that something halted its motion, doing its own work, or that the damping is so high that it can’t be ignored in your conservation of energy calculations. This question has been similarly asked and answered about a dozen times on this site. Oct 4 '21 at 18:39
• Yes, now I just realise and I fell so stupid. I was always thinking with a weight scale and there's should be a mechanism inside that when the spring reaches the maxium strech locks the system (and, therefore, i can't apply conservation of energy). Thank you so much Oct 4 '21 at 19:11

If you release the mass from $$\Delta y = 0$$ and do not stop it at its equilibrium point $$\Delta y = \frac {mg} k$$ then it accelerates until it reaches its equilibrium point and then decelerates until it is momentarily stationary at $$\Delta y = \frac {2mg} k$$ (because it has inertia it cannot stop at its equilibrium point, even though the net force on it here is zero). The mass then repeats the same motion in reverse. If there is no loss of energy then it will oscillate indefinitely between $$\Delta y = 0$$ and $$\Delta y = \frac {2mg} k$$ - this is SHM.
If you want the mass to be stationary at its equilibrium point $$\Delta y = \frac {mg} k$$ then you must remove kinetic energy from the mass to do this, either by bringing it to a hard stop at this point, or by making it fall slowly so that its velocity is always small. In either case energy is removed from the system, so you cannot use a conservation of energy argument.