Energy conservation of a spring

Question

The problem is as follows:

A spring is hung from the ceiling. A 0.450-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.150 m before momentarily coming to rest, after which it moves back upward. a) What is the spring constant of the spring?

I found this problem while helping a high school student with their physics homework. The hint I gave the student to solve the problem was to use conservation of energy, i.e.

$$ E_i = E_f $$ $$ (KE + PE_g + PE_e)_i = (KE + PE_g + PE_e)_f $$ $$ 0 + mgx + 0 = 0 + 0 + .5kx^2 $$ $$ => k = 58.8 N/m $$

While the student was doing this, I was looking for a different approach and see if we both can arrive at the same answer. My second approach was as follow:

When the block is momentarily resting, we know that the net force is zero such that

$$ F_{net} = F_e - F_g $$ $$ 0 = F_e - F_g $$ $$ => F_e = F_g $$ $$ kx = mg $$

The second approach gives an answer that is about half of that from the first approach. I know that the first approach yielded the correct answer, so what is the wrong assumption that I made in the second approach?

Answer 1

The incorrect assumption is here:

When the block is momentarily resting, we know that the net force is zero

The net force is zero when the acceleration is zero, not when the velocity is zero. In this problem that happens at the equilibrium position. But the momentarily resting position is below that.