How far should a weight of 640 g suspended on a spring with a stiffness of 400 N / m

How far should a weight of 640 g suspended on a spring with a stiffness of 400 N / m be taken down from the equilibrium position in order for it to pass the equilibrium position at a speed of 1 m / s?

Let us give the values from given in the SI system:
m = 640 g = 0.64 kg.

1. In the equilibrium position, the load has kinetic energy:
E = m * v² / 2
2. In the retracted position, the load has only potential energy:
Ep = k * x² / 2
3. According to the law of conservation of energy:
E = Ep
m * v² / 2 = k * x² / 2
Let us express the tension of the spring from this expression:
x² = m * v² * 2 / (2 * k)
x = √ (m * v² / k)
Let’s calculate the elongation of the spring:
x = √ (m * v² / k) = √ (0.64 * 1² / 400) = 0.04 m.
Answer: the spring needs to be stretched 0.04 m or 4 cm.