Solution:The kinetic energy of the spring is equal to its elastic potential energy, i.e. 1/2 mv² = 1/2 kx², when the spring is stretched some distance x from the equilibrium point and when its mass also has some velocity, v with which it is moving.
∵ 1/2 mv² = 1/2 kx²
⇒ 2 × 4² = 1000 × x²
⇒ x = 0.056 m = 5.5 cm.