Determine the stiffness of the spring, which lengthened by 3 cm under the action of a body with a mass of 600 g.

Since the force of elasticity balances the force of gravity acting on it from the load:
Fcont. = Fт.
Fcont. = k * ∆l, where k is the stiffness of the spring (N / m); ∆l is the amount of spring deformation (elongation, ∆l = 3 cm = 0.03 m).
Fт = m * g, where m is the mass of the load (m = 600 g = 0.6 kg), g is the acceleration of gravity (we take g = 10 m / s²).
Let us express and calculate the stiffness of the spring:
k * ∆l = m * g.
k = m * g / ∆l = 0.6 * 10 / 0.03 = 200 N / m.
Answer: The spring rate is 200 N / m.