A body weighing 600 g was suspended on a spring, it stretched by 5 cm. What is the stiffness of the spring.

The force of elasticity is balanced by the force of gravity, which acts on the suspended load, and the ratio is true:

Fcont. = Fт.

Fcont. = k * ∆l, where k is the stiffness of the spring (N / m); ∆l – elongation, the amount of deformation of the spring (∆l = 5 cm = 0.05 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 / 0.05 = 120 N / m.

Answer: The spring rate is 120 N / m.