The spring lengthened by 3 cm under the action of gravity acting on a body with a mass of 10 kg

The spring lengthened by 3 cm under the action of gravity acting on a body with a mass of 10 kg, what is the stiffness of the spring?

Given: L = 3 cm spring length increment.
m = 10 kg weight of the suspended load.
g = 10 m / s2.

Find: k – coefficient of spring stiffness.

Let’s use Hooke’s formula: F = k * x, where F is the weight of the load, equal to the product of the mass by the coefficient of free fall. According to Hooke’s law for longitudinal deformation, the deformation x of a body is proportional to its initial length L and the applied force F.

Let’s find the coefficient of spring stiffness, converting the length into meters.

k = F / x = 10 * 10 / 0.03 = 100: 3/100 = 100 * 100/3 ≈ 3333 n / m.

Answer: k ≈ 3333 n / m.