How much will the spring with a stiffness of 100 N / m lengthen under the action of a load weighing 100 g?

k = 100 N / m.

g = 10 N / kg.

m = 100 g = 0.1 kg.

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Under the action of the load, the spring will stretch, and the load will be in a state of equilibrium. Two forces will act on the load: the force of gravity m * g, directed vertically downward, and the force of elasticity of the spring Fпр, directed vertically upward. According to 1 Newton’s law, these forces will balance each other: Fcont = m * g.

The force of elasticity Fel, which arises in the spring, we express by Hooke’s law: Fel = k * x.

k * x = m * g.

x = m * g / k.

x = 0.1 kg * 10 N / kg / 100 N / m = 0.01 m.

Answer: under the action of the load, the spring will lengthen by x = 0.01 m.