# How long will a spring with a stiffness of 100 N / m stretch if a weight of 100 g is suspended from it?

k = 100 N / m.

g = 10 m / s2.

m = 100 g = 0.1 kg.

x -?

The condition for the balance of the load on the spring, according to Newton’s 1 law, is the balancing of all the forces that act on it. The body is acted upon vertically downward by the force of gravity Fт, vertically upwards by the elastic force of the spring Ftr: Ft = Ftr.

Let us express the force of gravity Ft by the formula: Ft = m * g.

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

The elongation of the spring x will be determined by the formula: x = m * g / k.

x = 0.1 kg * 10 m / s2 / 100 N / m = 0.01 m.

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

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