Find how much the 80H / m spring has lengthened under the 2 kg weight.

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

Fcont. = Fт.

Fcont. = k * ∆l, where k is the stiffness of the spring (k = 80 N / m); ∆l is the elongation of the spring, the amount of deformation.

Fт = m * g, where m is the mass of the load suspended from the spring (m = 2 kg), g is the acceleration of gravity (we take g = 10 m / s²).

Let us express and calculate the elongation of the spring:

k * ∆l = m * g.

∆l = m * g / k = 2 * 10/80 = 20/80 = 0.25 m = 25 cm.

Answer: The extension of the spring is 25 cm.

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