# A load was attached to a vertically suspended spring, the stiffness of which was 800 N / m.

A load was attached to a vertically suspended spring, the stiffness of which was 800 N / m. In this case, the spring lengthened by 10 cm. Find the mass of the load.

k = 800 N / m.

g = 10 m / s2.

x = 10 cm = 0.1 m.

m -?

Since the load is in a state of equilibrium, it means, according to Newton’s 1 law, the action of forces on it is compensated. The load is acted upon by the force of gravity Ft and the force of elasticity of the spring Fcont: Ft = Fcont.

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

We express the force of elasticity by Hooke’s law: Fel = k * x.

m * g = k * x.

m = k * x / g.

m = 800 N / m * 0.1 m / 10 m / s2 = 8 kg.

Answer: a load is attached to the vertical spring, the mass of which is m = 8 kg.

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