A load was suspended to a spring of stiffness of 200 Newtons per meter, calculate the mass

A load was suspended to a spring of stiffness of 200 Newtons per meter, calculate the mass of the load if the spring is stretched by 10 cm.

k = 200 N / m.

x = 10 cm = 0.1 m.

g = 10 m / s2.

m -?

The condition for finding the load at rest, according to Newton’s 1 law, is the balancing of all the forces that act on it. The load is acted upon by the force of gravity m * g, which is always directed vertically downward, and the elastic force of the stretched spring Ffr, directed vertically upward.

m * g = Fcont – the condition of the balance of the load.

We will express the force of elasticity Fcont according to Hooke’s law: Fcont = k * x, where k is the stiffness of the spring, x is the elongation of the spring.

The condition for the balance of the load will take the form: m * g = k * x.

m = k * x / g.

m = 200 N / m * 0.1 m / 10 m / s2 = 2 kg.

Answer: a load weighing m = 2 kg was suspended from the spring.