A load weighing 100 g, suspended on a spring with a stiffness of 20 N / m, makes vertical vibrations

A load weighing 100 g, suspended on a spring with a stiffness of 20 N / m, makes vertical vibrations. With what acceleration does the load move at the moment of time when the spring is stretched by 2 cm?

m = 100 g = 0.1 kg.

g = 10 m / s2.

k = 20 N / m.

x = 2 cm = 0.02 m.

a -?

When moving, two forces act on the load: the force of gravity Ft directed vertically downward, the elastic force of the spring Ffr, which is directed towards the balance of the spring.

Ft = m * g, Fcont = k * x.

When the elongation of the spring is x = 2 cm, the elastic force is directed vertically upwards.

Let’s write Newton’s 2 law for cargo in vector form: m * a = Fcont + Ft = k * x + m * g.

For projections 2 Newton’s law has the form: m * a = m * g – k * x.

a = (m * g – k * x) / m.

a = (0.1 kg * 10 m / s2 – 20 N / m * 0.02 m) / 0.1 kg = 6 m / s2.

When the load moves down, it is decelerated with acceleration a = 6 m / s2. When the load moves up, it accelerates with an acceleration a = 6 m / s2.

Answer: the acceleration of the body is a = 6 m / s2.