A bar weighing 800 g was attached to a spring with a stiffness of 40 N / m, stretched by 2 cm.

A bar weighing 800 g was attached to a spring with a stiffness of 40 N / m, stretched by 2 cm. Determine the acceleration that the bar will receive under the action of the spring force.

k = 40 N / m.
m = 800 g = 0.8 kg.
g = 10 m / s2.
x = 2 cm = 0.02 m.
a – ?
We express the acceleration of the load a according to Newton’s 2 law, the ratio of the resultant of all forces Fр that act on the body to the body mass m: a = Fр / m.
The spring is vertical.
The body is acted upon by the force of gravity m * g, directed vertically downward, and the force of elasticity Fel, directed vertically upward.
Their resultant Fр will be the difference: Fр = m * g – Fcont.
We will express the force of elasticity Fel by Hooke’s law: Fel = k * x.
a = (m * g – k * x) / m.
a = (0.8 kg * 10 m / s2 – 40 N / m * 0.02 m) / 0.8 kg = 9 m / s2.
The spring is horizontal.
The body is affected only by the elastic force Fel.
Fр = Fcont.
a = k * x m.
a = 40 N / m * 0.02 m / 0.8 kg = 1 m / s2.
Answer: the bar can have an acceleration a = 9 m / s2 when the spring is in a vertical position, and an acceleration a = 1 m / s2 when the spring is in a horizontal position.