A spring with a stiffness of 40 N / m is suspended A spring with a stiffness of 40 N / m is suspended vertically.

A spring with a stiffness of 40 N / m is suspended A spring with a stiffness of 40 N / m is suspended vertically. A body weighing 800 g is attached to the lower end. The spring is pulled down 15 cm and released. To what height will it rise after this body? Determine your maximum body speed?

The force of the action F of the spring stretched by the value x1 = 15 cm = 0.15 m:

F = k x1 = 40 × 0.15 = 6.0 H.

The spring force is directed vertically upward.

The gravity of a load with a mass of M = 800 g = 0.8 kg:

G = M g = 0.8 × 9.8 = 7.84 H.

The force of gravity is directed vertically downward. This means that the body will descend until the forces are balanced:

M g = k x2, whence: x2 = M g / k = 0.8 × 9.8 / 40 = 0.196 m.

This means that the body will drop another 46 mm.

The maximum speed of the body will be at point x2. From the law of conservation of energy:

k x1 ^ 2/2 + Мg (x2 – х1) = k x2 ^ 2/2 + МV ^ 2/2,

whence: V = [k (x1 ^ 2 − x2 ^ 2) / M + 2g (x2 – x1)] ^ 1/2 = 0.325 m / s.