When a load weighing 1 kg is suspended, the spring in equilibrium lengthened by 10 cm.

When a load weighing 1 kg is suspended, the spring in equilibrium lengthened by 10 cm.What is the maximum kinetic energy of the load when the load vibrates on a spring with an amplitude of 20 cm?

m = 1 kg.

x = 10 cm = 0.1 m.

g = 10 m / s2.

A = 20 cm = 0.2 m.

Ekmax -?

When the spring vibrates, it will have the maximum kinetic energy Ekmax when it passes the equilibrium position. According to the law of conservation of total mechanical energy, the entire potential energy of the spring, Епmax, when passing the equilibrium position, passed into the kinetic Епmax: Екmax = Епmax.

The maximum potential energy of the spring is determined by the formula: Enmax = k * A ^ 2/2, where k is the stiffness of the spring, A is the maximum deviation from the equilibrium position, which is called the amplitude.

Ekmax = k * A ^ 2/2.

The spring stiffness k is found from the condition of equilibrium of the load on the spring: m * g = k * x.

k = m * g / x.

k = 1 kg * 10 m / s2 / 0.1 m = 100 N / m.

Ekmax = 100 N / m * (0.2 m) ^ 2/2 = 2 J.

Answer: the maximum kinetic energy of the load during vibrations is Ekmax = 2 J when passing the equilibrium position.