A body weighing 200 g is suspended on a spring and performs harmonic vibrations. The highest body speed is 0.2 m / s

A body weighing 200 g is suspended on a spring and performs harmonic vibrations. The highest body speed is 0.2 m / s; the greatest deviation from the equilibrium position is 3 cm to determine the rigidity of the spring.

m = 200 g = 0.2 kg.

V = 0.2 m / s.

x = 3 cm = 0.03 m.

k -?

The body speed will be maximum at the moment it passes the equilibrium position. At this moment, the body has only kinetic energy Ek.

The body will have the greatest deviation x when it has only the potential energy En.

According to the law of conservation of total mechanical energy: Ek = En.

m * V ^ 2/2 = k * x ^ 2/2.

We find the stiffness of the spring by the formula: k = m * V ^ 2 / x ^ 2.

k = 0.2 kg * (0.2 m / s) ^ 2 / (0.03 m) ^ 2 = 8.88 N / m.

Answer: the spring rate is k = 8.88 N / m.