Question

A load of 0.5 kg is attached to a horizontally fixed 120 n / m rigidity spring, the spring was stretched by 4 cm and released. Which is equal to its velocity when the offset becomes equal to 1 cm.

Answer 1

Elastic potential energy of the spring turns into kinetic energy of the body:

$$\frac{kx_1^2}{2}=\frac{kx_2^2}{2}+\frac{mv^2}{2},\\\space\\ v=\sqrt{\frac{k(x_1^2-x_2^2)}{m}},\\\space\\ v=\sqrt{\frac{120(0.04^2-0.03^2)}{0.5}}=0.41\text{ m/s}.\\\space\\$$