One end of a horizontally located spring with a stiffness of k = 450H / m, compressed to l = 40mm

One end of a horizontally located spring with a stiffness of k = 450H / m, compressed to l = 40mm, is fixed. A block of mass m = 0.5 kg is leaning against the other end, resting on a smooth horizontal surface. What will be the speed modulus of the bar if the spring is released?

k = 450 H / m.
l = 40 mm = 0.04 m.
m = 0.5 kg.
V -?
When compressed, the spring has a potential energy En = k * l ^ 2/2, where k is the stiffness of the spring, l is the absolute elongation of the spring.
When the spring is released, all potential energy, according to the law of conservation of energy, is converted into kinetic Ek.
En = Ek.
The kinetic energy of the body, Ek, is determined by the formula: Ek = m * V ^ 2/2, where m is the mass of the body, V is the speed of the body.
k * l ^ 2/2 = m * V ^ 2/2.
V = √ (k * l ^ 2 / m) = l * √ (k / m).
V = 0.04 m * √ (450 N / m * 0.5 kg) = 0.6 m / s.
Answer: the speed of the bar will be V = 0.6 m / s.