A horizontally located weightless spring with a stiffness of 1000 is in an undeformed state; one end of it is fixed

A horizontally located weightless spring with a stiffness of 1000 is in an undeformed state; one end of it is fixed with the other touching a bar of 0.1 kg on the horizontal surface. The bar is shifted and compressed by the spring and released. How long is the spring compressed if, after releasing the bar, its speed becomes 1 m / s.

k = 1000 N / m.

m = 0.1 kg.

V = 1 m / s.

x -?

When the spring is compressed, the system has only the potential energy of the compressed spring, En, the value of which is expressed by the formula: En = k * x2 / 2, where k is the stiffness of the spring, x is the absolute deformation of the spring.

When the spring is completely unclenched, the body has only kinetic energy Ek, the value of which is expressed by the formula: Ek = m * V2 / 2, where m is the mass of the body, V is the speed of the body.

According to the law of conservation of total mechanical energy, all the potential energy of the spring has passed into the kinetic energy of the body: En = Ek.

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

x = V * √m / √k.

x = 1 m / s * √0.1 kg / √1000 N / m = 0.01 m.

Answer: the spring was compressed by x = 0.01 m.