The wooden block is evenly and rectilinearly moved with the help of a spring on a rough horizontal board.

The wooden block is evenly and rectilinearly moved with the help of a spring on a rough horizontal board. A sliding friction force equal to 0.4 N. acts on the bar, how much is the spring stretched, if its stiffness coefficient is 40 N

Ftr = 0.4 N.

k = 40 N / m.

x -?

The condition for the uniform rectilinear movement of a wooden bar on a rough surface, according to 1 Newton’s law, is the balancing of all the forces that act on it.

Vertically, the force of gravity m * g is balanced by the reaction force of the surface N: m * g = N.

Horizontally, the force F with which the bar is pulled is balanced by the friction force Ftr: F = Ftr.

The force F with which the bar is pulled is the elastic force of the spring, the value of which is expressed by Hooke’s law: F = k * x, where k is the stiffness of the spring, x is the elongation of the spring.

Ftr = k * x.

x = Ftr / k.

x = 0.4 N / 40 N / m = 0.01 m.

Answer: the spring will stretch x = 0.01 m.