A bar weighing 1.2 kg is evenly pulled over the table using a spring with a stiffness of 40N / m.

A bar weighing 1.2 kg is evenly pulled over the table using a spring with a stiffness of 40N / m. What is the elongation of the spring if the coefficient of friction between the bar and the table is 0.3?

m = 1.2 kg.

g = 10 m / s2.

k = 40 N / m.

μ = 0.3.

x -?

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

It is acted upon by 4 forces: F – the force with which they pull, m * g – gravity directed vertically downward, N – the reaction force of the table surface directed perpendicular to the plane upward, Ffr – friction force, along the table surface in the opposite direction of motion.

F + m * g + N + Ftr = 0 – the condition of uniform motion.

ОХ: F – Ftr = 0

OU: 0 = – m * g + N.

F = Ftr.

N = m * g.

Ftr = μ * N = μ * m * g.

F = k * x.

μ * m * g = k * x.

x = μ * m * g / k.

x = 0.3 * 1.2 kg * 10 m / s2 / 40 N / m = 0.09 m.

Answer: with a uniform movement of the bar, the spring elongation was x = 0.09 m.