A bar weighing 500 g slides evenly on a wooden platform under the action of a traction equal to 2.5 N

A bar weighing 500 g slides evenly on a wooden platform under the action of a traction equal to 2.5 N. What is the coefficient of friction of a bar against a tree?

m = 500 g = 0.5 kg.

g = 9.8 m / s2.

F = 2.5 N.

μ -?

Since the bar moves uniformly rectilinearly, then according to Newton’s 1 law, the action of forces on it is compensated.

In the horizontal direction, the force F with which the bar is pulled is compensated by the friction force Ff: F = Ff.

In the vertical direction, the force of gravity m * g is compensated by the reaction force of the support N: N = m * g.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.

F = μ * m * g.

μ = F / m * g.

μ = 2.5 N / 0.5 kg * 9.8 m / s2 = 0.51.

Answer: the coefficient of friction between the bar and the surface is μ = 0.51.