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

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

m = 500 g = 0.5 kg
F = 2.5 N
g = 10 m / s ^ 2
k -?
Since the bar slides uniformly rectilinearly according to Newton’s 1 law, the action of forces on it is compensated: the force with which the pull is compensated by the friction force of the bar on the surface F = Ffr, Ffr = k * N, where k is the coefficient of friction, N is the force with which the support acts on bar. N = m * g. The formula will look like F = k * m * g; k = F / m * g. Substitute the data k = 2.5N / 0.5kg * 10m / s ^ 2 = 0.5
Answer: coefficient of friction k = 0.5.