A bar with a mass of m = 5 kg moves along a horizontal plane. The coefficient of friction between

A bar with a mass of m = 5 kg moves along a horizontal plane. The coefficient of friction between the bar and the surface is u = 0.08. Determine the force F with which the bar moves.

m = 5 kg.
μ = 0.08.
g = 9.8 m / s ^ 2.
F -?
Since the bar moves uniformly, according to 1 Newton’s law, the force F with which the bar is pulled is equal to the friction force Ffr: F = Ffr.
The friction force Ffr is determined by the formula: Ffr = μ * N, where μ is the coefficient of friction between surfaces, N is the support reaction force.
The support reaction force N is determined by the formula: N = m * g.
F = μ * m * g.
F = 0.08 * 5 kg * 9.8 m / s ^ 2 = 3.92 N.
Answer: the bar is pulled with a force F = 3.92 N.