Under the action of a horizontal force of 5 N, a body weighing 1 kg. moves on a horizontal surface

Under the action of a horizontal force of 5 N, a body weighing 1 kg. moves on a horizontal surface with an acceleration of 2 m / s2. What is the coefficient of friction?

m = 1 kg.
F = 5 N.
a = 2 m / s ^ 2.
g = 10 m / s ^ 2.
μ -?
We write 2 Newton’s law in vector form: m * a = F + Ffr + m * g + N, where m is the body mass, a is the acceleration of the body, F is the force with which it is pulled horizontally, Ffr is the friction force, m * g – gravity, N – support reaction force.
ОХ: m * a = F – Ftr.
OU: 0 = m * g – N.
Ftr = F – m * a.
N = m * g.
The friction force Ffr is determined by the formula: Ffr = μ * N, where μ is the coefficient of friction.
Ftr = μ * m * g.
μ * m * g = F – m * a.
μ = (F – m * a) / m * g.
μ = (5 N – 1 kg * 2 m / s ^ 2) / 1 kg * 10 m / s ^ 2 = 0.3.
Answer: the coefficient of friction is μ = 0.3.