A truck weighing 5 tons starts to move with an acceleration of 1 m / s ^ 2

A truck weighing 5 tons starts to move with an acceleration of 1 m / s ^ 2. What traction force will the truck engine develop during this movement, if the coefficient of friction is 0.2.

m = 5 t = 5000 kg.
a = 1 m / s ^ 2.
μ = 0.2.
g = 9.8 m / s ^ 2.
Fт -?
The truck moves horizontally under the action of 2 forces: traction force Ft, which is directed in one direction, and friction force Ftr, which is directed in the opposite direction.
Let’s write 2 Newton’s law: m * a = Ft – Ftr.
Ft = m * a + Ftr.
The friction force Ffr is determined by the formula: Ffr = μ * m * g.
Fт = m * a + μ * m * g.
Ft = 5000 kg * 1 m / s ^ 2 + 0.2 * 5000 kg * 9.8 m / s ^ 2 = 14800 N.
Answer: the traction force of the truck engine is Ft = 14800 N.