# A body weighing 4 kg moves evenly along a horizontal plane. In this case, the thrust force is equal to 10 N

**A body weighing 4 kg moves evenly along a horizontal plane. In this case, the thrust force is equal to 10 N. What is the coefficient of friction?**

m = 4 kg.

g = 10 m / s ^ 2.

F = 10 N.

μ -?

According to 1 Newton’s law, a body moves uniformly in a straight line if the action of forces on it is compensated.

4 forces act on the body.

The force F, with which the body is pulled, is compensated by the friction force Ffr: F = Ftr.

The force of gravity m * g is compensated by the reaction force of the support N: m * g = N.

The friction force Ffr is determined by the formula: Ffr = μ * N, where μ is the coefficient of friction.

Ftr = μ * m * g.

μ = Ftr / m * g = F / m * g.

μ = 10 N / 4 kg * 10 m / s ^ 2 = 0.25.

Answer: the coefficient of friction between the body and the surface is μ = 0.2.