# On a body weighing 0.8 kg, lying on a horizontal surface, force F begins to act. The coefficient of friction

**On a body weighing 0.8 kg, lying on a horizontal surface, force F begins to act. The coefficient of friction between the body and the surface is 0.5. Determine the acceleration of the body if the modulus of force is 6 N.**

m = 0.8 kg.

g = 9.8 m / s2.

F = 6 N.

μ = 0.5.

a -?

According to 2 Newton’s law, the acceleration of a body a is directly proportional to the resultant of all forces F equal, which act on the body, and inversely proportional to the mass of the body m: a = Frav / m.

We find the resultant of all forces by the formula: Frav = F + Ftr + m * g + N.

ОХ: Frav = F – Ftr.

OU: 0 = m * g – N.

m * g = N.

Ftr = μ * N = μ * m * g.

Frav = F – μ * m * g.

The acceleration of the body a is found by the formula: a = (F – μ * m * g) / m.

a = (6 N – 0.5 * 0.8 kg * 9.8 m / s2) / 0.8 kg = 2.6 m / s2.

Answer: the body will move with acceleration a = 2.6 m / s2.