A body weighing 2 kg lying on a horizontal surface begins to act horizontal force F.

A body weighing 2 kg lying on a horizontal surface begins to act horizontal force F. The coefficient of friction between the body and the surface is 0.1. What is acceleration if the modulus of force is

m = 2 kg.

g = 10 m / s2.

μ = 0.1.

F1 = 0.5 N.

F2 = 2 N.

F3 = 2.5 N.

but – ?

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

Fr = F + Ftr + T + m * g.

ОХ: Fр = F – Fр.

OU: 0 = T – m * g.

T = m * g.

The sliding friction force Ftr is determined by the formula: Ftr = μ * T = μ * m * g.

If the horizontal force F is greater than the sliding friction force Ffr, then the body will move.

Ftr = 0.1 * 2 kg * 10 m / s2 = 2 N.

F1 <Ffr the body will be at rest a1 = 0 m / s2.

F2 = Ftr the body will be in a state of rest or uniform motion a2 = 0 m / s2.

a3 = F3 – μ * m * g / m.

a3 = 2.5 N – 0.1 * 2 kg * 10 m / s2 / 2 kg = 0.25 m / s2.

Answer: a1 = a2 = 0 m / s2, a3 = 0.25 m / s2.