On the horizontal surface of the table, under the action of 50 H directed along the surface of the table

On the horizontal surface of the table, under the action of 50 H directed along the surface of the table, a body with a mass of 7 kg moves. Friction coefficient 0.2. Find the acceleration of the body.

m = 7 kg.

g = 10 m / s2.

F = 50 N.

μ = 0.2.

a -?

4 forces act on the body during movement.

Let us write Newton’s 2 law in vector form: m * a = F + m * g + N + Ftr, where F is the force with which the load is pulled, m * g is the force of gravity, N is the surface reaction force, Ftr is the friction force.

ОХ: m * a = F – Ftr.

OU: 0 = – m * g + N.

a = (F – Ftr) / m.

N = m * g.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.

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

a = (50 N – 0.2 * 7 kg * 10 m / s2) / 7 kg = 5.1 m / s2.

Answer: the body on the table will move with acceleration a = 5.1 m / s2.