The body slides on a horizontal rough surface under the action of a horizontal force with a constant acceleration

The body slides on a horizontal rough surface under the action of a horizontal force with a constant acceleration of 8 m / s 2. The coefficient of friction of the body on the plane is 0.2, and the body weight is 5 kg. Determine the magnitude of the traction force applied to the body.

Given:

a = 8 m / s ^ 2 – body acceleration;

k = 0.2 – coefficient of friction;

m = 5 kg – body weight;

g = 9.8 m / s ^ 2 – acceleration of gravity

It is required to find Ft – the thrust force.

According to Newton’s second law:

Ft – Ftr = m * a, where Ftr = k * N = k * m * g (since the body moves horizontally, the reaction force of the support N is numerically equal to the force of gravity). Hence:

Fт – k * m * g = m * a

Ft = m * a + k * m * g = m * (a + k * g) = 5 * (8 + 0.2 * 9.8) = 49.8 N.

Answer: the pulling force is 49.8 Newtons.