On a horizontal surface there is a body with a mass of 2 kg and a horizontally directed force acts

On a horizontal surface there is a body with a mass of 2 kg and a horizontally directed force acts on it, the magnitude of which is 10N. Under the action of this force, the body moves according to the law according to the law: S = 7.5t + 2.5t ^ 2 (m). Determine the coefficient of friction between the body and this surface.

Given:

m = 2 kilograms – body weight;

F = 10 Newton – the force under which the body moves;

g = 10 m / s2 – free fall acceleration;

S = 7.5 * t + 2.5 * t ^ 2 – the equation of body motion.

It is required to determine k – the coefficient of friction between the body and the surface.

Let’s find the dependence of the speed on time:

v (t) = S ‘= (7.5 * t + 2.5 * t ^ 2)’ = 7.5 + 5 * t.

Let’s find the acceleration of the body:

a = v (t) ‘= (7.5 + 5 * t) = 5 m / s2.

Then, according to Newton’s second law:

F – F friction = m * a;

F friction = F – m * a = 10 – 2 * 5 = 10 – 10 = 0.

Since the friction force is 0, it means that the friction coefficient is also 0.

Answer: k = 0.