A small body slides along an inclined plane, the angle of inclination of which is 30, from a height of 1 m

A small body slides along an inclined plane, the angle of inclination of which is 30, from a height of 1 m and continues to move along a horizontal plane. the coefficient of friction between the body and the planes is 0.2. what distance S the body will travel along the horizontal plane.

h = 1 m.

g = 9.8 m / s2.

∠α = 30 °.

μ = 0.2.

Sg -?

Let’s find the body’s velocity V at the end of the inclined plane.

m * a = m * g * sinα – μ * m * g * cosα.

a = g * sinα – μ * g * cosα.

a = 9.8 m / s2 * sin30 ° – 0.2 * 9.8 m / s2 * cos30 ° = 3.2 m / s2.

We express the length of the inclined plane S by the formula: S = h / sinα = 1 m / 0.5 = 2 m.

V = √ (2 * a * S) = √ (2 * 3.2 m / s2 * 2 m) = 3.6 m / s

In horizontal motion, the body will move with acceleration a = μ * g.

ar = 0.2 * 9.8 m / s2 = 1.96 m / s2.

Sg = V2 / 2 * ar.

Sg = (3.6 m / s) 2/2 * 1.96 m / s2 = 3.3 m.

Answer: on a horizontal surface, the body will travel Sg = 3.3 m until it stops completely.