# A piece of stone weighing 18.5 kg slides at a speed of 6.72 m / s on a horizontal surface (friction coefficient 0.35)

**A piece of stone weighing 18.5 kg slides at a speed of 6.72 m / s on a horizontal surface (friction coefficient 0.35) and stops after 3.93 m under the influence of friction and some force F. Find the acceleration of the stone? Find the magnitude and direction of the force F?**

m = 18.5 kg. g = 9.8 m / s ^ 2. V0 = 6.72 m / s. V = 0 m / s. S = 3.93 m. Μ = 0.35. a -? F -?

Find the acceleration of the stone by the formula: a = (V ^ 2 – V0 ^ 2) / 2 * S. a = ((0 m / s) – (6.72 m / s) ^ 2) / 2 * 3.93 m = – 5.75 m / s ^ 2. The sign “-” means that the acceleration a directed in the opposite direction to the displacement S, the body is decelerated. Let us find the acceleration of stone a1 only under the action of the friction force. m * a1 = μ * m * g. a1 = μ * g. a1 = 0.35 * 9.8 m / s ^ 2 = 3.43 m / s ^ 2. a1 = 3.43 m / s ^ 2 <a = 5.75 m / s ^ 2, which means the force F, directed towards the action of the friction force, against the direction of motion. m * a = μ * m * g + F. F = m * a – μ * m * g = m * (a – μ * g). F = 18.5 kg * (5.75 m / s ^ 2 – 0.35 * 9.8 m / s ^ 2) = 42.92 N.

Answer: the acceleration of the body is a = 5.75 m / s ^ 2, a force acts on it in the opposite direction of motion F = 42.92 N.