A bar weighing 400 g moves under the action of a force of 2.4 N. Friction coefficient 0.3.

A bar weighing 400 g moves under the action of a force of 2.4 N. Friction coefficient 0.3. With what acceleration the bar moves.

The acceleration with which the bar should move (Newton’s second law):

ma = F – Ffr, where according to the condition m (mass of the bar) = 400 g or 0.4 kg, F (force acting on the bar) = 2.4 N, Ffr (friction force acting on the bar) = μ * m * g (μ is the coefficient of friction (μ = 0.1), g is the acceleration of gravity (g = 10 m / s2)).

ma = F – μ * m * g.

a = (F – μ * m * g) / m.

Let’s do the calculation:

a = (2.4 – 0.3 * 0.4 * 10) / 0.4 = (2.4 – 1.2) / 0.4 = 3 m / s2.

Answer: The acceleration of the movement of the bar is 3 m / s2.