A bar weighing 5 kg begins to move along a horizontal surface from a state of rest under the action of 40N

A bar weighing 5 kg begins to move along a horizontal surface from a state of rest under the action of 40N gravity directed at an angle of 45 degrees. Find its speed in 10 s if the sliding friction coefficient is 0.5.

m = 5 kg.

g = 10 m / s2.

F = 40 N.

∠α = 45 °.

V0 = 0 m / s.

μ = 0.5.

t = 10 s.

V -?

Let us write Newton’s 2 law in vector form: m * a = F + m * g + N + Ftr, where F is the force with which the load is pulled, m * g is the force of gravity, N is the surface reaction force, Ftr is the friction force.

ОХ: m * a = F * cosα – Ftr.

OU: 0 = F * sinα – m * g + N.

a = (F * cosα – Ftr) / m.

N = m * g – F * sinα.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * (m * g – F * sinα).

Find the acceleration of the bar by the formula: a = (F * cosα – μ * (m * g – F * sinα)) / m.

a = (40 N * cos45 ° – 0.5 * (5 kg * 10 m / s2 – 40 N * sin45 °)) / 5 kg = 3.5 m / s2.

The speed of the bar V is expressed by the formula: V = V0 + a * t.

V = 0 m / s + 3.5 m / s2 * 10 s = 35 m / s.

Answer: the speed of the bar will be V = 35 m / s.