A block weighing 1 kg is pulled evenly over a wooden board using a spring k = 100n / m.

A block weighing 1 kg is pulled evenly over a wooden board using a spring k = 100n / m. Find the extension of the spring if Mu = 0.5.

m = 1 kg.

g = 10 m / s2.

k = 100 N / m.

μ = 0.5.

x -?

Since the bar is pulled evenly, according to Newton’s 1 law, the action of forces on it is compensated: m * g + Ffr + Ffr + N = 0, where m * g is the force of gravity, Ffr is the friction force, Ffr is the force with which the bar is pulled , N is the reaction force of the wooden board.

ОХ: – Ftr + Fcont = 0.

OU: – m * g + N = 0.

N = m * g.

Ftr = μ * N = μ * m * g.

Fcont = Ftr.

Fupr = k * x.

μ * m * g = k * x.

x = μ * m * g / k.

x = 0.5 * 1 kg * 10 m / s2 / 100 N / m = 0.05 m.

Answer: with a uniform movement of the bar, the spring will lengthen by x = 0.05 m.