A block with a mass of m = 2 kg is moved with the help of a spring at a constant speed up the board

A block with a mass of m = 2 kg is moved with the help of a spring at a constant speed up the board, making an angle α = 30 with the horizon. Determine the stiffness k of the spring if it lengthened at the same time Δl = 5 cm. The axis of the spring is parallel to the board. Coefficient of friction between the bar and the board n = 0.2

m = 2 kg.

g = 10 m / s2.

∠α = 30 °.

Δl = 5 cm = 0.05 m.

n = 0.2.

k -?

Since the bar moves uniformly in a straight line, then, according to 1 Newton’s law, the action of all forces on it is compensated.

For the movement of a bar along an inclined plane 2, Newton’s law in vector form will have the form: 0 = Ft + m * g + N + Ftr, where Ft is the force with which the bar is pulled, m * g is the force of gravity, N is the surface reaction force inclined plane, Ffr – friction force.

ОХ: 0 = Fт – Fтр – m * g * sinα.

OU: 0 = – m * g * cosα + N.

Ft = Ftr + m * g * sinα.

N = m * g * cosα.

The friction force Ffr is expressed by the formula: Ffr = n * N = n * m * g * cosα.

Fт = k * Δl.

k * Δl = n * m * g * cosα + m * g * sinα.

k = m * g * (n * cosα + sinα) / Δl.

k = 2 kg * 10 m / s * (0.2 * cos30 ° + sin30 °) / 0.05 m = 338 N / m.

Answer: the spring rate is k = 338 N / m.