# A block weighing 5 kg is pulled along the table surface, grasping the dynamometer ring. In this case

**A block weighing 5 kg is pulled along the table surface, grasping the dynamometer ring. In this case, the acceleration of the body is equal to 0.5 m / s2. The spring rate is 200 N / m. Determine the tension of the spring. the coefficient of friction of the bar on the table is 0.05.**

m = 5 kg.

g = 10 m / s2.

a = 0.5 m / s2.

k = 200 N / m.

μ = 0.05.

x -?

Let’s write 2 Newton’s law in vector form: m * a = Fcont + Ftr + N + m * g.

ОХ: m * a = Fcont – Ftr.

OU: 0 = N – m * g.

Fcont = m * a + Ftr.

N = m * g.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.

We express the force of elasticity by Hooke’s law: Fel = k * x.

k * x = m * a + μ * m * g.

The tension of the spring x will be determined by the formula: x = (m * a + μ * m * g) / k.

x = (5 kg * 0.5 m / s + 0.05 * 5 kg * 10 m / s2) / 200 N / m = 0.025 m.

Answer: the tension of the spring was x = 0.025 m.