The bar is evenly pulled along a horizontal surface using a spring with a stiffness of 200 N / m

The bar is evenly pulled along a horizontal surface using a spring with a stiffness of 200 N / m, the elongation of which is 1.5 cm. The coefficient of sliding friction of the bar against the surface is 0.3. The determinant of these data is the mass of the bar.

k = 200 N / m.

x = 1.5 cm = 0.015 m.

g = 9.8 m / s2.

μ = 0.3.

m -?

According to Newton’s 1 law, a bar moves uniformly when the action of forces on it is compensated.

In the horizontal direction, the elastic force of the spring Ffr is compensated by the friction force Ffr: Ffr = Ffr.

In the vertical direction, the force of gravity m * g compensates for the reaction force of the support N: N = m * g.

Fupr = k * x.

Ftr = μ * N = μ * m * g.

k * x = μ * m * g.

m = k * x / μ * g.

m = 200 N / m * 0.015 m / 0.3 * 9.8 m / s2 = 1 kg.

Answer: the bar has a mass of m = 1 kg.