Using a dynamometer, the bar is pulled along the table with a force of 2.5 N.

Using a dynamometer, the bar is pulled along the table with a force of 2.5 N. Find the mass of the bar if the coefficient of friction is 0.2.

g = 10 m / s2.

F = 2.5 N.

μ = 0.2.

m -?

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

We will assume that the bar is pulled uniformly, which means that a = 0.

F + m * g + N + Ftr = 0.

ОХ: 0 = F – Ftr.

OU: 0 = – m * g + N.

F = Ftr.

N = m * g.

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

F = μ * m * g.

m = F / μ * g.

m = 2.5 N / 0.2 * 10 m / s2 = 1.25 kg.

Answer: the bar has a mass of m = 1.25 kg.