What is the distance evenly displaced on the floor with a weight of 100 kg, if the horizontal force applied

What is the distance evenly displaced on the floor with a weight of 100 kg, if the horizontal force applied to it did the work of 4.5 kJ? The coefficient of friction between the load and the floor is 0.15.

m = 100 kg.

g = 9.8 m / s2.

A = 4.5 kJ = 4500 J.

μ = 0.15.

S -?

The mechanical work A of the force F, which moves the load, is determined by the formula: A = F * S * cosα, where ∠α is the angle between the force F and the displacement S. Since the force is directed along the horizontal surface, then ∠α = 00.

cos00 = 1.

A = F * S.

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

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.

A = μ * m * g * S.

S = A / μ * m * g.

S = 4500 J / 0.15 * 100 kg * 9.8 m / s2 = 30.6 m.

Answer: the load was moved along the floor to a distance of S = 30.6 m.