A trolley weighing 200 kg moves with an acceleration of 4m / s ^ 2.

A trolley weighing 200 kg moves with an acceleration of 4m / s ^ 2. With what force does the worker push the trolley if the coefficient of friction is 0.6.

m = 200 kg.

g = 10 m / s2.

a = 4 m / s2.

μ = 0.6.

F -?

According to 2 Newton’s law, the acceleration of a body a is directly proportional to the resultant of all forces Fр and inversely proportional to the mass of the body m: a = Fр / m.

Fr = F + m * g + N + Ftr, where F is the force with which the worker pushes the trolley, m * g is the trolley’s gravity, N is the surface reaction force, Ftr is the friction force.

OH: m * a = F – Ftr.

OU: 0 = – m * g + N.

F = m * a + Ftr.

N = m * g.

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

F = m * a + μ * m * g = m * (a + μ * g).

We see that the force of the worker F goes to overcome the friction force Ffr and impart the acceleration a to the trolley.

F = 200 kg * (4 m / s2 + 0.6 * 10 m / s2) = 2000 N.

Answer: for the movement of the trolley, the worker applies a force F = 2000 N.