A bucket weighing 500 kg rises from the mine and in the first 10 s from the beginning

A bucket weighing 500 kg rises from the mine and in the first 10 s from the beginning of uniformly accelerated movement it passes 20 m. What is the tension force of the rope on which the bucket is lowered, if the acceleration is equal in magnitude to the acceleration when moving up?

m = 500 kg.

g = 9.8 m / s ^ 2.

t = 10 s.

S = 20 m.

F -?

Let us write Newton’s 2 law for a bucket in vector form: m * a = F + m * g, where m is the mass of the body, a is the acceleration of the body, F is the pulling force of the rope for which they are pulling, m * g is the force of gravity.

We express the acceleration of the body a from the formula: S = a * t ^ 2/2.

a = 2 * S / t ^ 2.

a = 2 * 20 m / (10 s) ^ 2 = 0.4 m / s ^ 2.

In projections on the vertical axis directed downward, 2 Newton’s law will take the form: m * a = m * g – F.

F = m * g – m * a = m * (g – a).

F = 500 kg * (9.8 m / s ^ 2 – 0.4 m / s ^ 2) = 4700 N.

Answer: the pulling force of the rope when lowering is F = 4700 N.