What is the acceleration of a 500 kg load, which is lowered with the help of a cable

What is the acceleration of a 500 kg load, which is lowered with the help of a cable, if the tension force of the cable is 4000 N?

m = 500 kg.

g = 9.8 m / s2.

N = 4000 N.

a -?

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

When lowering, two forces act on the load: the force of gravity m * g, directed vertically downward, and the tension force of the cable N, directed vertically upward.

We find the resultant forces Fr by the formula: Fpav = m * g – N.

Fomula for determining the acceleration of the body a will take the form: a = (m * g – N) / m.

a = (500 kg * 9.8 m / s2 – 4000 N) / 500 kg = 1.8 m / s2.

Answer: the acceleration of the load will be a = 1.8 m / s2.