A stone weighing 0.1 kg was thrown vertically upward at a speed of 10m / s and fell in the same

A stone weighing 0.1 kg was thrown vertically upward at a speed of 10m / s and fell in the same place at a speed of 8m / s. The work of the air resistance forces is?

m = 0.1 kg.

V1 = 10 m / s.

V2 = 8 m / s.

A -?

At the moment of throwing the stone, the total mechanical energy of the stone consists of kinetic Ek1 and potential energy En1: E1 = Ek1 + En1.

Kinetic Ek and potential En energy is determined by the formulas: Ek = m * V ^ 2/2, En = m * g * h.

E1 = m * V1 ^ 2/2 + m * g * h.

Let us express the total mechanical energy of the stone E2 at the moment of falling: E2 = m * V2 ^ 2/2 + m * g * h.

According to the law of conservation of energy, the change in the total mechanical energy went to overcome the work of the resistance force: A = E2 – E1.

A = m * V2 ^ 2/2 + m * g * h – m * V1 ^ 2/2 – m * g * h = m * (V2 ^ 2 – V1 ^ 2) / 2.

A = 0.1 kg * ((8 m / s) ^ 2 – (10 m / s) ^ 2) / 2 = – 1.8 J.

Answer: the work of the air resistance force was A = – 1.8 J.