From a helicopter descending uniformly at a speed of 2 m / s, a packet was dropped from a height of 50 m

From a helicopter descending uniformly at a speed of 2 m / s, a packet was dropped from a height of 50 m, which reached the Earth’s surface at a speed of 8 m / s. Package weight is 10 kg. Determine the work of the air resistance force acting on the package.

V0 = 2 m / s.

h = 50 m.

g = 9.8 m / s2.

m = 10 kg.

V = 8 m / s.

A -?

We express the work A of the resistance force by the formula: A = – Fcopr * h, where Fcopr is the air resistance force, h is the movement of the package.

The “-” sign indicates that the resistance force Fcopr is directed in the opposite direction of the packet movement.

m * a = m * g – Fc.

Fcopr = m * g – m * a = m * (g – a).

a = (V ^ 2 – V0 ^ 2) / 2 * h.

a = ((8 m / s) ^ 2 – (2 m / s) ^ 2) / 2 * 50 m = 0.6 m / s2.

Fcopr = 10 kg * (9.8 m / s2 – 0.6 m / s2) = 92 N.

A = – 92 N * 50 m = -4600 J.

Answer: the resistance force did the work A = – 4600 J.