A parachutist weighing 80 kg separated from the motionlessly hanging helicopter and, having covered a path of 200 m

A parachutist weighing 80 kg separated from the motionlessly hanging helicopter and, having covered a path of 200 m before the parachute was deployed, acquired a speed of 50 m / s. Find the work of the force of air resistance along the way.

m = 80 kg.

g = 10 m / s2.

S = 200 m.

V0 = 0 m / s.

V = 50 m / s.

Asopr -?

The work of the resistance force Asopr is expressed by the formula: Asopr = Fcopr * S * cos180 = – Fcopr * S.

Since the resistance force Fcopr is directed in the opposite direction of the parachutist’s movement S, then ∠α = 180, os180 = -1.

m * a = m * g – Fsopr – 2 Newton’s law for a parachutist.

Fcopr = m * g – m * a = m * (g – a).

The parachutist’s acceleration a is expressed by the formula: a = (V ^ 2 – V0 ^ 2) / 2 * S.

Since V0 = 0 m / s, then a = V ^ 2/2 * S.

The work of the resistance force will take the form: Asopr = – m * (g – V ^ 2/2 * S) * S.

Asopr = – 80 kg * (10 m / s2 – (50 m / s) ^ 2/2 * 200 m) * 200 m = – 60,000 J.

Answer: the work of the resistance force is Asopr = – 60,000 J.