The pilot takes the plane out of the dive. Aircraft speed 300 m / s. The radius of the trajectory is 1 km.

The pilot takes the plane out of the dive. Aircraft speed 300 m / s. The radius of the trajectory is 1 km. Pilot weight 70 kg. Determine the weight of the pilot at the bottom of the trajectory.

V = 300 m / s.

R = 1 km = 1000 m.

m = 70 kg.

g = 10 m / s2.

R – ?

When diving, 2 forces act on the pilot at the lowest point of the trajectory: gravity Ft directed vertically downward, force N of seat pressure on the pilot directed vertically upward.

m * a = m * g + N – 2 Newton’s law in vector form.

Let’s find the expression 2 of Newton’s law on the vertical axis directed vertically upward: m * a = – Fт + N.

N = m * a + Fт.

The gravity of the pilot Ft is determined by the formula: Ft = m * g.

N = m * a + m * g = m * (a + g).

According to Newton’s 3 Laws, the force N with which the seat presses on the pilot is equal to the force with which the pilot presses on the seat, that is, his weight P: N = P.

P = m * (a + g).

a = V ^ 2 / R.

P = m * (V ^ 2 / R + g).

P = 70 kg * ((300 m / s) ^ 2/1000 m + 10 m / s2) = 7000 N.

Answer: when the aircraft is withdrawn from their dive, the weight of the pilot will be P = 7000 N.