An airplane weighing 10 tons lands at a speed of 216 km / h and runs to a stop of 400 m.

An airplane weighing 10 tons lands at a speed of 216 km / h and runs to a stop of 400 m. Find the force of resistance to movement.

Given: m (aircraft mass) = 10 t (10 ^ 4 kg); V0 (speed at the beginning of landing) = 216 km / h (60 m / s); S (distance traveled by the plane to a stop) = 400 m.

1 way) Fc. = m * a = m * ((V ^ 2 – V0 ^ 2) / 2S) = m * ((0 – V0 ^ 2) / 2S) = -m * V0 ^ 2 / 2S = -104 * 60 ^ 2 / (2 * 400) = -45,000 N (45 kN).

2 way) Fsopr. = A / S = ΔEk / S = (mV ^ 2/2 – mV0 ^ 2/2) / S = (0 – mV0 ^ 2/2) / S = -m * V ^ 2 / 2S = -10 ^ 4 * 60 ^ 2 / (2 * 400) = -45,000 N (45 kN).

Answer: The force of resistance to the movement of the aircraft during landing was 45 kN.