An airplane weighing 5 tons moved horizontally at a speed of 360 km / h.

An airplane weighing 5 tons moved horizontally at a speed of 360 km / h. Then he climbed 2 km. At the same time, its speed became 200 km / h. Find the work that the engine took to lift the plane.

Given: m (aircraft mass) = 5 t = 5 * 10 ^ 3 kg; V0 (initial speed) = 360 km / h = 100 m / s; h (lifting height) = 2 km = 2 * 103 m; V1 (speed after ascent) = 200 km / h = 55.56 m / s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

To calculate the work done by the motor for lifting, we apply the formula: A = ΔE = E – E0 = m * V1 ^ 2/2 + m * g * h – m * V0 ^ 2/2 = m * (V1 ^ 2/2 + g * h – V0 ^ 2/2).

Calculation: A = 5 * 10 ^ 3 * (55.56 ^ 2/2 + 10 * 2 * 10 ^ 3 – 100 ^ 2/2) = 82.72 * 10 ^ 6 J (82.72 MJ).

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