A helicopter weighing 6 tons uniformly rises to a height of 100 m. What kind of work does the helicopter

A helicopter weighing 6 tons uniformly rises to a height of 100 m. What kind of work does the helicopter engine do in this case?

To find the value of the work performed by the engine of the specified helicopter during vertical takeoff, we use the formula: A = ΔEp = m * g * h.

Variables and constants: m is the mass of the specified helicopter (m = 6 t = 6 * 10 ^ 3 kg); g – acceleration due to gravity (g ≈ 10 m / s2); h – change in helicopter height (h = 100 m).

Let’s perform the calculation: A = m * g * h = 6 * 10 ^ 3 * 10 * 100 = 6 * 10 ^ 6 J.

Answer: When lifting 100 m, the engine of the specified helicopter will perform work of 6 MJ.