What work must be done to lift a load weighing 2 tons to a height of 300 dm?

These tasks: m (mass of the lifted load) = 2 t (in SI m = 2 * 103 kg); h (required lifting height) = 300 dm (in SI h = 30 m).

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

The work required to lift the load under consideration is equal to the change in its potential energy: A = ΔEp = m * g * h.

Let’s perform the calculation: A = 2 * 10 ^ 3 * 30 * 10 = 600 * 10 ^ 3 J = 600 kJ.

Answer: The required work is 600 kJ.