The crane lifts the load evenly at a speed of 0.4 m / s. What was the weight of the lifted load if the crane

The crane lifts the load evenly at a speed of 0.4 m / s. What was the weight of the lifted load if the crane performed 360 kJ in 45 seconds?

The useful work of the crane is:
A = Fd * h, where A is the work of the crane (A = 360 kJ = 360 * 10 ^ 3 J), Fd is the acting force (Fd = Ft (gravity) = m * g, m is the mass of the lifted load (kg) , g is the acceleration of gravity (g = 10 m / s ^ 2), h is the height of the lifting of the load (h = V * t, where V is the speed of lifting the load (V = 0.4 m / s), t is the lifting time (t = 45 s))).
Let us express and calculate the mass of the lifted load:
A = m * g * V * t; m = A / (g * V * t) = (360 * 10 ^ 3) / (10 * 0.4 * 45) = 2000 kg = 2 tons.
Answer: The mass of the lifted load is 2 tons.