The tower crane lifts in a horizontal position a steel beam 5 m long and 100

The tower crane lifts in a horizontal position a steel beam 5 m long and 100 cm2 to a height of 12 m. What useful work does the crane do?

Data: l (length of steel beam) = 5 m; S (cross-section of a steel beam) = 100 cm2 = 0.01 m2; h (height to which the tower crane lifted the steel beam) = 12 m.

Reference data: ρ (average density of steel) = 7800 kg / m3; g (acceleration due to gravity) = 10 m / s2.

The useful work that the tower crane has done will be equal to the change in the potential energy of the steel beam: A = Ep = m * g * h = ρ * V * g * h = ρ * l * S * g * h.

Calculation: A = 7800 * 5 * 0.01 * 10 * 12 = 46 800 J or 46.8 kJ.