The man took hold of the end of a homogeneous rod 2 m long and 100 kg in weight lying on the ground and slowly

The man took hold of the end of a homogeneous rod 2 m long and 100 kg in weight lying on the ground and slowly lifted this end to a height of 1 m. What work did he do?

Data: l (length of a homogeneous rod) = 2 m; m (rod mass) = 100 kg; one end of the rod was raised to a height of 1 m.

Reference values: g = 10 m / s2.

1) Since one end of the rod is raised to a height of 1 m, and the length of the entire rod is 2 m, its center of gravity has risen above the ground to a height of 0.5 m.

2) The work that the person has done will be equal to the change in the potential energy of the center of gravity of the rod: A = ΔEp = m * g * h = 100 * 10 * 0.5 = 500 J.

Answer: 4) The person has done the work of 500 J.