**Answer:**

Initial: bar power U₀

Final: bar power U = U₀ / 2

bar Kinetic energy K = U₀ / 2

These two bars are half the height of the initial bar

**Explanation:**

To know which graph is correct, let's discuss the solution to the problem

Initial mechanical energy

Em₀ = U₀ = m g H

The mechanical energy at the midpoint

Em₂ = K + U₂

As there is no friction, mechanical energy is conserved

Em₀ = Em₂

U₀ = K + U₂

K = U₀ - U₂

K = m g (H - y₂)

Indicates that position 2 corresponds to y₂ = H / 2

K = m g (H –H / 2)

K = ½ m g H

K = ½ Uo

Therefore the graph must be

Initial: bar power U₀

Final: bar power U = U₀ / 2

bar Kinetic energy K = U₀ / 2

These two bars are half the height of the initial bar