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