Answer:
for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.
Explanation:
Let us consider a ball falling from its maximum height
For a body falling from its maximum height to a point p
change in height = Δh
The potential energy decrease is then proportional to
ΔPE = mgΔh
where
ΔPE is the decrease in kinetic energy
m is the mass of the ball
g is acceleration due to gravity
Δh is the change in height
For a body falling from its maximum height, the increase change in velocity
Δv = u + 2gΔh (at maximum height u = 0)
where
u is the initial kinetic energy of the ball
Δv = 0 + 2gΔh
Δv = 2gΔh
The kinetic energy increases by
ΔKE = [tex]\frac{1}{2}[/tex]m(Δv)^2
but Δv = 2gΔh
therefore
ΔKE = [tex]\frac{1}{2}[/tex]m(2gΔh)^2 = 2m(gΔh)^2
comparing the increase in kinetic energy to the decrease in potential energy, we have
(2m(gΔh)^2)/(mgΔh) = 2gΔh
This means that for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.
