Question

How does the decrease in gravitational potential energy of a falling ball compare to its increase in kinetic energy? (Ignore air friction.)

Answer 1

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 = $\frac{1}{2}$m(Δv)^2

but Δv = 2gΔh

therefore

ΔKE = $\frac{1}{2}$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.