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

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

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Calantha 1 year 2021-07-31T10:57:53+00:00 1 Answers 30 views 0

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    2021-07-31T10:59:41+00:00

    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.

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