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What is it about GPE that makes it transfer to kinetic energy, why does it do this? What is GPE?

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    $\begingroup$ Two masses attract each other, leading to an acceleration which increases their velocity… $\endgroup$
    – Jon Custer
    Commented Feb 10, 2023 at 14:38
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Feb 10, 2023 at 14:52
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    $\begingroup$ By "GPE" do you mean the Gross–Pitaevskii equation? :P en.wikipedia.org/wiki/Gross%E2%80%93Pitaevskii_equation $\endgroup$
    – Quillo
    Commented Feb 10, 2023 at 16:25
  • $\begingroup$ @Quillo No: gravitational potential energy. $\endgroup$
    – J.G.
    Commented Feb 10, 2023 at 16:48

2 Answers 2

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Gravitational Potential Energy (GPE) is the amount of energy an object could have if it were to fall from that height. Thus - provided the gravitational acceleration is constant - gravitational potential energy only depends on mass and how far from the ground ($h=0$) the object is suspended, given by $U=mgh$. It is transfered to kinetic energy when droped from height h. This kinetic energy can be calculated from the equation $KE=\frac{1}{2}mv^2$. If you want to calculate the kinetic energy in terms of the gravitational force $mg$, you'll have to apply the Work-Energy Theorem, which states that the net work equals the change in kinetic energy; however, because the kinetic energy is initially $0$, it would just be the final kinetic energy, just before the object reaches $h=0$. The work is done by multiplying the force ($mg$) by the height, $h$. Hope this helps.

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As I understand it(your question), we can use conservation of mechanical energy to answer your question.

Consider a ball (of mass $m$)falling from a height $h$. After some time before the collision, the ball must be at height $h_{2}$ which is obviously less than the original height. It is clear from the above observation that: $$ mgh > mgh_{2} $$

This means that the GPE has decreased. To balance the decrement in GPE, Kinetic energy increases so that overall mechanical energy remains conserved. Since mass cannot change, velocity increases to increase the kinetic energy.

At the bottom most point(on the ground), GPE is $0$ which means all Potential energy had been converted to kinetic energy of the mass. This is why velocity is maximum till the time the object reaches the ground.

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