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Does an object's kinetic energy increase, decrease, or stay constant when it reaches terminal velocity while falling?

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    $\begingroup$ after reaching the terminal velocity $v_f$, the velocity does not change after that, so $E_k=1/2 m v_f^2$ staying constant. if this is what you mean. $\endgroup$
    – wonderich
    Jun 24 '14 at 2:02
  • $\begingroup$ Assuming your mass is non-changing this question is equivalent to "Does an object's speed increase, decrease, or stay constant when it reaches terminal velocity while falling". When you go from non-terminal velocity to terminal velocity, it increases. When you stay at non-terminal velocity, it's constant. (assuming constant terminal velocity) $\endgroup$
    – Cruncher
    Jun 24 '14 at 13:01
  • $\begingroup$ Understanding the feeling of this: It takes energy to push through the air. Terminal velocity is the velocity where all the energy gained from falling is spent pushing through the air resulting in no net gain. $\endgroup$
    – Phil
    Jun 24 '14 at 16:43
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Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes. Hence the speed and the kinetic energy of the falling body would both start to decrease after reaching the altitude where the terminal velocity is lower than the object's vertical velocity.

There's an excellent discussion about this subject in the following web-site: http://www.pdas.com/falling.html

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It stays the same, because it's dependent on the square of velocity. You're probably wondering where the extra energy goes because potential energy is falling. The answer is: it is dissipated into heat because of friction from the air or air resistance.

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Kinetic energy is dependent on square of velocity therefore constant.

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