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Say there is an object being dropped from a distance. I can calculate the GPE of that object using $GPE = mgh$. As the object falls, the gravitational potential energy is converted into kinetic energy. However, it's when the object reaches terminal velocity, its velocity is not changing, so there is no change in kinetic energy $KE = 0.5mv^2$, so where is the GPE being lost being transferred to? At terminal velocity, does an object have no GPE? (has it already all been transferred to Kinetic?)

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Without air, a body in free fall experiences a simple transfer from GPE to KE; KE and speed increase as long as GPE is decreasing.

With air present, the body still transfers energy from GPE to KE, however due to collisions with air molecules, energy is also simultaneously transferred from KE of the object to the air (as thermal energy, and as bulk KE of the air).

As speed increases, the drag force increases, and so does the rate of energy transfer to the air. At terminal velocity, the rate of energy transfer for those two processes (GPE ==> KE, and KE ==> air) are equal; GPE is transforming into KE, and KE is transforming into energy of the air at the same rate, which is why speed stops increasing.

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  • $\begingroup$ Thanks, that make sense now, I don't know why I overlooked drag $\endgroup$ – TheInfernalCow Dec 30 '15 at 17:38
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so where is the GPE being lost being transferred to?

Transferred to the air molecules hitting on it.

At terminal velocity, does an object have no GPE?

Yes, the whatever energy gained by GPE is being lost to the air molecules. The condition that must hold here that rate of energy gain by GPE is equal to the rate of energy lost to air molecules.

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  • $\begingroup$ In other words, air resistance is significant. $\endgroup$ – Jim Garrison Dec 31 '15 at 3:43

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