How is energy conserved when a dropped object hits the ground?

I know that when an object is held at a particular height above the Earth, it has a quantity of potential energy associated with it, but no kinetic energy. When the object is dropped, it loses potential energy and gains kinetic energy, thereby ensuring that energy is conserved. However, I've never understood what happens when the object finally strikes the ground; at this point, it has no potential and no kinetic energy. I'm sure some of it is released as heat and sound, but that can't account for all the energy, can it?How is energy conserved when a dropped object hits the ground?
If the collision between the object and the ground were perfectly elastic, you would find that the object would bounce back up, from the ground, to the initial height that it was dropped from. However, when the collision is inelastic (like the case you are referring to), energy is dissipated by three means: compression waves generated in the air (sound), deformation of the objects involved in the collision, and some heat release due to friction.





Energy is this case (as in every case) is conserved, the problem is that useful energy has been turned into very useless forms of energy - heat, sound, and deformed objects.How is energy conserved when a dropped object hits the ground?
Thanks for the great to the point answer; well done! Report Abuse

It can and does.

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