When an object is at rest on a smooth horizontal "surface" (floor) its potential energy $P_E$ equals zero and, as the object is at rest, its kinetic energy $K_E$ is also zero. When the same object is kept at some height $h$ it has $P_E=mgh$ and $K_E=0$. Now if you let it fall then when it reaches the ground $P_E=0$ and $K_E=\frac12mv^2$ but initially when the object was at rest it had $K_E=P_E=0$ so where does this $K_E$ go? Also $K_E$ as well as $P_E$ are frame dependent so what does $P_E+K_E=\mathrm{constant}$ really mean? (No nonconservative and external forces are acting on the "system")

  • $\begingroup$ Why was PE=0 initially? If you're letting the object fall from the height h, then initially PE=mgh. This has to be equated to KE at the ground. $\endgroup$
    – mavzolej
    Apr 16, 2019 at 4:32
  • 1
    $\begingroup$ if you're asking where does $K_e$ go when the object hits the ground, the energy is (not always) released as mechanical shock in the ground; see en.wikipedia.org/wiki/Coefficient_of_restitution $\endgroup$
    – user164843
    Apr 16, 2019 at 4:40
  • $\begingroup$ PE was zero initially as it was on the floor-zero PE level $\endgroup$
    – K. Jay
    Apr 16, 2019 at 5:00
  • 2
    $\begingroup$ Do you know about conservation of energy? your $GPE$ got converted to $K.E$. That's how your body got$ K.E$. Moreover you did work to raise the object to $h$ and hence you increased its $P.E$. $\endgroup$ Apr 16, 2019 at 5:40

4 Answers 4


If an object is kept on your palm over some height above the floor, there is no reason for you to talk about the energy of a system since it is just there. It is not moving, no obvious attraction to the ground (gravity cancels the normal force due to your palm).

Now, you have to apply the conservation of energy, when the system has no external force acting on it. If you chose the system as the earth and the object, the mutual force between the two counts as internal force. The normal force given by your palm counts as external force since it is from you.

So, if $K=0$ and $P=mgh$ assuming the earth's crust is the ground level for $h=0$. Basically you are assuming the object to be free from your palm when you apply this condition.


Let's consider what happens here in stages.

STAGE 1: Initial conditions (KE=0, PE=0)

Since macroscopic KE and PE are frame dependent, these initial conditions mean we are assigning a fixed point on the supporting surface near the object as our frame of reference. This is our reference frame for all stages.

STAGE 2: The acquisition of gravitational potential energy, mgh.

In order for the object to be at height h, an external agent (you perhaps) had to apply a force to raise the object off the ground and bring it to rest at height h. The work done by the external agent equals the increase in potential energy.

STAGE 3: Free fall (conservation of PE+KE).

As the object falls, gravity does positive work on the object giving it kinetic energy at the expense of its potential energy. The sum of KE and PE is constant (no air drag) at each point of the path up to the point of impact where all all the PE is now KE.

STAGE 4: Impact. What happens to the KE?

There are three possibilities or cases

Case 1: Completely inelastic collision.

In this case the object is brought to a complete stop (no rebound). The kinetic energy causes permanent deformation, friction, etc. on the object and/or surface (with a small amount of energy in sound). The temperature of the object and/or surface at the impact site is elevated. Heat transfer occurs from the impact site to the lower temperature surroundings.

Case 2: Partially elastic collision.

In this case the object bounces off the surface, but does not return to the height h. Part of the kinetic energy is lost, part recovered. Each subsequent bounce results in more KE lost until the object comes to rest on the surface.

Case 3: Completely elastic collision.

In this case the object rebounds and returns to height h. The process repeats. However, collisions at the macroscopic level are inelastic. Perfect elastic collisions are approached at the microscopic level, such as with the collisions of ideal gas molecules.

Hope this helps.


Short answer: the new "extra" KE comes from the work you did to rise the object.

Initially, object at rest on the ground. $PE=KE=0$.

Then, you say you rise it, until $PE=mgh$. But this is the key step. You cannot rise it unless you apply W=mgh. That work gets stored as potential energy.

Now, if you let it fall, $PE$ will become $KE$. It will be more than 0 because you have added energy during the process.


where does this $K_E$ go?

It is spent heating or crushing the surface.

  • Drop a tennis ball on a sandy beach and the $K_E$ is absorbed by the sand in order to displace the sand (work is done) and create a small "crater".
  • Drop a tennis ball on an asphalt road, and the $K_E$ is not absorbed but instead converted into elastic energy during the impact, which is converted back to $K_E$ in the next moment - and the ball jumps up again with a lot of $K_E$.

The $K_E$ could of course also be absorbed into the ball itself in a similar fashion, if it is soft or breakable.

what does $P_E+K_E=\text{constant}$ really mean?

If nothing else is involved, then their sum must be constant and never change. If their sum wasn't constant but suddenly was, say, higher than before, then one would be increasing more than the other would be decreasing.

That is impossible due to the the energy conservation law - if one energy type increases more than the other energy type decreases, then where does that extra energy come from?


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