When researching GPE and KE conversion, lots of websites say that as an object with GPE (Gravitational Potential Energy) falls, that GPE is converted into KE (Kinetic Energy), so there is less GPE and more KE. The problem I have with this principle is, what starts the conversion, because all the websites say "When the object falls", but there needs to be KE for it to fall, but for there to be KE, the object needs to convert that GPE into KE, which means the object needs KE to fall etc, etc.

  • 2
    $\begingroup$ While anything that already moves has kinetic energy, it is not true that objects need kinetic energy to start moving - it is perfectly acceptable to have vanishing velocity but non-vanishing acceleration. $\endgroup$
    – ACuriousMind
    Jan 31, 2015 at 15:14
  • $\begingroup$ @ACuriousMind So basically, the object with GPE will start moving as long as there is nothing with equal or a larger force going against it. If I'm understanding you correctly. $\endgroup$
    – Orfby
    Jan 31, 2015 at 15:50

4 Answers 4


The answer was in fact covered by the Curious Mind, but for you to see the process how the potential energy transforms into kinetic, here is an elementary elaboration.

The equation of motion in the gravitation field says that

$ \ (1) \ h_0 - h = \int _0^t v(t) \ \text d t $

Multiplying this equation by $mg$ which is constant

$ \ (2) \ E_P(0) - E_{P}(t) = m \int _0^t g \ v(t) \text d t $

Since $g = \text d v(t)/ \text d t$ we can write the expression under the integral in a more convenient way

$ \ (3) \ E_P(0) - E_P(t) = m \int _0^{v(t)} v(t) \text d v(t), $

where the lower integration limit is zero because the object starts from velocity zero. So, we get

$ \ (4) \ E_P(0) - E_P(t) = \frac {m v(t)^2}{2} - 0 = E_K(t).$

At $t = 0$ on the LHS you have zero, s.t. no kinetic energy on the RHS. As the LHS increases, the RHS will also increase.


Writing down the relation between KE and PE and taking the time derivative yields: $$\frac{m\dot{z}^2}{2} - mgz = 0 \Rightarrow$$ $$m\dot{z}\ddot{z} - mg\dot{z} = 0$$

If we assume $\dot{z}$ is non-zero, we conclude that $\dot{z} = g$. However, if $\dot{z}$ is truly zero, this equation says absolutely nothing about what the body should do. Indeed, a totally stationary body (that is not accelerating) in a gravitational field would not violate energy conservation. However in this case we just have to apply common sense to argue that classical physics is continuous and insert the $g$ manually (or just take the limit).


An object at rest will start moving whenever there is acceleration. And there is acceleration whenever there is non-zero net force, according to

$$\sum \vec F=m \vec a$$

As @ACuriousMind mentions in the comments, kinetic energy is a "result" of velocity, but velocity is not caused by kinetic energy.

When you (in the comments to the question) say

as long as there is nothing with equal or a larger force going against it

then remember that if there is something "with equal force against it" then the net force will be zero. If there is something "with larger force against it", then there will be acceleration since the net force is non-zero. But acceleration might be upwards (could be a strong wind blowing the falling object up preventing it from falling).


Your statement "but there needs to be KE for it to fall" is wrong. What it need is a force in that direction to fall. That force on it comes from the negative of the gradient(change with respect to distance in the maximum increase direction) of potential energy which is always present as long as the potential energy is changing with distance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.