When a ball falls from a high place, it experiences a gravitational force. Forces make objects accelerate (in this case, it is constantly increasing the velocity). Because $KE = \frac{1}{2}mv^2$ this should mean that Kinetic Energy should grow quadratically (please correct me if wrong) because of the increasing velocity right?

But also, $GPE=mgh$ where the potential energy is a linear equation. How can this happen? If energy has to be conserved wouldn't both equations have to change linearly?

Can you please explain how the $KE$, $GPE$ and conservation of energy are reconciled in this system? Could you also confirm the shape of the graph of $KE$ and $GPE$ against time?

(I had initially come up with this problem for electric fields but I think that it might've been easier to answer the question in terms of gravitational fields)


closed as off-topic by AccidentalFourierTransform, Jon Custer, Kyle Kanos, Aaron Stevens, stafusa Oct 22 '18 at 22:24

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  • 11
    $\begingroup$ Linear with respect to what? Linear with respect to time - no. With respect to velocity - no. With respect to the distance fallen - yes! $\endgroup$ – alephzero Oct 20 '18 at 9:40
  • $\begingroup$ I had the same doubt a while back. You think KE increases quadratically because of the square of velocity right.But in the formula of PE=mgh there is no v.Infact for an object dropped from rest the height the obj has fallen can be replaced by v^2/2g which gives a "quadratic" decrease in the PE. $\endgroup$ – A.R.K Oct 20 '18 at 14:53

You are mixing up two different SUVAT equations. The change of velocity with time is given by:

$$ v = u + at $$

So velocity increases linearly with time. However the change of velocity with distance is given by:

$$ v^2 = u^2 + 2as $$

So velocity increases as the square root of distance, not linearly with distance. That's why the kinetic energy increases linearly with distance. The kinetic energy does increase quadratically with time.

  • $\begingroup$ but then if KE increases quadratically with time, what shape must the GPE decrease with to ensure conservation of energy? $\endgroup$ – John Hon Oct 20 '18 at 9:49
  • $\begingroup$ @JohnHon GPE decreases (becomes more negative) linearly with distance and quadratically with time. This ensures that the sum of the GPE and KE remains constant. $\endgroup$ – John Rennie Oct 20 '18 at 10:01

Assume that something is dropped from a certain height and no air resistance etc.

Consider the following two graphs:

enter image description here

The left hand graph has height $h$ as the abscissa.
${\rm pe} = mg \cdot h +0$ and ${\rm ke} = - mg \cdot h + {\rm constant}$, both linear relationships.
The sum of the potential energy and the kinetic energy is constant for all heights.

The right hand graph has time $t$ as the abscissa.
Now the velocity $v$ does depend linearly on time $t$ with $v = gt$ but the relationship between kinetic energy and time is a quadratic ${\rm ke} = \frac 12 mg^2t^2$ as is the potential energy ${\rm pe} = {\rm constant} - \frac 12 mg^2t^2$.
Again the sum of the potential energy and the kinetic energy is constant for all times.


GPE will not decrease linear over time. This is because height will decrease at a greater rate over time as velocity increases due to acceleration (until either h is zero or terminal velocity is reached).

Otherwise, if in doubt; chart it and see.


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