Gravitational Potential Energy during a Free-Fall Impact

1. Jun 16, 2009

ptd-

http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html" [Broken]

Since the gravitational potential energy of the falling object at the point of impact is higher than the gravitational potential energy at the distance traveled after impact, where does the energy go? It is shown that the KE is removed due to the work done but I see no treatment of PE after impact.

More specifically, how does "the process of penetration" further decrease the gravitational potential energy if all the work went towards decreasing the kinetic energy?

Last edited by a moderator: May 4, 2017
2. Jun 16, 2009

diazona

Actually some of the work goes toward decreasing the potential energy as well. It looks like the page you're looking at doesn't take that into account.

Here's another explanation, hopefully a little more satisfying: conservation of energy says that
$$\Delta K + \Delta U - \Delta W = 0$$
that is, between any two points on the object's path, the change in kinetic energy plus the change in potential energy minus the work done by external nonconservative forces (those that have no associated potential) is equal to zero. So for example, between the top of the object's path (when it gets dropped) and when it hits the ground,
\begin{align*}\Delta K &= \frac{1}{2}mv^2\\ \Delta U &= -mgh\\ \Delta W &= 0\end{align*}
Between the impact and the point where the object finally stops,
\begin{align*}\Delta K &= -\frac{1}{2}mv^2\\ \Delta U &= -mgd\end{align*}
So the law of conservation of energy says that
$$\Delta W = -\frac{1}{2}mv^2 - mgd$$
which means that the impact force is
$$F = -\frac{\Delta W}{d} = \frac{mv^2}{2d} + mg$$
The website you were looking at seems to neglect the last term. (good catch! )

EDIT: Although they do say this:
which does sort of fix the problem, but the math done by those text boxes seems to be inconsistent with that statement.

3. Jun 16, 2009

Staff: Mentor

Note the disclaimer on that hyperphysics page:
Edit: diazona beat me to it

The math in the text boxes looks consistent with that to me.

4. Jun 16, 2009

ptd-

I believe the text boxes are inconsistent because they will give a different answer for average impact force.

For example, plug in m = 1 kg, h = 1 m, d = 1 m.

Diazona's equation would yield a value of 19.6 N, whereas text boxes 9.8 N.

Edit: it's actually consistent if you use h = 2 as including the "d" distance, but then the velocity/kinetic energy before impact text boxes are wrong.

Would it be appropriate to think of the extra term "mg" as the normal force of the surface acting against the falling object's self weight?

Curiously, is conservation of energy the only way to approach this? If i try to solve for the "average impact force" using Newton's second law I find that it also does not consider this extra term "mg".

F = ma, where a is the (assumed) constant decelerration required to bring the falling object to rest from the velocity at the point of impact, within the distance prescribed.

For our previous example:
m = 1 kg, h = 1 m, d = 1 m, v_impact = 4.427 m/s, v_rest = 0 m/s.
Average speed = 2.2135 m/s
Time to stop = 0.452 s
Acceleration = dv/dt = 4.427/.452 = 9.8 m/s^2
Upwards Force = 1*9.8 = 9.8 N.

Have I simply neglected to consider the normal force here? But if so, would not the self weight of the ground-penetrating object cancel that out?

Last edited: Jun 16, 2009
5. Jun 16, 2009

Staff: Mentor

Right: The speed/KE numbers "before impact" don't make much sense, since h includes the after impact distance. But the force numbers are correct.

The fact that the surface must overcome the object's weight to accelerate it adds the "mg" term.

Sure you can use Newton's 2nd law. Just realize that the impact force is not the only force acting on the object. The weight of the object provides the "extra" mg.

You neglected to consider the object's weight. If the upwards force equaled 9.8 N, there would be no net force on the object.

6. Jun 16, 2009

diazona

Yeah, that's what I meant by it being inconsistent.