Gravitational Potential Energy during a Free-Fall Impact

In summary: It gives the right value for the force, but the wrong values for the speed and energy.The fact that the surface must overcome the object's weight to accelerate it adds the "mg" term.Right, so the "mg" term is necessary for the acceleration of the object, but not for the calculation of the impact force. The impact force will be the same whether you take the weight into account or not.
  • #1
ptd-
6
0
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?
 
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  • #2
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
[tex]\Delta K + \Delta U - \Delta W = 0[/tex]
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,
[tex]\begin{align*}\Delta K &= \frac{1}{2}mv^2\\
\Delta U &= -mgh\\
\Delta W &= 0\end{align*}[/tex]
Between the impact and the point where the object finally stops,
[tex]\begin{align*}\Delta K &= -\frac{1}{2}mv^2\\
\Delta U &= -mgd\end{align*}[/tex]
So the law of conservation of energy says that
[tex]\Delta W = -\frac{1}{2}mv^2 - mgd[/tex]
which means that the impact force is
[tex]F = -\frac{\Delta W}{d} = \frac{mv^2}{2d} + mg[/tex]
The website you were looking at seems to neglect the last term. (good catch! :wink:)

EDIT: Although they do say this:
Note that the above calculation of impact force is accurate only if the height h includes the stopping distance, since the process of penetration is further decreasing its gravitational potential energy.
which does sort of fix the problem, but the math done by those text boxes seems to be inconsistent with that statement.
 
  • #3
Note the disclaimer on that hyperphysics page:
hyperphysics said:
Note that the above calculation of impact force is accurate only if the height h includes the stopping distance, since the process of penetration is further decreasing its gravitational potential energy.

Edit: diazona beat me to it

The math in the text boxes looks consistent with that to me.
 
  • #4
Thanks for your replies.


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?
 
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  • #5
ptd- said:
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.
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.

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


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".
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.

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?
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
Doc Al said:
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.
Yeah, that's what I meant by it being inconsistent.
 

1. What is gravitational potential energy during a free-fall impact?

Gravitational potential energy is the energy stored in an object when it is positioned at a certain height in a gravitational field. During a free-fall impact, this energy is converted into kinetic energy as the object falls towards the ground.

2. How is gravitational potential energy calculated during a free-fall impact?

The gravitational potential energy during a free-fall impact can be calculated using the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.

3. Does gravitational potential energy affect the speed of a falling object?

Yes, the higher the gravitational potential energy of an object, the faster it will fall. This is because the potential energy is converted into kinetic energy, which determines the speed of the object.

4. Can gravitational potential energy be negative?

Yes, gravitational potential energy can be negative if the object is positioned below the reference point, such as the ground. This indicates that the object is already in motion and does not need additional energy to start moving.

5. How is gravitational potential energy related to work?

Gravitational potential energy is related to work through the work-energy principle, which states that the change in an object's kinetic energy is equal to the net work done on it. In the case of a free-fall impact, the work done is equal to the change in the object's gravitational potential energy.

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