# Potential energy vs position graph w/ total mech. energy?

1. Jan 1, 2014

### Ascendant78

Ok, I've been trying to soak up the MIT physics material, but I am stuck on something and it is driving me crazy. They have several questions about graphs of potential energy vs. position, which also includes the total mechanical energy. I just can't fully wrap my mind around these graphs and the explanations they give about them are extremely convoluted. Questions 5-9 on here are what I'm talking about: http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/conservation-of-energy/mechanical-energy-and-the-simple-harmonic-oscillator/MIT8_01SC_quiz15.pdf

What is confusing me is how you determine which direction the object will head based on the graph alone? I understand that when potential decreases, kinetic will increase. However, for question 5 on that PDF, they indicate that the object would go to infinity. What I don't get is how you don't know that it doesn't just stop when it gets to the position that is directly on the mechanical energy line? What lets you know that the motion would stop there and change direction, or does it not even get there for some reason?

Also, what happens to the kinetic energy when the position goes to the point where potential is above the overall mechanical? Since kinetic is scalar, wouldn't it be impossible for it to become negative? If that's the case, then wouldn't a potential higher than the mechanical be impossible?

If someone could please make these types of graphs make sense to me, I'd really appreciate it.

2. Jan 1, 2014

### Simon Bridge

Use Newton's Laws. Lets say you have a potential energy function U(x), then the force for object at position x will be: $$\vec F = -\frac{d}{dx}U(x)\hat{\imath} = m\vec{a}$$ ... you'll have had that relation somewhere in the text.

What this bit of math is saying is that objects try to roll "downhill".

...it meets the total energy line, you see the slope of the function there is not zero
- so it experiences an unbalanced force there.
Sure it stops - but only for an instant - then what happens?

That is correct - a classical particle can never be in a position where it's total energy is greater than it's potential energy.
This should be clear by conservation of energy ... $E_{tot}=E_K+U$

It can help to think of the PE curve as a landscape and the object is a BB pellet rolling on it.
Where the PE curve meets the energy level is the maximum height the pellet may reach.

3. Jan 1, 2014

### Simon Bridge

Example: Lets say you have a potential energy function that looks like this:

$U(x)=kx^3$ - notice that $U(0)=0$.
For simplicity, let $K_0=K(x=0)$ - this will also be the total energy since $E_{tot}=K(x)+U(x)$

Thus $K(x)=K_0-kx^3$ with an initial velocity in the +x direction.
You can work it out - for a particle mass m: $\vec v=\sqrt{2K_0/m}\hat{\imath}$

As the particle travels to the right, it experiences a force opposing the motion (because the slope is positive that way and the particle is trying to "roll downhill").

Thus: it slows down - exchanging kinetic energy for potential energy.
The farthest the object can travel to the right is x=K_0^{1/3}, at which point it has exchanged all it's kinetic energy for potential energy - so it comes to rest.

The force there is $\vec F=-3kK_0^{2/3}\hat{\imath}=m\vec a$
Since K_0 is positive, the force, and therefore, the acceleration still points in the -x direction... so the particle, having come to rest, heads off, picking up speed, in the -x direction.

In the -x direction, there are no more barriers.
Thus the particle escapes to $-\infty$

In this example, the potential energy happened to be measured from x=0 ... but this is arbitrary.
This means that although the kinetic energy can never be negative, the PE can be as negative as you like and so can the total energy.

For the above system I set the total energy to be K_0 ... lets, instead, have a particle with total energy -1J ... what would this mean?

A x=0, the KE would have to be -1J which is impossible, to the particle would never reach as far to the right as the origin.

The closest to the origin the partical could get is where all the energy is potential energy, which is where
$kx^3=-1 \implies x = -1/\sqrt[3]{k}$
... which is to the left of the origin.
At that position, the kinetic energy is zero and the force points to the left.

4. Jan 2, 2014

### Ascendant78

Thanks for such a thorough response! You explained it all in a way that was crystal-clear, which is far more than I can say for the course materials. Though what they cover in their course is definitely more comprehensive than what I learned at my college, it is a pain trying to teach the unknown content to myself without a professor to ask for clarification on certain things like this.

You really helped make sense of this all for me. Thanks for taking the time to do so. I really do greatly appreciate it, as it would have kept running through the back of my head until I got this, lol.

5. Jan 2, 2014

### Simon Bridge

No worries - the lynchpin is the relationship between the force and the potential energy function.
The diagrams contain a slight irregularity: a and b should really be positions on the x axis rather than vague bits of the curve. However, lots of people label their graphs like that so you have to get used to it.

- if you look now at Q6, you'll see the total energy line is cut in three places ... considering the starting position of the particle, it can now never get away.
We say that the particle is "bound" to the potential well.
Can you see which answer best fits now?

Note: the answer to these questions is quite different in quantum mechanics :)

6. Jan 2, 2014

### Ascendant78

Yes, revisiting all the questions now after what you explained makes it easy to understand. As far as quantum mechanics, I am definitely looking forward to that.

7. Jan 2, 2014

### Simon Bridge

Well done.
Happy New Year BTW.