# SHowing that conservation of Energy is related to Newton's motion

1. Oct 14, 2013

### Emspak

1. The problem statement, all variables and given/known data

Given: $\frac{1}{2}m (\dot x)^2 + mg x = E$

Gravitational force is mg.

We need to show that by solving this DE that we can confirm that the conservation of energy correctly describes one-dimensional motion (the motion in a uniform field). That is, that the same motion is obtained as predicted by Newton' equation of motion.

3. The attempt at a solution

My attempt was as follows:

$\frac{1}{2}m\dot x + mg x = E$ so by moving things around a bit I can reduce this to

$(\dot x)^2 = \frac{2(E - mgx)}{m}$

and from there I can solve the DE:

$\dot x = \sqrt{\frac{2(E - mgx)}{m}} \Rightarrow \int dx =\int \sqrt{\frac{2(E - mgx)}{m}}dt$

from there I can move some variables some more:

$\frac{\sqrt{m}}{\sqrt{E-mgx}} \int dx = \int \sqrt{2}dt \Rightarrow \sqrt{m} \int \frac{dx}{{\sqrt{E-mgx}}} = \int \sqrt{2}dt$

trying a u substitution where $u=\sqrt{E-mgx}$ and $du = -mgdx$ I should have $\frac{\sqrt{m}}{mg} \int \frac{du}{{\sqrt{u}}} = \int \sqrt{2}dt$ which gets me to $\frac{\sqrt{m}}{mg} \sqrt{u} + c = \sqrt{2}t$

going back to what I substituted for u I have:
$\frac{\sqrt{m}}{mg} \sqrt{E-mgx} + c = \sqrt{2}t$

and when I do the algebra I get (after moving the c over and squaring both sides):

$\frac{m}{m^2g^2}{E-mgx} = {2}t^2 - 2\sqrt{2}c+ c^2$
$\frac{E}{mg^2}-\frac{x}{mg} = {2}t^2 - 2\sqrt{2}c+ c^2$
$-\frac{x}{mg} =-\frac{E}{mg^2}+ {2}t^2 - 2\sqrt{2}ct+ c^2$
$x(t) =\frac{E}{g}- 2mgt^2 + 2\sqrt{2}mgct- c^2mg$

I am unsure of the last step. I see something that looks like an equation of motion there -- and since $gt = v = \dot x$ I can see that term, and $gt^2 = x$. But I am not quite clear on what to do with the energy term.

I know, I know, don't just plug in formulas. But I presume that the prof gave us the instructions he did for a reason and it wasn't just to confuse us, though he has succeeded in that regard. :-)

Anyhow, any assistance is appreciated.

2. Oct 14, 2013

### tiny-tim

Hi Emspak!

hmm … you want …
sooo you want an equation with x'' in it

hint: what's the most obvious way of getting x'' out of the original equation?

3. Oct 14, 2013

But.. I did that, I came up with a solution for the DE. It's x(t). mg = force, int he solution I got, and I suppose I could rewrite the solution as $x(t) = \frac{E}{g}- 2m \ddot x t^2+ 2\sqrt{2}\dot x = c^2mg[/tex]. But I am still not sure about this. 4. Oct 14, 2013 ### Emspak But.. I did that, I came up with a solution for the DE. It's x(t). mg = force, int he solution I got, and I suppose I could rewrite the solution as [itex]x(t) = \frac{E}{g}- 2m \ddot x t^2+ 2\sqrt{2}\dot x - c^2mg$. But I am still not sure about this.

5. Oct 14, 2013

### Emspak

sorry for doubling up...

6. Oct 14, 2013

### tiny-tim

no, you started with x' and x, and you eliminated x' (by integrating)

try getting to something with x'' x' and x

7. Oct 14, 2013

### Emspak

is this what you are referring to? You're being a bit cryptic...

$x(t) = \frac{E}{g}- 2m \ddot x t^2+ 2\sqrt{2}\dot x - c^2m\ddot x$ which becomes

$x(t) = \frac{E}{g}- (2m-mc^2) \ddot x t^2+ 2\sqrt{2}\dot x$

$x(t) = \frac{E}{g}- m(2-c^2) \ddot x t^2+ 2\sqrt{2}\dot x$

EDIT: looking at this I can see the energy term could be moved over and you'd end up with Energy = equation of motion. Is that what we are after here ?

8. Oct 14, 2013

### tiny-tim

d/dt (1/2 mx'2) = mx'x''

9. Oct 14, 2013

### Emspak

(sigh) that isn't helping much, there are no x'x'' terms in the equation as I re-wrote it. I am not at all clear on how that derivative is helpful, unless you are alluding to something that should happen in the E term.

10. Oct 14, 2013

### tiny-tim

try differentiating the whole original equation …

11. Oct 14, 2013

### Emspak

How about this: when you talk about getting x out of the equation where in the derivation I did should that be happening? Did I do the whole bloody thing wrong or what? Or was I correct that there's a last step here? If I plug a KE equation in for E I would have $\frac{m(gt)^2}{g}$ there. But I suspect that is not what you are taking about doing.

12. Oct 14, 2013

### Emspak

OK, at that point I end up with $m \dot x \ddot x + mg \dot x = 0$, yes?

13. Oct 14, 2013

### tiny-tim

yup!

and now divide by x'

14. Oct 14, 2013

### Emspak

OK, we can do that because it would show up on both sides, right? $m \ddot x = -mg$. Now since g was the acceleration (earth's gravity) and we have $\ddot x$ in the same place we've just shown that the equation of motion works here, or at least that in a 1-D case in a field that it does, correct?

15. Oct 14, 2013

### tiny-tim

correct!

(and if you divide by m, you get x'' = -g, exactly as newton would have said)