# Show that potential energy is conserved

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1. Apr 19, 2017

### gelfand

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

potential energy function of :

$$U(x) = 4x^2 + 3$$

And have to

i) Work out the equation of motion

ii) Prove explicitly that the total energy is conserved

2. Relevant equations

$$F = \frac{dU}{dt}$$

3. The attempt at a solution

I would say that I have the force of

$$F = 8x$$

By differentiating the given potential energy function. I need to work out the
equation of motion, what I have an object with mass $m$.

So this means that I have

$$F = 8x = ma$$

Then I have that

$$a = \frac{8x}{m}$$

Is this an equation of motion? I mean, it's acceleration, or should I find for
$v(t)$ and $x(t)$ as well as this?

In which case I would have

$$v(t) = \int a(t) dt$$

Which in this case is found as (having the mass in the equation seems unusual?)

$$v(t) = v_0 + \frac{1}{2m}8x^2 = v_0 + \frac{4}{m} x^2$$

So then from this I have that

$$x(t) = x_0 + v_0t + \frac{4}{3m}x^3$$

And this would be all of the equations of motion for this 1D case?

Then I need to prove that energy is conserved here, and I've no idea how to go

I've not been given any frictional forces, so it seems like it's just a given
that I'm going to have

$$W + PE_0 + KE_0 = PE_f + KE_f + \text{Energy(Lost)}$$

Here I can remove work $W$ and the energy lost for

$$PE_0 + KE_0 = PE_f + KE_f$$

And I need to do something with these?

Potential energy - I have the potential energy function given as part of the
problem which is

$$U(x) = 4x^2 + 3$$

Then I can sub this into the energy expression as
$$4x_0^2 + 3 + KE_0 = 4x_f^2 + 3 + KE_f$$

Getting rid of the constants seems pretty harmless

$$4x_0^2 + KE_0 = 4x_f^2 + KE_f$$

Now I'm really not sure what I should do from here, sub in kinetic formulas of
$K = \frac{1}{2}mv^2$?

$$4x_0^2 + \frac{1}{2}mv_0^2 = 4x_f^2 + \frac{1}{2}mv_f^2$$

I'm not sure if I can arrange this to be 'nicer' in any way either, I'm purely
thinking in algebra at the moment though not physics :S

$$8(x_0^2 - x_f^2) = m(v_f^2 - v_0^2)$$

I'm not sure if differentiation should do anything nice here, but I really have
no idea what I'm doing with this.

Thanks

2. Apr 19, 2017

### haruspex

Dividing energy by time gives power, not force.

3. Apr 20, 2017

### gelfand

OK $F = - \frac{dU}{dx}$ sorry , i'm still unsure about the question

4. Apr 20, 2017

### haruspex

You got a=8x/m ok, but you cannot integrate that wrt t directly. The expression you got for v(t) was the integral wrt x (which just gets you back to U).

There is a useful trick for solving equations like $\ddot x=f(x)$. Multiply both sides by $\dot x$, then integrate dt.