Show that potential energy is conserved

AI Thread Summary
The discussion focuses on deriving the equation of motion from the given potential energy function, U(x) = 4x^2 + 3, and proving conservation of energy. The force is determined to be F = 8x, leading to the acceleration a = 8x/m. The user attempts to find velocity and position functions but struggles with the integration process and the relationship between kinetic and potential energy. Key insights include the need to multiply the equation of motion by velocity and integrate to establish energy conservation, emphasizing the importance of correctly applying physics principles. The conversation highlights common pitfalls in integrating motion equations and clarifies the correct approach to proving energy conservation.
gelfand
Messages
39
Reaction score
3

Homework Statement



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

Homework Equations

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

The Attempt at a Solution



I'm not too sure how to go about this.

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
about that.

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
 
Physics news on Phys.org
gelfand said:
2. Homework Equations ##F = \frac{dU}{dt}##
Dividing energy by time gives power, not force.
 
haruspex said:
Dividing energy by time gives power, not force.
OK ##F = - \frac{dU}{dx}## sorry , I'm still unsure about the question
 
gelfand said:
OK ##F = - \frac{dU}{dx}## sorry , I'm still unsure about the question
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top