Potential energy and Conservative Force

[Arman777]

Homework Statement

I have a force $\vec{F} = a_x\vec{i}+2a_y\vec{j}+3a_z\vec{k}$. Find the potential

Homework Equations

The Attempt at a Solution

Lets suppose
And we know that $\vec {F} = ∇U$

In this case I said that

$U_x=-\int F_xdx$
$U_y=-\int F_ydy$
$U_z=-\int F_zdz$
and then I said

$U= U_x+U_y+U_z$

Is this mathematically true ?

Normally I should have $U(x,y,z) = -\int F_xdx$ I guess..For example in this case I should have something like
$U(x,y,z) = -a_x^2+U(y,z)$
then I can say
$U(x,y,z) = \int (-a_x^2+ 2a_y)dy$ etc. Which one is correct ?

 

[haruspex]

Are a_x etc. constants? If so, what are their integrals wrt x, y, z?
You can always check your answer using the differential equation you quoted.

 

[Arman777]

Are a_x etc. constants? If so, what are their integrals wrt x, y, z?
You can always check your answer using the differential equation you quoted.

$a$ are constant. I find the answer its easy to find. I was asking about the mathematical part of it.Are both approaches are true ?

 

[PeroK]

Normally I should have $U(x,y,z) = -\int F_xdx$ I guess..For example in this case I should have something like
$U(x,y,z) = -a_x^2+U(y,z)$
then I can say
$U(x,y,z) = \int (-a_x^2+ 2a_y)dy$ etc. Which one is correct ?

You should never integrate a constant $a_x$ and get $\frac12 a_x^2$. That's a common mistake.

Instead, you have:

$\vec{F} = (a_x, 2a_y, 3a_z) = - \vec{\nabla}U(x, y, z) = -(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z})$

If we start with $-\frac{\partial U}{\partial x} = a_x$, then this gives:

$U(x, y, z) = -a_xx + f(y, z)$

Now we have:

$\frac{\partial U}{\partial y} = \frac{\partial f}{\partial y} = -2a_y$

Which leads to $f(y, z) = -2a_yy + g(z)$

And, finally we have:

$\frac{\partial U}{\partial z} = \frac{dg}{dz} = -3a_z$

Which gives $g(z) = -3a_zz + C$

Your first method is just a shortcut for this.

 

[Arman777]

You should never integrate a constant $a_x$ and get $\frac12 a_x^2$. That's a common mistake.

Instead, you have:

$\vec{F} = (a_x, 2a_y, 3a_z) = - \vec{\nabla}U(x, y, z) = -(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z})$

If we start with $-\frac{\partial U}{\partial x} = a_x$, then this gives:

$U(x, y, z) = -a_xx + f(y, z)$

Now we have:

$\frac{\partial U}{\partial y} = \frac{\partial f}{\partial y} = -2a_y$

Which leads to $f(y, z) = -2a_yy + g(z)$

And, finally we have:

$\frac{\partial U}{\partial z} = \frac{dg}{dz} = -3a_z$

Which gives $g(z) = -3a_zz + C$

Your first method is just a shortcut for this.

Yes sorry. Its a stupid mistake that I made...I see your point, thanks for clarifying it.