Potential energy and Conservative Force

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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 ?
 
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Answers and Replies

  • #2
haruspex
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Are ax etc. constants? If so, what are their integrals wrt x, y, z?
You can always check your answer using the differential equation you quoted.
 
  • #3
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Are ax 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 ?
 
  • #4
PeroK
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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.
 
  • #5
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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.
 

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