Question

help please. Rewriting a Potential Energy Function.

Answer 1

A proton with mass "m" moves in one dimension. The potential-energy function is\[U(x)=\frac{ \alpha }{ x^2 }-\frac{ \beta }{ x }\] where alpha and beta are positive constants. The proton is released from rest at \[x _{0}^2=\frac{ \alpha }{ \beta }\]
show that U(x) can be written as \[U(x)=\frac{ \alpha }{ x _{0}^2 }\left[ \left( \frac{ x _{0} }{ x } \right)^2-\frac{ x _{0} }{ x } \right]\]

Answer 2

if at the very least, please help me get on the right start. thank you to any who comment.

Answer 3

I must be making an error. I can't get that expression. The last term I only get a 1/x, not xo/x..

Answer 4

I just noticed that you can show what you need to if the initial condition is different.

Are you sure the condition isn't Xo = alpha/beta?

Answer 5

crap. yeah. there's not supposed to be second power there. i'm sorry. i was typing quickly before my circuits class.

Answer 6

it is \[x _{0}=\frac{ \alpha }{ \beta }\]

Answer 7

Okay, so it is as follows:

\[x_{o} = \frac{\alpha}{\beta} ;\beta=\frac{\alpha}{x_{o}}\]
\[U(x) = \frac{ \alpha }{ x^{2} }-\frac{\beta}{x}=\frac{\alpha}{x^{2}}-\frac{\alpha}{x_{o}x}=\frac{\alpha}{x_{o}^{2}}\left[ \frac{x_{o}^{2}}{x^{2}}-\frac{x_{o}}{x} \right]=\frac{\alpha}{x_{o}^{2}}\left[ \left( \frac{x_{o}}{x} \right)^{2}-\frac{x_{o}}{x} \right]\]

Answer 8

oh wow that's how i started this off in the first place. i saw that the final product didn't have a beta in it so i tried solving for beta in the initial condition but i got intimidated by the amount of arithmetic going on so i stopped and came here for help. anyway. thank you!

Answer 9

Not a problem.

Answer 10

ok now mastering physics is asking me to solve for U(x0) and to express it in terms of \[\alpha, x_{0}, m, x\]
i'm not so sure how to start this...

Answer 11

All you should have to do is replace any 'x' in the U(x) equation with 'x0'.

Things should cancel quite a bit. You're essentially solving for the initial energy, since it's the pot. energy at rest.

Answer 12

Hmm, I didn't peek at it when I gave my suggestion, but yes, the brackets go to 0. That's odd. I'll have to think about it for a few moments and see if anything comes to mind.

Answer 13

so.... uhhh... i just put in 0 into the answer.... and that was it... yeah.

Answer 14

lol well, mastering physics is still as stupid as i remember it to be

Answer 15

haha yeah. it's annoying. ok one more question and i'll shall leave you alone. it's asking for me to calculate v(x). could you just direct me on to the right path? i can take it from there. if so that would be the greatest. also thanks for the help. it's greatly appreciated.

Answer 16

Umm, I'm going to guess v(x) means velocity.

We know that v(x0) = 0, because it said that is when it was at rest.

If this is true, then at x=x0, U(x) would be the total energy. With no motion, there is no kinetic energy.

U(x0) should equal the total energy E at that initial condition.

E = U(x0) = 0

Now E= K + U ; K = 1/2 * m * v^2

0 = K + U(x)
K = - U(x)
1/2 * m * v^2 = -U(x)
v(x) = sqrt[-2U(x)/m]

\[v(x) = \sqrt{\frac{ -2 }{ m }\left( \frac{\alpha}{x_{0}^{2}} \right)\left[ \frac{ x_{0}^{2} }{ x^{2} }-\frac{x_{0}}{x} \right]}\]

If I haven't made any stupid mistakes, I think it makes sense to me.

Answer 17

thanks. i didn't make that connection there but after seeing the steps you did it's obvious and looks quite simple. now there's the next question, "For what value of x is the speed of the proton a maximum?", any insight on how to approach this?

Answer 18

i wonder if i should just rewrite v(x) and solve for x.

Answer 19

Calculus! To find the maximum of a function, take the derivative and set it equal to 0. Then solve for x and that will be the position with the max speed.

Answer 20

ok. i tried it and i got a crazy looking. result... oh yeah. x(0) is a constant. spoke to the teacher and got the same answer as take the derivative. but i'm not to sure of this. haha

Answer 21

so far I have \[\frac{ 1 }{ 2 }\left( \frac{ -2 \alpha }{ mx^2 }+\frac{ 2 \alpha }{ mx_{0}x } \right)^{-\frac{ 1 }{ 2 }}\left( \frac{ 4 \alpha }{ mx^3 }-\frac{ 2 \alpha }{ mx_{0}x^2 } \right)\] did i do this wrong or.... what?

Answer 22

It looks like you are right so far

Answer 23

yes, i have the same thing you have

Answer 24

Given:
\[\frac{dv(x)}{dx}=\frac{1}{2}\left( \frac{-2\alpha}{mx^{2}}+\frac{2\alpha}{mx_{0}x} \right) \left( \frac{4\alpha}{mx^{3}}-\frac{2\alpha}{mx_{0}x^{2}}\right)=0\]
Divide out the 1/2:
\[\frac{dv(x)}{dx}=\left( \frac{-2\alpha}{mx^{2}}+\frac{2\alpha}{mx_{0}x} \right) \left( \frac{4\alpha}{mx^{3}}-\frac{2\alpha}{mx_{0}x^{2}}\right)=0\]
First term solution:
\[\left( \frac{-2\alpha}{mx^{2}}+\frac{2\alpha}{mx_{0}x} \right) =0\]\[\left( \frac{2\alpha}{mx}\right)\left( \frac{-1}{x}+\frac{1}{x_{0}}\right) =0\]The first term is x = 0, which is trivial.\[\left( \frac{-1}{x}+\frac{1}{x_{0}}\right) =0\]This gives a solution of x = x0.

Second term solution:
\[\left( \frac{4\alpha}{mx^{3}}-\frac{2\alpha}{mx_{0}x^{2}}\right)=0\]\[\left( \frac{2\alpha}{mx^{2}}\right)\left( \frac{2}{x}-\frac{1}{x_{0}}\right)=0\]The first terms is x = 0, which is again trivial.\[\left( \frac{2}{x}-\frac{1}{x_{0}}\right)=0\]This gives a solution of x = 2(x0).

Solution:
We found two trivial cases of x = 0. We found one case of x = x0. We found one case of x = 2(x0).

x = 0 doesn't matter. x = x0 is the place the proton starts, so it doesn't matter.

That means that the max speed occurs at x = 2(x0).

Answer 25

wow. that's it right there. thank you again.