How to Graph Potential Energy from Force Fx Along the x-axis?

[voko]

sooo
U = -2x^2 +4x +k ]^x - ( (-2x^2 +4x +k)]^0.5)

right?

Compare this with #19.

 

[bestchemist]

Compare this with #19.

idk now... I'm really confused now :(

 

[HallsofIvy]

For this problem, where the force function is a "broken line", you don't really need "Calculus". The integral is the same as "area under the curve" so, here, all you need is the formula for area of a triangle.

For x between 0 and 0.5, F is given by F= 4x. The "area under the curve" from 0 to x is the area of a right triangle with legs of length "x" and "4x" so area (1/2)(x)(4x)= 2x^2.

Beyond x= 0.5, for x between 0.5 and x= 1, F= 4- 4x. The "area under the curve" from 1/2 to x is the area of a right triangle with legs of length "x- 0.5" and "4- 4x". The "area under the curve" from 1/2 to x is (1/2)(x- 0.5)(4- 4x)= (x- 1/2)(2- 2x)= -2x^2+ 3x- 1. The "area under the curve" from 0 to x, for x> 1/2, is the area from 0 to 1/2 which is 1/2, plus the area of the second triangle.

 

[bestchemist]

For this problem, where the force function is a "broken line", you don't really need "Calculus". The integral is the same as "area under the curve" so, here, all you need is the formula for area of a triangle.

For x between 0 and 0.5, F is given by F= 4x. The "area under the curve" from 0 to x is the area of a right triangle with legs of length "x" and "4x" so area (1/2)(x)(4x)= 2x^2.

Beyond x= 0.5, for x between 0.5 and x= 1, F= 4- 4x. The "area under the curve" from 1/2 to x is the area of a right triangle with legs of length "x- 0.5" and "4- 4x". The "area under the curve" from 1/2 to x is (1/2)(x- 0.5)(4- 4x)= (x- 1/2)(2- 2x)= -2x^2+ 3x- 1. The "area under the curve" from 0 to x, for x> 1/2, is the area from 0 to 1/2 which is 1/2, plus the area of the second triangle.

so from x = 0 to 0.5 I'll use 2x^2 to graph
anf for x = 0.5 to 1 I'll have to use -3x +1? since 2x^2 - (-2x^2 +3x -1) = -3x+1

 

[voko]

idk now... I'm really confused now :(

In #19: $U(x) = - \int_0^{0.5} a(x) dx - \int_{0.5}^x b(x) dx , \ 0.5 \le x < 1$.

In #30, you found what $\int_{0.5}^x b(x) dx$ is. You still need to compute $\int_0^{0.5} a(x) dx$ (which is the area of a triangle, as HallsofIvy remarked), and you need to mind the signs when plug all this into the formula for $U(x)$.

 

[bestchemist]

In #19: $U(x) = - \int_0^{0.5} a(x) dx - \int_{0.5}^x b(x) dx , \ 0.5 \le x < 1$.

In #30, you found what $\int_{0.5}^x b(x) dx$ is. You still need to compute $\int_0^{0.5} a(x) dx$ (which is the area of a triangle, as HallsofIvy remarked), and you need to mind the signs when plug all this into the formula for $U(x)$.

okay so..
from taking the integral of a and b I got
2x^2 -2x^2+4x+k +1/2-2-k

then U = 4x+5/2

I think I did it wrong again...:(

 

[voko]

okay so..
from taking the integral of a and b I got
2x^2 -2x^2+4x+k +1/2-2-k

then U = 4x+5/2

I think I did it wrong again...:(

You computed indefinite integrals correctly. But the formula needs definite integrals, and it is important to use correct limits, which you mostly ignore, except in #30, where you did use them - correctly.

And you need to mind the signs. Potential energy is defined with a minus sign, which, unfortunately for you, makes things more confusing than they already are. But you are on the right track, so don't give up just yet.

 

[bestchemist]

You computed indefinite integrals correctly. But the formula needs definite integrals, and it is important to use correct limits, which you mostly ignore, except in #30, where you did use them - correctly.

And you need to mind the signs. Potential energy is defined with a minus sign, which, unfortunately for you, makes things more confusing than they already are. But you are on the right track, so don't give up just yet.

so...

-2x^2 - ((-2x^2 +4x+k)]^x - (-2x^2 +4x-k)}^0.5)

does this look right??

 

[voko]

so...

-2x^2 - ((-2x^2 +4x+k)]^x - (-2x^2 +4x-k)}^0.5)

does this look right??

Almost. The first term should have been $-2x^2 |_{0}^{0.5}$ - see the difference?

 

[bestchemist]

Almost. The first term should have been $-2x^2 |_{0}^{0.5}$ - see the difference?

ohhh... yep I see it now... I didn't look at the limit like you said...
does the first term is -1/2 then since -2x^2 ]^0.5 = -1/2

how about from x -1 to 1 then... how will the function become constant since there are so many variable here

 

[voko]

ohhh... yep I see it now... I didn't look at the limit like you said...
does the first term is -1/2 then since -2x^2 ]^0.5 = -1/2

Correct. So what is the complete formula for $0.5 \le x < 1$?

how about from x -1 to 1 then... how will the function become constant since there are so many variable here

You mean "x > 1"?

You got the general formula in #20. Use it, paying attention to the limits and signs.

 

[bestchemist]

Correct. So what is the complete formula for $0.5 \le x < 1$?



You mean "x > 1"?

You got the general formula in #20. Use it, paying attention to the limits and signs.

okay... so If I did it right

U = 2x^2 -4x-2

and for x >1
I did the calculation and I'm so sure about it lol

U = -1

right?

 

[voko]

okay... so If I did it right

U = 2x^2 -4x-2

Not quite. The "-2" in the end is wrong. I think you got lost in the signs, check them.

An easy way to check whether you formula is correct: the potential energy is smooth, so the value at 0.5 must be the same calculated from either x < 0.5 or x > 0.5. Approaching from x < 0.5, the value is $-2x^2 = -0.5$; approaching it from x > 0.5, your formula gives $2x^2 - 4x - 2 = -3.5$, so it must be wrong.

and for x >1
I did the calculation and I'm so sure about it lol

U = -1

right?

Yes!

Note that this is the (negative) area of the entire force triangle, just like it should be.

 

[bestchemist]

Not quite. The "-2" in the end is wrong. I think you got lost in the signs, check them.

An easy way to check whether you formula is correct: the potential energy is smooth, so the value at 0.5 must be the same calculated from either x < 0.5 or x > 0.5. Approaching from x < 0.5, the value is $-2x^2 = -0.5$; approaching it from x > 0.5, your formula gives $2x^2 - 4x - 2 = -3.5$, so it must be wrong.



Yes!

Note that this is the (negative) area of the entire force triangle, just like it should be.

okie, so
U = -2x^2 -4x +2
right?
If x = 0.5, then U = -2(0.5)^2 -4(0.5)+2
U=-0.5

 

[voko]

okie, so
U = -2x^2 -4x +2
right?

Sorry, no. The formula is $- \int_0^{0.5} 4x dx - \int_{0.5}^x (4 - 4x) dx$, so this is $- \text{some number} + 2x^2 - 4x + \text{some other number}$. Note that it is $2x^2$, not $-2x^2$.

 

[bestchemist]

Sorry, no. The formula is $- \int_0^{0.5} 4x dx - \int_{0.5}^x (4 - 4x) dx$, so this is $- \text{some number} + 2x^2 - 4x + \text{some other number}$. Note that it is $2x^2$, not $-2x^2$.

I think I got it... lol

U = 2x^2 -6x +2

x = 0.5

U = 2(0.5)^2 -6(0.5)+2
U =-0.5

Right?

 

[voko]

Per #45, there must be -4x, not +6x.

 

[bestchemist]

Per #45, there must be -4x, not +6x.

I think I got it for real this time... lol sooo much confident lol

U = 2x^2 -4x +1

Right?

 

[voko]

Right!

 

[bestchemist]

Right!

yayyyyyyyyyyyy...
After a few days on this one problem... Finally got it lol
Thank you so much!

 

[PhanthomJay]

Sooo
∫F= 0
so U = 0x +k
so U =1?

looks good. Remember the check I mentioned, U at x = 1.1 is the Area under the force-displacement curve between x= 0 and x = 1.1. But you still need to graph it as the problem asks.