Mechanics question involving spring and hookes law.

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SUMMARY

The discussion revolves around the application of Hooke's Law and integration techniques to solve a mechanics problem involving a spring. The user, identified as "ly667," is attempting to derive the potential energy function V using the relationship between work done and kinetic energy change, specifically integrating the force function F. The integration by parts method is employed, with the user expressing uncertainty about the correctness of their calculations and the value of the lambda (λ) parameter, which is initially set to 4 but later suggested to remain as a variable.

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ly667
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My lamda value is 4.

What I have managed to get so far is:

dV/dx=-F

V=8/arctan(x/3)dx /=integrate sign!

integrate by parts

u=arctan(x/3), dv=dx
du=3/9+x^2dx, v=x

V=8/arctan(x/3)dx = 8[x.arctan(x/3)-/3x/9+x^2dx]

V= 8[x.arctan(x/3)-3/2log(9+x^2)+C]

Since V(0)=0

V= 8[x.arctan(x/3)-3/2log(9+x^2)+3/2log(9)]
V= 8[x.arctan(x/3)+3/2log(9/9+x^2)]


Not sure if this is correct, or where to go from here? Please help!
 

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Hi ly667. It looks like you are using the relation
Work Done = ∫F.dx = ∆K.E.[/size]​

and you will integrate between the limits of x=4 and x=0

I recommend that you get this written down at the start of the problem before you leap into the mechanics of problem solving, so that from the outset you can be confident of knowing precisely where you are headed. If you get into the habit of first laying down a good foundation, most of the time the solution will rise up almost Phoenix-like out of that foundation. :smile:

I'm not sure that your λ is 4, so let's keep it at λ.
V= 8[x.arctan(x/3)-3/2log(9+x^2)+C]
I don't quite get that. Can you do it again, this time keeping λ in the expression?
 

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