Mechanics question involving spring and hookes law.

  • Thread starter ly667
  • Start date
  • #1
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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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Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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Hi ly667. It looks like you are using the relation
Work Done = ∫F.dx = ∆K.E.

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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