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how does work done (energy stored) vary with extension of a wire ?hows the graph shape and why?

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#2

(Original post by

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**sarah99630**)...

W ∝ x^2, the sketch (from origin) is the +ve region (right half) of parabola y = ax^2 (a > 0). Ensure the curve's grad inc more gently than e.g) a.e^x.

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#3

Integrating

dF = (kx)dx

Giving a parabolic (x^2) curve for the variation of work done with displacement.

dF = (kx)dx

Giving a parabolic (x^2) curve for the variation of work done with displacement.

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#4

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Integrating dF = (kx) dx, giving a parabolic (x^2) curve for the variation of work done with displacement.

**Tskadem**)Integrating dF = (kx) dx, giving a parabolic (x^2) curve for the variation of work done with displacement.

To elaborate, we can assume a constant F acts to extend a wire by only dx, doing dW in the process:

dW =

**F.**

**dx**= F dx = kx dx (as

**F**and

**dx**are collinear)

So Int 1 dW = Int kx dx => W(x) = (k/2)x^2 + c

W(0) = 0 => c = 0, so W(x) = (k/2)x^2

OP doesn't need any of this tho.

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[QUOTE=Physics Enemy;73626654]* dW = kx dx, note x is extension here i.e) ≥ 0.

To elaborate, we can assume a constant F acts to extend a wire by only dx, doing dW in the process:

dW =

So Int 1 dW = Int kx dx => W(x) = (k/2)x^2 + c

W(0) = 0 => c = 0, so W(x) = (k/2)x^2

thanks! Is it like half of y=x^2 curve? Like only the positive part?

To elaborate, we can assume a constant F acts to extend a wire by only dx, doing dW in the process:

dW =

**F.****dx**= F dx = kx dx (as**F**and**dx**are collinear)So Int 1 dW = Int kx dx => W(x) = (k/2)x^2 + c

W(0) = 0 => c = 0, so W(x) = (k/2)x^2

thanks! Is it like half of y=x^2 curve? Like only the positive part?

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#6

(Original post by

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**sarah99630**)...

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(Original post by

Unsure why you quoted my 2nd post when my 1st post is addressed to you, which answers your Qs. Re: 2nd post, I said "OP doesn't need any of this".

**Physics Enemy**)Unsure why you quoted my 2nd post when my 1st post is addressed to you, which answers your Qs. Re: 2nd post, I said "OP doesn't need any of this".

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#8

(Original post by

how does work done (energy stored) vary with extension of a wire ?hows the graph shape and why?

**sarah99630**)how does work done (energy stored) vary with extension of a wire ?hows the graph shape and why?

The standard and “approved” way is

Energy stored = work done in stretching the wire

Since the restoring force in the spring

*F*is directly proportional to its extension

*x*(

*F*=

*kx*),

work done in stretching the wire = average force × extension

where the average force is ½

*F*

Energy stored = average force × extension

Energy stored = ½

*Fx*

Energy stored = ½(

*kx*)

*x*

Energy stored = ½

*kx*

^{2}

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(Original post by

I had seen some of postings but I would like to give some piece of advice in arriving the answer in exam. There are some examination boards which are very particular about how you start the “derivation”.

The standard and “approved” way is

Energy stored = work done in stretching the wire

Since the restoring force in the spring

work done in stretching the wire = average force × extension

where the average force is ½

Energy stored = average force × extension

Energy stored = ½

Energy stored = ½(

Energy stored = ½

**Eimmanuel**)I had seen some of postings but I would like to give some piece of advice in arriving the answer in exam. There are some examination boards which are very particular about how you start the “derivation”.

The standard and “approved” way is

Energy stored = work done in stretching the wire

Since the restoring force in the spring

*F*is directly proportional to its extension*x*(*F*=*kx*),work done in stretching the wire = average force × extension

where the average force is ½

*F*Energy stored = average force × extension

Energy stored = ½

*Fx*Energy stored = ½(

*kx*)*x*Energy stored = ½

*kx*^{2}
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