energy against extension graph shape? Watch

sarah99630
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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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Physics Enemy
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(Original post by sarah99630)
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W to extend wire by x equals the area under its F-x graph. As F = kx, it's the area of a triangle of height F and base x => 0.5Fx. So W = 0.5(kx)x = 0.5kx^2.

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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Tskadem
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Integrating

dF = (kx)dx

Giving a parabolic (x^2) curve for the variation of work done with displacement.
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Physics Enemy
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(Original post by Tskadem)
Integrating dF = (kx) dx, giving a parabolic (x^2) curve for the variation of work done with displacement.
* 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 = 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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sarah99630
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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 = 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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Physics Enemy
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(Original post by sarah99630)
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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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sarah99630
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(Original post by 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".
hahah im just clumsy, I asked a question and idk how I ended up quoting and even deleting some stuff lol, I needed to clarify the question,
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Eimmanuel
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(Original post by sarah99630)
how does work done (energy stored) vary with extension of a wire ?hows the graph shape and why?
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 ½F

Energy stored = average force × extension
Energy stored = ½Fx
Energy stored = ½(kx)x
Energy stored = ½kx2

Name:  energy_extension.JPG
Views: 53
Size:  51.5 KB
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sarah99630
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(Original post by 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 ½F

Energy stored = average force × extension
Energy stored = ½Fx
Energy stored = ½(kx)x
Energy stored = ½kx2

Name:  energy_extension.JPG
Views: 53
Size:  51.5 KB
Thanks a lot! I extremely appreciate your help.
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