energy against extension graph shape?Watch

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Thread starter 2 years ago
#1
how does work done (energy stored) vary with extension of a wire ?hows the graph shape and why?
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2 years ago
#2
(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. 0
2 years ago
#3
Integrating

dF = (kx)dx

Giving a parabolic (x^2) curve for the variation of work done with displacement.
1
2 years ago
#4
(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. 0
Thread starter 2 years ago
#5
[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?
0
2 years ago
#6
(Original post by sarah99630)
...
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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Thread starter 2 years ago
#7
(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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2 years ago
#8
(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 1
Thread starter 2 years ago
#9
(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 Thanks a lot! I extremely appreciate your help.
0
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