Understanding elastic potential energy

Click For Summary

Discussion Overview

The discussion revolves around the derivation of the formula for elastic potential energy, specifically the presence of the factor of 1/2 in the equation U = 1/2 Fx. Participants are exploring the theoretical underpinnings of this formula in the context of an experiment involving a catapulted trolley.

Discussion Character

  • Exploratory
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant inquires about the origin of the 1/2 factor in the formula for elastic potential energy, linking it to their coursework.
  • Another participant suggests that the work done is the average force multiplied by the distance moved in the direction of the force.
  • A different participant explains that the 1/2 factor arises from the Taylor expansion of the potential energy function, emphasizing that the equilibrium position leads to certain terms being zero.
  • One participant provides a mathematical derivation involving the force equation F = -kx and the calculation of work done for a variable force, leading to the conclusion that U = 1/2 k x^2.

Areas of Agreement / Disagreement

Participants present various explanations and derivations regarding the 1/2 factor, but there is no consensus on a single explanation, indicating that multiple views remain on the topic.

Contextual Notes

The discussion includes assumptions about the Taylor expansion and the conditions under which the formula is valid, but these assumptions are not fully explored or resolved.

Heijbroek
Messages
3
Reaction score
0
where does the formula 1/2 Fx come from - i need to find out where the half comes from for some coursework of mine :S cheers,:cool:
 
Physics news on Phys.org
What have you tried so far?
 
well its for an experiment i did - How does the speed of a catapulted trolley vary with distance pulled back? - as part of the task it says - word done or energy stored is 1/2Fx. THink why the 1/2 is in there - average? - yh tried a few sites didnt get v far...
 
Work done = average force times distance moved in direction of force.
 
It comes from the Taylor expansion of the potential energy function. We don't know what U(x) is, so we take the Taylor expansion of it around 0. You can add a constant to any potential energy function and it won't change the physics, so we can let the first term equal zero. If we define the equilibium position to be zero, then U'(0) must be 0 since it is in equilibrium. If it weren't zero, then it wouldn't be the equilibrium position. That leaves us with the third term, 1/2 kx^2, where k = U''(0). Terms beyond the third are ignored, since we assume that three terms of the Taylor expansion will be a good approximation for small displacements (which is was 1/2 kx^2 is valid for). That's where the 1/2 comes from.
 
F = -kx

Work = F*x
Work for variable force = integral (F dx)
W = integral (-kx dx) = -k integral(x dx) = -1/2k x^2

W = -1/2k x^2
U = -W = 1/2 k x^2
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 54 ·
2
Replies
54
Views
7K
Replies
6
Views
2K
  • · Replies 20 ·
Replies
20
Views
8K
  • · Replies 3 ·
Replies
3
Views
927
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K