Understanding elastic potential energy

Click For Summary
SUMMARY

The formula for elastic potential energy, represented as U = 1/2 k x^2, derives from the Taylor expansion of the potential energy function around the equilibrium position. The factor of 1/2 arises because work done is calculated as the average force times the distance moved in the direction of the force. In this context, the force F is defined as F = -kx, leading to the integral of the force over displacement resulting in the expression for work W = -1/2 k x^2. Thus, the elastic potential energy is equal to the negative work done, confirming U = 1/2 k x^2.

PREREQUISITES
  • Understanding of Taylor series expansion
  • Familiarity with Hooke's Law (F = -kx)
  • Basic knowledge of work-energy principles
  • Concept of equilibrium in physics
NEXT STEPS
  • Study the derivation of the Taylor series and its applications in physics
  • Explore Hooke's Law in detail and its implications in elastic systems
  • Investigate the relationship between work and energy in variable force scenarios
  • Learn about potential energy functions and their significance in mechanics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy concepts, as well as educators looking for clear explanations of elastic potential energy derivation.

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 20 ·
Replies
20
Views
8K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K