Elastic potential energy and natural length of a spring- relation?

Click For Summary
The spring constant is inversely proportional to the natural length of a spring, indicating that longer springs have lower spring constants. There is a notable change in extension as springs approach their breaking point. At the point of failure, both strain and force remain consistent across different springs. Understanding these relationships is crucial for analyzing spring behavior under stress. This highlights the importance of strain and force in the study of elastic potential energy.
Stormzy67
Messages
2
Reaction score
0
Homework Statement
The question states, a spring of length 0.5m is can be extended by 0.05m before fracturing, storing elastic energy ‘U’. Now, another spring of the same material but of half the length(0.25) is also stretched to its maximum extension.
What will be the new elastic energy in terms of ‘U’?
Relevant Equations
F=kx , E=1/2 kx^2, Young’s modulus?
I figured out that the spring constant is inversely proportional to the natural length, but there’s still an unknown change in a quantity( most likely extension).
 
Physics news on Phys.org
Is there something that’s the same for both springs at the point where they break?
 
The strain?
 
Stormzy67 said:
The strain?
Yes. Also, the force.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
3K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
29
Views
3K