Understanding Spring Compression: How Doubling It Affects Stored Energy

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Homework Help Overview

The discussion revolves around the relationship between the compression of a spring and the energy stored within it, specifically examining how doubling the compression affects the stored energy. The subject area is primarily focused on spring mechanics and potential energy calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the equation for a spring's potential energy and discuss the implications of doubling the compression. Some participants question the initial assumption regarding the factor by which energy increases, while others clarify the correct interpretation of the variables involved.

Discussion Status

The discussion is active, with participants providing insights into the correct formula for potential energy and questioning the assumptions made about the spring constant and compression. There is a mix of interpretations being explored, particularly regarding the relationship between compression and energy increase.

Contextual Notes

Participants are navigating the implications of changing the compression variable while considering the spring constant. There is an emphasis on ensuring the correct application of the potential energy formula in the context of the problem.

NewJersey
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If compression of a spring is doubled, the stored energy increased by a factor of?

I think that the stored energy is increased by a factor of 2.

Right?
 
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What is the equation for a spring's potential energy?
 
Fs1=1/2 kx^2 so if k = 25 v=4
= 200

Fs2= k=50 v=4
=400

so by factor of 2
 
Ah, you are doubling the wrong thing. k is the spring constant. The question is asking about changing x by a factor of 2.

[tex]F_{s1} = .5kx^2[/tex]

[tex]F_{s2} = .5k(2x)^2[/tex]

[tex]\frac{F_{s2}}{F_{s1}} = ?[/tex]
 

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