Why is the Kinetic Energy of a Spring (1/6)mv²?

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Homework Help Overview

The discussion revolves around understanding the kinetic energy of a spring, specifically why it is expressed as (1/6)mv² instead of the standard kinetic energy formula (1/2)mv². Participants are exploring the physical principles governing the behavior of springs in motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the application of the standard kinetic energy formula to a spring system, considering the fixed and free ends of the spring and the implications of pulling the free end. There is discussion about the velocity distribution along the spring and the nature of the motion involved.

Discussion Status

Some participants have offered insights into the physical situation of the spring and the dynamics at play, while others are still seeking clarification on how to reconcile the differing kinetic energy expressions. Multiple interpretations of the problem are being explored, particularly regarding the assumptions made about the spring's behavior.

Contextual Notes

There is an ongoing discussion about the idealization of the spring and the effects of internal oscillations due to sudden movements, as well as the need to consider a steady state in the analysis.

Differentiate it
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Homework Statement
Could someone explain the situation here?
Relevant Equations
(1/2)mv^2
Screenshot_2022-12-03-18-00-35-87_1ab726e7599468b75300a3cdb0d53113.jpg

I tried just using the formula for kinetic energy but that was apparently the wrong answer. The answer key says it's (1/6)mv². I don't understand how they got that answer. Could someone explain?
 
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Differentiate it said:
Yes, I tried
I think @BvU's point is that we need to see it!
 
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Differentiate it said:
Homework Statement:: Could someone explain the situation here?
Relevant Equations:: (1/2)mv^2

I tried just using the formula for kinetic energy
So, that would be correct for a particle, why do you think it doesn't work for a spring?
 
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You have to think about what is going on physically with this spring. One end is fixed and the other is free. The free end is suddenly pulled at a velocity ##v##. You are looking at it at time zero, the free end has a velocity but hasn't actually stretched yet, it is still length ##l##. What's going on with the rest of the spring at that instant?
 
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bob012345 said:
You have to think about what is going on physically with this spring. One end is fixed and the other is free. The free end is suddenly pulled at a velocity ##v##. You are looking at it at time zero, the free end has a velocity but hasn't actually stretched yet, it is still length ##l##. What's going on with the rest of the spring at that instant?
To add to what @bob012345 is hinting at, specifically think about the fixed end. What velocity is it moving at?
 
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bob012345 said:
The free end is suddenly pulled at a velocity v.
Not suddenly, I think. That would get you into complexities of internal oscillations.
 
  • #10
haruspex said:
Not suddenly, I think. That would get you into complexities of internal oscillations.
I looked at this problem as idealized. The free end of an unstretched spring is found to be moving at velocity ##v##. Suddenly was an unfortunate choice not because of what the actual dynamics of a real spring would be since this is an ideal spring, but because there is no need to involve time at all. This is a snapshot.
 
  • #11
bob012345 said:
since this is an ideal spring
An ideal spring is massless; this one isn't, which is why a sudden tug will likely induce internal oscillations.
 
  • #12
haruspex said:
An ideal spring is massless; this one isn't, which is why a sudden tug will likely induce internal oscillations.
Sure, but for this problem there is only an agent pulling the free end with uniform velocity. It must be assumed that there is also a velocity distribution set up along the spring and we don't need to worry about the details of how that came about.
 
  • #13
bob012345 said:
Sure, but for this problem there is only an agent pulling the free end with uniform velocity. It must be assumed that there is also a velocity distribution set up along the spring and we don't need to worry about the details of how that came about.
Quite so, we have to assume some sort of steady state, whatever that means in this context. Hence, no sudden movements.
 
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  • #14
Differentiate it said:
Homework Statement:: Could someone explain the situation here?
Relevant Equations:: (1/2)mv^2I tried just using the formula for kinetic energy but that was apparently the wrong answer. The answer key says it's (1/6)mv². I don't understand how they got that answer. Could someone explain?
@Differentiate it, have you used the hints we gave you to make progress with this problem?
 

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