How Does Spring Compression Vary with Different Methods of Loading?

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Homework Help Overview

The discussion revolves around the behavior of a vertical spring when subjected to different loading methods, specifically focusing on the compression of the spring when a block is either dropped or slowly lowered onto it. The subject area includes concepts from mechanics and spring dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to determine spring compression in two scenarios. They express uncertainty about the second scenario and seek clarification on how the loading method affects the outcome.

Discussion Status

Participants are actively engaging with the problem, with some providing hints and guidance. There is a recognition of the difference in behavior of the spring based on the method of loading, and the discussion is exploring the implications of net forces and Hooke's law.

Contextual Notes

Participants are considering the implications of static versus dynamic loading on the spring's behavior, and there is an acknowledgment of the need to understand the forces involved in each scenario. The original poster has expressed confusion regarding the second part of the problem, indicating a gap in understanding that is being addressed through discussion.

phazei
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A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0 kg block just above the spring, not quite touching it.

A)How far does the spring compress if you let go of the block suddenly?

B)How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?


A) This part was simple enough,
U(potential) = U(spring)
where the h in potential is the height from where the spring would compress to...
mg(deltaS) = 1/2 k*(deltaS)^2
plug #'s and solve for deltaS. 0.2m

B)I don't know where to start this any differently. 0.2m isn't the correct answer. I just don't know how this differs...
help?

Thanks
 
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Hint: If you can move your hand without disturbing it, what must be the net force on the block?
 
the net force must be 0.

So the force upward by the spring = mg

But how to i relate that to figure out what the displacement is?
 
How does the spring force depend on its displacement from equilibrium? (What is Hooke's law?)
 
oh...

mg=-kdeltaS
5(-9.81)=-490deltaS
:biggrin: :biggrin:


so they're different because it occilates when dropped and not when lowered, right?
 
phazei said:
oh...

mg=-kdeltaS
5(-9.81)=-490deltaS
Right! (Here's a tip: Don't get hung up on the signs. The minus sign in Hooke's law just means that the spring always exerts a force opposite to its displacement from equilbrium. And g is always a positive number, by the way.)


so they're different because it occilates when dropped and not when lowered, right?
Exactly right!
 

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