How Does Spring Compression Vary with Different Methods of Loading?

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The discussion focuses on how spring compression varies with different loading methods. When a 5.0 kg block is dropped onto a vertical spring with a spring constant of 490 N/m, it compresses the spring by 0.2 meters due to the sudden force of impact. In contrast, when the block is slowly lowered until it can rest on the spring without disturbance, the net force on the block is zero, leading to a different calculation for displacement. The key distinction lies in the oscillation caused by the sudden drop versus the static equilibrium when lowered. Understanding Hooke's law is crucial, as it explains the relationship between the spring force and its displacement from equilibrium.
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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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