block placed on spring

[phazei]

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

[Doc Al]

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

[phazei]

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?

[Doc Al]

How does the spring force depend on its displacement from equilibrium? (What is Hooke's law?)

[phazei]

oh...

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


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

[Doc Al]

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!