Solving for Spring Compression with a Vertical Spring and Hanging Block

[Anthonyphy2013]

Homework Statement

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.

The first question asks "How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?
The answer is easy...0.109m

the part I can't figure out is the second part of the question "How far does the spring compress if you let go of the block suddenly?


What is the pan's distance from the ceiling when the spring reaches its maximum length?

Homework Equations

1/2 kx^2 = 1/2mv^2 , mgh=Fsp=-kdelta(x)

The Attempt at a Solution

1st question : mgh=Fsp=-kdelta(x)
I got wrong for that question and I use the conservation of energy equation and I got it right . My question is how I know to use the energy eqaution on the first question and use net force to use the second question

 

[voko]

1st question : mgh=Fsp=-kdelta(x)

This equation does not make any sense dimensionally: you have units of energy one the left, units of force on the right, I have no idea units of what in the middle. What equation did you really use to solve #1?

 

[haruspex]

My question is how I know to use the energy eqaution on the first question and use net force to use the second question

Do you mean that the other way around?
The first question is a statics question. It concerns a system which is at rest, so you can use the statics equations of force. Yo cannot use energy because clearly energy has been lost.
In the second question, you can assume energy is conserved. But now things are dynamic (the system is not in equilibrium), so you cannot use static equilibrium equations.

 

[Anthonyphy2013]

#1 : I use static equilibrium , I use mg=-kdeltax and my answer is 0.1 m and the book said that as 0.2 and then I use energy conservation to solve #1 , which is 1/2 kx^2 = mgh and I got 0.2 which is matched with the solution on book.
Plus: mgh=Fsp=-kdelta(x) and the middle one is the restoring for of the ideal spring.

 

[haruspex]

#1 : I use static equilibrium , I use mg=-kdeltax and my answer is 0.1 m and the book said that as 0.2

then you are right and the book is wrong

and then I use energy conservation to solve #1 , which is 1/2 kx^2 = mgh and I got 0.2 which is matched with the solution on book.

You mean to solve #2, right?

Plus: mgh=Fsp=-kdelta(x) and the middle one is the restoring for of the ideal spring.

As voko pointed out, that equation makes no sense. mgh is energy, kΔx is a force. Fsp, I'm guessing, means the restoring force from the spring. If so, Fsp = -kΔx is correct, but (for #1) it should be mg = Fsp, not mgh.

 

[Anthonyphy2013]

that means # 1 is to consider the net force .My question for # 2 is to mean the dynamic equilibrium , how could that possible ?

 

[haruspex]

that means # 1 is to consider the net force .My question for # 2 is to mean the dynamic equilibrium , how could that possible ?

Sorry, I don't understand your question.

 

[gneill]

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.

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What is the pan's distance from the ceiling when the spring reaches its maximum length?

Isn't there a contradiction here? Either that or we need to know the floor to ceiling distance for the room (and the problem becomes trivial!).