Maximum displacement in mass spring system

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In a mass-spring system, a block suspended by an ideal spring experiences maximum displacement X when a constant force F is applied. The equilibrium condition is described by the equation F + mg = (X + a)K, where a is the initial extension due to the block's weight. The initial extension a can be calculated as a = mg/K. The discussion reveals a discrepancy between the calculated maximum displacement X = F/K and the book's assertion that X = 2F/K, prompting a deeper examination of the forces at play. Ultimately, the block can extend further due to inertia even after reaching equilibrium, indicating that the spring force equals the weight plus the external force at maximum displacement.
Fitz Watson
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Homework Statement


A block is suspended by an ideal spring of the force constant K. If the block is pulled down by applying a constant force F and if maximum displacement of the block from its initial position of rest is X then, find the value of X.

Homework Equations


mg + F = XK + K(mg/K)

The Attempt at a Solution



Let mass of block = m
Before applying F, the block is at rest. Let extension of spring here be a. So,
aK = mg
Hence, a = (mg/K) ... {1}

Now, after the block comes in equilibrium under F,

F + mg = (X+a) K ..... {2}

Solving {1} & {2} gives:
X = F/KBut, my book says that X = 2F/K
 
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Fitz Watson said:
after the block comes in equilibrium under F,
It does not say that it comes to equilibrium. The force is maintained as a constant.
 
haruspex said:
It does not say that it comes to equilibrium. The force is maintained as a constant.
But when it has reached maximum extension, doesn't it mean that now the spring force is equal to weight + external force applied. And that's why the block can't go further down
 
Fitz Watson said:
But when it has reached maximum extension, doesn't it mean that now the spring force is equal to weight + external force applied.
Not necessarily. Think about KE.
 
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haruspex said:
Not necessarily. Think about KE.
Now that I rethink, will it be like after forces becoming equal, block will go further down due to inertia?
 
Fitz Watson said:
Now that I rethink, will it be like after forces becoming equal, block will go further down due to inertia?
Yes.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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