Force applied to both sides of a spring. What is the net compression?

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A spring compressed from both ends by 100 Newtons with a spring constant of 1000 N/m results in a net compression of 10 cm, not 20 cm, due to the opposing forces acting on it. The confusion arises from the application of Newton's third law, where the force from one side acts as a counterforce, preventing additional compression. When discussing different scenarios, such as a spring compressed by 5 cm with a force of 50 N, the total force applied must be considered to determine the spring constant. The key takeaway is that the net compression is determined by the effective force acting on the spring, which is influenced by external constraints. Understanding these principles clarifies the behavior of springs under compression.
Hereformore
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Homework Statement


A spring is compressed from both ends by 100 Newtons and the spring constant is given as 1000N/m. What is the net compression distance/displacement (x)?

Homework Equations



F=-k*change(x)

The Attempt at a Solution


Initially, I thought that the net displacement would be 20cm since i thought 100 Newtons applied at each end would lead to 10cm compression on each end. 100 = -(1000)*x so x = 10cm. 10cm compression on both sides leads to net compression of 20cm.

But I was wrong, and it's because of Newton's third law? So the force applied by one arm effectively serves as a wall so that compression can take place.

For some reason this seems counterintuitive for me. But i understand that if there were no other hand or wall there (so force applied to only one end), then the spring would simply accelerate to the right. So there needs to be a counter there (whether a wall or hand). If there was a wall, then it would return equal and opposite force to that applied by the spring, hence 100N. So the same principle would apply if there was a hand applying the same force instead.

What confuses me i guess is that we're talking a spring here so its not like we're pushing a block against a wall. I didnt expect the force that the spring would apply to a wall or the second hand to be the same as if the spring were substituted for a block. Would it be okay to see it as, the force applied when the spring is compressed to its maximum amount per that force?

Another reason why I am confused is because in another problem, it says that a spring is compressed from both sides by 5cm, and a force of 50N is applied by both sides. Yet here, we actually do consider the combined force to find the spring constant. How is it that here the spring compresses yet in the case i described above it doesnt?
 
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If you push on one end of a spring, an equal but opposite force must react against the other end. You can't have one without the other, (assuming a static situation or a massless spring).

What matters is the total compression of the spring, i.e., how far one end is displaced relative to the other.
 
Hereformore said:
What confuses me i guess is that we're talking a spring here so its not like we're pushing a block against a wall. I didn't expect the force that the spring would apply to a wall or the second hand to be the same as if the spring were substituted for a block.
Why not?

Would it be okay to see it as, the force applied when the spring is compressed to its maximum amount per that force?
I'm not sure what you mean here. Are you asking about the force exerted by the spring on the wall or hand as it's being compressed?

Another reason why I am confused is because in another problem, it says that a spring is compressed from both sides by 5cm, and a force of 50N is applied by both sides. Yet here, we actually do consider the combined force to find the spring constant. How is it that here the spring compresses yet in the case i described above it doesnt?
Please provide more details.
 
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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