General Rule for Spring Compression?

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The discussion focuses on the physics of a block compressing a spring on a frictionless table. When the block of mass m compresses the spring with force constant k, it comes to rest, and the amount of compression x is determined by the work done by the spring. The key concept is that the spring exerts a force proportional to its compression, following Hooke's law, where k represents the spring constant. To find the work done by the spring, one must consider the force acting over the distance of compression. Understanding this relationship is essential for solving the problem effectively.
FrenchAtticus
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Though this is part of my homework I posted it here because it's a general rule that applies to all physics and doesn't contain specific values from my homework.

A block of mass m slides on a horizontal frictionless table with an initial speed. It then compresses a spring of force constant k and is brought to rest. How much is the spring compressed x from it's natural length.

*also* Is k the same K as in Force of spring = -Kx
 
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(Yes, the k in Hooke's law is lower case).

For this problem you must realize that as the block hits the spring, it moves (covers distance) against the force kx. As a result, the spring does work on the block. Now, how much work it takes to slow the block down as a function of compressed spring distance is essentially what they're asking in this question.

I can't tell you the formula, as that would be doing your homework for you. But...think about how much work the spring does for a small distance assuming the force is constant over that distance - and use that to figure out how much work is done for any compression length x.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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