One-sided force applied to spring

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SUMMARY

The discussion centers on the mechanics of a spring subjected to a constant force in a frictionless environment. The formula for the change in spring length is established as Δx = F/K, where F represents the applied force and K is the spring constant. It is clarified that the mass of the weight does not directly influence the spring's contraction; rather, the acceleration of the mass plays a critical role. A smaller mass accelerates more quickly, necessitating adjustments in the force application to maintain constant spring length.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with Hooke's Law and spring constants
  • Basic knowledge of force and mass relationships
  • Concept of acceleration in physics
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  • Study the implications of Hooke's Law in dynamic systems
  • Explore the effects of varying mass on spring dynamics
  • Learn about frictionless environments in classical mechanics
  • Investigate the relationship between force, mass, and acceleration in detail
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Physics students, mechanical engineers, and anyone interested in understanding the dynamics of springs under applied forces.

snoopies622
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In a place with no friction, I have a rocket that applies a constant force f to a weight of mass m, and between them is a spring with constant k. I assume that the mass of the spring is negligible.

As the three objects accelerate, by how much does the spring contract? My intuition tells me that the greater the mass of the weight, the more the spring contracts, but I don't see how this works out algebraically.
 
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Hi snoopies622,

Your intuition is wrong. The change in the spring length depends on the force F and the constant K as simply as:

Δx=F/K

when m in your problem is small, it accelerates faster , trying to make the length shorter, so you need to move the point of force faster to keep the length ( and force) constant.
 

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