Two blocks with spring (no friction), force applied: displacement?

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SUMMARY

The discussion centers on calculating the displacement of a spring connecting two blocks on a frictionless surface when a force is applied. A 15N force is exerted on a 3kg block connected to a 2kg block via a spring with a spring constant (K) of 140 N/m. The initial calculation of displacement using the formula F = -kx resulted in 0.107m, but further analysis revealed the need to account for the net force acting on the system, factoring in the gravitational forces of the blocks. The correct approach involves adjusting the applied force by subtracting the weight of the 2kg block, leading to a more accurate displacement calculation.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Basic knowledge of Newton's second law of motion
  • Familiarity with gravitational force calculations
  • Concept of net force in a system of connected objects
NEXT STEPS
  • Study the implications of Hooke's Law in dynamic systems
  • Learn about net force calculations in multi-body systems
  • Explore the effects of mass on spring displacement
  • Investigate the role of friction in spring systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of spring systems and force interactions in mechanics.

ddtozone
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Homework Statement



A 2kg block & 3kg block (left & right, respectively) are on a horizontal frictionless surface connected by a spring with K=140 N/m. A 15N force is applied to the 3kg block towards the right. How much does the spring stretch from its equilibrium length?


Homework Equations



F=-kx
others, maybe?

The Attempt at a Solution



I simply did force over constant, which is 15/140=0.107m. Is it this simple, or do you use the masses in some way?

Thank you!
 
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the gravitational force of the 3kg is going to be more than the 2 kg gravitational force. So, when you pull with 15N to the right, it will not act like a fixed spring constant. So, what you should do is say 3-2kg=1kg so the resultant will include 9.81 Newtons in terms of the spring being moved, not the entire system which includes the spring and the two blocks. So, I think the answer is 15N-9.81N then you multiply by its constant.
 
Oh ok, that makes more sense. But wouldn't you divide by the constant instead of multiplying by the constant, so the final answer would be N/(N/m)=m?
 

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