Hooke's Law with Series Springs

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SUMMARY

The discussion centers on Hooke's Law as applied to series springs, specifically the calculation of equivalent spring constant (Keq). The correct formula for Keq is established as Keq = (k1k2)/(k1+k2), where k1 and k2 are the spring constants of the individual springs. The forum participants confirm the derivation of the force equation F = -[(k1k2)/(k1+k2)]*x2, validating the user's approach to solving the problem.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of spring constants (k1, k2)
  • Familiarity with algebraic manipulation of equations
  • Basic concepts of series spring systems
NEXT STEPS
  • Study the derivation of the equivalent spring constant in parallel spring systems.
  • Learn about energy stored in springs and its relation to Hooke's Law.
  • Explore applications of Hooke's Law in real-world engineering problems.
  • Investigate the effects of damping in spring systems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as engineers and anyone interested in the principles of elasticity and spring systems.

maulucci
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Homework Statement



The problem is in the attachment

Homework Equations



F=-kx
k1x1=k2x1

The Attempt at a Solution



I got as my answer Keq= (k1k2)/(k1+k2)

i just wanted to see if i had the right idea or no.
 

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Welcome to PF;
i just wanted to see if i had the right idea or no.
... then you need to tell us the idea and not just the final result you got ;)
 
-k1x1=-k2x2=F
F=-keq*x
solving for x1
x1=[(k2)/(k1+k2)] +x2
putting into F=-k2x2+k2x1
F=x2 (-k1k2-k2^2+k2^2)/(k1+k2)
simplified
F=-[(k1k2)/(k1+k2)]*x2

then the Keq = (k1k2)/(k1+k2) ?
 

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