Elastic potential energy of spring

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SUMMARY

The discussion centers on the elastic potential energy of a spring with spring constant k and a mass M attached. It establishes that the sum of the elastic potential energy, represented by the equation E(x) = 1/2 kx², and the gravitational potential energy, given by -mgx, reaches a minimum at the equilibrium position x = mg/k. The equilibrium condition is derived from setting the net force to zero, leading to the conclusion that the minimum energy occurs when the derivative of the energy function is zero.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with gravitational potential energy concepts
  • Knowledge of calculus, specifically derivatives
  • Basic physics principles related to equilibrium
NEXT STEPS
  • Study the implications of Hooke's Law in mechanical systems
  • Learn how to derive energy conservation equations in physics
  • Explore the concept of equilibrium in dynamic systems
  • Investigate applications of potential energy in real-world scenarios
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the principles of energy conservation and equilibrium in spring systems.

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Homework Statement


A spring of spring constant k is suspended from the ceiling with a mass M attached to its lower end. The spring has negligible mass. Show that the sum of the elastic potential energy of the spring and the gravitational potential energy of the mass is a minimum at the equilibrium position.


Homework Equations





The Attempt at a Solution



So I think the basic setup would be:
(Assuming the datum is at the position when the spring doesn't have anything on it)

[tex]\frac{1}{2}kx^{2} - mgh = E(x)[/tex]
But the graph shows that energy decreases until a minimum and then increases. If x=0 is the equilibrium point (and thus minimum energy), why does that function continue to decrease? Any help would be appreciated.
 
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x=0 is not the equilibrium point...
 
the point of equilibrium: kx - mg = 0 --> x=mg/k

energy [tex]E(x)= 1/2kx^2 - mgx[/tex]
Minimum is where derivative is zero and that is exactly the fist equation.
 

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