Elastic potential energy of spring

AI Thread Summary
The discussion focuses on demonstrating that the sum of elastic potential energy and gravitational potential energy reaches a minimum at the equilibrium position of a spring-mass system. The energy equation is expressed as E(x) = 1/2 kx^2 - mgx, where k is the spring constant, m is the mass, and g is the acceleration due to gravity. The equilibrium position is identified as x = mg/k, where the forces balance (kx = mg). The minimum energy occurs where the derivative of the energy function equals zero, confirming that the equilibrium point is not at x=0 but rather at the calculated position. Understanding these concepts is crucial for solving related physics problems effectively.
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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)

\frac{1}{2}kx^{2} - mgh = E(x)
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 E(x)= 1/2kx^2 - mgx
Minimum is where derivative is zero and that is exactly the fist equation.
 
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