If a mass is attached to a spring(hanging from a wall) with spring constant $k$, causing an expansion $x$, then the potential energy stored by the spring is $\frac{1}{2}kx^2$. However, the potential energy stored by the spring is just the potential difference of the mass caused by its descent, isn't it? So why isn't the potential difference equal to $mgx$, which in turn is equal to $kx^2$, since $mg$=$kx$?
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However, the potential energy stored by the spring is just the potential difference of the mass caused by its descent, isn't it?
No, the two potential energies are quite independent of each other. Therefore it doesn't make sense to equate them.
- The potential energy of the mass due to its position in the
gravitational field of the earth is $mgx$.
This energy part is positive or negative, depending on whether mass is ascended ($x>0$) or descended ($x<0$). - The potential energy of the spring due its compression or expansion is $\frac{1}{2}kx^2$.
This energy part is positive both when the spring is compressed ($x<0$) and when it is expanded ($x>0$).
Rather than equating these two potential energies, it makes more sense to add them, i.e. to calculate the total potential energy. $$E_\text{pot}=mgx+\frac{1}{2}kx^2$$
Thomas Fritsch
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Assuming $x > 0$ lies in the upward direction, the spring is ideal (with no damping) and you release the mass from the spring's unstretched position ($x = 0$), it would keep oscillating up and down between $x = 0$ and $x = -2mg/k$, never settling at $x_0 = -mg/k$. This is because part of the initial gravitational potential energy is converted into kinetic energy.
In order for the mass to settle at $x = x_0$, there either needs to be a damping mechanism dissipating $\frac12mg|x_0|$ of energy as heat, or you have to manually intervene and do work equal to $-\frac12mg|x_0|$. In either scenario, mechanical energy is not conserved, so you cannot use conservation of energy to determine the equilibrium position.
Puk
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