If a horizontal mass on a spring is turned to a vertical position, then the spring is stretched by an amount \(x_{0}=m g / k\), giving it a new equilibrium position. For the vertical spring, the poten...If a horizontal mass on a spring is turned to a vertical position, then the spring is stretched by an amount \(x_{0}=m g / k\), giving it a new equilibrium position. For the vertical spring, the potential energy is still given by \(U=\frac{1}{2} k x^{2}\), but \(x\) in this case refers to the distance from the original (horizontal) equilibrium position.
