Exploring Energy of a Vertical Spring System

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SUMMARY

The discussion focuses on the energy dynamics of a vertical spring system with an attached mass, specifically how changing the reference point affects the energy equations. Two scenarios are analyzed: one where the zero point is at the unstretched position of the spring and another where it is at the new equilibrium point. The equations derived are mgh - 1/2kx^2 = 1/2mv^2 and mgh + 1/2kx^2 = 1/2mv^2, highlighting the impact of reference points on gravitational and elastic potential energy. The conversation emphasizes the importance of the reference point in calculating elastic restoring force.

PREREQUISITES
  • Understanding of classical mechanics principles, specifically energy conservation.
  • Familiarity with spring mechanics, including Hooke's Law.
  • Knowledge of gravitational potential energy calculations.
  • Basic proficiency in algebra for manipulating energy equations.
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems.
  • Explore Hooke's Law and its applications in various spring systems.
  • Learn about different reference points in physics and their implications on energy calculations.
  • Investigate the use of simulation tools like the provided applet for visualizing spring dynamics.
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the energy transformations in spring systems.

nothing123
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Hi,

If we have a vertical spring with a mass attached to it, how does the energy of the system work. Let's say we set the zero point as the point where the spring is unstretched (basically with no mass attached). Then at the top we have zero kinetic, zero spring potential as well as zero gravitational potential. As it passes through its new equilibrium point, it will have negative gravitational and positive elastic potential and kinetic energy so 1/2kx^2 + 1/2mv^2 - mgh which rearranged is mgh - 1/2kx^2 = 1/2mv^2. Now let's say we take the zero point to be where the new equilibrium point is. At the top, we have gravitational potential as well as spring potential. As it passes through the equilibrium point, it would only have kinetic. By conservation of energy, mgh + 1/2kx^2 = 1/2mv^2. Why are the two equations different if we change the reference point?

Thanks.
 
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zero point

Hi nothing123! :smile:
nothing123 said:
… Now let's say we take the zero point to be where the new equilibrium point is …

Your equations are completely confusing me (especially since h is related to x). :confused:

But if by "zero point", you mean for calculating the elastic restoring force, it has to be the "old" equilibrium point … that's how springs work! :smile:

(unless, I suppose, you take g out of the equation completely :rolleyes:)
 

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