Spring Elastic Energy
Calculates the elastic energy stored in a spring E=0.5·k·x² in joules from constant and deformation.
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Elastic potential energy: Ep = (1/2)·k·x²
When you pull a spring (or any elastic element) away from its natural length, it stores energy. That stored energy is given by Ep = (1/2)·k·x², in joules, where k is the spring constant in N/m and x is the deformation in meters. Notice that it's quadratic in x, so if you double the stretch you end up with four times the stored energy. Take a bow with k = 200 N/m drawn back 0.5 m: it holds Ep = 25 J, which becomes the arrow's kinetic energy the moment you let go. You get the formula by integrating Hooke's law force F = k·x over the displacement, and it holds as long as the spring stays within its elastic limit and doesn't deform permanently.
Applications
Competition bows and crossbows. Automotive suspension, where coil and leaf springs soak up road impacts. Engine valve springs, which open and close valves thousands of times a minute in every cylinder. Mechanical clocks store the winding energy in a mainspring. Then there are retractable pens, pinball plungers, trampolines and railway bumpers.
FAQ
Does the formula work for compression and stretching? It does. Since x² is always positive, you get the same Ep whether the spring is stretched or compressed by the same distance.
What happens beyond the elastic limit? The spring takes on a permanent deformation, and at that point Hooke's law (along with this formula) stops applying. Some of the energy ends up as plastic work instead.
How does k relate to spring stiffness? A larger k is a stiffer spring. You need more force to reach the same deformation, and for a given x it stores more energy.
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