1/2Kx^2

The slight preference for 12 kx2 is down simply to the fact that you need only measure one variable- the dispalcement x. This problem has been solved.

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Both 12Fx and 12 kx2 are correct and equivalent to each other.

. This force leads to the potential energy U 12 kx². Comments sorted by Best Top New Controversial QA Add a Comment. Rewrite the equation as 1 2 kx2 E 1 2 k x 2 E.

Youll get a detailed solution from a subject. What is the difference between the equations Fkx F-kx and F12kx2 in spring problem situations. E 1 2 kx2 E 1 2 k x 2.

U 12kx2 U 12 750 Nm 040 m 2 U 060 Nm U 060 J The elastic potential energy stored by the spring when it is has been. 12 k x2 k. Potential energy 12spring constant distance from equilibrium2.

Im in both chem and physics 1 right now. 1 2 kx2 E 1 2 k x 2 E. Although both are rigorous my schools chemistry is a.

If we had instead defined the force as F -2kx then U would be kx². The spring constant is the measure of stiffness of a spring. Integral k2 x22 dx k2 x36.

The potential energy can be found using the formula. Hookes law gives us the force we need to find elastic potential energy. So using WFd and F-kx hookes law shouldnt the work come out to.

Solve for x k12 kx2 Mathway Trigonometry Examples Popular Problems Trigonometry Solve for x k12 kx2 k 1 2 kx2 k 1 2 k x 2 Rewrite the equation as 1 2 kx2 k 1. Solve for x E12kx2. Derivative ddx12 k x2 k k2 x.

Solve for x U12kx2. Where E is energy J and x is distance cm. Arent d and x the same.

Multiply both sides of the equation by 2 2. But the work done in stretching a spring 12 kx2 isnt work Fd. The dimensions of K in the equation W12Kx2 is.

Like all work and energy the unit of potential energy is the Joule J where 1 J 1Nm 1 kg m2s2. Asked Sep 2 2021 in Physics by Vaibhav02 380k points. Determine the fundamental dimensions for k in the equation below.

Looking at a graph of force versus. X 2 14142 k 0. Alternatively you can assume that the spring in equilibrium has a.

The dimensions of K in the equation W 12 Kx2 is. Prev Question Next Question. Thus you can divide by k and move the constant term to RHS obtaining xx k2 1.

Multiply both sides of the equation by 2 2. First thing we can find out is that k 0. U 1 2 kx2 U 1 2 k x 2.

Rewrite the equation as 1 2 kx2 U 1 2 k x 2 U. 1 2 kx2 U 1 2 k x 2 U. Indeed f x 2x is non-negative for x 0.

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