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Why is my method for finding the spring constant incorrect?
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[QUOTE="Chestermiller, post: 6835999, member: 345636"] Here is may take on this from an energy perspective. The force balance equation on the mass in the upward direction is given by: $$-k(x-x_u)-mg=m\frac{dv}{dt}=mv\frac{dv}{dx}$$where ##x_u## is the unstretched length of the spring, x is the deformed length, and v is the upward velocity. If we integrate the equation with respect to x, we obtain: $$-k\left(\frac{(x-x_u)^2}{2}-\frac{(x_0-x_u)^2}{2}\right)-mg[(x-x_u)-(x_0-x_u)]=m\frac{v^2}{2}-m\frac{v_0^2}{2}$$where ##v_0## is the initial velocity of the mass when the initial length of the spring is ##x_0##. This energy conservation equation can be rewritten as $$E=k\frac{(x-x_u)^2}{2}+mg(x-x_u)+m\frac{v^2}{2}=k\frac{(x_0-x_u)^2}{2}+mg(x_0-x_u)+m\frac{v_0^2}{2}$$ In this representation, the datum of zero potential energy is conveniently situated at the top of the unstretched spring. The total energy E is the sum of stored elastic spring energy, mass potential energy, and mass kinetic energy, and is constant. We can use the principle of virtual work to determine the relationship between the equilibrium length of the spring and the spring constant. $$dE=(k(x_e-x_u)+mg)dx=0$$So, $$k=\frac{mg}{(x_u-x_e)}$$ [/QUOTE]
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Introductory Physics Homework Help
Why is my method for finding the spring constant incorrect?
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