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Albertgauss
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A massless spring of spring constant k = 13 Newtons per meter hangs purely vertically. A 20 Newton, 5(10-3) Coulombs charged particle is attached to the spring and released from rest. Besides all this, there is also a constant, external Electric Field of 9000 Newtons per Coulomb pointing upwards. See figure below. After the release point, and assuming the particle only travels downwards and vertical, what is the speed of the particle as it passes through the equilibrium point? See the attached figure called "Figure_for_Problem.jpg"
Conservation of energy, except that now we have a -qEx mixed in with the kinetic energy, mgy, and the spring potential energy.
I am only having some trouble knowing what the sign on -qEy should be, and/or which side of the conservation of energy this term goes on.
A few thoughts: since the charge is +Q, it has lower (more negative) potential energy at the Release Point. Thus, on the equilibrium side of the equation, I would put y = 0. On the “release” point of the equation, I would write in –qEy, where q is +, E is +, and y is +. Is this correct? I can't get the latex on this website to work on my computer, so I uploaded a jpeg of where I am so I can keep all my symbols and subscripts.
Homework Equations
Conservation of energy, except that now we have a -qEx mixed in with the kinetic energy, mgy, and the spring potential energy.
The Attempt at a Solution
I am only having some trouble knowing what the sign on -qEy should be, and/or which side of the conservation of energy this term goes on.
A few thoughts: since the charge is +Q, it has lower (more negative) potential energy at the Release Point. Thus, on the equilibrium side of the equation, I would put y = 0. On the “release” point of the equation, I would write in –qEy, where q is +, E is +, and y is +. Is this correct? I can't get the latex on this website to work on my computer, so I uploaded a jpeg of where I am so I can keep all my symbols and subscripts.
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