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cowmoo32
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I'm reviewing some problems from a few months ago and I remember getting hung on this before. A mass is dropped from a height h above a spring. I am given the deflection of the spring when the mass stops moving momentarily. The spring constant in the solution is given by mg/x. I understand how that works if the mass is placed on a spring, but why is the extra force of the mass falling not accounted for? Why am I incorrect in saying that all of the kinetic energy of the mass falling is converted into potential energy for the spring? I tried the latter and came up with a close answer, but not the same as the solution. The mass is momentarily at rest. This isn't after the spring has damped out and stopped moving. The end goal of the problem is to find the velocity of the mass at some smaller x once the spring starts pushing it back up during the first oscillation.
Drawing a free body diagram of the mass just above the spring and after it hits, F = mg for both of them, but in the second one the spring is pushing back. I know F=ma=kx and a is constant, but if a mass is dropped on a spring, the spring will compress more than it would if the mass was gently placed on top.
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Drawing a free body diagram of the mass just above the spring and after it hits, F = mg for both of them, but in the second one the spring is pushing back. I know F=ma=kx and a is constant, but if a mass is dropped on a spring, the spring will compress more than it would if the mass was gently placed on top.
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