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Maximum Spring Elongation for a Block on a Horizontal Ideal Spring
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[QUOTE="anthonych414, post: 4529873, member: 237568"] [h2]Homework Statement [/h2] A 2-kg block is attached to a horizonal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5m/s. What is the maximum elongation of the spring. [h2]Homework Equations[/h2] Conservation of mechanical energy, PEspring=0.5kx^2 KE=0.5mv^2 [h2]The Attempt at a Solution[/h2] It seems like a fairly simple question to me, applying conservation of mechanical energy at equilibrium and at maximum elongation we get 0 + 0.5kx^2 = 0 + 0.5mv^2, so x =√(mv^2/k) = 0.5 meters, however the solution says the answer should be 5 meters, can anyone tell me what I'm missing here? [/QUOTE]
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Maximum Spring Elongation for a Block on a Horizontal Ideal Spring
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