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Unformatted text preview: =? =? =? How should we proceed? Let’s try energy which buckets do we need KE g S KE g S mgh = 1 / 2 k ∆ x 2 = E is conserved ∆ x = h 1 / 2 k ∆ x 2 = mgh ∆ x= 2mg k Total fall distance A mass is held up at the springs relaxed length then released. Find all of the following: Distance the mass falls Height it returns to Amplitude Max velocity Max acceleration KE g S 1 / 2 k ∆ x 2 = ∆ x = 2mg k Instant of release =? =? =? ∆ x = h KE g S KE g S Returns to the starting position where v = 0 again A = mg k 1 / 4 PE max = 1 / 2 mv 2 W T T  W = ma k ∆ x  mg = ma 2mg  mg = ma a = g !!!!...
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This note was uploaded on 12/21/2010 for the course PHYS Physics 21 taught by Professor Larrywatson during the Fall '08 term at North Dakota.
 Fall '08
 LarryWatson
 Physics, Angular Momentum, Energy, Momentum, Simple Harmonic Motion

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