ch15-p092 - = + 4.0 = 1 2 + 1 2 . 2 2 E K U mv kx...

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1 2 2(4.0 J) 2 0.20 m. 22 0 0 N / m mm E Ek x x k = ¡ == = (b) Since 2 / 2 0.80 kg / 200 N / m 0.4 s , Tm k ππ == then the block completes 10/0.4 = 25 cycles during the specified interval. (c) The maximum kinetic energy is the total energy, 4.0 J. (d) This can be approached more than one way; we choose to use energy conservation: =+
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Unformatted text preview: = + 4.0 = 1 2 + 1 2 . 2 2 E K U mv kx Therefore, when x = 0.15 m, we find v = 2.1 m/s. 92. (a) Eq. 15-21 leads to...
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