9_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

9_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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Electrical Energy and Capacitance 43 16.9 (a) Use conservation of energy () se f i KE PE PE KE PE PE ++ =++ ( or ) ()() 0 KE PE PE ∆+ ∆ + ∆ = ( ) 0 KE ∆= since the block is at rest at both beginning and end. 2 max 1 0 2 s PE kx max x ( , where is the maximum stretch of the spring. ) max e PE W QE x = Thus, 2 max max 1 2 kx QE x +− = 00 , giving ( ) ( ) 65 5 . 0 01 0 Vm 0.500 m N m ×× = max 25 0 . 0 C 2 100 QE x k == ± ² ± ³ ±²±³ ±² (b) At equilibrium, Σ= Therefore, 0, or 0 e q FFF k xQ E − + = + = 1 0.250 m 2 eq max QE xx k = Note that when the block is released from rest, it overshoots the equilibrium
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