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Unformatted text preview: C HAPTER 13 Problem Solutions 13.1 (a) The force exerted on the block by the spring is ( 29 ( 29 160 N m 0.15 m 24 N s F kx =  = = + , or s F = 2 4 N d i r e c t e d t o w a r d e q u i l i b r i u m p o s i t i o n (b) From Newton's second law, the acceleration is 2 24 N m 60 0.40 kg s s F a m + = = = + = 2 m 60 to w a rd eq u ilib riu m p o sitio n s 13.7 (a) The spring constant of each band is 32 15 N 1.5 10 N m 1.0 10 m s F k x = = = Thus, when both bands are stretched 0.20 m, the total elastic potential energy is ( 29 ( 29 2 2 3 1 2 1.5 10 N m 0.20 m 2 s PE kx = = = 60 J (b) Conservation of mechanical energy gives ( 29 ( 29 s s f i KE PE KE PE + = + , or 2 1 60 J 2 mv + = + , so ( 293 2 60 J 50 10 kg v = = 4 9 m s 13.8 (a) 230 N 0.400 m max max F k x = = = 5 7 5 N m (b) ( 29 ( 29 2 2 1 1 575 N m 0.400 2 2 s work done PE kx = = = = 46.0 J 13.16 (a) 0 at KE x A = = , so 2 1 2 s E KE PE kA = + = + , or the total energy is ( 29 ( 29 2 2 1 1 250 N m 0.035 m 2 2 E kA = = = 0.15 J (b) The maximum speed occurs at the equilibrium position where s PE = . Thus, 2 1 2 max E mv = , or ( 29 2 250 N m 0.035 m 0.50 kg max E k v A m m = = = = 0 .7 8 m s (c) The acceleration is F kx a m m  = = . Thus, max a a = at max x x A =  =  . ( 29 ( 29 250 N m 0.035 m 0.50 kg max k A k a A m m  = = = = 2 18 m s 13.26 (a) At t = , ( 29 ( 29 0.30 m cos 0 x = = 0.30 m , and at 0.60 s t = , ( 29 ( 29 ( 29 ( 29 0.30 m cos rad s 0.60 s 0.30 m cos 0.20 rad 3 x = = = 0.24 m (b) ( 29 ( 29 0.30 m 1 max A x = = = 0.30 m (c) ( 29 0.30 m cos 3 x t = is of the form ( 29 cos x A t = with an angular frequency of rad s 3 = . Thus, 3 2 2 f = = = 1 Hz 6 (d) The period is 1 T f = = 6.0 s 13.30 (a) The height of the tower is almost the same as the length of the pendulum. From 2 T L g = , we obtain ( 29 ( 29 2 2 2 2 2 9.80 m s9....
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This note was uploaded on 06/27/2008 for the course PHYS 221 taught by Professor Ezell during the Summer '08 term at Campbell.
 Summer '08
 Ezell
 Acceleration, Force

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