05_InstSolManual_PDF_Part23

05_InstSolManual_PDF - 2 1 0.875 kg 2 5 2.08 kg m 5 kx g 5 1 357 N m 21 0.0572 m 2 9.80 m s 2 5 2.08 kg k 5 mg x 5 1 0.875 kg 21 9.80 m s 2 2

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5.53. Set Up: Take to be downward. Solve: (a) The versus graph is sketched in Figure 5.53a. (b) See Figure 5.53b. (c) at and as t increases. When See Figure 5.53c. (d) See Figure 5.53d. (e) graph has same shape as versus t . See Figure 5.53e. Figure 5.53 5.54. Set Up: so Solve: (a) (b) 5.55. Set Up: so x is the change in length of the spring. Solve: (a) (b) Reflect: The elongation of the vertical spring is proportional to the mass hung from it. So, the mass in (b) is 5.56. Set Up: where F is the pull on the strip or the force the strip exerts. Solve: (a) so (b) F 5 kx 5 1 40.0 N / m 21 0.0114 m 2 5 0.456 N k 5 F x 5 1.50 N 0.0375 m 5 40.0 N / m F 5 kx 0 F spr 0 5 F , 1 5.72 cm 2.40 cm
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Unformatted text preview: 2 1 0.875 kg 2 5 2.08 kg. m 5 kx g 5 1 357 N / m 21 0.0572 m 2 9.80 m / s 2 5 2.08 kg k 5 mg x 5 1 0.875 kg 21 9.80 m / s 2 2 0.0240 m 5 357 N / m mg 5 kx . F spr 5 mg m 5 kx g 5 1 327 N / m 21 0.0813 m 2 9.80 m / s 2 5 2.71 kg k 5 mg x 5 1 1.25 kg 21 9.80 m / s 2 2 0.0375 m 5 327 N / m mg 5 kx . F spr 5 mg ( a ) ( b ) F D v ( c ) t F net mg v v t F net mg ( e ) t a g ( d ) t v v t F net a 5 F net / m ; F net 5 0. v 5 v t , v S v t t 5 v 5 F net 5 2 F D 1 mg 5 2 D v 2 1 mg . v F D F D 5 D v 2 . 1 y Applications of Newton’s Laws 5-23...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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