05_InstSolManual_PDF_Part24

05_InstSolManual_PDF_Part24 - kx 2 5 mg 1 kx 3 5 2 mg x 3 5...

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5.57. Set Up: the weight of the object on the pan. and Solve: (a) so and (b) 5.58. Set Up: so Solve: (a) Yes, Hooke’s law says and this is true here since the graph of F versus x is a straight line. (b) k is the slope of the graph of F versus x : (c) 5.59. Set Up: Solve: (a) Yes. Hooke’s law says and this is true here since the graph of m versus x is a straight line. (b) The slope of the graph of m versus x is so and (c) Reflect: A 0.50 kg mass stretches the spring 20 cm, so a 2.35 kg mass stretches it 5.60. Set Up: Use coordinates with upward. Label the masses 1, 2, and 3 and call the amounts the springs are stretched and Each spring force is kx . Solve: (a) The three free-body diagrams are shown in Figure 5.60. Figure 5.60 (b) so so so The length of the springs, starting from the top one, are 14.4 cm, 13.6 cm and 12.8 cm. x 3 5 3 1 mg k 2 5 2.40 cm kx 1 5 mg 1 kx 2 5 3 mg x 2 5 2 1 mg k 2 5 1.60 cm
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Unformatted text preview: kx 2 5 mg 1 kx 3 5 2 mg x 3 5 mg k 5 1 6.40 kg 21 9.80 m / s 2 2 7.80 3 10 3 N / m 5 0.800 cm kx 3 5 mg x y a 5 mg kx 2 kx 1 x y a 5 mg kx 3 kx 2 x y a 5 mg kx 3 x 3 . x 2 , x 1 , 1 y 1 2.35 kg 0.50 kg 2 1 20 cm 2 . x 5 mg k 5 1 2.35 kg 21 9.80 m / s 2 2 24.5 N / m 5 0.94 m 5 94 cm k 5 1 2.50 kg / m 2 g 5 24.5 N / m k g 5 0.500 kg 0.200 m 5 2.50 kg / m k / g , m 5 1 k g 2 x mg 5 F spr 5 kx F 5 kx 5 1 150 N / m 21 0.17 m 2 5 25 N. k 5 15.0 N 0.100 m 5 150 N / m F 5 kx F 5 kx . F 5 2 F spr x 5 w k 5 0.973 N 0.93 N / cm 5 1.0 cm 5 10 mm. w 5 1 3.5 oz 2 1 0.278 N 1 oz 2 5 0.973 N. k 5 w x 5 0.278 N 0.30 cm 5 0.93 N / cm. w 5 k x F spr 5 k x , w 5 1 1.00 oz 2 1 1 lb 16 oz 21 4.448 N 1 lb 2 5 0.278 N. 1 lb 5 4.448 N. 1 lb 5 16 oz F spr 5 w , 5-24 Chapter 5...
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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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