04_InstSolManual_PDF_Part11

04_InstSolManual_PDF - 21 1058 m s 2 2 5 153 N g F x 5 ma x F S a x 5 v x 2 2 v x 2 2 1 x 2 x 2 5 1 46 m s 2 2 2 2 1 1.0 m 2 5 1058 m s 2 v x 2 5 v

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4.41. Set Up: Use a constant acceleration equation to find the stopping time and acceleration. Let be in the direction the bullet is traveling. is the force the wood exerts on the bullet. Solve: (a) The free-body diagram is given in Figure 4.41. Figure 4.41 (b) and gives (c) gives gives and 4.42. Set Up: Use coordinates where is in the direction the ball is thrown. Solve: (a) Solve for gives The free-body diagram for the ball during the pitch is shown in Figure 4.42a. The force is applied to the ball by the pitcher’s hand. gives Figure 4.42 (b) The free-body diagram after the ball leaves the hand is given in Figure 4.42b. The only force on the ball is the downward force of gravity. ( a ) ( b ) w x y w F a F 5 1 0.145 kg
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Unformatted text preview: 21 1058 m / s 2 2 5 153 N g F x 5 ma x F S a x 5 v x 2 2 v x 2 2 1 x 2 x 2 5 1 46 m / s 2 2 2 2 1 1.0 m 2 5 1058 m / s 2 v x 2 5 v x 2 1 2 a x 1 x 2 x 2 v x 5 46 m / s. v x 5 0, x 2 x 5 1.0 m, a x : 1 x F 5 2 ma x 5 2 1 1.80 3 10 2 3 kg 21 2 4.71 3 10 5 m / s 2 2 5 848 N 2 F 5 ma x g F x 5 ma x a x 5 v x 2 2 v x 2 2 1 x 2 x 2 5 2 1 350 m / s 2 2 2 1 0.130 m 2 5 2 4.71 3 10 5 m / s 2 v x 2 5 v x 2 1 2 a x 1 x 2 x 2 t 5 2 1 x 2 x 2 v x 1 v x 5 2 1 0.130 m 2 350 m / s 5 7.43 3 10 2 4 s. 1 x 2 x 2 5 1 v x 1 v x 2 2 t 1 x 2 x 2 5 0.130 m. v x 5 v x 5 350 m / s, x n y mg F v F S 1 x Newton’s Laws of Motion 4-11...
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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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