04_InstSolManual_PDF_Part3 - Solve v x 5 v x 1 a x t 5 1...

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and its components are shown in Figure 4.5. Figure 4.5 Reflect: Adding the forces as vectors gives a very different result from adding their magnitudes. 4.6. Set Up: Let be the direction of the force and acceleration. Solve: gives 4.7. Set Up: Let be the direction of the force. Use a constant acceleration equation to calculate the displacement of the probe. Solve: (a) (b) 4.8. Set Up: Take to be the direction in which the skater is moving initially. (comes to rest). Solve: so The only horizontal force on the skater is the friction force, so The force is 46.7 N, directed opposite to the motion of the skater. 4.9. Set Up: Take to be in the direction in which the cheetah moves. Solve: (a) so (b) The force is exerted on the cheetah by the ground. Reflect: The net force on the cheetah is in the same direction as the acceleration of the cheetah. 4.10. Set Up: Let be the direction of the force. Use to calculate the acceleration of the puck. Then use constant acceleration equations to find the speed and displacement of the puck for
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Unformatted text preview: Solve: v x 5 v x 1 a x t 5 1 1.56 m / s 2 21 2.00 s 2 5 3.12 m / s x 2 x 5 v x t 1 1 2 a x t 2 5 1 2 1 1.56 m / s 2 21 2.00 s 2 2 5 3.12 m v x 5 0. a x 5 g F x m 5 0.250 N 0.160 kg 5 1.56 m / s 2 . t 5 2.00 s. g F x 5 ma x g F x 5 0.250 N. 1 x F x 5 ma x 5 1 68 kg 21 10.05 m / s 2 2 5 680 N a x 5 v x 2 v x t 5 20.1 m / s 2 2.0 s 5 10.05 m / s 2 v x 5 v x 1 a x t v x 5 0. 1 x f x 5 ma x 5 1 68.5 kg 21 2 0.682 m / s 2 2 5 2 46.7 N. a x 5 v x 2 v x t 5 2 2.40 m / s 3.52 s 5 2 0.682 m / s 2 v x 5 v x 1 a x t v x 5 1 x x 2 x 5 v x t 1 1 2 a x t 2 5 1 2 1 4.31 m / s 2 21 10.0 s 2 2 5 215 m t 5 10.0 s. a x 5 4.31 m / s 2 , v x 5 0, a x 5 g F x m 5 140 N 32.5 kg 5 4.31 m / s 2 x 2 x g F x 5 140 N. 1 x m 5 g F x a x 5 48.0 N 3.00 m / s 2 5 16.0 kg. g F x 5 ma x g F x 5 48.0 N. 1 x u R R y y R x x O R S Newton’s Laws of Motion 4-3...
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