04_InstSolManual_PDF_Part13 - Newtons Laws of Motion 4-13(c...

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(c) She would think the scale reading is her weight, 1700 N. And she would therefore think her mass is Reflect: Since she has an upward acceleration the upward force n is greater than the downward force of gravity. 4.46. Set Up: From Problem 4.45, the shuttle and tool have an upward acceleration of Take to be upward. The mass of the tool is Solve: (a) The free-body diagram for the tool is shown in Figure 4.46. Figure 4.46 (b) gives and 4.47. Set Up: The car has mass Use coordinates where the axis is in the direction of the motion of the car. The friction force is then in the Since the car slows, its acceleration is opposite to is velocity. Solve: (a) The free-body diagram for the car is given in Figure 4.47. Figure 4.47 (b) The acceleration that slows the car is caused by the friction force. (c) (stops). gives gives so Reflect: The magnitude of the friction force between the tires and pavement determines the stopping distance. 4.48. Set Up: Use coordinates where is along the barrel, in the direction of the motion of the bullet. The
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Unformatted text preview: is along the barrel, in the direction of the motion of the bullet. The expanding gases in the barrel produce a horizontal force F on the bullet. Solve: (a) gives gives F 5 ma x 5 1 4.20 3 10 2 3 kg 21 1.03 3 10 6 m / s 2 2 5 4330 N g F x 5 ma x a x 5 v x 2 2 v x 2 2 1 x 2 x 2 5 1 965 m / s 2 2 2 2 1 0.450 m 2 5 1.03 3 10 6 m / s 2 v x 2 5 v x 2 1 2 a x 1 x 2 x 2 x 2 x 5 0.450 m. v x 5 965 m / s, v x 5 0, 1 x f 5 2 1 1122 kg 21 2 9.25 m / s 2 2 5 1.04 3 10 4 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 28.7 m / s 2 2 2 1 44.5 m 2 5 2 9.25 m / s 2 v x 2 5 v x 2 1 2 a x 1 x 2 x 2 v x 5 v x 5 28.7 m / s, x 2 x 5 44.5 m, f x y n mg v a 2 x-direction. 1 x m 5 w / g 5 1122 kg. T 5 m 1 g 1 a 2 5 1 12.8 kg 21 9.80 m / s 2 1 21.2 m / s 2 2 5 397 N T 2 mg 5 ma g F y 5 ma y x y T w a m 5 w / g 5 12.8 kg. 1 y 21.2 m / s 2 . w / g 5 173 kg. m 5 Newton’s Laws of Motion 4-13...
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