04_InstSolManual_PDF_Part15

04_InstSolManual_PDF_Part15 - upward acceleration F 5 m 1 g...

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4.50. Set Up: Let be downward. The instrument and the lander have the same acceleration. Solve: (a) so Since is negative, is upward. is in the direction opposite to when the speed is decreasing. (b) The free-body diagram is given in Figure 4.50. T is the scale reading. gives so Figure 4.50 (c) On earth, so On Mars, so (d) The apparent weight is the scale reading, 60.5 N. The apparent mass m is the apparent weight divided by so the apparent mass is 16.3 kg. 4.51. Set Up: Let be upward. At his maximum height, While he is in the air, Solve: (a) and gives (b) For the motion while he is pushing against the floor, and gives The acceleration is upward. (c) The free-body diagram while he is pushing against the floor is given in Figure 4.51. is the vertical force the floor applies to him. Figure 4.51 (d) gives and The force he applied to the ground has this same magnitude and is downward. Reflect: The ground must push upward on him with a force greater than his weight in order to give him an
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Unformatted text preview: upward acceleration. F 5 m 1 g 1 a 2 5 1 90.8 kg 21 9.80 m / s 2 1 16.2 m / s 2 2 5 2.36 3 10 3 N. F 2 mg 5 ma g F y 5 ma y x y F mg a F S a y 5 v y 2 v y t 5 4.85 m / s 2 0.300 s 5 16.2 m / s 2 . v y 5 v y 1 a y t 5 0.300 s. v y 5 4.85 m / s v y 5 0, v y 5 " 2 2 a y 1 y 2 y 2 5 " 2 2 1 2 9.80 m / s 2 21 1.2 m 2 5 4.85 m / s v y 2 5 v y 2 1 2 a y 1 y 2 y 2 y 2 y 5 1.2 m. a y 5 2 9.80 m / s 2 v y 5 0, m 5 w / g 5 90.8 kg a y 5 2 9.80 m / s 2 . v y 5 0. 1 y g M , T 5 6.52 w M . w M 5 mg M 5 9.28 N, T 5 2.47 w E . w E 5 mg E 5 24.5 N, x y T mg a m 1 g 2 a y 2 5 1 2.50 kg 21 3.71 m / s 2 1 20.5 m / s 2 2 5 60.5 N T 5 mg 2 T 5 ma y g F y 5 ma y v S a S a S a y a y 5 v y 2 v y t 5 444 m / s 2 5.36 3 10 3 m / s 240 s 5 2 20.5 m / s 2 . v y 5 v y 1 a y t t 5 240 s. v y 5 1600 km / h 5 444 m / s, v y 5 19,300 km / h 5 5.36 3 10 3 m / s, 1 y Newton’s Laws of Motion 4-15...
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