05_InstSolManual_PDF_Part33

05_InstSolManual_PDF_Part33 - v t2 5 1 6 m / s 2 62 kg 75...

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5.79. Set Up: If the truck accelerates to the right the friction force on the box is to the right, to try to prevent the box from sliding relative to the truck. The free-body diagram for the box is given in Figure 5.79. The maximum acceleration of the box occurs when has its maximum value, If the box doesn’t slide its acceleration equals the acceleration of the truck. Figure 5.79 Solve: gives so and gives Reflect: If the truck has a smaller acceleration it is still true that but now 5.80. Set Up: The free-body for the car is given in Figure 5.80. Figure 5.80 Solve: gives so (stops), gives He was speeding. 5.81. Set Up: Write the air resistance force as k has units of Solve: (a) At the terminal velocity, and without a parachute: with a parachute: (b) so is constant. and without: and with: Reflect: At the terminal velocity the air resistance force equals the weight, so a lighter person has a smaller terminal velocity.
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Unformatted text preview: v t2 5 1 6 m / s 2 62 kg 75 kg 5 5.5 m / s v t2 5 1 60 m / s 2 62 kg 75 kg 5 55 m / s v t2 5 v t1 m 2 m 1 . v t1 " m 1 5 v t2 " m 2 v t " m v t 5 mg D , D w 5 1 75 kg 21 9.80 m / s 2 2 1 6 m / s 2 2 5 20 kg / m D wo 5 mg v t 2 5 1 75 kg 21 9.80 m / s 2 2 1 60 m / s 2 2 5 0.20 kg / m D v t 2 5 mg F D 5 mg kg / m. F D 5 D v 2 . v x 5 " 2 2 a x 1 x 2 x 2 5 " 2 2 1 2 3.92 m / s 2 21 67.0 m 2 5 22.9 m / s 5 51 mph v x 2 5 v x 2 1 2 a x 1 x 2 x 2 x 2 x 5 219.75 ft 5 67.0 m. v x 5 a x 5 2m k g 5 2 3.92 m / s 2 2m k mg 5 ma x 2 f k 5 ma x . g F x 5 ma x x y n mg f k v f s , m s n . f s 5 ma , t 5 v x 2 v x a x 5 30.0 m / s 2 6.37 m / s 2 5 4.71 s v x 5 v x 1 a x t v x 5 30.0 m / s. v x 5 0, a 5 m s g 5 6.37 m / s 2 . m s mg 5 ma f s 5 ma g F x 5 ma x n 5 mg . x y mg n f s a f s 5 m s n . f s Applications of Newtons Laws 5-33...
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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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