This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: C HAPTER 4 4.1 (a) ( 29 ( 29 2 6.0 kg 2.0 m s F ma = = = 12 N . (b) 12 N 4.0 kg F a m = = = 2 3.0 m s . 4.3 ( 29 2000 lbs 4.448 N 2 tons 1 ton 1 lb w = = 4 2 10 N . 4.6 5 2 2 7 7.5 10 N 5.0 10 m s 1.5 10 kg F a m = = = , and i v v at = + gives 2 2 80 km h 0.278 m s 1 min 5.0 10 m s 1 km h 60 s i v v t a  = = = 7 .4 m in 4.8 The acceleration of the bullet is given by ( 29 ( 29 ( 29 2 2 2 320 m s 2 2 0.82 m i v v a x = = Then, ( 29 ( 29 ( 29 2 3 320 m s 5.0 10 kg 2 0.82 m F ma = = = 2 3 .1 1 0 N . 4.11 (a) From the second law, the acceleration of the boat is 2000 N 1800 N 1000 kg F a m = = = 2 0 .2 0 0 m s . (b) The distance moved is ( 29 ( 29 2 2 2 1 1 0.200 m s 10.0 s 2 2 i x v t at = + = + = 10.0 m . (c) The final velocity is ( 29 ( 29 2 0.200 m s 10.0 s i v v at = + = + = 2 .0 0 m s . 1 C H A P T E R 1 4.15 Since the burglar is held in equilibrium, the tension in the vertical cable equals the burglars weight of 600 N Now, consider the junction in the three cables: y F = , giving 2 sin 37.0 600 N T  = , or 2 T = 9 9 7 N i n t h e i n c l i n e d c a b l e . Also, x F = which yields 2 1 cos37.0 T T  = , or ( 29 1 997 N cos37.0 T = = 7 9 6 N i n t h e h o r i z o n t a l c a b l e 4.16 From x F = , 1 2 cos 40.0 cos 40.0 T T  = , or 1 2 T T = . Then, y F = gives ( 29 1 2 sin 40.0 100 N T  = , yielding 1 2 T T = = 77.8 N 4.23 The forces on the bucket are the tension in the rope and the weight of the bucket, ( 29 ( 29 2 5.0 kg 9.80 m s 49 N mg = = . Choose the positive direction upward and use the second law: y y F ma = ( 29 ( 29 2 49 N 5.0 kg 3.0 m s T = T = 64 N . 2 C H A P T E R 1 4.27 (a) The resultant external force acting on this system having a total mass of 6.0 kg is 42 N directed horizontally toward the right. Thus, the acceleration produced is 42 N 6.0 kg F a m = = = 2 7 . 0 m s h o r i z o n t a l l y t o t h e r i g h t . (b) Draw a free body diagram of the 3.0kg block and apply Newtons second law to the horizontal forces acting on this block: x x F ma = ( 29 ( 29 2 42 N 3.0 kg 7.0 m s T = , and therefore T = 21 N (c) The force accelerating the 2.0kg block is the force exerted on it by the 1.0kg block. Therefore, this force is given by: ( 29 ( 29 2 2.0 kg 7.0 m s F ma = = , or = F 1 4 N h o r i z o n t a l l y t o t h e r i g h t The velocity after moving 80.0 m up the incline is given by ( 29 ( 29 ( 29 2 2 = 2 = 0+2 0.390 m s 80.0 m80....
View
Full
Document
This note was uploaded on 06/27/2008 for the course PHYS 221 taught by Professor Ezell during the Summer '08 term at Campbell.
 Summer '08
 Ezell
 Work

Click to edit the document details