Phys 221 2nd Homework Solutions

# Phys 221 2nd Homework Solutions - C HAPTER 4 4.1 (a) ( 29 (...

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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 3.0-kg 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 2.0-kg block is the force exerted on it by the 1.0-kg 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....
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## This note was uploaded on 06/27/2008 for the course PHYS 221 taught by Professor Ezell during the Summer '08 term at Campbell.

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Phys 221 2nd Homework Solutions - C HAPTER 4 4.1 (a) ( 29 (...

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