Answers to the Conceptual Questions


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1 Chpter 3 Explining Motion 41 Physics on Your Own If the clss is not too lrge, tke them into freight elevtor to perform this exercise. This simple exercise is importnt if you re going to cover inertil forces lter. Sprott Contins demonstrtion (in Prt of rockets being lunched. Be sure to view this one before showing it: Depending on the level of your clss you my wnt to either omit it or sve it until Chpter 6 on Momentum. Video Encyclopedi #18 Rection Gliders #0 Cr on Rolling Bord #1 Fn Cr with Sil Computer Animtions Active Figure Animtions re vilble on the Multimedi Mnger Instructor s Resource CD. They re orgnized by textbook chpter, nd ech nimtion comes within shell tht provides informtion on how to use the nimtion, explortion ctivities, nd short quiz. Answers to the Conceptul Questions 1. The sncks re lso moving t 800 km/h west nd therefore pper to fll stright down.. The keys re lso moving sidewys t 400 m/s nd, by Newton s first lw, they keep moving sidewys. 3. We conclude tht there must be n unblnced force (tion opposing the cr's motion. 4. There must be n unblnced force opposing its motion. We conclude tht the surfce exerts tionl force on the book. 5. Becuse the velocity does not chnge, the force on ech cr must dd to zero. Therefore the two forces re the sme. 6. Becuse the velocity is constnt, the forces must dd to zero. The engine provides force equl to the sum of ll the opposing forces such s tion nd ir resistnce. 7. Becuse of its inerti. When the cr ccelertes bckwrd, the tssel tends to remin t its originl speed. 8. Becuse of its inerti, the wter does not move with the fur. 9. Becuse of your inerti. When the cr stops suddenly, your hed (nd body stys t its originl speed until your hed hits the windshield. 10. The hedrests re most effective in rer end collision. The hedrests mke the hed move forwrd with the rest of the body. 11. The lrge inerti of the nvil mens tht it does not move much when the hmmer hits. 1. The lrge cutting bord hs lrge inerti nd therefore smll ccelertions. 13. No, inerti refers more generlly to how hrd it is to chnge n object s motion, whether it is speeding up or slowing down. 14. If you pply the sme force to ech one nd they hve the sme ccelertion, they hve the sme inerti. 15. No, the force is vector sum nd will not necessrily point in the direction of ny of the individul forces. 16. The force is not the dominnt force; it is the vector sum of ll the forces. 17. If the forces ct in opposite directions, the force is 50 newtons. If they ct in the sme direction, the force is 130 N. 18. Yes, the force resulting from these two forces cn hve ny vlue between zero nd 1400 newtons. 19. Initilly, the force ws zero. Incresing the pplied force mkes the force nonzero. The wgon now ccelertes (speeds up indefinitely. 0. Initilly, the force ws zero. Removing the books reduces the force of tion mking the force nonzero. The crte now ccelertes (speeds up indefinitely. 1. The ccelertion lwys points in the sme direction s the force. Therefore, it will point due est in
2 4 Chpter 3 Explining Motion both cses.. The ccelertion lwys points in the sme direction s the force. Therefore, it will point upwrd in both cses. 3. You re slowing down while moving downwrd so your ccelertion is up. The force must lso be up. 4. You re slowing down while moving upwrd so your ccelertion is down. The force must lso be down. 5. The ccelertion doubles. 6. The ccelertion is hlved. 7. As ccelertion is inversely proportionl to mss, reducing the mss by fctor of two will double the ccelertion to 4 (meter per second per second. 8. Becuse the cr s re identicl, they hve the sme mss. In order to hve the sme ccelertion, ech cr must experience the sme force. 9. Her weight chnges becuse it is force of ttrction between her nd the Moon wheres her mss, which is mesure of her quntity of mtter, doesn't chnge. 30. Kilogrm is mss unit. 31. Six cns of pop must hve 6 times the weight of single cn. 3. Becuse the weight is proportionl to the ccelertion due to grvity, the weight is onesixth s lrge on the Moon. 33. If the skier is moving to the right, her freebody digrm would be: N snow,skier 34. If the cr is moving to the right, its freebody digrm would be: fwind,cr W Erth,cr Nrod,cr f rod,cr f snow,skier W Erth,skier 35. If there is no ir resistnce, the two would fll with the sme ccelertion. 36. Aristotle would sy tht hevier objects nturlly fll fster; Glileo would ttribute the difference to ir resistnce. 37. The ccelertion is zero becuse the upwrd force of resistnce due to the sop is equl to the downwrd force of grvity. 38. The ccelertion is zero becuse the upwrd force of ir resistnce is equl to the downwrd force of grvity. 39. The ccelertion of both people is zero, so the forces must be zero. 40. The ccelertion of both objects is zero, so the forces must be zero. 41. Newton's first lw is lwys vlid. There must be other forces opposing the tion so tht the force is zero. 4. Newton's second lw is lwys vlid. The tionl force must be dded to the other forces to find the force. 43. The tionl force is equl to 400 newtons becuse the force is zero. 44. The kiic tionl force must be 10 newtons becuse the force is zero when moving t constnt speed. The mximum sttic tionl force is lwys greter thn the sliding tionl force nd so it is possible tht the jr would sty t rest. 45. According to Newton's first lw the force must be zero if the ccelertion is zero. If the tionl force is 50 newtons, the pplied force must lso be 50 newtons (in the opposite direction.
3 Chpter 3 Explining Motion The ccelertion is zero so the force must lso be zero by Newton s first lw. 47. They re equl nd opposite by Newton s third lw. 48. They re equl nd opposite by Newton s third lw. 49. By Newton's third lw, the force exerted on the Erth by the Sun must be equl (nd oppositely directed to the force exerted on the Sun by the Erth. 50. By Newton s third lw, the two forces must be equl in size. 51. Becuse the pple is not ccelerting, the force must be zero. 5. The force must be equl to its weight of 4 newtons. 53. According to Newton's third lw there is rection force cting on the cnnon. 54. According to Newton's third lw there is rection force cting on the tennis rcquet due to its interction with the bll. This force is opposite the rcquet s velocity nd therefore cuses the rcquet to slow down. 55. The forces tht llow you to wlk cross room re the tionl forces of the floor on your feet. 56. The engine provides downwrd force on its pistons tht rottes the driving xle. The xle rottes the wheels tht push bckwrd on the rod's surfce. The rod, in rection to this force, pushes forwrd on the cr. 57. The bll exerts n upwrd force on the Moon equl to the size of the force the Moon exerts on the bll, nmely 40 newtons. 58. Ceiling on string & string on ceiling; string on bll & bll on string; bll on erth & erth on bll. Ech pir is equl by Newton's third lw. The force of the ceiling on the string nd the force of the bll on the string re the sme by Newton's second lw. Likewise for the force of the string on the bll nd the force of Erth on the bll. Therefore, ll of the forces re equl in size. 59. The forces ct on different objects. The tionl force of the ground on the horse's hoofs llows the horse to move the crt. 60. Newton s thirdlw forces lwys ct on different objects, so they cn never cncel. The force of the door on Gry does not ffect the door s motion. 61. The upwrd force of the tble on the cn must be equl nd opposite the weight by Newton s second lw becuse the force is zero. 6. The rection force to the grvittionl force on the mouse is the upwrd grvittionl pull of the mouse on Erth. Answers to the Exercises 1. 8 N + 6 N = 14 N b 8 N 6 N = N 8 N + 6 N = 10 N or use scle drwing c ( (. 1 N + 5 N = 17 N b 1 N 5 N = 7 N in the direction of the 8N force 1 N + 5 N = 13 N or use scle drwing c ( ( N 300 N = 50 N in the direction of the 300N force 3N + 4N = 5N 4. ( (
4 44 Chpter 3 Explining Motion F 1800 N = = 3ms m 600 kg F 4000 N = = ms m 000 kg F 9000 N = = 900,000 m s m 0.01 kg F 6000 N = = 0.1 m s m 60,000 kg 9. F m ( ( 10. F m ( ( 11. = = 60 kg 3 m s = 180 N = = 0 kg 5 m s = 100 N 10 N = = = 1.67 m s m 6kg 1. F m ( ( = = 100 kg 3 m s = 3600 N F 300 N 13. m = = 75 kg 4ms 4 N 14. m = = = 60 kg 0.4 m s 15. F W mg ( ( 16. ( ( = = = 0.5 kg 10 m s = 5 N down F = W = mg = = 30 N = = = 4ms m 80 kg F 555 N = = 3.7 m s m 150 kg 19. F m ( ( 1 kg 10 m s 10 N down; = 10 m/s down = = 4 kg 3 m s = 7 N Therefore, F = F + F = 7 N + 90 N = 16 N pp t 0. F m ( ( = = 10 kg 3 m s = 30 N Therefore, F = F + F = 30 N + 50 N = 80 N pp t 1. F m ( ( = = 40 kg 5 m s = 00 N F = F F = 10 N 00 N = 10 N. F m ( ( = = 60 kg m s = 10 N F = F + W = 10 N N = 70 N scle 3. Frope Terry mterryterry ( ( Chris, = = 75 kg m s = 150 N F 150 N = = = 3ms m 50 kg rope, Chris Chris
5 Chpter 3 Explining Motion F m ( ( F = = 5 kg m s = 50 N, = Fdughter, mother = 50 N by the 3rd lw mother dughter 50 N mother = = 1ms 50 kg Answers to the Problems in Problem Solving F = 50 N + 50 N = 70.7 N northwest 1. ( ( F = 4 N + 9 N 6 N = 5 N 37 south of est. ( ( F = 40 N + 60 N = 7.1 N upwrd to the left t 56.3 to the horizontl 3. ( ( 1 30 N 4. θ = tn = N 5. F m ( ( 6. F m ( ( = = 400 kg 3 m s = 700 N est = = 100 kg m s = 400 N 7. = = = m 90 kg 88 N 9.8 m s to the right 8. = = = m 90 kg 150 N 1.67 m s downwrd 3. N 9. m = = = 8kg 0.4 m s 100 N 10. m = = = 66.7 kg 1.5 m s m = mb r 30 m/s m 3; red one hs smller mss m = = 10 m s = b b r r c 9 m s m t t m c c mt mc t 3 m s b = = = 0. kg = 0.6 kg = F 4N v 8ms = = 4ms t = = m 1kg 4ms = s F m 10 N v 8 m s ms t 60 kg m s = = = = = = v f = 8 m s in the opposite direction 15. The ccelertion is up nd therefore the upwrd force is greter thn 300 N. b The crte is moving downwrd but slowing down, which mens the ccelertion is up. c The ccelertion should be in the sme direction s the force. 4s
6 46 Chpter 3 Explining Motion v 30 m s 0 m s = = = 000 kg = 3000 N t 0 s 16. F m m ( 17. W mg ( ( Mss = 1 kg; = = 1 kg 3.7 m s = 44.4 N 18. Mss = 40 kg; weight = ( 39 N / 6 = 65.3 N 19. Fs mg m m( g ( ( 0. Fs mg m m( g ( ( = + = + = 85 kg 9.8 m s + 1. m s = 935 N = + = + = 70 kg 9.8 m s 0.8 m s = 630 N 1m/s F W Fs m mg g = = = = 600 N = 61. N 9.8 m/s 1. ( 1m/s F Fs W m mg g = = = = 900 N = 91.8 lb 9.8 m/s ( N floor, womn = 700 N f ir, mn = 400 N W Erth, womn = 700 N W Erth, mn = 800 N T cord, light = 14 N N floor, block = 196 N f floor, block = 100 N F hnd, block = 400 N F hnd, light = 30 N W Erth, light = 10 N W Erth, block = 196 N 7. N mg m m( g ( ( = + = + = 100 kg 9.8 m s + 1 m s = 1080 N ( F = F = µ N = N = 43 N pp
7 Chpter 3 Explining Motion N mg m m( g ( ( = = = 100 kg 9.8 m s 1.5 m s = 830 N ( F = F = µ N = N = 33 N pp 9. F = F = 650 N (becuse = pp ( ( mg F 90 kg 9.8 m s 500 N 4.4 m s = = = = m m 90 kg 31. F = m = ( ( = Fir mg F ( ( 3. F = m = ( ( = 70 kg 7 m s 490 N = = 70 kg 9.8 m s 490 N = 196 N 5.4 kg 1.5 m s 8.1 N F = Fpp F = 1 N 8.1 N = 3.9 N 33. F Fpp m ( ( = = 160 N 0 kg 3 m s = 100 N µ = F F 100 N 0.51 N = mg = = 34. pp F = m + F = m + µ mg ( 0 kg( 9.8 m s ( ( ( ( ( = 00 kg 1.5 m/s kg 9.8 m/s = 300 N N = 496 N F pp / dog = 496 N/10 = 49.6 N
46 ROTATIONAL MOTION
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