Unit 2 Review, Pages 208 215 Understanding 28. (a) 2. 3. 4 .
Unit 2 Review, pages 208 215Knowledge1. (b)2. (a)3. (a)4. (b)5. (a)6. (a)7. (b)8. (d)9. (c)10. (a)11. (c)12. (a)13. (a)14. False. Friction is a force that acts to resist themotion or attempted motion of an object.15. False. In the nucleus of atoms, the strongnuclear force holds together the protons andneutrons.16. False. Gravity is the only fundamental forcethat does not have a repulsive effect.17. True18. False. A force of 10 N that acts on a box with a2mass of 2.5 kg will accelerate it at a rate of 4 m/sas long as there are no other forces acting on thebox.19. False. Steven Hawking considers Newton to bethe greatest figure in mathematical physics and thePrincipia is his greatest work.20. True21. False. On the Moon, only your weight willdecrease.22. False. For a flat surface, the static frictionalforce can be determined by multiplying thecoefficient of static friction by the normal force ofan object.23. True24. True25. (a) (iv)(b) (ii)(c) (vi)(d) (v)(e) (i)(f) (iii)26. From least amount of inertia to most amount ofinertia: proton, atom, pencil, baseball, hockeyplayer, motorcycle, car27. From least amount of kinetic friction to mostamount of kinetic friction: ice on ice, Teflon onTeflon, wood on dry snow, rubber on concreteCopyright 2011 Nelson Education Ltd.Understanding28. (a)(b)29. Answers may vary. Sample answer:A system diagram is a simple sketch of the objectsinvolved in the problem and is used to show howeach object is interacting with others. A free-bodydiagram is a basic vector diagram that shows all ofthe forces acting on a single object. It is used todetermine the net force acting on an object and tohelp set up force equations.30. (a)Unit 2: ForcesU2-3
(b)35. (a) According to Newton’s second law, anobject will accelerate in the direction of the netforce with a magnitude given by a (c) Answers may vary. Sample answer:There are no differences between the FBDs inparts (a) and (b). The direction of each arrowdepends on the direction you choose to startdrawing a FBD. If you choose right for thedirection of the pulling force in (a) and you chooseright for the direction of the pushing force in (b),you will end up with the same force diagram forthe two different situations.31. Choose force northward as positive. So forcesouthward is negative.Fnet 42 000 N ( ! 1200 N)Fnet 40 800 NThe net horizontal force on the plane is 40 800 N[northward].32. According to Newton’s first law, when anobject is at rest, the net force on the object must bezero. So the normal force pushing upward on thebook must be equal to the force of gravity pullingdownward for the book to remain at rest.33. Answers may vary. Sample answer:The ball is in a horizontal motion when it is in theballistic cart. Since there are no horizontal forcesacting on the ball, it continues this horizontalmotion following Newton’s first law. When theball is fired upward, it rises and falls due to theshooting force and the force of gravity in thevertical direction, but at the same time continues tomove horizontally. As a result, the ball follows anarched path.34. Choose right as positive. So left is negative.Since the rope does not break and the students arestationary, Fnet 0.!!!!!Fnet FR1 FR2 FL1 FL20 95 N 87 N (!104 N) FL2FL2 !78 N!FL2 78 N [left]Fnetm. If twocars are pushed by an equivalent net force, the onethat has less mass will accelerate faster.(b) As more boxes are added to the cart, the massof the cart will increase. If the person continues topull with a constant force, the acceleration of thecart will decrease.!36. (a) Given: m 68 kg; a 2.4 m/s2 [forwards]!Required: FnetAnalysis: Use the equation Fnet ma to calculateFnet. Choose forwards as positive. So backwards isnegative.Solution:Fnet ma (68 kg)( 2.4 m/s 2 )Fnet 160 NStatement: The net force acting on the sprinter is160 N [forwards].!(b) Given: m 0.425 kg; a 9.8 m/s2 [down]!Required: FnetAnalysis: Use the equation Fnet ma to calculateFnet. Choose up as positive. So down is negative.Solution:Fnet ma (0.425 kg)(-9.8 m/s 2 )Fnet !4.2 NStatement: The net force acting on the baseball is4.2 N [down].!37. (a) Given: m 2200 kg; Fnet 4500 N [N]!Required: aAnalysis: Use the equation a Fnetmto calculate a.Choose north as positive. So south is negative.Solution:Fnetm 4500 N 2200 kga a 2.0 m/s 2!a 2.0 m/s 2 [N]Statement: The acceleration of the car is22.0 m/s [N].The fourth student on the left is pulling with aforce of 78 N [left].Copyright 2011 Nelson Education Ltd.Unit 2: ForcesU2-4
(b) Given: m 71.2 kg; Fnet 245 N [up]Required: aAnalysis: Use the equation a Fnetmto calculate a.Choose up as positive. So down is negative.Solution:Fa netm 245 N 71.2 kga 3.44 m/s 2!a 3.44 m/s 2 [up]Fnet Fg ! FTm2 a m2 g ! FT (Equation 2)Add the equations to solve for a.m1a m2 a FT m2 g ! FT(m1 m2 )a m2 gm2 ga m1 m2 Statement: The acceleration of the skydiver is23.44 m/s [up].38. Given: m 0.175 kg; a 1.3 m/s2!Required: FnetAnalysis: The frictional force is the net forceacting on the puck. Use the equation Fnet ma tocalculate Fnet. Choose forwards as positive. Sobackwards is negative. Since the puck is slowingdown, the acceleration should be negative(backwards).Solution:Fnet ma (0.175 kg)( ! 1.3 m/s 2 )Fnet !0.23 N!Required: aFnetmto calculate a.Choose north as positive. So south is negative.Solution:Fnetm!8.0 " 10 3 N 0.145 kga a !5.5 " 10 4 m/s 2!a 5.5 " 10 4 m/s 2 [S]Statement: The acceleration of the ball is425.5 10 m/s [S].40. Choose right as positive. So left is negative.(a) For the cart,Fnet FTm1a FT (Equation 1)Copyright 2011 Nelson Education Ltd.(0.20 kg)( 9.8 m/s 2 )0.4 kg 0.20 kga 3.3 m/s 2The acceleration of the cart is 3.3 m/s2 [right].(b) For the cart, Fnet FT ! Ffm1a FT ! Ff (equation 1)For the hanging object, Fnet Fg ! FTm2 a m2 g ! FT (equation 2)Add the equations to solve for a.m1a m2 a FT ! Ff m2 g ! FT(m1 m2 )a m2 g ! Ffa Statement: The frictional force acting on the puckis 0.23 N [backwards].!39. Given: m 0.145 kg; Fnet 8.0 103 N [S]Analysis: Use the equation a For the hanging object,m2 g ! Ffm1 m2(0.20 kg)( 9.8 m/s 2 ) ! 0.10 N0.4 kg 0.20 kga 3.1 m/s 2The acceleration of the cart is 3.1 m/s2 [right].(c) Use equation 2 in part (b).m2 a m2 g ! FTFT m2 g ! m2 a m2 (g ! a) (0.20 kg)(9.8 m/s 2 ! 3.1 m/s 2 )FT 1.3 NThe magnitude of the tension in the string is 1.3 N.41. (a) The force of the student pushing on theskateboard causes it to accelerate in that direction.According to Newton’s third law, the skateboardexerts a reaction force of equal magnitude pushingback on the student, causing him to accelerateaway from the skateboard.(b) When the person leans against the wall with acertain amount of force, the wall exerts a reactionforce of equal magnitude but in opposite directionon the person. The net force acting on the person iszero and the wall is anchored to the ground so theyboth remain stationary.Unit 2: ForcesU2-5
(c) As the ball rolls forwards and hits the group ofballs with a force, the group of balls exerts areaction force of equal magnitude but in oppositedirection on the ball, causing it to roll backwards.42. (a) Given: m 80 kg; Fa 68 N; Ff 30 NRequired: a!Analysis: Choose forwards as positive. Sobackwards is negative. Use the equationFnet Fa Ff to find a.Solution:Fnet Fa Ffma Fa Ff(80 kg)a 68 N (!30 N)38 Na 80 kga 0.48 m/s 2!a 0.48 m/s 2 [forward]Statement: The acceleration of the player is20.48 m/s [forwards], off the boards.(b) The boards do not move because they aremassive and anchored to the ground. Sincea Fnetm, for a large value of m, the value a will bevery small. The force that the player pushes with isso small compared to the mass of the boards thattheir acceleration is not noticeable.43. (a) Given: Fnet 0 N; m 0.50 kg,2g –9.8 m/sRequired: FTAnalysis: . The forces on the weight are thetension pulling it upward and the gravity pulling itdownward. Add all the vertical forces:Fnet FT Fg . Choose up as positive. So down isnegative. Since the helicopter is stationary,Fnet 0.Solution:Fnet FT Fg0 FT mgFT !mg !(0.50 kg)(!9.8 m/s 2 )FT 4.9 NStatement: The tension in the string is 4.9 N.2(b) Given: m 0.50 kg; g –9.8 m/s ;2a 0.80 m/sRequired: FTAnalysis: In this situation, Fnet ma . Choose upas positive. So down is negative.Copyright 2011 Nelson Education Ltd.Solution:Fnet FT Fgma FT mgFT ma ! mg (0.50 kg)( 0.80 m/s 2 ) ! (0.50 kg)(!9.8 m/s 2 )FT 5.3 NStatement: The tension in the string is 5.3 N.2(c) Given: m 0.50 kg; g –9.8 m/s ;2a –0.92 m/sRequired: FTAnalysis: In this situation, Fnet ma . Choose upas positive. So down is negative.Solution:Fnet FT Fgma FT mgFT ma ! mg (0.50 kg)(!0.92 m/s 2 ) ! (0.50 kg)(!9.8 m/s 2 )FT 4.4 NStatement: The tension in the string is 4.4 N.44. (a) The force of gravity and air resistance arethe forces acting on the brick as it falls.(b) In the instant the brick is initially dropped, theforce of gravity acts more strongly.(c) As the brick reaches its terminal speed, theforce of gravity and air resistance are equal inmagnitude and acting in opposite directions.2(d) Given: m 0.22 kg; g –9.8 m/sRequired: FairAnalysis: Since the net force on the brick is zero,at terminal speed, Fair Fg.Solution:Fair Fg mg (0.22 kg)(9.8 m/s 2 )Fair 2.2 NStatement: The force of air resistance is 2.2 N.45. (a) Answers may vary. Sample answer:One way to measure gravitational field strength isto measure the weight of a known mass and dividethe weight by the mass. A second way is tomeasure the acceleration of an object falling froma known height.Unit 2: ForcesU2-6
(b) Use the equation Fg mg to determine g.Fg mgg Fgm24.475 N 2.50 kgg 9.79 N/kgThe gravitational field strength acting on the objectis 9.79 N/kg.(c) This altitude is probably above sea levelbecause the gravitational field strength is lowerthan that at sea level.46. (a) Mass is the quantity of matter in an object.Weight is a measure of the force of gravity actingon an object.(b) It is possible to change the mass of an objectby adding or removing material from that object.(c) To change the weight of an object, but not themass, move the object to a location with a differentgravitational field strength.(d) The magnitudes of an object’s mass and weightwill be equal when the gravitational field strengthequals 1 N/kg.47. (a) If both cylinders hit the ground at the sametime, the cylinder with the greater mass has alarger cross-sectional area.(b) If they have the same cross-sectional area, thecylinder with the greater mass will hit the groundfirst.(c) If both cylinders were dropped in a vacuum,there is no air resistance and they would hit theground at the same time.48. Draw a FBD of the chandelier.49. (a) The forces acting on the rock are the forceof gravity, the normal force, the applied force, andstatic friction.(b)2(c) Given: m 210 kg; g –9.8 m/s!Required: FNAnalysis: Consider the vertical forces acting onthe rock. Choose up as positive. So down isnegative.Solution: Since the net force on the rock is zero,FN Fg 0FN !Fg !mg !(210 kg)(!9.8 m/s 2 )FN 2100 NStatement: The normal force acting on the rock is2100 N [up].50. (a) Given: m 5.0 kg; FSmax 29.89 NRequired: µSAnalysis: Use the equation µS FSmaxFNto find µS.Solution:µS FSmaxFNFSmaxmg28.89 N(5.0 kg)(9.8 m/s 2 )µS 0.61 The chandelier is at rest. So the net force on thechandelier is zero.Choose up as positive. So down is negative.FN Fg Fa 0FN (3.2 kg)(!9.8 m/s 2 ) 53 N 0FN !22 NThe normal force acting on the chandelier is22 N [down].Copyright 2011 Nelson Education Ltd.Statement: The coefficient of static friction is0.61.(b) The two materials involved in this interactionare leather and oak.51. (a) Given: FN 52 N; µS 0.1Required: FSmaxAnalysis: The initial force required is themaximum static friction. Use the equationFSmax µS FN to calculate FSmax .Unit 2: ForcesU2-7
Solution:Analysis: At the time the runner starts from rest,Fnet FSmax . Use the equations FSmax µS FN (0.1)(52 N)FSmax 5.2 NFnet ma to find a. Choose forwards as positive.FSmax µS FNStatement: The magnitude of the initial horizontalforce required is 5.2 N.(b) Given: FN 52 N; µK 0.03Required: FKAnalysis: The force required is the kinetic friction.Use the equation FK µ K FN to calculate FK.Solution:FK µK FN (0.03)(52 N)FK 1.6 NStatement: The force required to maintain the hutsliding at a constant speed is 1.6 N.52. (a) Given: m 12 kg; FSmax 47 NRequired: µSAnalysis: Use the equation µ S FSmaxFNto find µS.Solution:µS FSmaxFNFSmaxmg47 N(12 kg)(9.8 m/s 2 )µS 0.40 Statement: The coefficient of static friction is0.40.(b) Given: m 12 kg; Fa 47 N; a 1.1 m/s2Required: µKAnalysis: Use the equations Fnet FK Fa andFK µK FN to find µK. Choose forwards as positive.So backwards is negative.Solution:Fnet FK Fama ! µK FN 47 Nma ! µK mg 47 N(12 kg)(1.1 m/s 2 ) ! µK (12 kg)(9.8 m/s 2 ) 47 NµK 0.29Statement: The coefficient of kinetic friction is0.29.53. Given: m 72 kg; µS 0.79!Required: aCopyright 2011 Nelson Education Ltd.So backwards is negative.Solution:FSmax µS FNFnet µS mgma µS mga µS g (0.79)(9.8 m/s 2 )a 7.7 m/s 2!a 7.7 m/s 2 [forward]Statement: The maximum acceleration of the2runner is 7.7 m/s [forwards].54. Given: m 6.2 kg; a 0.50 m/s2; µK 0.24Required: FaAnalysis: Use the equations Fnet FK Fa andFK µK FN to find Fa.Solution: Choose forwards as positive. Sobackwards is negative.Fnet FK Fama ! µK FN FaFa ma µK FN ma µK mg m(a µK g) (6.2 kg)[0.50 m/s 2 (0.24)(9.8 m/s 2 )]Fa 18 NStatement: The magnitude of the force that theman pulls is 18 N.55. Given: m 22 kg; vi 5.2 m/s; µK 0.44Required: ΔdAnalysis: First use the equations Fnet FK andFK µK FN to find the acceleration of the box. Thenuse the equation v f 2 v i 2 2a!d to calculate thedistance travelled. Since the box comes to rest,vf 0 m/s.Solution:Fnet FKma µK FNma µK mga µK g (0.44)(9.8 m/s 2 )]a 4.31 m/s 2 (one extra digit carried)Unit 2: ForcesU2-8
Now calculate the distance travelled. Chooseforwards as positive. So backwards is negative.vf 2 vi 2 2a!d0 vi 2 2a!dvi 2 "2a!dvi 2"2a( 5.2 m/s)2 "2("4.31 m/s 2 )!d 3.1 m!d Statement: The box slides 3.1 m before it comesto rest.56. Answers may vary. Sample answers:(a) One safety feature is the seat belt. A seat belt isdesigned to keep a person in an optimal crashposition in a car seat to prevent the person fromaccelerating toward the dashboard or windshieldduring a car accident. Another safety feature is thecrumple zone. The crumple zones of a car aredesigned to crush during an accident. The crushingaction increases the time it takes to stop the carduring a collision, reducing both the accelerationand the forces acting on the people inside the car.(b) During a crash test, an engineer uses a dummyto accurately simulate what will happen to a personduring a car accident. The information gatheredincludes the accelerations of different parts of thebody (like the head and torso) usingaccelerometers, how much the chest getscompressed during the crash using a motionsensor, and other forces acting on the body usingload sensors all over the body.(c) In the photograph, the crumple zone in the frontof the car is crushed while the car test dummystays inside in the car seat uninjured.57. Answers may vary. Sample answer:When a person sits, the hip has to support theupper body’s weight and the hip slides quite oftenon different seat surfaces. So the material used inhip replacement must be strong, flexible butsturdy, wear resistant, and durable for the artificialhip to look good in its shape, function properly,and last longer.58. Answers may vary. Students’ answers shouldexplain their findings about snow tires. Samplestudent answer: The tread pattern on snow tires isspecifically designed to dig into the snow andcreate more friction. They are also made from asofter rubber that allows the tires to flex in thewinter and grip the road. Snow tires cannot be lefton all year because the soft rubber will wear a lotCopyright 2011 Nelson Education Ltd.more easily during the summer and ruin theirtraction for the winter.Analysis and Application59. (a) The ball should go up to a height of 25 cmbecause of inertia. The amount of inertia the ballcarries down the first ramp is the same as that itwill carry up the second ramp.(b) After rolling back, the ball should once againroll up to a height of 25 cm.(c) This cannot happen in real world because offriction. It causes the ball to slow down and loseinertia.60. Given: v2 32 m/s; Δt 0.42 s; m 0.22 kgRequired: FnetAnalysis: First use the equation v 2 v1 a!t tocalculate the acceleration. Then use Fnet ma tocalculate the average force on the potatoes.Solution: At the beginning, the velocity of thepotatoes is 0 m/s.v2 v1 a!tv2 a!tva 2!t32 m/s 0.42 sa 76.2 m/s 2 (one extra digit carried)Now calculate the average force.Fnet ma (0.22 kg)(76.2 m/s 2 )Fnet 17 NStatement: The average force the potato gunexerts on the potatoes is 17 N.61. Given: m 71 kg; v1 3.4 m/s; v2 6.7 m/s;Δt 2.8 sRequired: FnetAnalysis: First use the equation v 2 v1 a!t tocalculate the acceleration. Then use Fnet ma tocalculate the net force on the runner.Solution:v2 v1 a!tv " v1a 2!t6.7 m/s " 3.4 m/s 2.8 sa 1.18 m/s 2 (one extra digit carried)Unit 2: ForcesU2-9
Now calculate the average force.Fnet ma (71 kg)(1.18 m/s )2Fnet 84 NStatement: The net force the runner experiences is84 N.62. (a) Given: mT 9.0 kg 3.0 kg 12.0 kg;Ff 18 N 6.0 N 24 N; Fa 30.0 N!Required: aAnalysis: We can treat the two weights as onesingle object. Use the equations Fnet Fa Ff andFnet ma to calculate the acceleration. Chooseforwards as positive. So backwards is negative.Solution:Fnet Fa Ffma Fa Ff(12.0 kg)a 30.0 N (!24 N)a 0.50 m/s 2!a 0.50 m/s 2 [forward]Statement: The acceleration of the weights during2the first 5.0 s is 0.50 m/s [forwards].(b) If the applied force is removed, the onlyhorizontal force acting on the weights is friction,which is in the opposite direction of their motion.So the weight will slow down and stop.(c) Answers may vary. Sample answer:The two weights will move the same distancebecause the only force acting on the weights afterthe initial push is the frictional force, which isproportional to the normal force.Ff µ K FNma µ K mga µK gThe acceleration depends only on the coefficient offriction and the force of gravity, which are thesame for both weights. Thus, both weights slowdown at the same rate and travel the same distance.63. Calculate the net force on the students and useFnet ma to calculate the acceleration. Chooseright as positive. So left is negative.Fnet FR1 FR2 FL1 FL2 55 N 65 N (!58 N) (!70 N)Fnet !8 NThe total mass of the students is:mT 60 kg 62 kg 59 kg 64 kg 245 kgCopyright 2011 Nelson Education Ltd.Now calculate the acceleration.Fnet mT a!8 N (245 kg)a!8 Na 245 kga !0.03 m/s 2The students accelerate to the left with a2magnitude of 0.03 m/s .64. (a) Given: m 220 kg; Δv 5.0 km/h;Δt 8.0 sRequired: Fdog!vto calculate the!tacceleration. Then use Fnet ma to find Fdog.Analysis: First use a Solution: Convert the velocity to SI metric units.!km ! 1 h ! 1 min ! 1000 m 5.0 km/h # 5.0h &% #" 60 min &% #" 60 s &% #" 1 km &%" 1.389 m/s (two extra digits carried)Then calculate the acceleration.!v!t1.389 m/s 8.0 sa a 0.1736 m/s 2 (two extra digits carried)Calculate the average applied force.Fnet ma (220 kg)(0.1736 m/s 2 )Fnet 38.19 N (two extra digits carried)For each dog,38.19 N9 4.2 NFdog FdogStatement: The average force applied by each dogis 4.2 N.(b) The frictional force equals the total forceapplied by the dogs.Given:
(0.425kg)-9.8m/s2) F net !4.2N Statement: The net force acting on the baseball is 4.2 N [down]. 37. (a) Given: m 2200 kg; ! F net 4500 N [N] Required:! a Analysis: aUse the equation m F net to calculate a. Choose north as positive. So south is negative. Solution: a F net m 4500N 2200kg a 2.0m/s2! a 2.0m/s2 [N] Statement: The .
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contents page 2 fuel consumption pages 3-6 fiat 500 pages 7-10 fiat 500c pages 11-13 fiat 500 dolcevita pages 14-16 fiat 500 120th anniversary pages 17-21 fiat 500x pages 22-24 fiat 500x 120th anniversary pages 25-27 fiat 500x s-design pages 28-31 fiat 500l pages 32-35 fiat 500l 120th anniversary pages 36-39 tipo hatchback pages 40-43 tipo station wagon pages 44-47 tipo s-design
Pipe Fittings. pages 32-37. Unpolished Fittings. pages 74-80. Polished Fittings. pages 64-73. European Fittings. pages 81-85. Filters / Strainers. pages 111-117. Custom Fabrications. pages 109. Swivels. pages 140-141. Instrumentation. pages 118-133. Air Fittings. pages 162-170. High Press. Quick Disc. pages 171-179. Check Valves. pages 214-222 .
