Forces inOne Dimension - Staff

CHAPTER

4 Forces inOne Dimension— — — —

J -,-f.-”

Practice Problems4.1 Force and Motion

pages 87—95page 89For each of the following situations, specify the system

and draw a motion diagram and a free-body dia

gram. Label all forces with their agents, and indicate

the direction of the acceleration and of the net force.

Draw vectors of appropriate lengths.

1. A flowerpot falls freely from a windowsill.

(Ignore any forces due to air resistance.)

+y

a Fnet{

FEarth s mass on flowerpot

2. A sky diver falls downward through the air

at constant velocity. (The air exerts anupward force on the person.)

3. A cable pulls a crate at a constant speedacross a horizontal surface. The surface provides a force that resists the crate’s motion.

Ffriion 4 • on crateon crate Fnet = 0

4. A rope lifts a bucket at a constant speed.(Ignore air resistance.)

-l-y

Sys$m

5. A rope lowers a bucket at a constant speed.

(Ignore air resistance.)

— — —— System

— — %

V V V V

Systemji

I

______

II “.l

a=0

vt

vj

vj+y

— — — — —— v 4 Fair resistance

on diver

vTa=OfFnetO

SystI — — — — — — VT‘FEarths mass

on diver

a=

Frope on bucket

0 FnetO

FEahs mass on bucket

System

v

v

vJ

v

a=

Frope on bucket

0 Fnet=O

FEah’s mass on bucket

Physics: Principles and Problems Solutions Manual 61

Chapter 4 continued

page 936. Two horizontal forces, 225 N and 165 N,

are exerted on a canoe. If these forces areapplied in the same direction, find the nethorizontal force on the canoe.

Fnet = 225 N + 165 N = 390X102N

in the direction of the two forces

7 If the same two forces as in the previousproblem are exerted on the canoe in opposite directions, what is the net horizontalforce on the canoe? Be sure to indicate thedirection of the net force.

Fnet 225 N —165 N = 6.0X101 N

in the direction of the larger force

8. Three confused sleigh dogs are trying topull a sled across the Alaskan snow. Alutiapulls east with a force of 35 N, Seward alsopulls east but with a force of 42 N, and bigKodiak pulls west with a force of 53 N.What is the net force on the sled?Identify east as positive and the sled asthe system.

Fnet = FAlutla on sled + Fseward on sled —

Fkodlak on sled

= 35 N + 42 N — 53 N= 24 N

Fnet = 24 N east

Section Review4.1 Force and Motion

pages 87—95page 95

9. Force Identify each of the following aseither a, b, or C: weight, mass, inertia, thepush of a hand, thrust, resistance, air resistance, spring force, and acceleration.a. a contact force

b. a field force

C. not a force

weight (b), mass (C), inertia (C), push ofa hand (a), thrust (a), resistance (a),air resistance (a), spring force (a),acceleration (C)

10. Inertia Can you feel the inertia of a pencil?Of a book? If you can, describe how.Yes, you can feel the inertia of eitherobject by using your hand to give eitherobject an acceleration; that is, try tochange the objects velocity.

11. Free-Body Diagram Draw a free-body diagram of a bag of sugar being lifted by yourhand at a constant speed. Specifically identifythe system. Label all forces with their agentsand make the arrows the correct lengths.

Fhand on bag

12. Direction of Velodty If you push a bookin the forward direction, does this mean itsvelocity has to be forward?No, it could be moving backward andyou would be reducing that velocity.

13. Free-Body Diagram Draw a free-body diagram of a water bucket being lifted by a ropeat a decreasing speed. Specifically identifythe system. Label all forces with their agentsand make the arrows the correct lengths.

“1I Sugar

LII,I,

‘ k-

1 —————

System

a=O

FEah’s mass on bag

II

Frope on bucket

FEahs mass on bucketSystem

62 Solutions Manual Physics: Principles and Problems

Chapter 4 continued

14. Critical Thinking A force of 1 N is theonly force exerted on a block, and the acceleration of the block is measured. When thesame force is the only force exerted on asecond block, the acceleration is three timesas large. What can you condude about themasses of the two blocks?

Because m = F/a and the forces are thesame, the mass of the second block isone-third the mass of the first block.

Practice Problems4.2 Using Newton’s Laws

pages 96—101

You place a watermelon on a spring scale atthe supermarket. If the mass of the watermelon is 4.0 kg what is the reading on the scale?

The scale reads the weight of the watermelon:

Fg = mg = (4.0 kg)(9.80 mis2) = 39 N

16. Kamaria is learning how to ice-skate. Shewants her mother to pull her along so thatshe has an acceleration of 0.80 rn/s2. IfKamaria’s mass is 27.2 kg, with what forcedoes her mother need to pull her? (Neglectany resistance between the ice andKamaria’s skates.)

Fnet = ma = (27.2 kg)(0.80 mis2) = 22 N

17. Tam and Reiko simultaneously grab a0.75-kg piece of rope and begin tugging onit in opposite directions. If Taru pulls witha force of 16.0 N and the rope acceleratesaway from her at 1.25 rn/s2,with whatforce is Reiko pulling?

Identify Reiko’s direction as positiveand the rope as the system.

Fnet = FRleko on rope — FTaru on rope = ma

FRIeko on rope = ma + FTa on rope

= (0.75 kg)(1.25 m/s2)+

18. In Figure 4-8, the block has a mass of1.2 kg and the sphere has a mass of 3.0 kg.What are the readings on the two scales?(Neglect the masses of the scales.)

Bottom scale: Identify the sphere as thesystem and up as positive.

Fnet = Fscaie on sphere

FEaI.Jls mass on sphere = ma = 0

Fscaie on sphere = FEaIihs mass on sphere

= m5pg

= (3.0 kg)(9.80 mis2)

= 29 N

Top scale: identify the block as the system and up as positive.

Fnet = Ftop scale on block

Fbouom scale on block —

FEafths mass on block

Ftop scale on block = Fbollom scale on block +

FEahs mass on block

= Fbottom scale on block +

= 29 N + (1.2 kg)(9.80 m/s2)

= 41 N

1

page 9715.

a

E8

II8

• Figure 4-8

= ma = 0

16.0 N

= 17 N


Chapter 4 continued

page 100 c. It speeds up at 2.00 rn/s2 while moving19. On Earth, a scale shows that you weigh downward.

585 N.Accelerating downward,

a. What is your mass?so a = —2.00 mIs2The scale reads 585 N. Since there is

no acceleration, your weight equals Fscaie = Fnet + F9

the downward force of gravity: = ma + mgF9mg

=m(a+g)F9 — 585 N

= (75.0 kg)(—2.00 rn/s2 += 59.7 kgso m= — 9.80

9.80 mIs2)b What would the scale read on the Moon= 585 N(g = 1.60 m/s2)?

d. It moves downward at constant speed.On the moon, g changes:Constant speed, soFg = mgMOOfla 0 and Fnet = 0

= (59.7 kg)(1 .60 mIs2)Fscaie = Fg = mg

= 95.5 N= (75.0 kg)(9.80 mis2)

20. Use the results from Example Problem 2 to = 735 Nanswer questions about a scale in an eleva- e. It slows to a stop at a constant magnitor on Earth. What force would be exerted tude of acceleration.by the scale on a person in the following

Constant acceleration = a, thoughsituations?the sign of a depends on the direca. The elevator moves at constant speed. tlon of the motion that is ending.

Constant speed, so a = 0 andFscaie = Fnet ÷ Fg

Fnet=O.=ma+mg

Fscaie = F9= (75.0 kg)(a) ÷ 0

8= mg = (75.0 kg)(9.80 mIs2)

(75.0 kg)(9.80 mis2)a= 735 N

= (75.0 kg)(a) + 735 Nb. It slows at 2.00 rn/s2 while moving

upward.

3.Slowing while moving upward, so9,

a —2.00 m/s2

aFscaie = Fnet + F9

=ma+mg3=m(a+g)

= (75.0 kg)(—2.00 m/s2 +,

9.80 m/s2)

=585N


Chapter 4 continued

Section Review4.2 Using Newton’s Laws

pages 96—101

Lunar Gravity Compare the force holdinga 10.0-kg rock on Earth and on the Moon.The acceleration due to gravity on theMoon is 1.62 rn/s2.

To hold the rock on Earth:

Fnet = FEaI.tJl on rock — Fhold on rock = 0

Fhold on rock = FEah on rock =

= (10.0 kg)(9.80 mis2)

= 98.0 N

To hold the rock on the Moon:

Fnet = FMoon on rock Fhold on rock = 0

Fhold on rock = FMOOfl on rock mg00fl

= (10.0 kg)(1 .62 mIs2)

= 16.2 N

22. Real and Apparent Weight You take a ridein a fast elevator to the top of a tall buildingand ride back down while standing on abathroom scale. During which. parts of theride will your apparent and real weights bethe same? During which parts will yourapparent weight be less than your realweight? More than your real weight? Sketchfree-body diagrams to support your answers.

Apparent weight and real weight are thesame when you are traveling either up ordown at a constant velocity. Apparentweight is less than real weight when theelevator Is slowing while rising or speeding up while descending. Apparentweight is greater when speeding upwhile rising or slowing while goingdown.

Fg

apparent weight> real weight

23. Acceleration Tecle, with a mass of 65.0 kg,is standing by the boards at the side of anice-skating rink. He pushes off the boardswith a force of 9.0 N. What is his resultingacceleration?

Identity Tecle as the system and thedirection away from the boards aspositive. The ice can be treated as aresistance-free surface.

Fnet = Fboards on Tecle ma

a = Fboardn Tecle

9.0 N— 65.0 kg

1= 0.14 mIs2 away from the boards

page 10121.

Constant Velocity

Fscaie

F9

apparent weight = real weight

Slowing While Rising!Speeding Up While Descending

Fscaie

Fg

apparent weight < real weight

Speeding Up While Rising!Slowing While Descending

B8z

00

I

II


a=

— 9.3N—9.BON1 kg

= —0.5 m/s2

26. Acceleration A sky diver falls at a constantspeed in the spread-eagle position. After heopens his parachute, is the sky diver accelerating? If so, in which direction? Explainyour answer using Newton’s laws.

Yes, for a while the diver is acceleratingupward because there is an additional

Chapter 4 continued

24. Motion of an Elevator You are riding inan elevator holding a spring scale with a1-kg mass suspended from it. You look atthe scale and see that it reads 9.3 N. What,if anything can you conclude about theelevator’s motion at that time?

It the elevator is stationary or movingat a constant velocity, the scale shouldread 9.80 N. Because the scale reads alighter weight, the elevator must beaccelerating downward. To find theexact acceleration: identify up as positive and the 1-kg mass as the system.

Fnet = Fscaie on 1 kg —

FEahs mass on 1 kg = ma

Fscaie on 1 kg — FEarths mass on 1 kam

= 0.5 rn/s2 downward

25. Mass Marcos is playing tug-of-war with hiscat using a stuffed toy. At one instant duringthe game, Marcos pulls on the toy with aforce of 22 N, the cat pulls in the oppositedirection with a force of 19.5 N, and the toyexperiences an acceleration of 6.25 rn/s2.What is the mass of the toy?

Identify the toy as the system and thedirection toward his cat as the positivedirection.

Fnet = FMarcos on toy — Fcat on toy = ma

m= FMarcos on toy — F on toy

22 N — 19.5 N= 6.25 m/s2

= 0.40 kg

upward force due to air resistance on theparachute. The upward accelerationcauses the driver’s downward velocity todecrease. Newton’s second law says thata net force in a certain direction willresult in an acceleration In that direction(F = ma).

27. Critical Thinking You have a job at a meatwarehouse loading inventory onto trucks forshipment to grocery stores. Each truck has aweight limit of 10,000 N of cargo. You pusheach crate of meat along a low-resistanceroller belt to a scale and weigh it beforemoving it onto the truck. However, rightafter you weigh a 1000-N crate, the scalebreaks. Describe a way in which you couldapply Newton’s laws to figure out theapproximate masses of the remaining crates.

Answers may vary. One possible answeris the following: You can neglect resistance If you do all your maneuvering onthe roller belt. Because you know theweight of the 1000 N crate, you can useit as your standard. Pull on the 1000 Ncrate with a particular force for 1 s,estimate its velocity, and calculate theacceleration that your force gave to It.Next, pull on a crate of unknown masswith as close to the same force as youcan for 1 s. Estimate the crate’s velocityand calculate the acceleration yourforce gave to it. The force you pulledwith on each crate will be the net forcein each case.

Fnet 1000-N crate = Fnet unknown crate

(1000 N)(al000.Ncrate) = (muflk)(aunk)

— (1000 N)(al,N crate)mUflk—

aUflk

$1

(


Chapter 4 continued

Practice Problems4.3 Interaction Forces

pages 102—107page 10428. You lift a relatively light bowling ball with

your hand, accelerating it upward. What arethe forces on the ball? What forces does theball exert? What objects are these forcesexerted on?

The forces on the ball are the force ofyour hand and the gravitational force ofEarth’s mass. The ball exerts a force onyour hand and a gravitational force onEarth. All these forces are exerted onyour hand, on the ball, or on Earth.

29. A brick falls from a construction scaffold.Identify any forces acting on the brick. Alsoidentify any forces that the brick exerts andthe objects on which these forces are exerted.(Air resistance may be ignored.)

The only force acting on the brick is thegravitational attraction of Earth’s mass.The brick exerts an equal and oppositeforce on Earth.

30. You toss a ball up in the air. Draw a free-body diagram for the ball while it is stillmoving upward. Identify any forces actingon the ball. Also identify any forces that theball exerts and the objects on which theseforces are exerted.

I ma on ball

The only force acting on the ball is theforce of Earth’s mass on the ball, whenignoring air resistance. The ball exertsan equal and opposite force on Earth.

31. A suitcase sits on a stationary airport luggagecart, as in Figure 4-13. Draw a free-bodydiagram for each object and specifically indicate any interaction pairs between the two.

Fsurtace on cart

FEarth’smasson cart

on cart

You are helping to repair a roof by loadingequipment into a bucket that workers hoistto the rooftop. If the rope is guaranteed notto break as long as the tension does notexceed 450 N and you fill the bucket until ithas a mass of 42 kg, what is the greatestacceleration that the workers can give thebucket as they pull it to the roof?

Identify the bucket as the system andup as positive.

Fnet = ope on bucket —

= ma

a= Frope on bucket — FEarthe mass on bucket

m

= Frope on bucket — mgm

= 450 N — (42 kg)(9.80 mis2)42kg

= 0.91 m/s2

• Figure 4-13

CartSuitcase

Fcart on suitcase

FEarths masson suitcase

page 10632.

S

aS8

CC

I

0

8

FEahs mass on bucket


Diego and Mika are trying to fix a tire onDiego’s cai but they are having trouble getting the tire loose. When they pull togethei,Mika with a force of 23 N and Diego with aforce of 31 N, they just barely get the tire tobudge. What is the magnitude of the strengthof the force between the lire and the wheel?

Identify the tire as the system and thedirection of pulling as positive.

Fnet = Fwheei on tire — FMjka on tire —

FDiego on tire

= ma = 0

Fwhl on tire = FMika on tire + FDiego on tire

= 23 N + 31 N

= 54 N

Section Review4.3 Interaction Forces

pages 102—107page 10734. Force Hold a book motionless in your

hand in the air. Identify each force and itsinteraction pair on the book.

The forces on the book are downwardforce of gravity due to the mass ofEarth and the upward force of the hand.The force of the book on Earth and theforce of the book on the hand are theother halves of the interaction pairs.

35. Force Lower the book from problem 34 atinaeasing speed. Do any of the forces or theirinteraction-pair partners change? Explain.

Yes, the force of the hand on the bookbecomes smaller so there is a downward acceleration. The force of thebook also becomes smaller; you canfeel that. The interaction pair partnersremain the same.

36. Tension A block hangs from the ceiling bya massless rope. A second block is attachedto the first block and hangs below it onanother piece of massless rope. If each ofthe two blocks has a mass of 5.0 kg, what isthe tension in each rope?

For the bottom rope with the positivedirection upward:

Fnet = Fbottom rope on bottom biock

FE1h.S mass on bottom block

For the top rope, with the positivedirection upward:

Fnet = Ft0 rope on top block

Fboflom rope on top block —

FEaI.thS mass on top block

37. Tension If the bottom block in problem36 has a mass of 3.0 kg and the tension inthe top rope is 63.0 N, calculate the tension in the bottom rope and the mass ofthe top block.

For the bottom rope with the positivedirection upward:

Fnet = Fboffom rope on bottom block —

FErnIhIS mass on bottom block

Chapter 4 continued

33.

= ma = 0

Fbottom rope on bottom block

= mass on bottom block

=mg

= (5.0 kg)(9.80 m/s2)

= 49 N

= ma = 0

Ft0 rope on top block

= FEa,.ths mass on top block +

Fbottom rope on top block

= mg + Fbottom rope on top block

= (5.0 kg)(9.80 m/s2) + 49 N

= 98 N

(

4

= ma = 0

Fbottom rope on bottom block

= FEahJs mass on bottom block

= (3.0 kg)(9.80 mis2)

= 29 N


Chapter 4 continued

For the top mass with the positivedirection upward:

Fnet = Ftop rope on top block —


FEahs mass on top block

63.0 N — 29 N= 9.80 mIs2

= 3.5 kg

38. Normal Force Poloma hands a 13-kg boxto 61-kg Stephanie, who stands on a platform. What is the normal force exerted bythe platform on Stephanie?

Identify Stephanie as the system andpositive to be upward.

Fnet Fplafform on Stephanie —

Fbox on Stephanie —

FEaI.ths mass on Stephanie

Fplafform on Stephanie

= Fbox on Stephanie +

FEahs mass on Stephanie

= m0g+ mstephaflleg

= (13 kg)(9.80 mis2) + (61 kg)(9.80 mis2)

= 7.3X102N

39. CritIcal Thinldng A curtain prevents twotug-of-war teams from seeing each other.One team ties its end of the rope to a tree.If the other team pulls with a 500-N force,what is the tension? Explain.

The tension would be 500 N. The ropeis in equilibrium, so the net force on Itis zero. The team and the tree exertequal forces in opposite directions.

Mastering Conceptspage 11241. A physics book is motionless on the top of

a table. If you give it a hard push with yourhand, it slides across the table and slowlycomes to a stop. Use Newton’s laws toanswer the following questions. (4.1)

a. Why does the book remain motionlessbefore the force of your hand is applied?

An object at rest tends to stay atrest if no outside force acts on it.

b. Why does the book begin to move whenyour hand pushes hard enough on it?

The force from your hand is greaterthan any opposing force such as friction. With a net force on it, the bookslides in the direction of the net force.

c. Under what conditions would the bookremain in motion at a constant speed?

The book would remain In motion ifthe net force acting on It Is zero.

42. Cycling Why do you have to push harderon the pedals of a single-speed bicyde tostart it moving than to keep it moving at aconstant velocity? (4.1)

A large force is required to accelerate themass of the bicycle and rider. Once thedesired constant velocity is reached, amuch smaller force is sufficient to overcome the ever-present frictional forces.

Chapter AssessmentConcept Mappingpage 11240. Complete the following concept map using

the following term and symbols: normal,

I J FTFg.

Ii= ma = 0

FEaI.ths mass on top block = mg

= Ftop rope on top block


m=

op rope on top block — Fboftom rope on top blockg


Chapter 4 continued

43. Suppose that the acceleration of an object iszero. Does this mean that there are noforces acting on it? Give an example supporting your answer. (4.2)

No, it only means the forces acting on itare balanced and the net force is zero.For example, a book on a table is notmoving but the force of gravity pullsdown on it and the normal force of thetable pushes up on it and these forcesare balanced.

44. Basketball When a basketball player dribbles a ball, it falls to the floor and bouncesup. Is a force required to make it bounce?Why? If a force is needed, what is the agentinvolved? (4.2)

Yes, its velocity changed direction;thus, it was accelerated and a force isrequired to accelerate the basketball.The agent is the floor.

45. Before a sky diver opens her parachute, shemay be falling at a velocity higher than theterminal velocity that she will have after theparachute opens. (4.2)

a. Describe what happens to her velocityas she opens the parachute.

Because the force of air resistancesuddenly becomes larger, thevelocity of the diver drops suddenly.

b. Describe the sky diver’s velocity fromwhen her parachute has been open for atime until she is about to land.

The force of air resistance and thegravitational force are equal. Their’sum is zero, so there is no longer any‘acceleration. The sky diver continuesdownward at a constant velocity.

46. If your textbook is in equilibrium, what canyou say about the forces acting on it? (4.2)

If the book is in equilibrium, the netforce is zero. The forces acting on thebook are balanced.

47. A rock is dropped from a bridge into a valley. Earth pulls on the rock and acceleratesit downward. According to Newton’s thirdlaw, the rock must also be pulling on Earth,yet Earth does not seem to accelerate.Explain. (4.3)

The rock does pull on Earth, but Earth’senormous mass would undergo only aminute acceleration as a result of sucha small force. This acceleration wouldgo undetected.

48. Ramon pushes on a bed that has beenpushed against a wall, as in Figure 4-17.Draw a free-body diagram for the bed andidentify all the forces acting on it. Make aseparate list of all the forces that the bedapplies to other objects. (4.3)

Fflooron bed

• Figure 4-17

Fwaii on bed FRamon on bed

Fg

Forces that bed applies to otherobjects:

Fbed on Ramon’ Fbed on Earth’ Fbed on floor’

Fbed on wall

c)

0

z


Chapter 4 continued

49. Figure 4-18 shows a block in four differentsituations. Rank them according to the magnitude of the normal force between theblock and the surface, greatest to least.Specifically indicate any ties. (4.3)

• Figure 4-iS

from left to right: second > fourth>third > first

50. Explain why the tension in a massless ropeis constant throughout it. (4.3)

If you draw a free-body diagram for anypoint on the rope, there will be two tension forces acting in opposite directions. Fnet = — Fdown = ma = 0(because it is massless). Therefore,

= Fdowfl. According to Newton’sthird law, the force that the adjoiningpiece of rope exerts on this point isequal and opposite to the force that thispoint exerts on it, so the force must beconstant throughout.

51. A bird sits on top of a statue of Einstein.Draw free-body diagrams for the bird andthe statue. Specifically indicate any interaction pairs between the two diagrams. (4.3)

52. Baseball A slugger swings his bat and hitsa baseball pitched to him. Draw free-bodydiagrams for the baseball and the bat atthe moment of contact. Specifically indicate any interaction pairs between the twodiagrams. (4.3)

I I

FbaIl on bat Fbafter on bat Fbat on ball

Applying Conceptspages 112—11353. Whiplash If you are in a car that is struck

from behind, you can receive a serious neckinjury called whiplash.

a. Using Newton’s laws, explain whathappens to cause such an injury.The car is suddenly acceleratedforward. The seat accelerates yourbody, but your neck has to accelerate your head. This can hurt yourneck muscles.

b. How does a headrest reduce whiplash?The headrest pushes on your head,accelerating it in the same directionas the car.

54. Space Should astronauts choose pencilswith hard or soft lead for making notes inspace? Explain.

A soft lead pencil would work betterbecause It would require less force tomake a mark on the paper. The magnitude of the interaction force pair couldpush the astronaut away from the paper.

l.

0.E

(3

000

x

00

0

4

Fstatue on bird

Fg, bird

FEah on statue

Fbird on statue

Fg, statue


Chapter 4 continued

55. When you look at the label of the productin Figure 4-19 to get an idea of how muchthe box contains, does it tell you its mass,weight, or both? Would you need to makeany changes to this label to make it correctfor consumption on the Moon?

I Figure 4-19

The ounces tell you the weight in Englishunits. The grams tell you the mass inmetric units. The label would need toread “2 oz” to be correct on the Moon.The grams would remain unchanged.

56. From the top of a tall building, you droptwo table-tennis balls, one filled with airand the other with water. Both experienceair resistance as they fall. Which ball reaches terminal velocity first? Do both hit theground at the same time?

The lighter, air-filled table tennis ballreaches terminal velocity first. Its massis less for the same shape and size, sothe friction force of upward air resistance becomes equal to the downwardforce of mg sooner. Because the forceof gravity on the water-filled table-tennisball (more mass) is larger, its terminalvelocity is larger, and it strikes thegrouhd first.

57. It can be said that 1 kg equals 2.2 lb. Whatdoes this statement mean? What would bethe proper way of making the comparison?

It means that on Earth’s surface, theweight of 1 kg Is equivalent to 2.2 lb.You should compare masses to massesand weights to weights. Thus 9.8 Nequals 2.2 lb.

58. You toss a ball straight up into the air.

a. Draw a free-body diagram for the ball atthree points during its motion: on theway up, at the very top, and on the waydown. Specifically identify the forcesacting on the ball and their agents.

On the way up

I mass on ball

I mass on ball

I mass on bail

b. What is the velocity of the ball at thevery top of the motion?

o mIs

c. What is the acceleration of the ball atthis same point?

Because the only force acting on itis the gravitational attraction ofEarth, a = 9.80 m/s2.

What is the net force acting on a 1.0-kg ballin free-fall?

Fnet = F9 = mg

jáigm ‘ia* OtLØ

OR ROTIMctuin in

120z(340g) @

At the top

On the way down

Mastering Problems4.1 Force and Motion

page 113

Level 159.

I0.

0

‘I= (1.0 kg)(9.80 mIs2)

= 9.8 N


Chapter 4 continued

60. Skating Joyce and Efua are skating. Joycepushes Efua, whose mass is 40.0-kg, with aforce of 5.0 N. What is Efua’s resultingacceleration?

F = ma

a=Fm

5.0 N— 40.0kg

= 0.12 mIs2

61. A car of mass 2300 kg slows down at a rateof 3.0 rn/s2 when approaching a stop sign.What is the magnitude of the net forcecausing it to slow down?

F = ma

= (2300 kg)(3.0 mis2)

= 6.9X103N

62. BreakIng the Wishbone AfterThanksgiving, Kevin and Gamal use theturkey’s wishbone to make a wish. If Kevinpulls on it with a force 0.17 N larger than theforce Gamal pulls with in the opposite direction, and the wishbone has a mass of 13 g,what is the wishbone’s initial acceleration?

a=Lm

0.17 N0.013 kg

= 13 mIs2

4.2 Using Newton’s Lawspages 113-114

Level 163. What is your weight in newtons?

Fg = mg = (9.80 mIs2)(m)

Answers will vary.

64. Motorcycle Your new motorcyde weighs2450 N. What is its mass in kilograms?

65. Three objects are dropped simultaneouslyfrom the top of a tall building: a shot put,an air-filled balloon, and a basketball.a. Rank the objects in the order in which

they will reachterminal velocity, fromfirst to last.

balloon, basketball, shot putb. Rank the objects according to the order

in which they will reach the ground,from first to last.

Shot put, basketball, balloonc. What is the relationship between your

answers to parts a and b?They are inverses of each other.

66. What is the weight in pounds of a 100.0-Nwooden shipping case?

(100.0 N)(918N )()= 22 lb

67. You place a 7.50-kg television on a springscale. If the scale reads 78.4 N, what is theacceleration due to gravity at that location?Fg=mg

m

78.4 N— 7.50 kg

Drag Radng A 873-kg (1930-ib) dragster,starting from rest, attains a speed of26.3 m/s (58.9 mph) in 0.59 s.a. Find the average acceleration of the

dragster during this time interval.

a iiv

(26.3 m/s — 0.0 m/s)0.59 s

= 45 mis2

b. What is the magnitude of the average netforce on the dragster during this time?F = ma

= (873 kg)(45 mis2)

= 3.9X104N

Eax

CC

II85

E1

= 10.5 mIs2

Level 268.

F9=mg

F 2450Ng — 9.80m/s2

= 2.50X102N


Chapter 4 continued

c. Assume that the driver has a mass of c. It slows to a stop while moving down-

68 kg. What horizontal force does the ward with a constant acceleration.

seat exert on the driver? Depends on the magnitude of the

F = ma = (68 kg)(45 mis2) acceleration.

= 3.1X103N Fscaie = Fg + Fsiowing

= mg + ma69. Assume that a scale is in an elevator on

Earth. What force would the scale exert on = m(g + a)

a 53-kg person standing on it during the = (53 kg)(9.80 mis2 + a)following situations?

a. The elevator moves up at a constant 70. A grocery sack can withstand a maximum

speed. of 230 N before it rips. Will a bag holding15 kg of groceries that is lifted from the

Fscaie = Fgcheckout counter at an acceleration of

= mg 7.0 rn/s2 hold?

= (53 kg)(9.80 mis2) Use Newton’s second law Fnet = ma.

5.2X1 02 N If Fgroceries> 230, then the bag rips.

b. It slows at 2.0 rn/s2 while moving Fgrocerjes = mgrocerjesr.rl +upward.

Slows while moving up or speedsup while moving down, = mgroceries(agrerjes + 9)

Fscaie = F9 + Iowing = (15 kg)(7.0 rn/s2 + 9.80 mIs2)

=mg+ma =250N

= m(g + a) The bag does not hold.

= (53 kg)(9.80 iii/s2— 2.0 mIs2) 71. A 0.50-kg guinea pig is lifted up from the

= 4.1 Xl 02 N ground. What is the smallest force needed.3to lift it? Describe its resulting motion.

c. It speeds up at 2.0 rn/s2 while movingSdownward. F = F9

Slows while moving up or speeds = mg a

up while moving down,= (0.50 kg)(9.80 mis2)

Fscaie = F9 + Fspeedlng= 4.9 N

5.

= mg + ma It would move at a constant speed.

=m(g+a)Level 3 a

= (53 kg)(9.80 rn/s2 — 2.0 mIs2) 72. Astronomy On the surface of Mercury, the

= 4.1X102N gravitational acceleration is 0.38 times itsS

value on Earth.d. It moves downward at a constant speed.

a. What would a 6.0-kg mass weigh onFscaie = F9

Mercury?= mg F9 = mg(0.38)

= (53 kg)(9.80 mis2)= (6.0 kg)(9.80 m/s2)(0.38) 1%

= 5.2x102N= 22 N


Chapter 4 continued

b. If the gravitational acceleration on thesurface of Pluto is 0.08 times that ofMercury, what would a 7.0-kg massweigh on Pluto?

F9 = mg(0.38)(O.08)

= (7.0 kg)(9.80 mIs2)(O.38)(O.08)

= 2.1 N

73. A 65-kg diver jumps off of a 10.0-rn tower.

a. Find the diver’s velocity when he hitsthe water.

V2 = + 2gd

V1 = 0 mIs

sov=\/i

= V2(9.80 mIs2)(10.0 m)

= 14.0 mIs

b. The diver comes to a stop 2.0 m belowthe surface. Find the net force exerted bythe water.

v12=v12+2ad

—v2= 0, so a

=

and F= ma

- -my122d

= —(65 kg)(14.0 rn/s)22(2.0 m)

= —3.2X103N

74. Car Racing A race car has a mass of 710 kg.It starts from rest and travels 40.0 m in 3.0 s.The car is uniformly accelerated during theentire time. What net force is exerted on it?

d= vot+()a12

Since v0 = 0,

2da = -- and F = ma, so

2mdF=—

4.3 Interaction Fonespage 114

Level 175. A 6.0-kg block rests on top of a 7.0-kg

block, which rests on a horizontal table.a. What is the force (magnitude and direc

tion) exerted by the 7.0-kg block on the6.0-kg block?

Fnet = N — mg

FN = F7..kg block on 6-kg block

=mg

= (6.0 kg)(9.80 mis2)

= 59 N; the direction is upward.b. What is the force (magnitude and direc

tion) exerted by the 6.0-kg block on the7.0-kg block?

equal and opposite to that in part a;therefore, 59 N downward

76. Rain A raindrop, with mass 2.45 mg, fallsto the ground. As it is falling, what magnitude of force does it exert on Earth?

alndrop on Earth= F9

=mg

= (0.00245 kg)(9.80 mis2)

= 2.40X102N

77. A 90.0-kg man and a 55-kg man have atug-of-war. The 90.0-kg man pulls on therope such that the 55-kg man accelerates at0.025 rn/s2.What force does the rope exerton the 90.0-kg man?

same in magnitude as the force therope exerts on the 55-kg man:F = ma = (55 kg)(0.025 mis2) = 1.4 N

— (2)(71 0 kg)(40.0 m)— (3.0 s)2

= 6.3X103 N

‘3

8

ci

0C0

C

ci0

8


Chapter 4 continued

Male lions and human sprinters can bothaccelerate at about 10.0 rn/s2. If a typicallion weighs 170 kg and a typical sprinterweighs 75 kg, what is the difference in theforce exerted on the ground during a racebetween these two species?

Use Newton’s second law, Fnet = ma.

The difference between

F,1 and Fhuman is

Fiicrn — Fhuman

= m110a110— mhumanahuman

= (170 kg)(l0.0 mis2) —

(75 kg)(l 0.0 mIs2)

= 9.5X102N

79. A 4500-kg helicopter accelerates upward at2.0 rn/s2.What lift force is exerted by theair on the propellers?

ma = Fnet = Fappi + F9 = Fappi + mg

SO Fappi = ma — mg = m(a—

g)

= (4500 kg)((2.Om/s2)

(—9.8 mIs2))

= 5.3X104N

Three blocks are stacked on top of oneanother, as in Figure 4-20. The top blockhas a mass of 4.6 kg, the middle one has amass of 1.2 kg, and the bottom one has amass of 3.7 kg. Identify and calculate any.normal forces between the objects.

Fnet = Fmlddle block on top block —

FEahs mass on top block

= ma 0

Fmlddle block on top block

= FEaIihs mass on top block

= mtop block9

= (4.6 kg)(9.80 mis2)

=45 N

The normal force Is between the bottomand middle block; the middle block is

the system; upward is positive.

Fnet = Fboltom block on middle block —

iop block on middle block —

FEahls mass on middle block

= ma = 0

op block on middle block

= Fmlddle block on top block

Fbottom block on middle block

= Fmiddle block on top block +

FEahs mass on middle block

= Fmiddle block on top block +

mmlddle block9

= 45 N + (1.2 kg)(9.80 mis2)

= 57 N

The normal force is between the bottomblock and the surface; the bottom blockis the system; upward is positive.

Fnet = Fsuace on bottom block —

Fmiddle block on bottom block —

FEahs mass on bottom block

= ma = 0

The normal force is between the topand middle blocks; the top block is thesystem; upward is positive.

(

Level 278.

Level 380.

I

0

C.

000

0

C

C

U Figure 4-20


Chapter 4 continued

Mixed Reviewpages 114—115

The dragster in problem 68 completed a402.3-rn (0.2500-mi) run in 4.936 s. If thecar had a constant acceleration, what was itsacceleration and final velocity?

= + v1t+ iat2

c11 = v1 = 0, so

2dfa — --

— (2)(402.3 m)(4.936 S)2

= 33.02 m/s2

dt=dh+ -(V-- vt

c11 = V1 = 0, so

2d1t= r

= (2)(402.3 m)4.936 S

= 163.0 mIs

Jet A 2.75X 106-N catapult jet plane isready for takeoff. If the jet’s engines supplya constant thrust of 6.35X 106 N, how muchrunway will it need to reach its minimumtakeoff speed of 285 km/h?

Vt = (285 km/h)(1 000 m/km)(3O,s)

— (1YFthrust1 vfm 2

— 2A m )\Fthrust)

_(12m\ 2 /1ihrust

IFv 2_Li f\g

\2) hrust

—

(79.2 mIs)2(275Xt0621)

—(i-) 6.35X106N

=139m

83. The dragster in problem 68 aossed the finishline going 126.6 rn/s. Does the assumptionof constant acceleration hold true? Whatother piece of evidence could you use todetermine if the acceleration was constant?

126.6 mIs is slower than found Inproblem 81, so the acceleration cannot

be constant. Further, the accelerationin the first 0.59 s was 45 mIs2, not33.02 mIs2.

Fsuace on bottom block

= Fmlddle block on bottom block +

FEa..ths mass on bottom block

= Fmlddle block on bottom block ÷

mbonom l,Iockg

= 57 N + (3.7 kg)(9.80 mis2)

= 93 N

Level 181.

hrust = ma, so a =

Fg—mg

m=.g

= v1 + atand v1 = 0, so

a

= (Fth,us,

vfmFthrust

d= + v1t+ at2

= = 0,so

d at2

Level 282.

‘4= 79.2 m/s


Suppose a 65-kg boy and a 45-kg girl use a

massless rope in a tug-of-war on an icy,

resistance-free surface as in Figure 4-21.

If the acceleration of the girl toward the

boy is 3.0 rn/s2, find the magnitude of theacceleration of the boy toward the girl.

and a1= —rn2a2

= —(45 kg)(3.0 m/s2)(65 kg)

= —2.1 mIs2

85. Space Station Pratish weighs 588 N and isweightless in a space station. If she pushesoff the wall with a vertical acceleration of3.00 rn/s2, determine the force exerted bythe wall during her push off.

Use Newton’s second law to obtainPratish’s mass, mPratish. Use Newton’s

third law FA = —FB mAaA = —mBaB.

Fmpratish

=

Fwaii on Pratlsh Fpratlsh on wall

= mpratlshaPratlsh

— Fgapratlsh

g

(588 N)(3.00 m/s2)— 9.80 m/s2

86. Baseball As a baseball is being caught, itsspeed goes from 30.0 rn/s to 0.0 rn/s inabout 0.0050 s. The mass of the baseball is0.145 kg.

a. What is the baseball’s acceleration?v4-vltf — t1

= 0.0 m/s — 30.0 mIs0.0050 s — 0.0 s

= —6.OX io mis2

b. What are the magnitude and directionof the force acting on it?

F = ma

= (0.145 kg)(—6.0x103m/s2)

= —8.7X102N

(opposite direction of the velocity ofthe ball)

C. What are the magnitude and directionof the force acting on the player whocaught it?

Same magnitude, opposite direction(in direction of velocity of ball)

Air Hockey An air-hockey table works bypumping air through thousands of tinyholes in a table to support light pucks. Thisallows the pucks to move around on cushions of air with very little resistance. One ofthese pucks has a mass of 0.25 kg and ispushed along by a 12.0-N force for 9.0 s.

a. What is the puck’s acceleration?

F= ma

m

— 12.ON— 0.25 kg

= 48 m/s2

b. What is the puck’s final velocity?

v1=v1+at

V1 = 0,so V = at

= (48 mIs2)(9.0 s)

= 4.3X102rn/s {

Chapter 4 continued

84.

(

• Figure 4-21

F1,2 = —F1,2,so m1a1 = —m2a2

Level 387.

a

000

It

a

a

a0rI

1.80X102N


A student stands on a bathroom scale in anelevator at rest on the 64th floor of a building. The scale reads 836 N.

a. As the elevator moves up, the scale reading inaeases to 936 N. Find the acceleration of the elevator.

Fnet = F9 + Feievator

Feievator = Fnet — Fg = ma

Fm

=so

a=FnetFg

g

— g(F — F9)

Fg

— (9.80 mIs2)(963 N — 836 N)836 N

= 1.17 mIs2

b. As the elevator approaches the 74thfloor, the scale reading drops to 782 N.What is the acceleration of the elevator?

Fnet = Fg + Feievator

Feievator = Fnet — F9 = ma

Fm = --, so

a=FnetFq

g

— g(F — Fg)Fg

= (9.80 mIs2)(782 N — 836 N)836 N

= —0.633 m/s2

c. Using your results from parts a and b,explain which change in velocity, starting or stopping, takes the longer time.

Stopping, because the magnitude ofthe acceleration is less and

a

89. Weather Balloon The instruments attachedto a weather balloon in Figure 4-22 have amass of 5.0 kg. The balloon is released andexerts an upward force of 98 N on theinstruments.

a. What is the acceleration of the balloonand instruments?

Fnet = Fappi + Fg

= Fappi + mg

= 98 N + (5.0 kg)(—9.80 mis2)

= +49 N (up)

a=

+49 N5.0 kg

= +9.8 m/s2

b. After the balloon has accelerated for10.0 s, the instruments are released.What is the velocity of the instrumentsat the moment of their release?

v = at

= (+9.8 m/s2)(10.0 s)

= +98 m/s (up)

c. What net force acts on the instrumentsafter their release?

just the instrument weight, —49 N(down)

Chapter 4 continued

88.

98 N

5.0 kg

• Figure 4-22

‘3x

ci

000

0

ci0

8


The velocity becomes negative afterit passes through zero. Thus, use

= V1 + gt, where V =0, or

—vg

— —(+98m1s)— (—9.80 m/s2)

90. When a horizontal force of 4.5 N acts on ablock on a resistance-free surface, it produces an acceleration of 2.5 rn/s2. Supposea second 4.0-kg block is dropped onto thefirst. What is the magnitude of the acceleration of the combination if the same forcecontinues to act? Assume that the secondblock does not slide on the first block.

F = mfjrst blockalnitlal

Fmflrst block

— aInitial

F = mbOh blocksatinal

= (mflrst block + msecond block)aflnal

1=na flfjS block + msecond block

FF

a+ msecond block

Initial

4.5 N4.5N 40k2.5m/s2+ . g

= 0.78 mIs2

91. Two blocks, masses 4.3 kg and 5.4 kg, arepushed across a frictionless surface by ahorizontal force of 22.5 N, as shown inFigure 4-23.

U Figure 4-23

a. What is the acceleration of the blocks?

Identify the two blocks together asthe system, and right as positive.

Fnet = ma, and m = m1 + m2

a= Fm1 + m2

22.5 N— 4.3kg+5.4kg

= 2.3 mIs2 to the right

b. What is the force of the 4.3-kg block onthe 5.4-kg block?

Identify the 5.4-kg block as thesystem and right as positive.

Fnet = F4.3..kg block on 5.4-kg block

= ma

= (5.4 kg)(2.3 mis2)

= 12 N to the right

C. What is the force of the 5.4-kg block onthe 4.3-kg block?

According to Newton’s third law,this should be equal and opposite tothe force found in part b, so theforce is 12 N to the left.

Chapter 4 continued

d. When does the direction of the instruments’ velocity first become downward?

= 1.0X101 s after release

(

(


Chapter 4 continued

92. Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied togetherwith a massless rope as in Figure 4-24. This rope is strung over a massless, resis

tance-free pulley. The blocks are released from rest. Find the following.

a. the tension in the rope

b. the acceleration of the blocksHint: you will need to solve two simultaneous equations.

3.0kg

5.0 kg

• Figure 4-24

Equation 1 comes from a free-body diagram for the 5.0-kg block. Down is

positive.

Fnet = FEarths mass on 5.0-kg block — Frope on 5.0-kg block = m5.0-kg blOCka (1)

Equation 2 comes from a free-body diagram for the 3.0-kg block. Up is

positive.

Fnet = Frope on 3.0-kg block — FEarths mass on 3.0-kg block = m3.0kg blOCka (2)

The forces of the rope on each block will have the same magnitude,

because the tension is constant throughout the rope. Call this force T.

FEarths mass on 5.0-kg block — T = m5.0.kg blOCka (1)

T FEarths mass on 3.0-kg block = m3.okg blOCka (2)

Solve equation 2 for Tand plug into equation 1:

mS.0.kg blOCka = Earth’s mass on 5.0-kg block —

Earth’s mass on 3.0-kg block b,00ka

a= Frth mass on 5.0-kg block — mass on 3.0-kg block

m50- block +73.0.kg block

= (“s.O.kg block — m30-k9blOCk)g

S m3.0-kg block + m5.0-kg block

— (5.0 kg — 3.0 kg)(9.80 m/s2)

3.0 kg + 5.0 kg

8= 2.4 m/s2


Chapter 4 continued

Solve equation 2 for T:

T = FEa,.ths mass on 3.0-kg block +

m3kg blOCka

= m3JJ..kO blOCkg + m3.0..kg blOCka

= m31 biock(g + a)

= (3.0 kg)(9.80 mis2 + 2.4 mis2)

= 37 N

Thinking Criticallypages 115—11693. Formulate Models A 2.0-kg mass, m,,

and a 3.0-kg mass, mB, are connected to alightweight cord that passes over a friction-less pulley. The pulley only changes thedirection of the force exerted by the rope.The hanging masses are free to move.Choose coordinate systems for the twomasses with the positive direction beingup for mA and down for m.

a. Create a pictorial model.

I.b. Create a physical model with motion

and free-body diagrams.

I.c. What is the acceleration of the smaller

mass?

ma = Fnet where m is the total mass

being accelerated.

For mA, mAa = FT — mg

For mB, mBa = — FT + m8g

FT = mg — mBa = m(g — a)

Substituting into the equation for

mA gives

mAa = mg — m8a — mg

or (mA + mB)a = (mB fl7)g

I m— mATherefore a = I

mA + mB,

— (3.0 kg — 2.0 kg— \ 2.0 kg + 3.0 kg

(9.80 mIs2)

= 2.0 mis2 upward

94. Use Models Suppose that the masses inproblem 93 are now 1.00 kg and 4.00 kg.Find the acceleration of the larger mass.

1mB—rna=i Alg

\ m + mB /

i4.00 kg — 1.00 kg= I 1.00 kg + 4.00 kg

)(9.8o m/s2)

= 5.88 mis2 downward

95. Infer The force exerted on a 0.145-kgbaseball by a bat changes from 0.0 N to1.0 X N in 0.0010 s, then drops back tozero in the same amount of time. The baseball was going toward the bat at 25 rn/s.

a. Draw a graph of force versus time. Whatis the average force exerted on the ballby the bat?

za)I.,I0

LI

= (-)(1.OX1O4 N)

= 5.0X103N(

j m8gTime (ms)

ave— j peak


Chapter 4 continued

b. What is the acceleration of the ball?

a= Fnet = 5.0X103N

m 0.145kg

= 3.4X104mis2

c. What is the final velocity of the ball,assuming that it reverses direction?

v=v1+at

—25 mIs + (3.4X104m/s2)

(0.0020 s)

= 43 mIs

96. Observe and Infer Three blocks that areconnected by massless strings are pulledalong a frictionless surface by a horizontalforce, as shown in Figure 4-25.

FT1 FT2 6.0 kg

:• Figure 4-25

a. What is the acceleration of each block?

Since they all move together, theacceleration is the same for all3 blocks.

F = ma(m1 + m2 + m3)a

Fa— m1-l-m2+m3

b. What are the tension forces in each ofthe strings?

Hint: Draw a separate free-body diagramfor each block.

Fnet = ma

F— FT2 = m3a

FT2 — FT1 = m2a

FT1 = m1a

97. Critique Using the Example Problems inthis chapter as models, write a solution tothe following problem. A block of mass3.46 kg is suspended from two verticalropes attached to the ceiling. What is thetension in each rope?

1 Analyze and Sketch the Problem

Draw free-body diagrams for the blockand choose upward to be positiveSolve for the Unknown

Known:

mbIOCk = 3.46 kg

Unknown:

Fropei on block = Frope2 on block = ?

2 Solve for the Unknown

Use Newton’s second law to find thetension in the ropes

Fnet =2Fropei on block —

FT2 = m2a + FT1

= (4.0 kg)(3.0 mis2) + 6.0 N

= 18 N

TT

Fg

Ec3

ci

U

CC0

ci

C

ci

36 N2.0 kg + 4.0 kg + 6.0 kg

= 3.0 mIs2

FEarths mass on block

= ma = 0

F— FEah•s mass on block

ropel on block — 2

— mgFropei on block

— 2

= (3.46 kg)(9.80 m/s2)2

= 17.0 N

= (2.0 kg)(3.0 mis2)

= 6.0 N


Chapter 4 continued

3 Evaluate the Answer

• Are the units correct? N is the correctunit for a tension, since it is a force.

• Does the sign make sense? The positive sign Indicates that the tension is

- pulling upwards.

Is the magnitude realistic? We wouldexpect the magnitude to be on thesame order as the block’s weight.

98. Think Critically Because of your physicsknowledge, you are serving as a scientificconsultant for a new science-fiction TVseries about space exploration. In episode 3,the heroine, Misty Moonglow, has beenasked to be the first person to ride in a newinterplanetary transport for use in our solarsystem. She wants to be sure that the transport actually takes her to the planet she issupposed to be going to, so she needs totake a testing device along with her tomeasure the force of gravity when shearrives. The script writers don’t want her tojust drop an object, because it will be hardto depict different accelerations of fallingobjects on TV. They think they’d like something involving a scale. It is your job todesign a quick experiment Misty can conduct involving a scale to determine whichplanet in our solar system she has arrivedon. Describe the experiment and includewhat the results would be for Pluto(g = 0.30 m/s2), which is where she is supposed to go, and Mercury (g = 3.70 m/s2),which is where she actually ends up.

Answers will vary. Here is one possibleanswer: She should take a knownmass, say 5.00-kg, with her and place iton the scale. Since the gravitationalforce depends upon the local acceleration due to gravity, the scale will read adifferent number of newtons, dependingon which planet she is on. The following analysis shows how to figure outwhat the scale would read on a givenplanet:

Identify the mass as the system andupward as positive.

Fnet = Fscaie on mass — Fg = ma = 0

Fscaie on mass = Fg

Fscaie on mass = mg

Pluto: Fscaie on mass

= (5.00 kg)(0.30 m/s2)

= 1.5 N

Mercury: Fscaie on mass

= (5.00 kg)(3.7 mis2)

= 19 N

99. Apply Concepts Develop a CBL lab, usinga motion detector, that graphs the distancea free-falling object moves over equal intervals of time. Also graph velocity versus time.Compare and contrast your graphs. Usingyour velocity graph, determine the acceleration. Does it equal g?

Student labs will vary with equipmentavailable and designs. p-f graphs andv-f graphs should reflect uniform acceleration. The acceleration should beclose to g.

Writing in Physicspage 116100. Research Newton’s contributions to physics

and write a one-page summary. Do you thinkhis three laws of motion were his greatestaccomplishments? Explain why or why not.Answers will vary. Newton’s contributions should include his work on lightand color, telescopes, astronomy, lawsof motion, gravity, and perhapscalculus. One argument in favor of histhree laws of motion being his greatestaccomplishments Is that mechanics isbased on the foundation of these laws.His advances in the understanding ofthe concept of gravity may be suggested as his greatest accomplishmentinstead of his three laws of motion.

(

(


Answers will vary. Newton’s first law of

motion involves an object whose net

forces are zero. It the object is at rest, it

remains at rest; if it is in motion, it will

continue to move in the same direction

at a constant velocity. Only a force act

ing on an object at rest can cause it to

move. Likewise, only a force acting on

an object in motion can cause it to

change its direction or speed. The twocases (object at rest, object in motion)

could be viewed as two different frames

of reference. This law can be demon

strated, but it cannot be proven.

102. Physicists classify all forces into four funda

mental categories: gravitational, electro

magnetic, strong nudear, and weak nuclear.

Investigate these four forces and describe

the situations in which they are found.

Answers will vary. The strong nuclear

force has a very short range and is what

holds protons and neutrons together In

the nucleus of an atom. The weak

nuclear force is much less strong than

the strong nuclear force and is involved

in radioactive decay. The electromag

netic force is involved in holding atoms

and molecules together and is based on

the attraction of opposite charges.

Gravity is a long-range force between

two or more masses.

Cumulative Reviewpage 116103. Cross-Country Skiing Your friend is

training for a cross-country skiing race, and

you and some other friends have agreed to

provide him with food and water along his

training route. It is a bitterly cold day, so

none of you wants to wait outside longer

than you have to. Taro, whose house is the

stop before yours, calls you at 8:25 A.M. to

tell you that the skier just passed his house

and is planning to move at an average

speed of 8.0 km/h. If it is 5.2 km from

Taro’s house to yours, when should you

expect the skier to pass your house?

(Chapter 2)

d= vt,ort=

d= 5.2 km = 5.2X103m

v = (8.0 kmlh)( 1km X= 2.2 mIs

5.2X1Om— 2.2 mIs

= 2.4X103S

= 39 mm

The skier should pass your house at

8:25 + 0:39 = 9:04 A.M.

104. Figure 4-26 is a position-time graph of

the motion of two cars on a road.

(Chapter 3)

• Figure 4-26

a. At what time(s) does one car pass the

other?

3 5, 8 s

b. Which car is moving faster at 7.0 s?

car A

c. At what time(s) do the cars have the

same velocity?

5s

d. Over what time interval is car B speed

ing up all the time?

none

e. Over what time interval is car B slow

ing down all the time?

— 3s to lOs

Chapter 4 continued

101. Review, analyze, and critique Newton’s

first law. Can we prove this law? Explain.

Be sure to consider the role of resistance.

Ea)‘-IC

2 4 6 8

Time (s)

0U

U

000

>0

U

0

0

U

‘I

I.8


Chapter 4 continued

105 Refer. to Figure 4-26 to find theinstantaneous speed for the following:(Chapter 3)

a. carBat2.Os

o mIs

b. carBat9.Os

—0 mIs

c. carAat2.Os

—lmIs

2. It takes the glider 1.3 s to pass through asecond gate. What is the distance betweenthe two gates?

d=d+ v1t+ at2

Let d = position of first gate = 0.0 m

d = 0.0 m + (0.25 m/s)(1 .3 s) +

(..)(o.8o mIs2)(1.3 )2

= 1.Om

Challenge Problempage 100An air-track glider passes through a photoelectricgate at an initial speed of 0.25 rn/s. As it passesthrough the gate, a constant force of 0.40 N isapplied to the glider in the same direction as itsmotion. The glider has a mass of 0.50 kg.

1. What is the acceleration of the glider?

F = ma

m

0.40 N— 0.50 kg

= 0.80 mis2

3. The 0.40-N force is applied by means of astring attached to the glider. The other endof the string passes over a resistance-freepulley and is attached to a hanging mass,m. Howbigism?

Fg = mmassg

m -mass— g

0.40 N= 9.80 n’Js2

= 4.1X102kg

4. Derive an expression for the tension, T, inthe string as a function of the mass, M, ofthe glider, the mass, m, of the hangingmass, and g.

T= mg= Ma


Forces inOne Dimension - Staff

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