Sm PDF Chapter4

7/28/2019 Sm PDF Chapter4

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Chapter 4

The Laws of Motio

Quick Quizzes

1. (a) True. Motion requires no force. Newtons first law says an object in motion continuesto move by itself in the absence of external forces.

(b) False. It is possible for forces to act on an object with no resulting motion if the forces

are balanced.2. (a) True. If a single force acts on an object, it must accelerate. From Newton's second law,

= ma FG

G

, and a single force must represent a non-zero net force.

(b) True. If an object accelerates, at least one force must act on it.

(c) False. If an object has no acceleration, you cannot conclude that no forces act on it. Inthis case, you can only say that the net force on the object is zero.

3. False. If the object begins at rest or is moving with a velocity with only an x component,the net force in the x direction causes the object to move in the x direction. In any othercase, however, the motion of the object involves velocity components in directions other

than x. Thus, the direction of the velocity vector is not generally along the x axis. What wecan say with confidence is that a net force in the x direction causes the object to accelerate inthe x direction.

4. (a). Because the value of g is smaller on the Moon than on the Earth, more mass of goldwould be required to represent 1 newton of weight on the Moon. Thus, your friend on theMoon is richer, by about a factor of 6!

5. (c) and (d). Newtons third law states that the car and truck will experience equalmagnitude (but oppositely directed) forces. Newtons second law states that accelerationis inversely proportional to mass when the force is constant. Thus, the lower mass vehicle(the car) will experience the greater acceleration.

6. (c). The scale is in equilibrium in both situations, so it experiences a net force of zero.Because each person pulls with a force F and there is no acceleration, each person is inequilibrium. Therefore, the tension in the ropes must be equal to F. In case (i), the personon the right pulls with force F on a spring mounted rigidly to a brick wall. The resultingtension F in the rope causes the scale to read a force F. In case (ii), the person on the leftcan be modeled as simply holding the rope tightly while the person on the right pulls.Thus, the person on the left is doing the same thing that the wall does in case (i). Theresulting scale reading is the same whether there is a wall or a person holding the left sideof the scale.

99


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100 CHAPTER 4

7. (c). The tension in the rope has a vertical component that supports part of the total weightof the child and sled. Thus, the upward normal force exerted by the ground is less than thetotal weight.

8. (b). Friction forces are always parallel to the surfaces in contact, which, in this case, are thewall and the cover of the book. This tells us that the friction force is either upward or

downward. Because the tendency of the book is to fall due to gravity, the friction forcemust be in the upward direction.

9. (b). The static friction force between the bottom surface of the crate and the surface of thetruck bed is the net horizontal force on the crate that causes it to accelerate. It is in thesame direction as the acceleration, to the east.

10. (b). It is easier to attach the rope and pull. In this case, there is a component of yourapplied force that is upward. This reduces the normal force between the sled and thesnow. In turn, this reduces the friction force between the sled and the snow, making iteasier to move. If you push from behind, with a force with a downward component, thenormal force is larger, the friction force is larger, and the sled is harder to move.


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The Laws of Motion 101

Answers to Even Numbered Conceptual Questions

2. If the car is traveling at constant velocity, it has zero acceleration. Hence, the resultantforce acting on it is zero.

4. The force causing the ball to rebound upward is the normal force exerted on the ball bythe floor.

6. w = mg and g decreases with altitude. Thus, to get a good buy, purchase it in Denver. If gold was sold by mass, it would not matter where you bought it.

8. If it has a large mass, it will take a large force to alter its motion even when floating inspace. Thus, to avoid injuring himself, he should push it gently toward the storagecompartment.

10. The net force acting on the object decreases as the resistive force increases. Eventually, theresistive force becomes equal to the weight of the object, and the net force goes to zero. Inthis condition, the object stops accelerating, and the velocity stays constant. The rock has

reached its terminal velocity.12. The barbell always exerts a downward force on the lifter equal in magnitude to the

upward force that she exerts on the barbell. Since the lifter is in equilibrium, themagnitude of the upward force exerted on her by the scale (that is, the scale reading)equals the sum of her weight and the downward force exerted by the barbell. As the

barbell goes through the bottom of the cycle and is being lifted upward, the scale readingexceeds the combined weights of the lifter and the barbell. At the top of the motion and asthe barbell is allowed to move back downward, the scale reading is less than the combinedweights. If the barbell is moving upward, the lifter can declare she has thrown it just byletting go of it for a moment. Thus, the case is included in the previous answer.

14. While the engines operate, their total upward thrust exceeds the weight of the rocket, andthe rocket experiences a net upward force. This net force causes the upward velocity of therocket to increase in magnitude (speed). The upward thrust of the engines is constant, butthe remaining mass of the rocket (and hence, the downward gravitational force or weight)decreases as the rocket consumes its fuel. Thus, there is an increasing net upward forceacting on a diminishing mass. This yields an acceleration that increases in time.

16. The trucks skidding distance can be shown to be20

2 k

vx

g = where k is the coefficient of

kinetic friction and is the initial velocity of the truck. This equation demonstrates thatthe mass of the truck does not affect the skidding distance, but halving the velocity willdecrease the skidding distance by a factor of four.

0v

18. Because the mass of the truck is decreasing, the acceleration will increase.


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102 CHAPTER 4

20.

(a)

Rur

ur

mgur

mg

fur

Rur

ur

n

(c)

Tur

(b)

ur

mg

Rur

In the free-body diagrams give above, R

JG

represents a force due to air resistance, T

JG

is aforce due to the thrust of the rocket engine, nJG

is a normal force, fG

is a friction force, andthe forces labeled are gravitational forces.mg

G


5/44


Answers to Even Numbered Problems

2. 25 N

4.

2

1.7 10 N

6. 7.4 min

8. 23.1 10 N

10. (a) The gravitational force for S is greater than F(b) The time of fall for S is less than F(c) The times are equal (d) The total force on S is greater than F

12. (a) 799 N at 8.77 to the right of forward direction(b) 20.266 m s in the direction of the resultant force

14. 21.59 m s at 65.2 N of E

16. 77.8 N in each wire

18. , 6121.7 10 N

20. 31.04 10 N rearward

22. (a) sinmg

T

= (b) 1.79 N

24. (a) 1.5 m (b) 1.4 m

26. 24.43 m s up the incline, 53.7 N

28. 13 N down the incline

30. 26.53 m s , 32.7 N

32. (a) (b) T w> T w= (c) T w< (d) , Yes (e) 1. , Yes (f) , Yes41.85 10 N 447 10 N 41.25 10 N

34. (a) 36.8 N (b) 22.45 m s (c) 1.23 m

36. (a) 0 (b) 20.70 m s

38. (a) 21.20 m s (b) 0.122 (c) 45.0 m

40. (a) 55.2 (b) 167 N


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104 CHAPTER 4

42. 3.17 s

44. (a) 20.366 m s (b) 21.29 m s down the incline

46. (a) 98.6 m (b) 16.4 m

48. (a) 20.125 m s (b) (c) 0.23539.7 N

50. (a) 18.5 N (b) 25.8 N

52. (a) 2.13 s (b) 1.67 m

54. 22.6 m s

56. 21.5 N

58. (a) 50.0 N (b) 0.500 (c) 25.0 N

60. 0.814

62. (a) 21.63 m s(b) 57.2 N tension in string connecting 5-kg and 4-kg, 24.5 N tension in string

connecting 4-kg and 3-kg

64. (b) 9.8 N (c) 20.58 m s

66. upward31.18 10 N

68. 20.69 m s

70. 173 lb

72. 23 m s

74. (a) (b) 37.25 10 N 24.57 m s

76. 104 N

78. 6.00 cm

80. 35.10 10 N


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Problem Solutions

4.1 (a) ( )( )26.0 kg 2.0 m s 12 Nx xF ma = = =

(b) 212 N 3.0 m s4.0 kg

xx

Fm

=a = =

4.2 From v v , the acceleration given to the football is0 at= +

20 10 m s 0 50 m s0.20 sav

v va

t= = =

nd2

(

Then, from Newtons law, we find

)( )2g 50 m s 25 Na= = =av avF m 0.50 k

4.3 ( ) 42000 lbs 4.448 N2 tons 2 10 N1 ton 1 lb

w = =

4.4 ( ) 24.448 N38 lbs 1.7 10 N1 lb

w = =

4.5 The weight of the bag of sugar on Earth is ( ) 4.448 N5.00 lbs 22.2 N1 lbE E

w mg = = = . If

M g is the free-fall acceleration on the surface of the Moon, the ratio of the weight of an

object on the Moon to its weight when on Earth is M M ME E E

mg gww mg g

= = , so M M EE

gw w

g

=

.

Hence, the weight of the bag of sugar on the Moon is ( ) 1 3.71 N6

= 22.2 N Mw = . On

Jupiter, its weight would be ( )( )22.2 J J EE

g g

= =

N 2.64 58.7 N=w w

The mass is the same at all three locations, and is given by

( )( )2

5.00 lb 4.448 N lb2.27 kg

9.80 m sE

E

wm g

= = =


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106 CHAPTER 4

4.6 5

27

7.5 10 N 5.0 10 m s1.5 10 kg

Fa

m = = =

0v v at= +

2 , and

gives

02 2

80 km h 0 0.278 m s 1 min 7.4 min5.0 10 m s 1 km h 60 s

v va

= = = t

4.7 Summing the forces on the plane shown gives

( )( )210 N 0.20 kg 2.0 m sxF F f f = = =

From which, 9.6 N f =

f

a = 2.0 m/s 2

F = 10 Nur

4.8 The acceleration of the bullet is given by( )

( )( )

22 20 320 m s 0

2 2 0.82 mv v

ax

= =

Then, ( ) ( )( )2

3 2320 m s5.0 10 kg 3.1 10 N2 0.82 m

F ma = = =

4.9 The vertical acceleration of the salmon as it goes from 0 3.0 m s yv = (underwater) to6.0 m s

yv = (after moving upward 1.0 m or 2/3 of its body length) is

( ) ( )( )

2 22 20 26.0 m s 3.0 m s 13.5 m s

2 2 1.00 m y y

y

v va

y

= = =

Applying Newtons second law to the vertical leap of this salmon having a mass of , we find61 kg

y y yF ma F mg ma = =

( ) ( )

or 32 2m m9.8 1.4 10 Ns s + = 61 kg 13.5 yF m a g= + =


9/44


4.10 (a) The sphere has a larger mass than the feather. Hence, the sphere experiences alarger gravitational force gF mg= than does the feather.

(b) The time of fall is less for the sphere than for the feather. This is because airresistance affects the motion of the feather more than that of the sphere.

(c) In a vacuum, the time of fall is the same for the sphere and the feather. In theabsence of air resistance, both objects have the free-fall acceleration g.

(d) In a vacuum, the total force on the sphere is greater than that on the feather. In theabsence of air resistance, the total force is just the gravitational force, and the sphereweighs more than the feather.

4.11 (a) From the second law, the acceleration of the boat is

22 000 N 1 800 N 0.200 m s1 000 kg

Fam

= = =

(b) The distance moved is

( )( )22 20 1 10 0.200 m s 10.0 s 10.0 m2 2x v t at = + = + =

(c) The final velocity is ( )( )20 0 0.200 m s 10.0 s 2.00 m sv v at= + = + =

4.12 (a) Choose the positive y-axis in the forward direction. Weresolve the forces into their components as

Force x -component y-component400 N 200 N 346 N450 N 78.1 N 443 N

Resultant 122 NxF = 790 N yF =

30

10 F 2 =

4 5 0 N F 1

= 4 0 0 N

+x

+ y

f = 1 800 NF = 2 000 N

+ y+x

F

ur

fur

The magnitude and direction of the resultant force is

( ) ( )22 799 NR x yF F F= + = , 1tan 8.77x y

FF

= =

to right of y-axis.

Thus, 799 N at 8.77 to the right of the forward directionR = FG


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108 CHAPTER 4

(b) The acceleration is in the same direction as FRG

and has magnitude

2799 N 0.266 m s3 000 kg

RFam

= = =

4.13 Starting with v and falling 30 m to the ground, the velocity of the ball just before ithits is

0 0 y =

( )( )2 20 2 9.80 m s 30 m 24 m s y ya y= = + =

0 yv

1 0 2v v +

On the rebound, the ball has = after a displacement 20 m y = + . Its velocity as itleft the ground must have been

( )( )2 22 0 2 9.80 m s 20 m 20 m s y ya y= + = + = +2v v

Thus, the average acceleration of the ball during the 2.0-ms contact with the ground was

( ) 4 2av 3

20 m s 24 m s2.2 10 m s

2.0 10 sa

+ = +

2 1v v

t= =

(

The resultant force acting on the ball during this time interval must have been

)( )4 2 4kg 2.2 10 m s 1.1 10 NF m + = + 0.50a= =

or 41.1 10= N upwardFG

4.14 Since the two forces are perpendicular to each other, their resultant is:

( ) ( )2 2180 N 390 N 430 NRF = + = , at 1390 Ntan 65.2 N of E180 N

= =

Thus, 2430 N 1.59 m s270 kg

RFam

= = = or 21.59 m s at 65.2 N of E= aG


11/44


4.15 Since the burglar is held in equilibrium, the tension in thevertical cable equals the burglars weight of 600 N

0

Now, consider the junction in the three cables:

, giving

or

0 yF = 2 sin 37.0 600 NT =

2 997T =

0xF =

(

N in the inclined cable

2 1cos37.0 0 =

Also, which yields T T

or )1 99T 7 N cos rizontal cable37.0 796 N in the ho= =

4.16 From , T T

or

Then, gives

0xF =

1 2T T =

0 yF =

1 2cos40.0 cos 40.0 0 =

( )12 sin 40.0 100 NT 0=

yielding 1 2 77.8 N= =T T

4.17 From , T T

or

0xF =

(

1 2cos30.0 cos60.0 0 =

)2 13T T

(

1.7=

0 yF =

(1)

Then becomes

)1 130.0 1.73 sin 60.0 150T T + sin N 0=

which gives 1 75.0 N in the right side= cableT

Finally, Equation (1) above gives 2 130T = N in the left side cable

x37.0T1

w = 600 N

yT2ur

ur

100 N

T2

x

y

40.0

T1

40.0

ur ur

150 N

T2

x

y

60.0

T1

30.0

ur

ur


12/44

110 CHAPTER 4

4.18 If the hip exerts no force on the leg, the system must bein equilibrium with the three forces shown in the free-

body diagram.

Thus becomes0xF =

( )2 cosw 110 N cos40 =

(

(1)

From , we find0 yF =

)2 sinw 220 N 110 N sin 40 =

( )

(2)

Dividing Equation (2) by Equation (1) yields

( )1 220 N 110 Nn110 N cos

sin40 6140

= =

ta

Then, from either Equation (1) or (2), 22 1.7 10 Nw =

4.19 Free-body diagrams of the two blocks are shownat the right. Note that each block experiences adownward gravitational force 98.0 N gF mg= = .Also, each has the same upward acceleration asthe elevator, 22.00 m s ya = +

Applying Newtons 2 nd law to the lower block :

y y lower g yF ma T F ma = =

or ( )( )2m s 118 N=98.0lower g yT F ma= + =

N+ 10.0 kg 2.00+

Then, applying the 2 nd law to the upper block

y y upper lower g yF m = a T T F ma =

or ( )( )210.0 kg 2.00 m s 236 N+ =118 N 98.0 N+a = +upper lower g yT T F m= + +

220 N

1 1 0 N

40

w 2

ax

yur

ur

T lower

Upper Blockm = 10.00 kg

ur

Tupper

ur

F g

Lower Blockm = 10.00 kg

ur

T lower

ur

F g


13/44


4.20 The resultant force exerted on the boat by the people is( ) 32 600 N cos30.0 1.04 10 N = in the forward direction. If the boat moves with

constant velocity, the total force acting on it must be zero. Hence, the resistive forceexerted on the boat by the water must be

31.04 10 N in the rear= fG

ward direction

4.21 and1.00 kgm = 9.80 Nmg =

1tan 0.458

yF

in

0.200 m25.0 m

= =

0=

T mg

2 s

Require that ,

=

9.80 N 613 N2sin

= =T

4.22 (a) An explanation proceeding from fundamental physicalprinciples will be best for the parents and for you.Consider forces on the bit of string touching the weighthanger as shown in the free-body diagram:

Horizontal Forces: 0 cos 0xF F T = + =0 sin 0 yF mg T

Vertical Forces: = + =

T T

a a0.200 m

25.0 m25.0 m

urur

mgur

T

F

mg

q

ur

ur

ur

You need only the equation for the vertical forces to find that the tension in the

string is given bysinmg

T

= . The force the child feels gets smaller, changing from T

to T cos , while the counterweight hangs on the string. On the other hand, the kitedoes not notice what you are doing and the tension in the main part of the stringstays constant. You do not need a level, since you learned in physics lab to sight to ahorizontal line in a building. Share with the parents your estimate of theexperimental uncertainty, which you make by thinking critically about themeasurement, by repeating trials, practicing in advance and looking for variationsand improvements in technique, including using other observers. You will then beglad to have the parents themselves repeat your measurements.

(b)( )( )20.132 kg 9.80 m s

1.79 Nsin sin 46.3mg

=

T = =


14/44

112 CHAPTER 4

4.23 The forces on the bucket are the tension in the rope and the weightof the bucket, ( )( )25.0 kg 9.80 m s 49 Nmg = = . Choose the positivedirection upward and use the second law:

y yF =ma

(

)( )2N 5.0 kg 3.0 m s =49T

64 NT = mg

T

a

ur

ur

ur

F = 10 N4.24 (a) From the second law, we find the acceleration as

210 N 0.33 m s30 kg

Fam

= = =

To find the distance moved, we use

( )( )22 20 1 10 0.33 m s 3.0 s 1.5 m2 2x v t at = + = + =

(b) If the shopper places her ( )30 N 3.1 kg child in the cart, the new acceleration will be2s

10 N0.30 m

33 kgtotal

Fam

= = = , and the new distance traveled in 3.0 s will be

( )( )2

3.0 s 1.4=21

0 0.30 m s m2x = +

4.25 (a) The average acceleration is given by

20av

5.00 m s 20.0 m s3.75 m s

4.00 sv v

at

= = =

(

The average force is found from the second law as

)( )

2 3av av 2000 kg 3.75 m s 7.50 10 NF ma= = =

(b) The distance traveled is:

( ) ( )av5.00 m s 20.0 m s

4.00 s 50.0 m2

x v t+ = = =


15/44


4.26 Let m , , and1 10.0 kg= 2 5.00 kgm = 40.0 = .Applying the second law to each object gives

(1)

and

1 1m a m g=

2m a T

T

2 sinm g = (2)

Adding these equations yields

1 2

1 2

sina g

m m

+

(

m m= , or

) ( )2 2in40 m s 4.43 m s= =10.0a

(

kg 5.00 kg s15.0 kg

.09.80

Then, Equation (1) yields

) ( )( ) 29. m s 53.7 N = 1T m g= 10.0 kga = 80 4.43

q

T

ur

m1

T

a

m1gur

ur

ur

a

m2g

n

m2

ur

ur

ur

4.27 (a) The resultant external force acting on this system having a total mass of 6.0 kg is42 N directed horizontally toward the right. Thus, the acceleration produced is

242 N 7.0 m s horizontally to the right6.0 kg

Fa

m= = =

(b) Draw a free body diagram of the 3.0-kg block and apply Newtons second law tothe horizontal forces acting on this block:

x xF ma = ( )( )242 N 3.0 kg 7.0 m sT = , and therefore 21 NT =

(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:

( )( )22.0 kg 7.0 m sF ma= = , or 14 N horizontally to the right=FG


16/44

114 CHAPTER 4

4.28 The acceleration of the mass down the incline is given by

20

12

x v t at = + , or ( )210.80 m 0 0.50 s2

a= +

This gives2

6.4 m sa =

Thus, the force down the incline is ( )( )22.0 kg 6.4 m s 13 NF ma= = =

4.29 Choose the positive x axis to be up the incline.

Then,

which gives

( )sin18.5x xF ma T mg ma= = x

( ) ( )2 2140 Nsin18.5 9.80 m s sin18.5 0.390 m s40.0 kgxT a gm

= = =

( )

The velocity after moving 80.0 m up the incline is given by

( )( )2 20= 2 = 0+2 0.390 m s 80.0 m 7.90 m sxv v a x+ =

4.30 First consider the block moving along thehorizontal. The only force in the directionof movement is T . Thus,

x xF ma = ( )5.00 kgT =

(

a (1)

Next consider the block which movesvertically. The forces on it are the tensionT and its weight, 98.0 N.

y yF ma = )98.0 N 10.0 = kgT a (2)

+ y

T

mg = 98.0 N

(b)

10.0 kg

ur

(a)

T

+x

5.00 kgur

Note that both blocks must have the same magnitude of acceleration. Equations (1) and(2) can be solved simultaneously to give.

26.53 m sa = , and 32.7 NT =


17/44


4.31 Taking forward as the positive direction, the acceleration that the braking force gives thetrain is

62

61.87 10 N 0.358 m s

5.22 10 kgF

am

= = =

(a) The velocity of the train at 30.0 st = is then

0.447 mh 1.61 km

( )( )20 skm90.0 0.358 m s 30.0 s 14.2 m shv v at = + = + =

(b) During this time, the displacement of the train is

( ) ( )( )22 21 10 2 20.447 m skm90.0 30.0 s 0.358 m s 30.0 sh 1.61 km hx v t at = + = +

or 588 mx =

4.32 (a) When the acceleration is upward, the total upward force T must exceed the total

downward force ( )( )2 41 500 kg 9.80 m s 1.47 10 N g= = = w m

(b) When the velocity is constant, the acceleration is zero. The total upward force T andthe total downward force w must be equal in magnitude .

(c) If the acceleration is directed downward, the total downward force w mustexceed the total upward force T.

(d) ( )( )2 21 500 kg 9.80 m s 2.50 m s 1.85 10 N y y yF ma T mg ma = + = + = 4 =

Yes T w>, .

(e) ( )( )2 41 500 kg 9.80 m s 0 1.47 10 N y y yF ma T mg ma = + = + = =

Yes T w=, .

(f) ( )( )2 21 500 kg 9.80 m s 1.50 m s 1.25 10 N y y yF ma T mg ma = + = = 4 =

Yes T w


18/44

116 CHAPTER 4

4.33 Trailer300 kg

T

Car1000 kg

F

n T w T

T

n c w cq

Rcar

F

n c

ur ur

ur ur ur

ur

ur

ur

urur

Choose the + x direction to be horizontal and forward with the + y vertical and upward.The common acceleration of the car and trailer then has components of

22.15 m s and 0x ya a= + = .

(a) The net force on the car is horizontal and given by

( ) ( )( )2 31000 kg 2.15 m s 2.15 10 N forwardx car xcarF F T m a = = = =

(b) The net force on the trailer is also horizontal and given by

( ) ( )( )2300 kg 2.15 m s 645 N forwardx trailer xtrailerF T m a = + = = =

(c) Consider the free-body diagrams of the car and trailer. The only horizontal forceacting on the trailer is 645 N forwardT = , and this is exerted on the trailer by thecar. Newtons third law then states that the force the trailer exerts on the car is

645 N toward the rear

(d) The road exerts two forces on the car. These are and cF n

3 310 N

shown in the free-bodydiagram of the car.

From part (a),

Also,

2.15 10 N 2.80F T = + = +

( ) 0c car yw m a = = c cn w= = y ccarF n = , so

The resultant force exerted on the car by the road is then

310 N9.80carm g =

( ) ( )232.80 10 N 9.80 10n = + 2 410 N2 2car cR F= + 3 N 1.02=

at ( )1 1tan tan 3.51 74.1cnF

= = = above the horizontal and forward. Newtons

third law then states that the resultant force exerted on the road by the car is

4.02 10 N at 74.1 below the hori 1 zontal and rearward


19/44


4.34 First, consider the 3.00-kg rising mass.The forces on it are the tension, T , andits weight, 29.4 N. With the upwarddirection as positive, the second law becomes

( )29.4 N 3.00 kgT a =

(

(1)

The forces on the falling 5.00-kg massare its weight and T , and itsacceleration has the same magnitude asthat of the rising mass. Choosing thepositive direction down for this mass,gives

)49 N 5.00 kgT = a (2)

Equations (1) and (2) can be solved simultaneously to give

w2 = 49.0 N

T

+

FallingMass

m2 = 5.00 kg

ur

RisingMass

T

w1 = 29.4 N

+

m1 = 3.00 kg

ur

(a) the tension as 36.8 NT =

(b) and the acceleration as 22.45 m sa =

(c) Consider the 3.00-kg mass. We have

( )( )22 20 1 10 2.45 m s 1.00 s 1.23 m2 2 y y y v t a t = + = + =

4.35 When the block is on the verge of moving, the static friction force has a magnitude( )maxs s s f f n = =

75

.

Since equilibrium still exists and the applied force is 75 N, we have

orN 0x sF f = = ( )max 75 Ns f =

In this case, the normal force is just the weight of the crate, or n mg= . Thus, the

coefficient of static friction is

( ) ( )( )( )

maxs

max2 0.38m s

s s f f n mg

= = = =75 N20 kg 9.80


20/44

118 CHAPTER 4

After motion exists, the friction force is that of kinetic friction, k k f n =

Since the crate moves with constant velocity when the applied force is 60 N, we find thator . Therefore, the coefficient of kinetic friction is60 N 0x k F f = = 60 Nk f =

( )( )2N 0.310 m s g

=6020 kg 9.8

=k k k f f n m = =

4.36 (a) The static friction force attempting to prevent motion may reach a maximum valueof

( ) ( )( )( )21 1max 0.50 10 kg 9.80 m s 49 Ns s s f n m g = = = =

This exceeds the force attempting to move the system, the weight of . Hence, the

system remains at rest and the acceleration is

2m

0a =

(b) Once motion begins, the friction force retarding the motion is 1 1k k k f n m g = =

2

.This is less than the force trying to move the system, weight of m . Hence, thesystem gains speed at the rate

( )( )22 11 2

4.0 kg 0.30 10 kg 9.80 m s4.0 kg 10 kg

k net

total

m g m gFa

m m m = = =

+ +20.70 m s=

4.37 (a) Since the crate has constant velocity, 0x y

a a= = .

Applying Newtons second law:

, orcos20.0 0x k F F f ma = = =x ( )300 Nk f cos 20.0 282 N= =

and , orsin 20.0 0 yF n F w = =

( ) 310 N= 300 N sin 20.0 1000 Nn = + 1.10

The coefficient of friction is then 3N

0.25610 N =282

1.10k f

n= = k


21/44


(b) In this case,

so

The friction force now becomes

sin 20.0 0 yF n F w = + =

sin 20.0 897 NF =n w=

( )( )0.256 897 N 230 Nk k f n = = =

Therefore, cos20.0x k F F f ma = = x xw

a g

=

and the acceleration is

( ) ( ) ( )2 230 N 9.80 m sco 0.509 m sN

a =300 N coss20.0 k F f g

w

= = 20.0 2

1000

4.38 (a) 206.00 m s 12.0 m s

1.20 m s5.00 s

x xx

v va

t= = =

(b) From Newtons second law, , orx k x k xF f ma f ma = = =

mg

.

The normal force exerted on the puck by the ice is n = , so the coefficient of friction is

( )( )

2

2

1.20 m s

9.80 m sk

k

m f n m

= = 0.122=

(c) ( ) ( )0av6.00 m s 12.0 m s

5.00 s 45.0 m

2 2

x xx

v vx v t t

++ = = = =

4.39 When the load on the verge of slidingforward on the bed of the slowingtruck, the static friction force has itsmaximum value

( )max =s s s lo f n m =

x x s loaF ma

ad g

This single horizontal force must give the load an acceleration equal to that the truck.

Thus, d load truck m g m a =

truck sa g

=

If slipping is to be avoided, the maximum allowable rearward acceleration of the truck isseen to be =

( ) ( )

and the minimum stopping distance will be

2 20 0x x

s

v va gmin max

02 2truck

x

= =

fsur

ur

mloadg

nur


22/44

120 CHAPTER 4

(a) If 0 12 m s and 0.500x sv = = , then ( )( )

( )( )

2

min 2

12.0 m s14.7 m

2 0.500 9.80 m sx = =

(b) Examining the calculation of Part (a) shows that neither mass is necessary

4.40 20.0 kg, 35.0 N, 196 Nm F mg= = =

(a) Since the velocity is constant,

cos 0xF F f = = , or

20.0 Ncos35.0 N

f F

= = 0.571= , 55.2 =

(b) sin 0 yF n F mg = + = , so

( )sin 196n mg F = = 35.0sin 55.2 N 167 N =

4.41 The normal force acting on the crate is given bycosn F mg = +

mg. The net force tending to move the crate down

the incline is sin s f

sinmg

, where is the force of staticfriction between the crate and the incline. If the crate is inequilibrium, then

s f

0s f = , so that sins g f F =

But, we also know ( )cosmgs s s f n F = +

Therefore, we may write ( )sinmg cosmgs F + , or

( )( )200 kg 9.80 m s 32.1 N=sins

sin30.30

g = 5.00

cosF m

cos 35.0 3.

q

nF

f

mgur

ur

ur

ur

mg

F n

fs

q = 35.0

ur

ur

ur

ur


23/44


4.42 In the vertical direction, we have

from which, n

Therefore,

300 N (400 N)sin 35.2 0n

531 N=

(

=

)( )= 0.570 531 Nk f n 302 N = =

(

From applying the second law to the horizontal motion, we have

) ( )400 N cos 35.2 302 N 30.6 = kg xa , yielding 20.798 m sxa =

Then, from 2012x x

x v t a t = + , we have ( 2 210 0.798 m s2 t= + )4.00 m , which gives3.t = 17 s

4.43 (a) The object will fall so that( )

, ormg bv

ma mg bv am

= =

where the downward direction is taken as positive.

Equilibrium ( )0a = is reached when

( )( )250 kg 9.80 m s33 m s

15 kg sterminalmg

v vb

= = = =

35.24 0 0 N 300 N

n

fm = 30.6 kg

ur

ur

m

f = bv

mg

v

ur

ur

ur

ur

(b) If the initial velocity is less than 33 m/s, then a 0 and 33 m/s is the largestvelocity attained by the object. On the other hand, if the initial velocity is greater than 33 m/s, then a 0 and 33 m/s is the smallest velocity attained by the object.Note also that if the initial velocity is 33 m/s, then a = 0 and the object continuesfalling with a constant speed of 33 m/s.


24/44

122 CHAPTER 4

4.44 (a) Find the normal force on the 25.0 kg box:nJG

( )80.0 N s 25.0 245 N 0 yF n = + =

211 Nn =

(

in

or

Now find the friction force, f, as

)0.300 211 63.4 Nk f n N = = =

x

From the second law, we have F ma =

(

, or

) ( )80.0 N cos25.0 N= 25.0 kg a 63.4 which yields 20.366 m sa =

(b) When the box is on the incline,

( ) ( )80.0 N sin 25.0 245 N cos10.0 0 yF n = + =

207 Nn =

(

giving

The friction force is

)0.300 207 N 62.2 Nk f n = = =

(

The net force parallel to the incline is

) ( )80.0 N cos 25.0 245 N sin 10.0 62.2 NxF = =-32.3 N

Thus, 232.3 N

1.29 m s25.0 kg

xFam

= = = , or 21.29 m s down the incline

x

y

F = 8 0. 0

N

25.0

mg = 245 N

n

f

ur

ur

x

y

F = 8 0. 0 N

25.0

2 4 5 N

n

f

10.0

ur

ur

f Tm

1g

n = m1g

a

m1

ur

ur ur

ur

ur

ur

Ta

m2g

m2ur

ur

ur

4.45 The acceleration of the system is found from

20

12 y

y v t at = + , or ( )2100 m 0 1.20 s2

a= +1.

which gives2

1.39 m sa =


25/44


Using the free body diagram of m , the second law gives2

( )( ) ( )( )2 25.00 kg 9.80 m s .00 kg 1.39 m sT

42.1 NT =

(

5 =

or

Then applying the second law to the horizontal motion of m 1

)( )242.1 N 10.0 kg 1. m s f = 28.2 N f 39 , or =

Since , we have1 98.0 Nn m g= =28.2 N 0.28

N= 7

98.0k

f n

= =

4.46 (a) The force of friction is found as ( )k k f n mg = =

Choose the positive direction of the x-axis in the direction of motion and apply thesecond law. We have , orx x

f a

m k f ma g

= = =

From ( )2 20 2a= + v v , withx 00, 50.0 km hv v 13.9 m s= = = , we find

( ) ( )( )2 k g x 2

0 13.9 m s= + , or ( )2

s g

13.9 m2 k

x

= (1)

With 0.100k = , this gives 98.6 mx =

(b) With 0.600k = , Equation (1) above gives 16.4 mx =


26/44

124 CHAPTER 4

4.47 (a) 2 201 102 2x x

x v t a t a t = + = + gives: ( ) ( )( )

222

2 2.00 m21.78 m s

1.50 sxx

at= = =

(b) Considering forces parallel to the incline, the secondlaw yields

( ) ( )( )229.4 N sin 30.0 3.00 kg 1.78 m sx k F f = =

9.37 Nk f =

(

or

Perpendicular to the plane, we have equilibrium, so

)29.4 N cos 30.0 0 yF n = = 25.5 Nn, or =

Then, k9.37 N= = = 0.36825.5 N

k f

n

(c) From part (b) above, 9.37 Nk f =

(d) Finally, ( )2 20 2 xa xv v= +

( )

gives

( )( )2 22 1.78 m s 2.00 m 2.67 m s+ =0 2 0xv v a x= + =

4.48 (a) Both objects start from rest and have accelerations of the same magnitude, a. Thismagnitude can be determined by applying 210 2 y y y v t a t = + to the motion of :1m

( )( )

222

2 1.00 m20.125 m s

4.00s y

at= = =

(b) Consider the free-body diagram of and apply Newtons 2nd law:1m

( )1 1 y yF ma T m g m = = +

(

a

or ) ( )(1 4.00 kg 9.80 mT m g a= + = s )2 20.125 m s 39.7 N+ =

30.0

30.0

w = mg = 29.4 N

n f k

= m k n

+y

+x

ur

T

m1g

m1ur

ur


27/44


(c) Considering the free-body diagram of : 2m

2 cos 0 y yF ma n m g = = or n m2 cos g =

so ( )( )2 cos9.00 kg 9.8n 0 67.6 Nm s 40.0= =

( )2 2msinx xF ma m g T f =

(

k = a+

Then )2 sink f m g =

(

a T

or ) ( )2 2s 39.7 N 15.9 N 9.00 kg 9.80k f m sin 40.0 0.125 m= s =

The coefficient of kinetic friction is 15.9 N 0.23567.6 N

= = =k f nk

T

w = 118 N

+ y12.0 kg

ur

37.0

w = 68.6 N

T

f

n

+x

7.00 kg

ur

ur

ur

Tur

m 2

+ y

+ x m2g

ur

nur

fk ur

q

4.49 First, taking downward as positive, applythe second law to the 12.0 kg block:

( )118 N 12.0 kg yF T = =

(

a (1)

For the 7.00 kg block, we have

)68.6 N cos37.0 54.8 Nn =

(

= , and

)( )0.250 54.8 N 1k f n 3.7 N = = =

Taking up the incline as the positive direction and applying the second law to the 7.00kg block gives ( ) ( )68.6 N sin 37.0 7.00 kgxF T f = = a , or

( )13.7 N 41.3 N 7.00 kgT a= + + (2)

Solving Equations (1) and (2) simultaneously yields 23.30 m sa .=


28/44

126 CHAPTER 4

4.50 When the minimum force F is used, the block tends toslide down the incline so the friction force,

G

sfG

is directedup the incline.

While the block is in equilibrium, we have

( )cos60.0 19.6 N sin60 0F F = + =.0x s f (1)

and

( )sin60.0 19.6 N cos60 0 yF n F = =.0 (2)

fs

F

mg = 19.6 N

n60.0

60.0

+x+ y

m

ur

ur

ur

For minimum F (impending motion), ( ) ( )max 0.300s s s f f n = = = n (3)

Equation (2) gives (4)0.866 9.80 Nn F= +

(a) Equation (3) becomes: 0.260 2.94 Ns f F= +

94 N 17.0 N 0 =

, so Equation (1) gives

, or0.500 0.260 2.F F+ + 18.5 NF =

(b) Finally, Equation (4) gives the normal force 25.8 Nn =

4.51

n ground = w/2 = 85.0 lb

F2

+x

+ y

w = 170 lb

Free-Body Diagram of Person

22.0 22.0

F1ur ur n tip

f

F =

4 5

. 8 l

b

Free-Body Diagram of Crutch Tip

22.0

+ y

+x

ur

ur

(

From the free-body diagram of the person,

) ( )1 2sin 22.0 sin 22.0 0xF F F = = 1 2, which gives or F F F= =

2 cos22.0 85.0 lbs 170 lbs 0 yF F = + = 45.8 lb

Then, yields F =


29/44


(a) Now consider the free-body diagram of a crutch tip.

( )45.8 lb sin 22.0 0xF f = = , or 17.2 lb f =

( )45.8 lb cos22.0 y tipF n = 0= , which gives 42.5 lbtipn =

For minimum coefficient of friction, the crutch tip will be on the verge of slipping,

so ( )maxs s tip f f n = = and17.242.5

lb 0.404lb

=stip

f n

= =

(b) As found above, the compression force in each crutch is

1 2 45.8 lbF F F= = =

4.52 (a) First, draw a free-body diagram (Fig. 1) of the top block. Since and,10, 19.6 N ya n= =

( )( )1 .88 Nk f n 0.300 19.6 N 5 = =

xF m =

=

T a

gives ( )10.0

or

N 5.88 N = 2.00 kg T a

2m s2.06T a = (for top block)

Top Block2.00 kg

n1 = 19.6 N

mg = 19.6 N

F

f = mkn1

Figure 1

ur

Figure 2

Bottom Block8.00 kg

Mg

fn1 n2

ur

ur urur

Now draw a free-body diagram (Fig. 2) of the bottom block and observe that x BF Ma = gives

( )5.88 N 8.00 kg= = B f a , or

20.735 m sBa = . (for the bottom block)

In time t, the distance each block moves (starting fromrest) is

( )2 2s t 21 1.03 m2T T d a t= = , and

( )2 2s t 21 0.368 m2B Bd a t= =


30/44

128 CHAPTER 4

For the top block to reach the rightedge of the bottom block, it is necessary(See Fig. 3.) that

, orT Bd d= +L

( ) ( )2 2 2 21.03 m s 0.368 m s 3.00 mt t= + which gives 2.13 st = Figure 3

L

dT dB

(b) From above, ( )( )22 21 0.368 m s 2.13 s2B Bt= =d a = 1.67 m

4.53 (a)

2.0 kg

ax = a

20 N

ur

T1

n 1ur

fk ur

3.0 kg

ax = + a

fk ur

ur

T1ur

T2

30 N

n 1ur

n 2ur

10 kg a y = a

ur

T2

98 N

(b) For the 10-kg object:( )( )2 98 N 10 kg y yF ma T a = = or ( )2 98 N 10 kg=

1 20 Nn

T a (1)

For the 2.0-kg object: or1 20 N y yF ma n = 0= =

so ( ) ( )1 0.30 20 N 6.0 Nk k f n = = =

Also, ( )( )16.0 N 2.0 kgx xF ma T a = = or ( )1 6.0 N 2.0 kg= +T a (2)

Finally, for the 3.0-kg object:

( )( )2 1 6.0 N 3.0 kgx xF ma T T a = = + or ( )2 1 6.0 N 3.0 = + kgT T (3)

Substituting Equations (1) and (2) into Equation (3) yields:

a

( ) ( ) ( )98 N 10 kg 6.0 N 2.0 kg 6.0 Na a = 3.0+ kg a or 2s86 N 5.7 m

15 kga = =

(c) Substituting the computed value for the magnitude of the acceleration intoEquations (1) and (2) gives: ( )( )21 6.0 N 2.0 kg 5.7 m s 17 N= + =T and ( )( )2m s 41 N=2 98 N 10 kg 5.7= T


31/44


32/44

130 CHAPTER 4

4.56 The acceleration of the ball is found from

( )( )

( )

22 220 20.0 m s 0 133 m s

2 1.50 m2v v

a y

= = =

From the second law, y yF F w ma = = , so

( )( )2133 m s 21.5 N yF w ma= + = =1.47 N 0.150 kg+

F

w = 1.47 N

m = 0.150 kg

ur

4.57 On the level surface, the normal force exerted on the sled by the ice equals the totalweight, or n = 600 N. Thus, the friction force is

( )( )0.050 600 N 30 Nk f n = = =

Hence, the second law yields x xF f ma = = , or

( )( )2 230 N 9.80 m s 0.49 m s600 Nx

f f a

m w g

= = = =

The distance the sled travels on the level surface before coming to rest is

( )( )

22 20

2

0 7.0 m s50 m

2 2 0.49 m sx x

x

v vx

a

= = =

4.58 (a) For the suspended block, 50.0 N 0 yF T = = , so the tension in the rope is. Then, considering the horizontal forces on the 100-N block, we find

, or

50.0 NT =

xF T = 0s f = 50.0 N f T = =s

(b) If the system is on the verge of slipping, ( )maxs s s f f n = = . Therefore,

the required coefficient of friction is 50.0 N 0.500100 N

ss

f n

= = =


33/44


(c) If 0.250k = , then the friction force acting on the 100-N block is

( )( )0.250 100 N 25.0 Nk k f n = = =

F T

Since the system is to move with constant velocity, the net horizontal force on the

100-N block must be zero, or 25.0 N 0x k f T = 25.0 N=

= =

0 y

. The required tension inthe rope is . Now, considering the forces acting on the suspended blockwhen it moves with constant velocity,

T F T w = = , giving the required weight of

this block as 25.0 Nw T = =

4.59 (a) The force that accelerates the box is the friction force between the box and thetruck bed.

(b) The maximum acceleration the truck can have before the box slides is found by

considering the maximum static friction force the truck bed can exert on the box:

( ) ( )maxs s s f n = =

(

mg

Thus, from the second law,

) ( ) ( )( )2 2maxmax 0.300 9.80 m s 2.94 m ss s s f mg

a gm m

= = = = =

4.60 Consider the vertical forces acting on the block:

( )85.0 N sin 55.0 39.2 N 0 y yF n = = =

30.4 Nn =

(

ma ,

so the normal force is

Now, consider the horizontal forces:

) ( )( )26.00 m s85.0 N cos55.0 4.00 kgx k xF f ma = = =

(

or )85.0 N cos 55.0 24.0 N 24.8 Nk f = =

The coefficient of kinetic friction is then 24.k k f n

= = 8 N 0.81430.4 N

=

4.00 kg

n

mg = 39.2 N55.0

85.0 N

fk

ur

ur


34/44

132 CHAPTER 4

4.61 When an object of mass m is on this frictionless incline, the only force acting parallel tothe incline is the parallel component of weight, sinmg directed down the incline. Theacceleration is then

( )2 2sin sin 9.80 m s si 5.62 m smgFa gm m

n35.0

= = = = =

directed down the incline.

(a) The time for the sled projected up the incline to come to rest is given by

02

0 5.00 m s0.890 s

5.62 m sv v

ta

= = =

The distance the sled travels up the incline in this time is

( )0av 0 5.00 m s 0.890 s 2.22 m2 2v vs v t t ++ = = = =

(b) The time required for the first sled to return to the bottom of the incline is the sameas the time needed to go up, that is, t 0.890 s= . In this time, the second sled musttravel down the entire 10.0 m length of the incline. The needed initial velocity is

found from 2012

s v t at = + as

( )( )2s 0.89020

5.62 m s10.0 m 8.74 m s2 0.890 s

s atvt

= = =

or 8.74 m s down the incline

4.62 Let m m . Let T be the tension in the string between , and the tension in the string between .

1 2 35.00 kg, 4.00 kg, and 3.00 kgm= = =

1 2andm m 2T 1

2 3andm m


35/44


(a) We may apply Newtons second law to each of the masses.

for m1: (1)

for m : (2)

for m : (3)

Adding these equations yields

1 1 1m a T m g=

2 2m a T m g= +

3 3m a m g T =

2

3

2 T

2

1

( ) ( )1 2 3 1 2 3m a m m m g+ + = + +m m , so

( ) 229.80 m s 1.63 m s=1 21 2

m mm m

++ +

3

3

2.0012.0 kg

ma g

m += =

kg

(b) From Equation (1), ( ) ( )( )21 1 5.00 kg 11.4 m s 57.2 Na g= + = =T m , and

from Equation (3), ( ) ( )( )22 3 3.00 kg 8.17 m s 24.5 NT m g a= = =

4.63 (a) Free-body diagrams for the two blocks are given atthe right. The coefficient of kinetic friction foraluminum on steel is 1 0.47 = while that for copperon steel is 2 0.36 = . Since 0 ya = for each block,

and . Thus,1n w= 1 2 2n w cos30.0= ( )1 1 1 0.47 f n 19.6 N 9.21 N = = =

and ( )2 2 2 0.36 f n 58.8 N cos30.0 18.3 N = = =

For the aluminum block:( )1 1 x xa T = T

(or f m a= + F m = f ma+

giving )9.21 N 2= +

(

.00 kgT (1)

For the copper block:

a

) ( )18.x xa T 58.8(

N sin 30.0 3 N 6= .00 kgF m = a or )11.1 aN 6 =

(

.00 kgT (2)

Substituting Equation (1) into Equation (2) gives

) ( )11.1 N 6.00 kg=9.21 N 2.00 kg a a or 20.232 m s= =1.86 N8.00 kg

a

Aluminum2.00 kg

ur

T

n 1ur

aur

f1ur

w1 = 19.6 N

ur

Tn 2ur

aur

f2ur

w2 = 58.8 N30

C o p p e r 6 .0 0 k g +x

(b) From Equation (1) above, ( )( )29.21 N 2.00 kg 0.232 m s 9.68 N= + =T


36/44

134 CHAPTER 4

4.64 (a) Force diagrams for penguin and sled are shown. The primed forces are reactionforces for the corresponding unprimed forces.

(forceon Pulling

Agent)

Earth

w = 49 N w2 = 98 N

Earth

n 1

T F

f1

f1

f2

n1 n 2

f2 n 2

w w 2

T W A L L

F = 45 N

ur

ur

ur

ur

ur ur

ur

ur ur

ur

ur

ur

ur

(b) The weight of the penguin is 49 N, and hence the normal force exerted on him bythe sled, , is also 49 N. Thus, the friction force acting on the penguin is:1n

( )1 1 0.20 49 N 9.8 Nk f n = = =

1 f

Since the penguin is in equilibrium, the tension in the cord attached to the wall andthe friction force must be equal: 9.8 NT =

(c) The normal force exerted on the sled by the Earth is the weight of the penguin(49 N) plus the weight of the sled (98 N). Thus, the net normal force, n equals147 N, and the friction force between sled and ground is:

2

( )2 2 0.20 147 N 29.4 Nk f n = = =

(

Applying the second law to the horizontal motion of the sled gives:

)1 245 N 10 kg f f a = or 20.58 m sa =

f1

ur

Figure 1

a

50 Nm1

n 1

m1g

m2

n2

f2 m2g

ur

ur ur

urur

ur

4.65 Figure 1 is a free-body diagram for the systemconsisting of both blocks. The friction forces are

( )1 1 1k k f n m = = g and ( )2 k f m = 2 g . For thissystem, the tension in the connecting rope is an

internal force and is not included in secondlaw calculations. The second law gives( )1 250 NxF f = 1 2m m a+ f = , which reduces to

1 2

50 Na

m m k g = +

(1)


37/44


Figure 2 gives a free-body diagram of alone. Forthis system, the tension is an external force andmust be included in the second law. We find:

, or

1m

1 1xF T f m = =

(

a

)1T m a= + k g (2)

m1

n1

m1g

T

Figure 2

ur

ur

ur

ur

a

(a) If the surface is frictionless, 0k = . Then, Equation (1) gives

2

1 2

50 N 50 N .7 m30 kg

am m

= =+ 0 1

= s

and Equation (2) yields ( )( )21.7 m s 0 17 N= +10 kgT =

(b) If 0.10k = , Equation (1) gives the acceleration as

( )( )2 20.10 9.80 m s 0.69 m s= =50 N30 kga

(

while Equation (2) gives the tension as

) ( )( )2 20.69 m s 0.10 9.80 m s 17 N = + 10 kgT =

4.66 Before he enters the water, the diver is in free-fall with an acceleration of 29.80 m s

downward. Taking downward as the positive direction, his velocity when he reaches thewater is given by

( ) ( )( )2 20 2 0 2 9.80 m s 10.0 m 14.0 m sv v a y= + = + =

His average acceleration during the 2.00 s after he enters the water is

( ) 20av

0 14.0 m s7.00 m s

2.00 sv v

at

= = =

av av , or yF F mg ma = + =

(

Continuing to take downward as the positive direction, the average upward force by thewater is found as

) ( )( )2 2av av 70.0 kg 7.00 m s 9.80 m sF m a g = = 31.18 10 N=

or 3av 1.18 10 N upwardF =


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136 CHAPTER 4

4.67 We shall choose the positive direction to be to the right and call the forces exerted byeach of the people F

G

and FG

. Thus, when pulling in the same direction, Newtonssecond law becomes

1 2

( )( )21 2 kg 1 m sF F+ = 200 .52 , or 1 2 304 NF F+ = (1)

When pulling in opposite directions,

( )( )21 2 kg 18 m sF F = 200 0.5 , or 1 2 104 NF F = (2)

Solving simultaneously, we find: 1F 100= N , and 2F = 204 N

4.68 In the vertical direction, we have

, orcos 4.0 0 yF T mg = = cos4.0mgT =

In the horizontal direction, the second law becomes:

, sosin4.0xF T ma = =

20.69 m s =sin4.0 tan4.0T a gm

= =

T

4.0+ y

+x

mgur

ur

4.69 The magnitude of the acceleration is2

2.00 m s=a for all three blocks and applyingNewtons second law to the 10.0-kg block gives

( )( ) ( )( )2 2110.0 kg 9.80 m s 10.0 kg 2.00 m sT = , or 1 78.0 NT =

Applying the second law to the 5.00-kg block gives:

( )( ) ( )( )2 2m s1 2 5.00 kg 9.80 m s 5.00 kg 2.00k T T =

1 78.0 N=

With , this simplifies to:T ( )2 68.0 N- 49.0 NT k = (1)


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For the 3.00-kg block, the second law gives

With

2 sin25.0k T n mg ma =

2 2s , and cos25.0n mg= 3.00 kg, 2.00 m s , 9.80 mm a g= = =

(

, this reduces to:

)2 26.6 N 18.4 Nk T = (2)

Solving Equations (1) and (2) simultaneously, and using the value of T from above, wefind that

1

(a) 1 N78.0T = , 2 35.9 NT = , and (b) 0.656k =

4.70 The scale simply reads the magnitude of the normal force exerted on the student by theseat. The seat is parallel to the track, and hence inclined at 30.0 to the horizontal. Thus,the magnitude of this normal force and the scale reading is

( )cos 200 lb cos30.0 173 lbn mg = = =

q

T

m

+x

+ y

a

mgq

ur

ur

ur

4.71 Choose the positive x axis to be down the incline and the y axis perpendicular to this as shown in the free-bodydiagram of the toy. The acceleration of the toy then hascomponents of

230.0 m s0, and 5.00 m s6.00 s

x y x

va a

t+= = = = +

Applying the second law to the toy gives:

(a) sinx xF mg ma = ,=2

1 12

5.00 m ssin sin 30.7

9.80 m sxa

g

= = =

and

(b) cos 0 y yF T mg ma = , or = =

( )( )29.80 m s cos 30.7 0.843 N =cos 0.100 kgT mg = =


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138 CHAPTER 4

4.72 Taking the downward direction as positive, applying the second law to the fallingperson yields y yF mg f ma = = , or

2 2100 N 8.6 m s80 kg y

a g

= 9.80 m s f m

= =

(

Then, )2 20 y yv v 2 ya y= +

(

gives the velocity just before hitting the net as

) ( )( )2 22 8.6 m s 30 m 23+ =0 2 0 y y ya y= + = m sv v

4.73 The acceleration the car has as it is coming to a stop is

( )

( )

( )

22 220 0 35 m s 0.61 m s

2 2 1000 m

v va

x

= = =

Thus, the magnitude of the total retarding force acting on the car is

( )2 228800 N 0.61 m s 5.5 10 N9.80 m sw

F m a a g

= = = =

4.74 (a) In the vertical direction, we have

( )8000 N sin65.0 0 y yF w = = =

( )

ma

so 38000 N sin65.0 7.25 10 Nw = = w = mg

8000 N

65.0

urur

(b) Along the horizontal, the second law yields

( )8000 N cos65.0x xw

F m g

= = =

(

xa a , or

) ( )( )2 23

9.80 m s 8000 N cos65.08000 N cos65.04.57m s

7.25 10 Nx g

aw

= = =


41/44


4.75 First, we will compute the needed accelerations:

(1) Before it starts to move: 0 ya =

(2) During the first 0.80 s: 0 21.2 m s 0

1.5 m s0.80 s

y y y

v va

t

= = =

(3) While moving at constant velocity: 0 ya =

(4) During the last 1.5 s: 0 20 1.2 m s

0.80 m s1.5 s

y y y

v va

t

= = =

Applying Newtons second law to the vertical motion of the man gives:

y yF n mg ma = = , or ( ) yn m g a= +

(a) When 0a = , y ( )( )2 272 kg 9.80 m s 0 7.1 10 Nn = + =

(b) When 2m s y 1.5a = , 28.1 10 Nn =

(c) When 0a = , y 27.1 10 Nn =

(d) When 20.80 m s ya = , 26.5 10 Nn =

4.76 Consider the two free-bodydiagrams, one of the penguin aloneand one of the combined systemconsisting of penguin plus sled.

The normal force exerted on thepenguin by the sled is

and the normal force exerted on the combined system by the ground is

The penguin is accelerated forward by the static friction force exerted on it by the sled.When the penguin is on the verge of slipping, this acceleration is

1 1 1n w m g= =

2 total totn w m= =

(

130 Nal g =

) ( ) ( )( )1 1 2 2maxmax1 1

0.700 9.80 m s 6.86 m ss s f m g

a gm m

= = = = =

n 1

w1 = 70.0 N

Ff1

f2

wtotal = 130 N

n 2urur

ur

ur

ur


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140 CHAPTER 4

Since the penguin does not slip on the sled, the combined system must have the sameacceleration as the penguin. Hence, applying the second law to the combined systemgives , or2 mx totF F f m a = = axal

( )2 maxtotal k F f m a = + =

( )(

max

total

total

ww a

g

+

)

This yields ( )22130 N 6.86 m s 104 N

9.80 m s

= + =

0.100 130 NF

4.77 Since the board is in equilibrium, 0xF = and we see thatthe normal forces must have the same magnitudes on bothsides of the board. Also, if the minimum normal forces(compression forces) are being applied, the board is on the

verge of slipping and the friction force on each side is( )s smax f f n = =

The board is also in equilibrium in the vertical direction,so

2 0, or2w

w f = = yF f =

The minimum compression force needed is then

( )95. 72.0 N2 2s s f w = = = 5 N0.663 =n

w = 95.5 N

f fn nur ur

ur ur

4.78 The friction force exerted on the mug by the moving tablecloth is the only horizontalforce the mug experiences during this process. Thus, the horizontal acceleration of themug will be

20.100 N 0.500 m s0.200 kg

k mug

mug

f a

m= = =

The cloth and the mug both start from rest ( 0 0xv = ) at time 0t = . Then, at time , the

horizontal displacements of the mug and cloth are given by

0t >210 2x xx v t a t+ = as:

( ) ( )2 2 2120 0.500 m s 0.250mugx t = + = 2m s t and ( ) ( )2 2 2 2120 3.00 m s 1.50 m sclothx t = + = t


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In order for the edge of the cloth to slip under the mug, it isnecessary that , or0.300 mcloth mugx x = +

( ) ( )2 2 2 2s 0.250 m st t= +

( )

1.50 m 0.300 m

The elapsed time when this occurs is

20.300 m 0.490s

1.50 0.250 m st = =

At this time, the mug has moved a distance of ( )( )20.250 m s 0.490 s 6mugx = =2 2.00 10 m 6.00 cm =

4.79 (a) Consider the first free-bodydiagram in which Chris and thechair treated as a combined system.The weight of this system is

, and its mass is480 Ntotalw =49.0 kgtotaltotal

wm

g= =

2

Taking upward as positive, theacceleration of this system is foundfrom the second law as

y total total yF T = w = m a

Thus,2

0.408 m s= +500 N 480

49.0 kg ya

=

Nor

2

0.408 m s upward

wtotal = 320 N + 160 N

T = 250 N

T = 250 N

a

n

wChris = 320 N

T = 250 N

ur

ur

(b) The downward force that Chris exerts on the chair has the same magnitude as theupward normal force exerted on Chris by the chair. This is found from the free- body diagram of Chris alone as

Hence,

, y Chris Chris y Chris y ChrisF T n w m a n m a w T = + = = +

( )22320 N

0.408 m s 320 N 250 N 83.3 N9.80 m s

n

= + =


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142 CHAPTER 4

4.80 Let R represent the horizontal force of air resistance. Since thehelicopter and bucket move at constant velocity,

JG

0x ya a= = .The second law then gives:

cos 40.0 0 yF T mg = = ,

or cos40.0

mg=

sin40.0R T

T

Also, , or

sin 40.0 0xF T R = = =

Thus, ( )( )2 34 10 Ntan 40.0 620 kgR mg= = 9.80 m s tan 0.0 5.10 =

4.81 Consider two free-body diagrams, one of thecement bag and one of the junction of the threewires as given at the right.

Using the first diagram:

(1)

Then, using the second free-body diagram:

30 yF T = = w

11 1 2

2

coscos 0 orcos

T T T 12 20 cosxF T

=

2 20 sin yF T

= =

1 1 3 2 2sin 0 or sinT T T

(2)

and 1 1sinT 3T + = + = = (3)

w

T

R

w = mg

40.0

+ y

+xur

ur

urur

q 1

ur

T3

ur

T1q 2

ur

T2

Cement

ur

T3

ur

(a) Substituting Equations (1) and (2) into Equation (3) yields

( )1 2 1 1 2sin cos sin cos cosT w 2 + =

Making use of the trigonometric identity ( )1 2 2 1 1sin sin cos sin cos 2 + = +

( )

, the

above expression reduces to: 211 2

cossin

wT

=+

(b) If w 1 2325 N, while 10.0 and 25.0 ,

( )

= = we find that:= ( )

( )2

11 2

325 N cos 25.0cos514 N

sin sin 10.0 25.0wT

= = =+ +

( )1 12

2

514 N cos10.0cos 558 Ncos cos25.0

T T

= = =

and 3 325T w= =

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