Hallyday Fundamentals of Physics 8E Student Solution Manual

STU DENT SOLUTIONS MAN UAL J. Richard Christman Professor Emeritus U.,S. Coast Guard Academy

FUNDAMENTALS OF PHYSICS Eighth Edition

David Halliday (Jnivers iQ of P itts burgh

Robert Resnick Rens s elaer

Polytechnic Institute

Jearl Walker Cleveland State Univers iQ

John Wiley & Sons, Inc.

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2008 John Wiley

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PREFACE

This solutions manual is designed for use with the textbook Fundamentals of Physics, eighth edition, by David Halliday, Robert Resnick, and Jearl Walker. Its primary puqpose is to show students by example how to solve various types of problems given at the ends of chapters in the text.

Most of the solutions start from definitions or fundamental relationships and the final equation is derived. This technique highlights the fundamentals and at the same time gives students the opportunity to review the mathematical steps required to obtain a solution. The mere plugging of numbers into equations derived in the text is avoided for the most part. We hope students will learn to examine any assumptions that are made in setting up and solving each problem. Problems in this manual were selected by Jearl Walker. Their solutions are the responsibility of the author alone.

The author is extremely grateful to Geraldine Osnato, who oversaw this project, and to her capable assistant Aly Rentrop. For their help and encouragement, special thanks go to the good people of Wiley who saw this manual through production. The author is especially thankful for the dedicated work of Karen Christman, who carefully read and coffected an earlier version of this manual. He is also grateful for the encouragement and strong support of his wife, Mary Ellen Christman. J. Richard Christman Professor Emeritus IJ.S. Coast Guard Academy New London, CT 06320

TABLE

OF CONTENTS

Chapterl .., . .. . .... . o. .. . I Chapter?. .i . . . . . . . . . . . . . .4 Chapter 3 . . . . . . . . . . . 0 . . . . . 10 Chaptet 4 . .. . . . . . . . . . . . . . .I4 Chaptgr5 . . . . . . . . . . . . . . . . .21 Chapter6. . . . . . . . . . . . . . . . .28

ChapterT o.......... o.... .37 ChapterS . . . . . . . . . . . . . . . . .42

Chapter9. Chapter Chapter Chapter Chapter Chapter Chaptgr Chapter Chaptgr Chapter Chaptgr

...............

50 10 . . . . . . . . . . . . . . . . 58

11

.......... r... ..63

12.. .. . .. . .. .. .. ..71 13 ... o... . .. . ... ..77 14 .. . . . . r . . . . . . . . . 84

15 . . . . . . . . . . 16.. .. ... . .. 17 . . . . . . . . . . 18 . . . . . . . . . . 19 . . . , . . . . . . Chaptgr20. . . . . . . . . .

. . . . . .

. . . . o 89 . . .. .95 . . . , . 101 . . . r . 109 . . . . . 115 . . . . . 122

Chapter2l ............... a I28 Chaptet 22. . . . . . . . . . . . . . . . 134

Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

23

..

.. ..

. .140

24

o.o.o..146

25

.......154

26

. . . . . . . 159

27

28 29 30 31

32 33

34 35

36 37 38 39

40

4l 42 43

44

.......162

......

.170

. . . . . . .175 . . . . . . . 193 . . . . . . , l9l

....,..199

...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ...... ......

.205 .213 .221 .229 .235 .239 .243 .247 .251 .254 .260 .264

Chapter

L

1yd

- (0.9144mX106

pmlm)

- 9.144 x 105 pm.

3

use the given conversion factors. (a) The distance d

d,

in rods is

- 4.0 furlongs -

(4.0 furlongsX2Ol . 168 m/furlong)

5.0292mf rod

(b) The distance in chains is d

- 4.Lfurrongs -

(4'0 furlongsX20l ' 168 m/furlong) 20.L7 mlchain

:

4|chains

.

I(a) The circumference of a sphere of radius R is given by 2r R. Substitute R 106mX10-tk^lm) should obtain 4.00

-

(6.37 x

x 104km.

(b) The surface area of a sphere is given by 4trR2, so the surface area of Earth is 4n(6.37 x 103

k*)'

(c) The volume of a sphere is given by (4nlrR3, so the volume of Earth is G"13X6.37 x 103 k*)3 _ 1.08 x 1012 km3. t7 None of the clocks advance by exactly 24h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. If the clock reading jumps around from one 24-h period to another, it cannot be coffected since it would impossible to tell what the coffection should be. The followittg table gives the coffections (in seconds) that must be applied to the reading on each clock for each 24-h period. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning.

Chapter

I

CLOCK

A

-Mon.

Mon. -Tues.

Sun.

-Thurs.

-Sat

-15

-15

Wed.

-Wed.

-16

-16

-3

+5

-15 -10

-17

B

+5

+6

C

-58

-58

-s8

-s8

-58

+67 +70

D E

+67 +55

Fri.

Thurs. -Fri.

Tues.

+67 +2

+67 +20

+67 +10

-7 -58 +67 +10

Clocks C and D are the most consistent. For each clock the same coffection must be applied for each period. The coffection for clock C is less than the coffection for clock D, so we judge clock C to be the best and clock D to be the next best. The coffection that must be applied to clock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to *10 s, for clock E it is in the range from -70 s to -2s. After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest range. From best the worst, the ranking of the clocks is C, D, A, B, E. 21

(a) Convert grams to kilograms and cubic centimeters to cubic meters: 1 g 1cm3 - (1 x To-2 m)3

rs-(1

e)(#) (,%) \t

_ I x 10-' kg and

:rx1o3kg

(b) Divide the mass (in kilograms) of the water by the time (in seconds) taken to drain it. The mass is the product of the volume of water and its density: M - (5700m3X1 x 103 kg/nt') 5.70 x 106kg. The time is t - (10.0hX3600s/h) - 3.60 x 104 s, so the mass flow rate n is R

M- t

5.70 x 106kg1 58 kg/s 3.60 x 104 s

"

3s (a) The amount of fuel she believes she needs is (750mi)l(0mif gal): 18.8gal. This is actually the number of IJ.K. gallons she needs although she believes it is the number of IJ.S. gallons.

(b) The ratio of the U.K. gallon to the U.S. gallon is (4.545963 IL)l(3.785 3060L) The number of U.S. gallons she actually needs is (18.8

:

1.201.

IJ.K. galx 1.201 U.S gallu.K. gal) : 22.5 IJ.S. gal .

39

The volume of a cord of wood is V - (8 ftx 4 tt)(4 ft) : I28ft3. IJse l ft ft)3 - 3.62m3. Thus 1.0 m3 of wood coffesponds Appendix D) to obtain V - I28ft3X0 .3048 ^l to (l 13.62) cord - 0 .28 cord.

2

Chapter

I

4T

(a) The difference in the total amount between 73 freight tons and 73 displacement tons is (8 barrel

bulk/freight ton)(73 freight ton)

- (7 barrel bulk/displacement : 73 barrel bulk

ton)(73 displacementton)

.

Now

l banel bulk -

0.141 5m3

bushel):

- (0.!41 5m3)(28.378U.S.

4.01

5U.S. bushel ,

SO

- (73 barrel bulk)(4.01 5 U.S. bushellbarcel bulk) -

73barceI bulk

293

U.S. bushel

.

(b) The difference in the total amount between 73 register tons and 73 displacement tons is (z}barrel bulk/register ton)(7 3 register ton)

:

(7 barrel bulk/displacement ton)(73 displacementton) 949 barrel

bulk.

Thus g4:gbarrel bulk

- (949barrel bulk)(4.01 5U.S. bushellbarcel bulk):3810U.S. bushel

.

45

m-1x10-3km, 0.12

ATJ

lmin.

57

(a) We want to convert parsecs to astronomical units. The distance between two points on a circle of radius r is d - 2r stn 0 12, where 0 is the angle subtended by the radtal lines to the points. See the figure to the right. Thus r - d,lz sin 0 12 and

l pc where

!" -

t

^",., I" 12) -

2 sin(

2.06

x

105

/-"

AU,

- (2.78 x 10-4)' was used. Finally 1 AU - (1 AU) 1Q.06 x 105 AU lpc) : 4.9 x 10-u p,

.,

"

012

(I13600)"

.

(b) A light year is (1.86

x

10s

mrlsxl.0 yX3 65.3 daly)Q|hld)(3600 s/h) - 5 .87 x

1012

mt

and

l AuJ :

92'9 x

l.5g L'J\) x 1o-' ly. },9^6 Tt,; 5.87xlot2millyChapter

I

Chapter 2 1 (a) The average velocity during any time interval is the displacement during that interval divided by the duration of the interval: ?)av' - Lr I Lt, where Lr is the displacement and Lt is the time interval. In this case the interval is divided into two parts. During the first part the displacement is Lr 1 : 40 km and the time interval is

Ltr

:

During the second part the displacement

(40km) (30

k*/h) -1.33h.

is Lrz: 40km and the time intenral is

Ltz-(60(4gq) l*ttr - o '67h ' Both displacements arc in the same direction, so the total displacement is Lr 40km +40km:80km. The total time interval is Lt: Lt1 * Ltz- 1.33h+0.67h:2.00h. The average velocity is (80km) : nl

'avg

(2.0 h)

40k*/h.

(b) The average speed is the total distance traveled divided by the time. In this case the total distance is the magnitude of the total displacement, so the average speed is 40 km fh. (c) Assume the automobile passes the origin at r g0 : time t 0. Then its coordinate as a function of Gm) time is as shown as the solid lines on the graph 60 to the right. The average velocity -- --Q- --Jis the slope -- -r - of 40

the dotted

line.

20

1.0

0.5

t (h)

1.5

2.0

5-

Substitute,inturn, t- I,2,3, and4sintotheexpression meters and t is in seconds: (a)

(b) (c)

(d)

r(t):3t-4*+#,where

r isin

r(Is): (3^lsxl s) - (mls2xl r)2 + (l^11311 s;3 - 0 r(zs) : (3mlsx2 s) - (4mls\(2s)2 + Qmls';12 s)' - -2m r(3s): (3^lsx3 s) r(4s): (3mlsx4s) -

(4mls2x3

s)2

+ Omlr311:

r13

-0

(4mls2x4s)2 + Q^lr3;1+s;3

(e) The displacement during an interval is the coordinate at the end of the interval minus the coordinate at the beginning. For the interval from t -- 0 to t - 4 s, the displacement is Lr - r(4s) - tr(0) - I2m- 0 - +12m. The displacement is in the positive r direction.

4

Chapter 2

(0 The average velocity during an interval is defined as the displacement over the interval divided by the duration of the interval: uavs- L, I Lt. For the interval from t- 2s to t-4s the displacement is Lr - r(4 s) r(2 s) - I2m (-2m) - I4m and the time intenral is

Lt:

4s

-

2s

:

2s. Thus

Lr l4m 'uavg:E-7m/s. (d) The solid curye on the graph to the right ^ r r2.o shows the coordrnate r as a function of time. (m) The slope of the dotted line is the average velocitybetween t- 2.0s and t- 4.0s.

9.0 6.0 3.0 0.0

-3.0 19

If ur is the velocity at the beginning of a time interval (at time t)

and u2 is the velocity at the end (at tz), then the average acceleration in the interval is given by eav' Take h : 0, u1 : 18 m/s, t2 - 2.4 s, and u2

- 18 m/s 2.4s

m/s auus -30

t -zo ^lr'

.

The negative sign indicates that the acceleration is opposite to the original direction of travel. 25

(a) Solve u- us* at for t: t - (u - uo) f a. Substitute u-0.1(3.0 x 108 mls):3.0 u0 : 0, and e,:9.8 The result is t - 3.06 x 106 s. This is !,zmonths. ^lr'. (b) Evaluate r 4.6 x 1013 m.

x

I07

ml

105

mls,

s,

27

Solveu2 -u3+2a(r-r0 fora. TakeffO:0. Then a-(r'-ril|2tr. Use'u0:1.50x u - 5.70 x 106m/s, and r - 1.0cm - 0.010m. The result is (5.70 x 106m/s)2 - (1.50 x 105m/s)2 a_ - 1.62x l0r, ^/r, . 2(0.010m) 33 (a) Take Ug

-

frs- 0, and solve r-

56.0k*/h

-

15.55 m/s, and

&-

ust

t

* *ot'

for

a: a_

2(r

ust)f

- 2.00 s. The result is

2lzq.0m- (15.55m1sx2.00s)]

1

t2. Substitute r-

rr ^

24.0m,

r^Z

Chapter

2

5

The negative sign indicates that the acceleration is opposite the direction of motion of the car. The car is slowing down.

(b) Evaluate

u

(30.3 km lh). 45

(a) Take the A axis to be positive in the upward direction and take t - 0 and A - 0 at the point from which the wrench was dropped. If h is the height from which it was dropped, then the ground is at A : -h. Solve u2 - uzo + 2gh for h:

hSubstitute uo : 0,

1)

1)

:

uo

2s

: -24mls, and g - g.8 m/s2: h-

(b) Solve

ul

u2

(24mls)2

:29.4m.

2(9.8 m/s2)

- gt for t:

t-

(to

-u)

24mls 9.8

-2.45s.

^l12

(c)

t 0

t

(s) 2

a-10

(-)-zo

0

(s) 2

u -10 (m/s)

'

'_20

-30

-30

t (s)

The acceleration is constant until the wrench hits the ground: a- -g.8 m/s2. Its graph is as shown on the right.

0

a-5 (-is2)

-10 _15

47

(a) At the highest point the velocity of the ball is instantaneously zero. Take the A axis to be in u2 upward, set u and solve for u$ us- \EA. Substitute g

6

Chapter 2

A:50mtoget

- 31 m/s. - uot t7t'

2(9.8^ls2x50m)

(b) It will be in the atr until A-0 again. Solve A for t. Since A : solutions aret:0 and t 2uo I g . Rej ect the first and accept the second: t-_ .L

2uo 2(31 m/s) -

I

9.8

(c)

the two

-6.4s.

^lt' u40

a60

(n/s)

(m) 40

20

20

0

_20

68

t

(s)

-40

t (s)

The acceleration is constant while the ball is in

flight:

a,

0

on the right.

a-5

(*/r')

-10 _15

49

(a) Take the A axis to be upward and place the origin on the ground, under the balloon. Since the package is dropped, its initial velocity is the same as the velocity of the balloon, +lTm/s. The initial coordinate of the package is ao: 80 m; when it hits the ground its coordinate is zero. Solve a : Uo + uot - *gt' for t:

t-uo+ I

9.8

where the positive solution was used. package was dropped.

(b) Use 't):7)0

- gt: l}mls -

;-i-

L

ll

r

:J.-TD-c

g.g m/s2 V (9.8 mls2)2 negative value for t coffesponds to a time before the

^lt'

A

(9.8 mls2x5.4s)

-

-.41

m/s. Its speed is 4Imf

s.

51

The speed of the boat is given by u6 _ d lt, where d is the distance of the boat from the bridge when the key is dropped (IZm) andt is the time the key takes in falling. To calculate t, put the Chapter 2

origin of the coordinate system at the point where the key is dropped and take the y axis to be positive in the upward direction. Take the time to be zero at the instant the key is dropped. You want to compute the time t when A - -45 m. Since the initial velocity of the key is zera, the coordinate of the key is given by A - -+g*. Thus

2(-45 m)

-3.03s.

This means

U6:

12m 3.03

s

- 4.0 m/s

.

-I5 First find the velocity of the ball just before it hits the ground. During contact with the ground its average acceleration is given by

a'avg

Lu N

,

where Lu is the change in its velocity durittg contact and

Lt is the time of contact.

To find the velocity just before contact take the A axis to be positive in the upward direction and put the origin at the point where the ball is dropped. Take the time t to be zero when it is dropped. The ball hits the ground when A - -15.0m. Its velocity then is found fromu2- -\ga, SO

- -l

-ze.g^ls2x-15.0m)

: -17.rmls.

The negative sign is used since the ball is traveling downward at the time of contact. The average acceleration during contact with the ground is

- (- 17 .I m/s) *avs0 zo.o x 1o-3 s rt:

857

^lt2

The positive sign indicates it is upward. 89

The velocity at time t is given by integration. use the condition that 2.5(2.0 s)2 - 7.0 m/s. The velocity

u: f adt: f S.\tdt:2.5t2+C, where e is aconstantof u- +ITmlsatt- 2.0stoobtain C:u-2.5t2 - ITmlsatt - 4.0 s * 7.0^lsr2.5t2 - 7.0^ls*2.5(4.0 r)2 : 47 m/s.

9l (a) First convert the final velocity to meters per second: 1) : (60 k*/hx1000 m/km)1Q600 s/h) 16.7mf s. The average acceleration is ult - (16.7^ls)/(5.4s):3.1

(b) Since the initial velocity is zero,the distance traveled is r

8

Chapter 2

-

Lot'

-

^lr'.

Ltl .L ^ls2x5 .4 r)2

-

- 45 m.

(c) Solve n

- *ot'

for

t.

The result is 2(0.25

103

m)

-

13 s.

97

The driving time before the change in speed limit was t6 - Lr luo, where Lr is the distance and u6 is the original speed limit. The driving time after the change is to - A^r lro, where I)q is the new speed limit. The time saved is

|

t6- te- Lr(*;)

-(700km)(0.62ramilkm)( milh \ss

I

)-r2h. L'L lh)

)

65 mi

This is about t h and TZmrn. 99

Let t be the time to reach the highest point and us be the initial velocity. The velocity at the highest point is zero, so 0 - us - gt and us- gt. Thus H- uot i7t': gtz ig* where the substitution was made for ns. Let H2 be the second height. It is given by H2 : LgQt)2 - 2gt2 4H. The balls must be thrown to four times the original height. I07

(a) Suppose the iceboat has coordinate Ut at time h and coordinate Az at time t2. If a is the acceleration of the iceboat and tre is its velocity att - 0, thenAr: ?rltt+Lot? andy2: u0t2**t7. Solve these simultaneously for a and us. The results are 2(azh

-

t1t2(t2

aftz)

- tt)

ug:ffi

and

h - 2.0s and tz: 3.0s. The graph indicates that at: 16m and Uz - 27 m. These values a, - 2.0^lt2 and us - 6.0m/s. (b) The velocity of the iceboat at t - 3.0s is ,u: uo* at:6.0 mls * (2.0m1s2x3.0s) - L}mf s. (c) The coordinate at the end of 3.0 s is az - 27 m. The coordinate at the end of 6.0 s is Uz: uotz+ |"*3- (6.0^lsx6.0s) + l(2.0m1s2x6.0r)' - 72m. The distance traveled during the Take

yield

second 3.0-s interval is Uz

-

Uz:72m

-

27

m-

45m.

Chapter 2

Chapter

3

1

(a)use

a:

lmt

to obtain

e,

:

(-25.0 m)2 + (+40.0 m)2

:

(b) The tangent of the angle between the vector and the positive tano

aa

- clr-

47

r

.2m.

axis is

4o'o m

-25.0 m

The inverse tangent is -58.0o or -58.0" + 180' - I22". The first angle has a positive cosine and a negative sine. It is not correct. The second angle has a negative cosine and a positive sine. It is correct for a vector with a negative r component and a positive A component. 3

The

r

component is given by

by a,a

ar -

(7 .3

m) cos 250o

-- -2.5 m and the A component is given

the components can also be computed using ar - -Q.3 m) cos 70" and e,a - -Q,3 m) sin 70o . It is also 20" from the negative A axis, so you might also use a,n - -Q.3 ) sin 20" and a,a : -(7.3 m) cos 20". These expressions give the same results. 7

(a) The magnitude of the displacement is the distance from one corner to the diametrically opposite corner: d - \/ (3.00 m)2 + (3 .70 m)2 + (4.30 m)2 see this, look at the diagram of the room, with the displacement vector shown. The length of the diagonal across the floor, under the displacement vector, is given by the Pythagorean theorem: L- tM, where (. is the length and u is the width of the room. Now this diagonal and the room height form a right triangle with the displacement vector as the hypotenuse, so the length of the displacement vector is given by

d-{L2+h2(b), (c), and (d) The displacement vector is along the straight line from the beginning to the end point of the trip. Since a straight line is the shortest distance between two points the length of the path cannot be less than the magnitude of the displacement. It can be greater, however. The fly might, for example, crawl along the edges of the room. Its displacement would be the same but the path length would be (. + ut * h" The path length is the same as the magnitude of the displacement if the fly flies along the displacement vector.

10

Chapter 3

(e) Take the r axis to be out of the page, the y axis to be to the right, and the z axis to be upward. Then the r component of the displacement is 'u) : 3.70 m, the A component of the displacement is 4.30m, and the z component is 3.00m. Thus i- (3.70m)i +(4.30m)j+(3.00m)t. You may write an equally correct answer by interchanging the length, width, and height. (f) Suppose the path of the fly is as shown by the dotted lines on the upper diagram. Pretend there is a hinge where the front wall of the room joins the floor and lay the wall down as shown on the lower diagram. The shortest walking distance between the /l-t. lower left back of the room and the upper right / ..t ./ \/ front corner is the dotted straight line shown on the diagram. Its length is

(

T

, umtn

(3.70m

* 3.00 m)2 + (4.30 m)2 -

7

.96m

w .

h

2

d,+6. Then r, : an*b* 10m. Thus r-: (-9.0m)i*(10m)j.

(a) Let

i-

(b) The magnitude of the resultant is

4.0m-

13m

- -g.0m andrr: &y*b, - 3.0m +7.0m -

r : ^ lr2 + ,2 -

Y'*''a

(-9.0m)2 + (10m)2 _

13 m.

(c) The angle 0 between the resultant and the positive r axis is given by tan? - ,a lr* (10 m)l(-9.0m) - -1.1. 0 is either -48o or I32". The first angle has a positive cosine and a negative sine while the second angle has a negative cosine and positive sine. Since the n component of the resultant is negative and the A component is positive, 0

-

I32o

.

t7 (a) and (b) The vector d has a magnitude 10.0m and makes the angle 30o with the positive n axis, so its components are a,n- (10.0m)cos30o - 8.67m and aa: (10.0m)sin30o - 5.00m. The vector 6'hur a magnitude of 10.0m and makes an angle of 135o with the positive n axis, so its components are b*- (10.0m)cos 135o - -7.07 m and ba _ (10.0m) sin 135o The components of the sum are rn: a* * b*- 8.67 m - 7.07 m- 1.60m and ra: ea * ba:

5.0m*7.07

m- 12.1 m.

(c) The magnitude of

iis r - ym:

(1.60m)2 + (12.I

-

I2.2m.

(d) The tangent of the angle 0 between i and the positive r axis is given by tan? _ rr lr* (lz.Lm)l(l.60m) : 7.56. e is either 82.5o or 262.5o. The first angle has a positive cosine and a positive sine and so is the correct answer. Chapter 3

11

39

: a*b* * oab, r a"br,

Since ab cos Q

cos

d:

a*b*+a,abr+&"b, ab

:

The magnitudes of the vectors given in the problem are e, (3.0)2+(3.0)2+(3.0)2 :3.7. (2.0)2 + (1.0)2 + (3.0)2 The angle between them is found from

b-

cos

and the angle is

d:22o

Q:

(3.0X2.0) + (3.0X1.0) + (3.0X3.0) _ 0 .926 (5.2)(3.7)

.

43

(a) and (b) The vector d, is along the r axis, so its r component is ar component is zero. (c) and (d) The r component of 6 is b* - b cos 0 -- (4.00 m) cos 30.0o component is ba - b sin 0 - (4.00 m) sin 30.0o - 2.00 m. (e) and (f) The r component of c*is cr -- ccos(0 + 90")- (10.0m)cos I20" - -5.00m and the A component is ca: csin(9 + 90o): (10.0m)sin I20o - 8.66m. (g) and (h) In terms of components cr : pa* * qb* and ca : paa + eba. Solve these equations simultaneously for p and q. The result is

P:ffi-- bac* b*ca

and

(3.46 mX8 .66m)

q:ffi-- arca -(3.00 aac*

-

(2.00

mX-5.00 m)

mXS .66 m)

--6'67

-4'34'

47 (a) The scalar product is

d . b - abcos d

:

(10X6.0) cos 60o

:

30

(b) The magnitude of the vector product is

ld*61

- ab sin d:

(10X6.0) sin 60o

: 52

.

51

Take the r axis to run west to east and the A axis to run south to north, with the origin at the starting point. Let f,dest - (90.0 k-) j be the position of the destination and r'r the position of the sailor after the first leg of his journey and iz be the remaining displacement

12

Chapter

j

required to complete the journey. The total journey is the vector sum ?'dest : r'I + iz and ri,,, iz - (90 km)i (50.0 km) i.

n:

-

of the two parts,

so

-

The magnitude of the remaining trip is T2

'?.*

+ r?a

(50.0 km)2 + (50.0 km)2

_

103

km

.

The tangent of the angle with the positiye r direction is tan d : r2a lrr* - (90.0 km) 160.0 km) 1.80. The angle is either 60.9' or 180'+ 60.9o :24Io. Since the sailor must sail northwest to reach his destination the correct angle is 24Io . This is equivalent to 60.9" north of west.

7l According to the problem statement to obtain 2A-2.0i+ 8.0j and then

i+ E - 6.0i+ 1.0j and A- E - -4.0i +7.0j. Add these A- 1.0i++.0j. The magnitude of A'is (1.0)2 + (4.0)2

Chapter 3

13

Chapter

4

7

The average velocity is the total displacement divided by the time interval. The total displacement i is the sum of three displacements, each calcul ated as the product of a velocity and a time interval. The first has a magnitude of (60.0 km lh)(40.0 min) l$0.0 min lh) direction is east. If we take the r axis to be toward the east and the A axis to be toward the north, then this displacement is i1 - (40.0 km) i. The second displacement has a magnitude of (60.0 km lh)(20.0 min) 1rc0.0 min lh) direction is 50.0o east of north, so it may be written

iz - (20.0km) sin50.0'i + (20.0km)cos 50.0"i -

(15.3

:

20.0

km. Its

km)i* (12.9k*)i.

The third displacement has a magnitude of (60.0 km lh)(50.0 min) l(60.0 min lh) _ 50.0 km. Its direction is west, so the displacement may be written fi - (-50 km) i. The total displacement is

r-: r-r+ i2+ i, - (40.0km)i+ (15.3km)i+ (I2.9km)j - (50k*)i

- (5.3 km) i + (12.9 km) j

.

The total time for the trip is 40 min * 20 mtn + 50 min interval to obtain an average velocity of du,,e - (2.9k^lh) i + (7 .05k*/h)i. The magnitude of the average velocity is lr-uu*l

:

-7.6km/s

and the angle O it makes with the positive

r

axis satisfies

tanQ:'!:.^l,n - 2.43 2.9 k-/h The angle is d

.

: 68o.

11

(a) The velocity is the derivative of the position vector with respect to time: 6

- *(i. 4*i +,t) :

8,i + r

in meters per second for t in seconds. (b) The acceleration is the derivative of the velocity with respect to time:

d-* (r,i.o) :8i 14

Chapter 4

in meters per second

squared.

17

(a) The velocity of the particle at any time t is given by d - 6o + dt, where do is the initial velocity and d is the acceleration. The r component is 'trr:'uyrl a,nt - 3.00 mls - (1.00 mls2)t and the g component rs uy - 'tr1a* aat - -(0.500 mlsz)t When the particle reaches its maximum r coordinate 'trr:0. This means 3.00 mls - (1.00 mls\t - 0 or t - 3.00s. The A component of the velocity atthis time is rra - (-0.500 mls2x3.00s): -1.50m/s. Thus 6 - (-1.50 mls)j. (b) The coordinates of the particle at any time t are r- nyrt + *.o*t' and A - uyat + *,oot' . At t - 3.00 s their values are

r:(3.00

and a

Thus

r':

(4.50 m)

i

-

mlsx3.00s)

:

- )ft00 mls2x3.00s)2 - 4.50m

I

-;(o.5oo m/s2x3.oo

s)2

_

-2.25m

.

(2.25m)i.

29

(a) Take the y axis to be upward and the r axis to be horizontal. Place the origin at the point where the diver leaves the platforrn. The cornponents of the diver's initial velocity are uyn - 3.00m/s and uya - 0. At t - 0.800 s the horizontal distance of the diver from the platforrn is :x : 'uy*t - (2.00 mlsx0.800 s) : 1.60 m. (b) The driver's a coordinate is a- -+g*- -+(9.8 mls2x0.800s)2 - -3. 13m. The distance above the water surface is 10.0m - 3.13 m - 6.86m. (c) The driver strikes the water when A: -10.0m. The time he strikes is

2(and the horizontal distance from the platform is

10.0 m)

r:1)0rt -

- 1.43 s (2.00

mlsxl

.43 s): 2.86m.

31

(a) Since the projectile is released its initial velocity is the same as the velocity of the plane at the time of release. Take the A axis to be upward and the r axis to be horizontal. Place the origin at the point of release and take the time to be zero at release. Let r and y (: -730 m) be the coordinates of the point on the ground where the projectile hits and let t be the time when it hits. Then A

where 0o

:

-uot

cos

1. 96 1gt'

,,

- 53.0o. This equation gives y+ ug: \'/ -T-f cos*gt, 0g

-730 m + |(g.go mls2x5.oo

(b) The horizontal distance traveled is

s)2

:202m/s.

(5.00 s) cos(53.0o)

r:

uotsin0s

- (202*1sX5.00

s)

sin(53.0")

:

806m.

Chapter 4

15

(c) and (d) The n component of the velocity is 't)r

:

't)0

sin 06

- (202^ls) sin(53.0o) :

161

m/s

and the A component is

ua:

-uscos 06 -

gt:

-(202mls) cos(53o) - (9.80 mls2x5.00

s)

: -17Imf s

.

39

Take the A axis to be upward and the r axis to the horizontal. Place the origin at the firing point, let the time be zero at firing, and let 0o be the firing angle. If the target is a distance d, away, then its coordinates are r- dandA:0. The kinematic equations are d:uotcos0s and 0 - uotsin 0o - Lgt'. Elimrnate t and solve for de. The first equation gives t - dlurcosgs. This expression is substituted into the second equation to obtain 2rB sin 0s cos 9s - gd,- 0. [Jse the trigonometric identity sin 9s cos ds _ | sin(zlil to obtain ufr sin (200 - gd,or

' 4ufi

sin(200):

Q'Smfl2)(45'] m) (460 mls)z

- 2.12x

10-3

.

The firing angle is 0o: 0.0606o. If the gun is aimed at a point a distance (. above the target, then tan2o: (, 1d, or (, - dtan?s - (45.7 m)tan0.0606' - 0.0484m - 4.84cm. 47

You want to know how high the ball is from the ground when its horizontal distance from home plate is 97.5 m. To calculate this quantity you need to know the components of the initial velocity of the ball. [Jse the range information. Put the origin at the point where the ball is hit, take the y axis to be upward and the r axis to be horizontal . If r (: 107 m) and y (- 0) are the coordinates of the ball when it lands, then tr : uyrt and 0 - uyat - *St', where t is the time of flight of the ball. The second equation gives t - 2uoa I g and this is substituted into the first equation. Use 1)0n - uya, which is true since the initial angle is 0o : 45o . The result is r : 2u\a I g. Thus

u,a

-

gr

(9.8 mls2xl07m) :22.9mf

2

s

Now take r and A to be the coordinates when the ball is at the fence. Again r a: uyat-Lgt'. The time to reach the fence is given by t - rluo* - (97.5m)lQ2,9mf s) : 4.26s. When this is substituted into the second equation the result is

1. a: utat - ,gt' - (22.9^lsx4 .26s) -

)e.8^ls2x4

.26s)2:8.63m.

:

Since the ball started I.22m above the ground, it is 8.63 m + 1,.22m 9.85 m above the ground when it gets to the fence and it is 9.85 m 7 .32m - 2.53 m above the top of the fence. It goes over the fence.

-

l6

Chapter 4

51

Take the A axis to be upward and the r axis to be horizontal. Place the origin at the point where the ball is kicked, or the ground, and take the time to be zero at the instant it is kicked. r and A are the coordinates of ball at the goal post. You want to find the kicking angle 0s so that A- 3.44m when r and A into the second the result is

y:

gn2

rtan0s

2r'ocos2 gs

You may solve this by trial and effor: systematically try values of 0g until you find the two that satisfy the equation. A little manipulation, however, will give you an algebraic solutioll. Use the trigonometric identity 1/ cos'

1 ar2

tft

0o:

.

tan2 os

1

+

tan2 0s

to obtain

I -^^2

- ntanoo + + ;TY

o'

This is a quadratic equation for tanfls. To simplify writing the solutior, let c - ig*'lr| ifq.B0 mls2x50*)t lQ5mls)z - 19.6m. Then the quadratic equation becomes ctanz 0s r tan?o + y + c : 0. It has the solution tan 0o

r* 50m

-

+ 4(A + c)c

*

(50 m)2

-

4(3.44 m + 19.6 mX 19.6 m)

2(19.6 m)

The two solutions are tan?s and 0s on the goal post. s3

Let h be the height of a step and u be the width. To hit step n, the ball must fall a distance nh and travel horizontally a distance between (n - l)u and nu. Take the origin of a coordinate system to be at the point where the ball leaves the top of the stairway. Take the U axis to be positive in the upward direction and the r axis to be horizontal. The coordinates of the ball at time t are given by n the level of step n:

The

r

coordinate then is

n-uor

(r.szmls)

2n(0.203 m)

- (0.30e n)\fr,. Chapter 4

t7

Try values of n until you find one for which r lu is less than n but greater than n 1. For n nlw - 2. 15. This is also greater than rL. For n- 3, r_0.535m and rlu - 2.64. This is less than n and greater than n - 1. The ball hits the third step. 67

To calcul ate the centripetal acceleration of the stone you need to know its speed while it is being whirled around. This the same as its initial speed when it flies off. [Jse the kinematic equations of projectile motion to find that speed. Take the A axis to be upward and the r axis to be horrzontal. Place the origin at the point where the stone leaves its circular orbit and take the time to be zeto when this occurs. Then the coordinates of the stone when it is a projectile are given by r- uot and A_ -+g*. It hits the ground when r- 10m and A- -2.0m. Note that the initial velocity is horizontal. Solve the second equation for the time: t : {:di. Substitute this expression into the first equation and solve for uoi

ug:r

_9 2y

(10m)

2(-2.0m)

The magnitude of the centripetal acceleration is

a,:

u2

15.7

m/s

lr - (15.7*1il2

.

l(1

.5m):

160

^/

,'.

73

(a) Take the positive r direction to be to the east and the positive A direction to be to the north. The velocity of ship A is given by d

t

: _rr")LT]

: -t? knots) sin 4s")r i +

r

(z'knots)

cos 4s " r

i

I1

and the velocity of ship B is given by dn

""|;;:li : -tQ8 : _ll; ;LT:] I _rrllnff

knots) sin 40'r

i

- tQ8 knots) cos 40"r i

The velocity of ship A relative to ship B is

6tn:6A-6s - (1 .0 knots) i + 1f 8. knots) j

.

The magnitude is

uln: (b)The angle 0 that

u2tnr*'Io, 6te

-

tan-1

This direction is 1.5o east of north. Chapter 4

knots)z + (38.4 knots)2

makes with the positive 0

18

(1 .0

UAB A UAB

r

r

axis is

, 38.4 -. knots - tan-I 1.0 knots -

88.5o

- 38.4 knots

(c) The time t for the separation to become d is given by mile, t - (l60nautical miles) lQS. knots): 4.2h. (d) Ship B will be 1.5o west of south, relative to ship A.

t

d

lr

tu.

Since a knot is a nautical

75

Relative to the cur the velocity of the snowflakes has a vertical component a horizontal component of 50 km/h tan 0 - unlu, - (13.9^ls)/(8.0 mls): I.74. The angle is 60o.

of 8.0 m/s and

77

fall vertic aIIy relative to the train, the honzontal component of the velocity of a raindrop LS u1" - 30^ls, the same as the speed of the train. If u, is the vertical component of the velocity and 0 is the angle between the direction of motion and the vertical, then tan 0 - uh/ur. Thus 't)u:'uhf tan? - (30 mls)ltan70" - 10.9mf s. The speed of a raindrop is u Since the raindrops

(30 mls)2 + (10.9m1s)z

-

32mf

G*,?:

s.

91

(a) Take the positive A axis to be downward and place the origin at the firing point. Then the A coordinate of the bullet is given by A - i7t'. If t is the time of flight and A is the distance the bullet hits below the target, then 2(0.019 m)

:

6.3 x 10-2 s

(b) The mvzzle velocity is the initial velocity of the bullet. It is hortzontal. If distance to the target, then r : uOt and Ug

-yt

30m 6.3

x

10-2 s

r

is the honzontal

- 4.8 x 102 mls

t07 (a) Use A : uyat - *gt' and 'na - 't)ga - gt, where the origin is at the point where the ball is hit, the positive y direction is upward, and uya is the vertical component of the initial velocity. At the highest point ua:0, so uya - gt and a: *gt': ifg.8^ls2x3.0r)t - 44m.

(b) Set the time to zero when the ball is at its highest point. The vertical component of the initial velocity is then zero and the ball's initral y coordinate ts 44m. The ball reaches the fence at time t- 2.5 s. Then its height above the ground is a: Uo_ Lst' :44m- itg.8^ls\(2.5 s)t (c) Since the ball takes 5.5 s to travel the horizontal distance of 97 .5 m to the fence, the horizontal component of the initial velocity is 1)0r: Q7.5m)l(5.5 s) - 17.7 mf s. Since the ball took 3.0 s to rise from the ground to its highest point it must take the same time, 3.0s to fall from the highest point to the ground. Thus it hits the ground 0.50 s after clearing the fence. The point where it hits is (17 .7 s) : 8.9 m. ^1sX0.50 Chapter 4

T9

111

:

(a) The position vector of the particle is given by i 6ot + |dtz, where ds is the velocity at time t - 0 and d is the acceleration. The r component of this equation is r : ?Jont* +e*tT. Sinc a uo*

0 this becomes r

2rfa* \l zesm)l(4.0 *.o*t' . The solution for t is t^ls') : The a coordinate then is A uyat + *ort' - (8 .0 s) + *tz.o^1s2x3.81 s)2 : 45 m. ^1sX3.81 (b) The r component of the velocity is 'ur: rr0* + e*t - (4.0 mls2x3.81s) - 15.2mls and the a component is rra: uya + Lort: 8.0 mls * (2.0m1s2x3.81 s) - 15.6mf s. The speed is ?r: l-j (15 .2^ls)2 + (15 .6mls)2 : 22m/s. --

-

-

lr'"*r;

12t (a) and (b) Take the r axis to be from west to east and the A axis to be from south to north. Sum the two displacements from A to the resting place. The first is Lfi j sin 37o) : (60 km) i+1+S km) j and the second is Liz- -(65 k*)i. The sum is Ar - (60 km) i(20k*)i. The magnitude of the total displacement is Lr the tangent of the angle it makes with the east is tan? - (-20 km)/(60 km) - -0 .33. The angle is 18" south of east. (c) and (d) The total time for the trip and rest is 50h+ 35h+ 5.0h:90h, so the magnitude of the ayeruge velocity is (63 km) lQ0 h) - 0.70 kmlh. The average velocity is in the same direction as the displacement, l8o south of east.

(e) The average speed is the distance traveled divided by the elapsed time. The distance is 75 km + 65 km - 140 km, so the average speed is (140 k*) lQ0 h) : 1.5 km/h. (0 and (g) The camel has I20h 90 h - 30 h to get from the resting place to B. If Lin is the displacement of B from A and A4.r, is the displacement of the resting place from A, the displacement of the camel during this time is Lr'B- Lr-rest: (90km)i-(60km)i- (-20k*)j:

(30k-)i+1zok-)i.Themagnitudeofthedisp1acementis-36kmand

the magnitude of the average velocity must be (36 km) lQO h) _ average velocity must make with the east is given by tan Q angle is 34" north of east.

20

Chapter 4

I.zk*/h. The angle 0 that the

Chapter

5

5-

Label the two forces Ft and Fr. According to Newton's second law, F, + F, - rnd, - Fr. In unit vector notation Ft - (20.0 N) i and d - -(1

2^lrt;1ritr

30o) i

-

(I2mlr';1.os 30o)j

- -6.0*lrt) i -

(10.

Fr:

md,

a^lr') j

Thus Fz

- (z.0ks)(-6.o^lrt)i +(2.0ks)(-1l. mlrt)i - e0.0N)i - (-32N)i -

(2rr.Di

F2: @(-32N)2 + (-21 X;z - 3gN. The angle that F2makes with the positive r axis is given by tan? - FzalFr* - (21 N) lQ2N) - 0.656. The angle is either 33" or 33o + 180' - 213o. Since both the r and A components are negative the coffect result is 2I3". You could also take the angle to be 180o - 213o - -I47o. (b) and (c) The magnitude of F2 is

13

In all three cases the scale is not accelerating, which means that the two cords exert forces of equal magnitude on it. The scale reads the magnitude of either of these forces. In each case the magnitude of the tension force of the cord attached to the salami must be the same as the magnitude of the weight of the salami. You know this because the salami is not accelerating. Thus the scale reading is ffig,where mtsthe mass of the salami. Its value is (11.0kgX9 .8*lr2) 108N.

t9 (a) The free-body diagram is shown in Fig. 5 - 16 of the text. Since the acceleration of the block is zero, the components of the Newton's second law equation yield T - mg sin 0:0 and

l7/r,'

mg cos 0 (8.5 kexg .8^lrtl rin 30"

: 42 N.

(b) Solve the second equation for ,F^/: Fry

-

mgcos 0

- (8.5kg)(9.8^lrt;ros30o :72N.

(c) When the string is cut it no longer exerts a force on the block and the block accelerates . The r component of the second law becomes -mgsin e - me,, so a,: - gsin 0 - -(9.8^lr';ritt30o -4.9mf s2. The negative sign indicates the acceleration is down the plane. 25

According to Newton's second law F - ma, where F is the magnitude of the force, a" is the magnitude of the acceleration, and m is the mass. The acceleration can be found using the equations for constant-acceleration motion. Solve u : lJ1 * at for a: e, : u lt The final velocity Chapter

5

2l

:

is u - (1600 korlhx1000m/km)1p600 s/h) : 444mls, so o (444mls)/(1.8 s) the magnitude of the force is Ir - (500 kgQaT I .2 x 105 N.

-

^lst)

: 247

^lr'

and

29

The acceleration of the electron is vertical and for all practical puqposes the only force acting on it is the electric force. The force of gravity is much smaller. Take the fr axis to be in the direction of the initial velocity and the A axis to be in the direction of the electrical force. Place the origin at the initial position of the electron. Since the force and acceleration are constant theappropriateequations arer_ uot and a- Lot': ifgl*)*, where F- ma, wasusedto substitute for the acceleration a. The time taken by the electron to travel a distance n (: 30 mm) horizontally is t - r luo and its deflection in the direction of the force is

1F

a: ;/.m

4.5 x 1 g-t6 y 9.11 x 10-3t kg )(

(#)':;(

30x 10-3 m 1.2

x

107

,a2

m/s )

35

The free-body diagram is shown at the right. F* is the normal force of the plane on the block and mj ts the force of gravity on the block. Take the positive r axis to be down the plane, in the direction of the acceleration, and the positive A axis to be in the direction of the normal force. The r component of Newton's second law is then mg sin 0 - ma) so the acceleration is a : g stn 0 .

(a) Place the origin at the bottom of the plane. The equations for motion alongthe r axis are r: uot+)atz and 'u: n0*at. The block stops when 'u : 0. According to the second equation, this is at the time t

3) *)"(+)

fr : 'tJ.( -u0

tq .B^lrt; |Ltn '

-us f a. The coordinate when it stops is

2--1

6

2 a,

A'

(-3.s0 mls)2 t_ 2

-

ritt 32.0" ]

2 gstn1

: -t.t8m

(b) The time is

t (c) Now set

r:

Ug

-3.50 m/s

Ug

g srn? (9.g mlrt;,in 32.0"

0 and solve tr : uot + \atz for

,

t:::

2uo a

t.

0.674s.

The result is

2uo

2(-3.50 m/s)

g stn?

(9.8 mlrt; ritt 32.0o

-

1.35 s.

The velocity is u

22

-

uol_

Chapter 5

at:

uo* gtsrnl

-

-3.50 mls + (9.8 mls2xl.35 s)sin 32o - 3.50 mls,

as expected since there is no

friction. The velocity is down the plane.

45

The free-body dtagrams for the links are drawn below. The force affrows are not to scale.

F:onz{ 4onr{ 'll-\rrrJl-\-'rr-rll

fLon:{

fton+{

FA

ltl

'o'

Ior

lltrl

t

ltl

I

mgy

mgililr'

I

Link 2

Link

on'

Link 3

msll

F3

Link

4

on+ msll F+ Link

on

s

5

(a) The links are numbered from bottom to top. The forces on the bottom link are the force of gravity md, downward, and the force F, onr of hnk2, upward. Take the positive direction to be upward. Then Newton's second law for this link is Fz on 1 - mg : ma. Thus

Fz onl : m(a+ g)

- (0.100kgX2.50 mlr'+

9.8 mlr')

-

I.23N.

(b) The forces on the second link arc the force of gravity md, downward, the force Fr on z of tink I, downward, and the force Fz onz of link 3, upward. According to Newton's third law Ft onz has the same magnitude as Fzonr. Newton's second law for the second link is F3 onz -

fl

on2

- mg :

mQ,,

SO

Fz onz- m(&+ g)+ Fr onz- (0.100kg)(2.50 mlr'+9.8 mls2)+ 1.23N - 2.46N, where Newton's third law was used to substitute the value of F2on

I for Ft

onz.

(c) The forces on the third link are the force of gravity mj, downward, the force Fz on z of link 2, downward, and the force F+ on z of link 4, upward. Newton's second law for this link is

F+onZ- F2on3 -mg:TnA, F+on3

SO

- m(a+ g)+ Fzon3 -

(0.100NX2.50

mlr'+9.8 mls2)+ 2.46N - 3.69N,

where Newton's third law was used to substitute the value of F3 on2 for F2 on3. (d) The forces on the fourth link are the force of gravity mj, downward, the force Ft on + of link 3, downward, and the force Fs on + of link 5, upward. Newton's second law for this link is FS on +

-

F3

Fs

on

on+

4

- mg - mA, SO

- m(a+ g)+ Ft on4 -

(0.100kgX2.50

mlr'+

9.8 mls2) + 3.6gN

- 4.92N,

where Newton's third law was used to substitute the value of Faon 3 for F3 on4. (e) The forces on the top link arc the force of gravrty mrt, downward, the force Fq link 4, downward, and the applied force F, upward. Newton's second law for the top F - F+ on S - mg - nLa, SO

s of link is

on

F -m(a+ g)+ Fqon5 - (0.100k9(2.50 mlrt+9.8 mls2)+ 4.92N- 6.15N, where Newton's third law as used to substitute the value of Fs

on

4 for F+

on s.

Chapter 5

23

(0 Each link has the same mass and the same acceleration, of them:

Fnet

-

ma

-

(0.100 kg(z.50 mls2)

-

so the same net force acts on each

0.25 N.

s3

(a) The free-body dragrams are shown to the right.

F is the applied

force and f is the force of block 1 on block 2. Note that F is applie,C only to block 1 and that block 2 exerts the force - i on block 1. Newton's third law has thereby been taken into account. Newton's second law for block 1 is F f - TrL1a, where o, is the acceleration. The second law for block 2 is f -- TrL2e,. Since the blocks move together they have the same acceleration and the same symbol is used in both equations. Use the second equation to obtain an expression for a: a - f l*r. Substitute into the first equation to get F - f : TTLI f l*r. Solve for f :

n Fmz Q.2NXl.2kg) 'f - m1 * TTL2- z.3kg + r.zk|(b) is

If F is applied to blo ck2 instead of block

1' 1 N '

llrr

1, the force of contact

ti-_ Fml _ Q.2NX2.3 kg) a' TrLl I mz 2.3 kg + l.2kg

.t

(c) The acceleration of the blocks is the same in the two cases. Since the contact force f is the only horizontal force on one of the blocks it must be just right to give that block the same acceleration as the block to which F is applied. In the second case the contact force accelerates ---.1N. a more massive block than in the first, so it must be larger. 57

(a) Take the positive direction to be upward for both the monkey and the package. Suppose the monkey pulls downward on the rope with a force of magnitude F. According to Newton's third law, the rope pulls upward on the monkey with a force of the same magnitude, so Newton's second law for the monkey is F - mrng - mrnarn, where mrn is the mass of the monkey and e,rn is its acceleration. Since the rope is massless F is the tension in the rope. The rope pulls upward on the package with a force of magnitude F, so Newton's second law for the package is F + Flr - mpg : mpap, where mp is the mass of the package, a,p is its acceleration, and Fx is the normal force of the ground on it.

Now suppose F is the minimum force required to lift the package. Then F'lr - 0 and e,,p - 0. According to the second law equation for the package, this means F - mpg. Substitute wpg for F in the second law equation for the monkey, then solve for am. You should obtain aTn

24

Chapter 5

: F - rrlrng: ,rr-,

(mo

- m,n)g (15 kg *-,

10kg)(9.8

lo

An - _ r kg ^lr\ - +'Y mf s

2 '

(b) Newton's second law equations are F -mpg: ffipap for the package and F -rrLrng : TrL,n,arn for the monkey. If the acceleration of the package is downward, then the acceleration of the monkey is upward, so o??" - -ap. Solve the first equation for -F: p - mp(g + op) : ffip(g - arn). Substitute the result into the second equation and solve for arni

anL-

- *'')g mp*Tftm.

(TY

/

15kg+10kg

(c) The result is positive, indicating that the acceleration of the monkey is upward.

(d) Solve the second law equation for the package to obtain

p-

Trl,p(g

-

a,nr): (15kgX9.8mf

s2

-2.0*ls2) -

120N

61

The forces on the balloon arc the force of gravity mj, down , and the force of the atr Fo, up. Take the positive direction to be up. When the mass is M (before the ballast is thrown out) the acceleration is downward and Newton's second law is Fo M g - - M a. After the ballast is thrown out the mass is lVI wL, where m is the mass of the ballast, and the acceleration is upward. Newton's second law is Fo (M m)g - (M m)a. The first equation gives Fo: M(g a) and the second gives M(g a) (M m)g - (M m)a. Solve for m:

m:ZMal(g + a). 73

r Fcosd f

Take the

axis to be horizontal and positive in the direction that the crate slides. Then

acceleration (the only nonvanishing component). In part (a) the acceleration is e,r :

In part (b)

m:

Wlg

F cos? - f

m

_- (3

nr

-

n n -^^ 1^2 - 125N-u't+m/s 310kg

(450N)cos38"

10N) lQ.8^ls')

Fcos0-f

m

-

A

'

31.6kg and

(450N)cos38o 125N

,1

..r

r2

36.1 kg

79

F be the maglitude of the force, a4 (: L2.0mls2) be the acceleration of object I, and a,2 (:3.30 mlt'; be the acceleration of obj ect2. According to Newton's second law the masses are rrLl: Flol and TrL2- Ff e,2. (a) The acceleration of an object of mass rTL2 - mr is (L- F - +'O m/ S ' TrL2 or u, I2.0 - Tft1 Let

Chapter

5

25

(b) The acceleration of an object of mass TtLl 1 m2 is

A'-

F Tft2

* mr

F

_ (F

lor) + (F lo)

eta,z

(12.0

a1 a aZ

12.0

^ls2x3.3o

mlr'

*lt') - 2.6mf

+ 3.3 0 mf

s2

s2

9l (a) Both pieces arc station dry, so you know that the net force on each of them is zero. The forces on the bottom piece are the downward force of gravity, with magnitude Tftzg, and the upward tension force of the bottom cord, with magnitude T6. Since the net force is zero,

T6:

TTL27

:

(4.5kgX9

.8^lr2) - 44N.

(b) The forces on the top piece are the downward force of gravity, with magnitude r(Ltg, the downward tension force of the bottom cord, with magnitude Ta, and the upward force tension of the top cord, with magnitude ?7. Since the net force is zero, T1

:

Tb

t mrg :44 N + (3.5 kg)(9.8 mlst) - 78 N

.

(c) The forces on the bottom piece are the downward force of gravity, with magnitude rrLsg, and the upward tension force of the middle cord, with magnitude Trn Since the net force is zeto, T,n

:

TTLsg

:

(5.5 kgXg .8

^ls') -

54

N.

(d) The forces on the top piece are the downward force of gravity, with magnitude w4g, the upward tension force of the top cord, with magnitude T7 (: I99 N), and the downward tension force of the middle cord, with magnitude Trn Since the net force is zero)

T,n: Tt -

Tft3g

:

IggN

-

(4.8 kg)(9

.8^ls2)

-

152N.

9s

(a) According to Newton's second law the magnitude of the net force on the rider is (60.0kg)(3.0 mls') - 1.80 x 102N.

It -

ma,

:

(9) Take the t force to__b. thl vector sum of the force of the motorcycle and the force of Earth: t Fnet: Fr, + Fn. Thus Frn: 4r.t - Fs. Now the net force is parallel to the ramp and therefore makes the angle 0 (: 10') with the horrzontal, so Fnt_ (F cos 0)i+(lrsin 0)j, where the r axis is taken to be horrzonta| and,the A axLS is taken to be vefitcal. The force of Earth is F" - -mgj, so F,- - (F cos il?+ (F sin o + milj.Thus F,n*

- It cos 0 - (1.80 x I02 N) cos 10o

and

Frna

- (1.80 x 102) sin 10o + (60.0kg)(g.g mls2) - 6.19 x 102 N.

The magnitude of the force of the motorcycle is TA

-r rn

26

Chapter 5

t02 N)2

r02 N)2

-6.44 x

102N.

99

The free-body diagrams for the two boxes are shown below.

Flrr Fxz

Here T is the tension in the cord, FxTr is the norrnal force of the left incline on box 1, and FNz is the norrnal force of the right incline on box 2. Different coordinate system are used for the two boxes but the positive r direction are chosen so that the accelerations of the boxes have the same sign. The r component of Newton's second law for box 1 gives T - TTLI7 stn01 : TTL1& and the r component of the law for box 2 gives mzg stn 02 - T : TTL27. These equations are solved simultaneously for T. The result is

T:

mtmzg (sin \ ?TL1 * mZ

91^

* sin oz)

(3.0 kgx2.0 kgX9.8 m/s2) 3.0 kg +

2.0kg

(sin 30o + sin

60.)

:

16

N

101

for the two tins are shown on the right. T is the tension in the cord and Fy is the normal force of the incline on tin 1. The positive n direction for tin I is chosen to be down the incline and the positive r direction for tin 2 is choFree-body diagrams

T

to be downward. The sign of the accelerations of the two tins arc both then positive. Newton's second law for tin 1 gives T + Tft1g stn {3 - ma, and for tin 2 gives mzg T F - TrL2cL. The second equation is sen

solvedforT,withtheresultT-

TrL2@_

\r

a)-F-(2.0kgX9.8^lrt-5.5m1s2)-6.0N-2.6N.

The first Newton's law equation is solved for sin p, with the result sin

C:mra-T mts

(1.0kgX9.8 m/s)

The angle is I7o

Chapter

5

27

Chapter

5

1

(a) The free-body diagram for the bureau is shown on the right. F is the applied force, "f- i, the force of friction, fi" is the normal force of the flooq and mj is the force of gravity. Take the r axis to be honzontal and the A axis to be vertical. Assume the bureau does not move and write the Newton's second law equations. The r componentis F- f :0 andthe A componentis Fl/ -rng:0. The force of friction is then equal in magnitude to the applied

force: f force of gravity: ,Fl/ - mg. As F increases, f increases until f : ltrrFx. Then the bureau starts to move. The minimum force that must be applied to start the bureau moving is

It - FrFx : (b) The equation for

F F

ltr,mg

- (0 .45)(45 kgX9.8 mls2) - 2.0 x I02 N

is the same but the mass is now 45 kg

- F,mg - (0 .45)(28 kgX9.8 mlst) -

-

1.2

1,7

kg

.

- 28 kg. Thus

x T02 N

.

3 (a) The free-body dragram for the crate is shown on the right. F is the force of the person on the crate,,f-ir the force of friction, Fr it the noffnal force of the floor, and mj ts the force of gravity. The magnitude of the force of friction is given by f - l.rrcFw, where ltr* is the coefficient of kinetic friction. The vertical component of Newton's second law is used to find the normal force. Since the vertical component of the acceleration is zero, F'l/ - mg - 0 and Fl/ - mg. Thus

f:

F;Fw

:

ltkmg

- (0.35X55 kgX9.8 mls2) - l.g x I02 N

.

(b) IJse the horizontal component of Newton's second law to find the acceleration. F-f:ffia, (F - f) (220N - 18eN)

a:

28

Chapter 6

m

-

55kg

-

0.5

6mf

s2

Since

13

(a) The free-body diagram for the crate is shown on the right. f is the tension force of the rope on the crate, Fn is the norrnal force of the floor on the crate, rnfi rs the force of gravity, and ,i ir the force of friction. Take the r axis to be horizontal on the right and the A axis to be vertically upward. Assume the crate is motionless. The r component of Newton's second law is then 7 cos e - f : 0 and the A component is Tsin 0+F^/ *mg:0, where 0 (: 15") is the angle between the rope and the horizontal. The first equation gives / - T cos I and the second gives Fx - mg - T sin 9. If the crate is to remain at rest, / must be less than lr"Fx, or T cos? < the tension force is sufficient to just start the crate moving T cos 0 for T: T'

:

F'mg cos 0 + p,,

stn0

lrr(mg

7

sin d). Solve

cos 15o + 0.50 sin 15o

(b) The second law equations for the moving crate are 7 cos 0 f : ma and -Fl/ + T sin 0 mg:0. Now f f - p,n(mg 7 sin 0). This expression is substituted for f in the first equation to obtain ? cos 0 - pn(mg - T sin 0) : ffia, so the acceleration is

a:

T(cos 0 +

p,7"

sin 0)

-ltt

*

9

Its numerical value ls e,

=

-

(0.35X9

.8^ls2)

-

1.3

^lr'

23

The free-body diagrams for block B and for the knot just above block ,4 arc shown on the right. T1 is the magnitude of the tension force of the rope pulling on blo ck B , Tz is the magnitude of the tension force of the other rope, f is the magnitude of the force of friction exerted by the horizontal surface on block B, FAr is the magnitude of the norrnal force exerted by the surface on block B, We is the weight of block A, and W n is the weight of block B . 0 (- 30") is the angle between the second rope and the horizontal.

For each object take the r axis to be horizontal and the A axis to be vertical. The fr component of Newton's second law for block B is then T1 f :0 and the A component is Fy Wn:0. The r aomponent of Newton's second law for the knot is T2 eos 0 - Tr - 0 and the A component is Tzsin? Wt:0. Eliminate the tension forces and find expressions for f and F11r in terms Chapter 6

29

of Wt and Wn, then select Wt so f : F"Fx. The second Newton's law equation gives Fyr - Ws immediately. The third gives Tz - Tr I cos 9. Substitute this expression into the

fourth equation to obtain T1 to obtain f - Waf tan?. For the blocks to remain stationary Waltan? < Solve for We:

We: FrWntanT -

(0.25)(711N) tan3Oo

-

f

must be less than

1.0

F"Fx or

x I02 N.

27

(a) The free-body dragrams for the two blocks are shown on the right. T is the magnitude of the tension force of the string , Fx t is the magnitude of the normal force on block A, Fw s is the magnitude of the nonnal force on blo ck B , f a is the magnitude of the friction force on block A, f e is the magnitude of the friction force on block B, TTLy is the mass of block A, and TTLs is the mass of block B . 0 is the angle of the incline (30'). We have assumed that the incline goes down from right to left and that block A is leading. It is the 3.6-N block. For each block take the r axis to be down the plane and the A axis to be in the direction of the normal force. For block A the r component of Newton's second law is

mtgsin0- fa-T:TrLAa,g and the A component is

Fnt-Tftagcos0:0.

Here a,s is the acceleration of the block. The magnitude of the frictional force is

ft : FkAFxt :

FkAmtgcos 0,

where Fxa- mtg cos9, from the second equation, is substituted. Fnt is the coefficient of kinetic friction for block A. When the expression for f a is substituted into the first equation the result is Tftsg sin 0

The same analysis applied to block

-

B

Trlpg sin 0

-

FntThsg cos 0 - T

- ?rLga,n

.

p

.

leads to

FnsTrlyg cos 0 + T

-

rTL

Ba,

We must first find out if the rope is taut or slack. Assume the blocks are not joined by a rope and calcul ate the acceleration of each. If the acceleration of A is greater than the acceleration of B,

then the rope is taut when it is attached. If the acceleration of B is greater than the acceleration of A, then even when the rope is attached B gains speed at a greater rate than A and the rope is slack. 30

Chapter 6

Set

T:0

in the equation you derived above and solve for as and a,p. The results arc

a,A- g(sin0

-

ltp7cos

0): (9.8 mlrt;1ritt30o -

0.10cos30")

coS

0): (9.8 mlr';1ritr30o -

0

: 4.05^lt'

and

a,s

- g(sin0 -

ltn6

.20cos30o): 3.20mlrt.

We have learned that when the blocks are joined, the rope is taut, the tension force is not zero, and the two blocks have the same acceleration.

Now go back to mAg sin 0 - Fntmeg cos 0 -T - TrL4a, and ThBgsrn) - FnsmBgcos 0 +T TTL6a, where a has been substituted for both a,s and a,s. Solve the first expression for T, substitute the result into the second, and solve for a. The result is

a:

gsin g

ll'naml* H"nnm6 g cos a^

-

rYL4

- (9 .g^lrt; - 3.5 ^lrt

,i,r

+

mB

30o '\'

N)l (9.g 3.6N+7.2N l\-/ mlrr; ,os 30o N) + (0'20X7'2

[(0'10X3'6

L

.

Strictly speakirg, values of the masses rather than weights should be substituted, but the factor g cancels from the numerator and denominator. (b) IJse ffLsg sin 0 - Fneme7 cos 0 - T

T

- rTLAa to find the tension force of the rope:

- mAg sin 0 - FntTTLsgcos e - TtLAa, - (3.6 N) sin 30o - (0. 10X3.6 N) cos 30o -

(3.6

N/9.8 mls2x3 .49 mlst)

- 0.21 N

.

35

Let the magnitude of the frictional force be au, where e, : 70 N . s/m. Take the direction of the boat's motion to be positive. Newton's second law is then -aa : m du ldt Thus

fJu, du: u where us is the velocity at time zero and evaluated, with the result

1)

: ,ol2 and solve for f

/

u is the velocity at time t. The integrals

-uat tn*-

Take

mJo

can be

-A.

:

t -m

a

rn2

loookg

- 70N.s/m hz - g.9s.

49

(a) At the highest point the seat pushes up on the student with a force of magnitude F'^/ (: 556 N). Earth pulls down with a force of magnitude W (: 667 N). The seat is pushing up with a force that is smaller than the student's weight in magnitude. The student feels light at the highest point. Chapter

6

31

(b) When the student is at the highest point, the net force toward the center of the circular orbit is W - Fy and, according to Newton's second law, this must be mu'lR, where u is the speed of the student and R is theradius of the orbit. Thus mu2f R-W - f^i:667N- 556N- 111].{. The force of the seat when the student is at the lowest point is upward, so the net force toward the center of the circle is F^i W and lr^/ W - mu2 I R. Solve for Fn:

,.,/,/-

667N- 77gN.

rytW:ltlN+

(c) At the highest point W F^/ : mu2 f R, so f^i mu2 lR increases by a factor of 4, to 444N. Then F^/ :667 N - 444N

:

(d) At the lowest point W + Fl/ m'u'I R, so F^/ 1.11 mu2 I R is still 444 N, F^r 667 N + 444N

:

-

- 223NI.

-W-mu2fR,so-F1/ x 103 N.

53

The free-body diagram for the plane is shown on the right" F is the magnitude of the lift on the wings and m LS the mass of the plane. Since the wings are tilted by 40" to the honzontal and the lift force is perpendicular to the wings, the angle 0 is 50o. The center of the circular orbit is to the right of the plane, the dashed line along r being a portion of the radius. Take the r axis to be to the right and the A axis to be upward. Then the r component of Newton's second law is F cos g - mu2 I R and the A mg - 0, where R is the radius of the orbit. The component is F sin I first equation gives It - mu2 I Rcos 0 and when this is substituted into the second, (*r'lR)tan? - mg results. Solve for R:

tr

n12

RThe speed of the plane is

R

_

'tr

I

tanl.

:480 kmfh

033 mls)2 9.8 mls'

-

tansoo

133

mls,

so

: 2.2 x 103 m

.

59

(a) The free-body dragram for the ball is shown on the

right.

f,

is the tension force of the upper string , ip is the tension force of the lower string, and m is the mass of the ball. Note that the tension force of the upper string is greater than the tension force of the lower string. It must balance the downward pull of gravity and the force of the lower string. 32

Chapter 6

r

axis to be to the left, toward the center of the circular orbit, and the A axis to be upward. Since the magnitude of the acceleration is a -- ,2 I R, the r component of Newton's second law is Take the

Tucosg+

Tacos0-ry,

where u is the speed of the ball and R is the radius of its orbit. The A component is

2,, sin 0

- Tzsrn? -

mg

- 0.

The second equation gives the tension force of the lower string: Tt triangle is equilateral 0 - 30o. Thus

Tr: 35 N -

(1'3a kgX9"9 m/s2) srn 30o

(b) The net force is radially inward and has magnitude 8.74N) cos 30"

- 8.74N Fnet, str

- 37.9 N.

(c) [Jse Fn"t,str:

nr,uz

f R. The radius of the orbit RFn"t,

1.3

4kg

0

. Since the

.

- (7, + TD cos 0

is [(1.70m) l2)Jtan30"

(1.a7 mX37.eN)

str

: Tu - mg f stn

- 6.45 mls

-

1.47

m. Thus

.

65

The first sentence of the problem statement tells us that the maximum force of static friction between the two block is f ,,^u*:12N. When the force F is applied the only horizontal force on the upper block is the frictional force of the lower block, which has magnitude f and is in the forward direction. According to Newton's third law the upper block exerts a force of magnitude f on the lower block and this force is in the rearward direction. The net force on the lower block is F - f . Since the blocks move together their accelerations are the same. Newton's second law for the upper block gives f - Tftt& and the second law for the lower block gives F - f : TTL6a, where cr is the common acceleration. The first equation gives a - f l*t Use this to substitute for a, in the second equation and obtain F - f : (mal^t)f .Thus

Ii-(r*e)mt/ f \ If f

has its maximum value then F has its maximum value, So the maximum force that can be applied with the block moving together is

F-

(t.ffi)

The acceleration is then LL:

f

lzN):27N

l2N ._ ,)

TrLl 4.0 N

Chapter

6

33

77

(a) The force of friction is the only horizontal force on the bicycle and provides the centripetal force need for the bicycle to round the circle. The magnitude of this force is f : mu2 f r, where m is the mass of the bicycle and rider together, u is the speed of the bicycle, and r is the radius of the circle. Thus : (85.0 kgXg.oo ml s)2

-

f

,n

25.0m (b) In addition to the frictional force the road also pushes up with a nonnal force that is equal in magnitude to the weight of the bicycle and rider together. The magnitude of this force is .F1r/ - mg: (85.0kg)(9.8 The frictional and normal forces are pe{pendicular to

^lt')-833N.

each other, so the magnitude of the net force of the road on the bicycle is 4r"t vL - , E' Va' (275 N)2 + (833 N)2

* ** -

- 877 N.

81

The free-body diagrams are shown on the right. T is the tension in the cord, F* is the norrnal force of the incline on block^ A, F* u is the normal force of the platfonn on block B, I is the angle that the incline makes with the horizontal (which is also the angle between the normal force and the vertic aI), and ,f-'ir the frictional force of the platforrn on block B . The r axis for each block is also shown.

r

component of Newton's second law for block A gives mgsinO component of the second law for block B is T f Fxn - TTLsg - 0.Note that the blocks have the same acceleration.

The

T-

Trlye,, the fr

The magnitude of the frictional force LS ptrFxs: ltkTftsg,where ?rLpg was substituted for Fxn, and the r component for B becomes T - Fnmng - TTLsa,. The equations mgsin 0 - T - TLna and T - Fnmng: TTLsa are solved simultaneously for T and a. The results are (4.0 kgX2.0 kg)(sin 30o +

+ t _ TftlTrLB(sin 0 Fil *^+r",

rF

and

g: e,- Masin 0 - Fnrns ms+ mB r

0.50)

10ro

rJl\ ' \r 1

(4.0kg) sin3Oo - (0.50x2.0kg) ., , r 2 :1.6m/s 4.0k9 + 2.0k9

8s (a) If u is the speed of the car, m ts its mass, and r is the radius of the curve, then the magnitude of the frictional force on the tires of the car must be f : muz f r or else the cat does not negotiate the curve. Since m - W I g, where W is the weight of the car,

r. wu2 a' gr JL-J./rLz\lWl\.

34

Chapter 6

(10.7

x

103

N(13.4m1s)2 .

(9.g mls2x61.0 m)

,_,1

1

(b) The norrnal force of the road on the car is l7^/

friction is

-

W

and the maximum possible force of static

.f", -u* - FrFx force that is required is less than the maximum possible, the car successfully rounds the curve.

9t Let F be the magnitude of the applied force and f be the magnitude of the frictional force. Assume the cabinet does not move. Then its acceleration is zero and, according to Newton's second law, F - f . The norrnal force is F1,' maximum force of static friction is Fr,max if F is less than 378 N the cabinet does not move and the frictional force is f greater than 378 N, then the cabinet does move and the frictional force is f (0.56X556 N) - 311 N.

f : 222I\. (b) The cabinet does not move and f : 334 N. (c) The cabinet does move and f : 311 N. (d) The cabinet does move and f : 311 N. (a) The cabinet does not move and

(e) The cabinet moves in attempts (c) and (d). 99

(a) The free-body diagram for the block is shown on the right. The magnitude of the frictional force is denoted by f , the magnitude -. Fl'/ of the normal force is denoted by lrl/ , and the angle between the incline and the horizontal is denoted by 0. Since the block is sliditg down the incline the frictional force is up the incline. The positive /0 fr r direction is taken to be down the incline. For the block when it is sliding with constant velocity the n component of Newton's second law gives mg sin 0 f - 0 and the 'A component gives mg cos 0 -Fl/ - 0. The second equation gives F1/ - mg cos9, so the magnitude of the frictional force is f obtain mg sin 0 - Tftuamg cos 0 : 0. Thus the coefficient of kinetic friction is Fp - tan 0 . When the block is sliditrg up the incline the frictional force has the same magnitude but is directed down the plane. The r component of the second law equation becomes mg sin 0 + Fnmg cos e rrLa, where a, is the acceleration of the block. Thus a - (sin 0 + [L7"cos0)g - 29 sind, where tan? was substituted for Fn and tan 0 - sin 0 f cos 0 was used.

If d is the displacement of an object with constant

acceleration a, us is its initial speed and u is its final speed, then u2 ulr: Zad. Set u equal to zero and a, equal to 29 stn? and obtain d - -ufil2n -- -ufrlag sin9. The negative sign indicates that the displacement is up the plane. (b) Since the coefficient of static friction is greater than the coefficient of kinetic friction the maximum possible static frictional force is greater than the actual frictional force and the block remains at rest once it stops. Chapter 6

3s

105

The box is subjected to two horizontal forces: the applied force of the worker, with magnitude F, and the frictional force, with magnitude f .Newton's second law gives F - f : ma) where m is the mass of the box and a is the magnitude of its acceleration. The magnitude of the frictional force is f the normal force of the floor. In this case F1y - mg and f becomes F - Ltnmg - me' so ltt - (lr - ma) l*g.

Let u be the final speed of the box and d, be the distance it moves. Then u2 a,- u2 lza- (1.0 mlilz l2(1 .4m):0.3 57mf s2. The coefficient of kinetic friction is ltt

36

Chapter 6

-

F - ma m9

-

(85

N-

(a0kgX0.357 m/s2)

(40 kgX9.8 m/s2)

0.18

-

Zad, and

Chapter

7

3

(a) Use Eq. 2-16: u2 - u|+Zar, where us is the initial velocity, u is the final velocity, displacement, and a is the acceleration. This equation yields

r

is the

(b) The initial kinetic energy is

K,;,-

i*rt

The final kinetic energy is

iO.67 x 10-27 kg)(2.9 x r07 ^ls)2:6.9 x 10-13J. The change in kinetic energy is LK :6.9 x 10-13 J - 4.8 x 10-13 J :2.I x 10-13 J.

Ky

- **r':

t7 (a) Let F be the magnitude of the force exerted by the cable on the astronaut. The force of the cable is upward and the force of gravity is mg is downward. Furtherrnore, the acceleration of the astronaut is glI0, upward. According to Newton's second law, F mg: mgl10, so

F done by

F is

Wp: Fd -

ILmgd,-

11(72 kgX9.8 m/s2X15 10

10

m)

- l.16 x 104J.

(b) The force of gravity has magnitude mg and is opposite in direction to the Since cos 180"

displacement.

- - 1, it does work ws

(c) The net work done is W :1.16x 104J- 1.06x 10aJ - 1.1x 103 J. Since the astronaut started from rest the work-kinetic energy theorem tells us that this must be her final kinetic energy.

(d) Since

K - i*r'her final speed is

u-

2K m

2(lJ x 103 J)

- 5.3 m/s

.

t9 (a) Let F be the magnitude of the force of the cord on the block. This force is upward, while the force of gravity, with magnitude M g, is downward. The acceleration is g 14, down. Take the downward direction to be positive. Then Newton's second law is Mg F- Mgl4,

soF Wp Chapter

7

37

(b) The force of gravity is in the same direction as the displacement, so it does work Ws

:

M gd.

(c) The net work done on the block is Wr: -3MSdl4+ Mgd - Il[gd,l4. Since the block stafts from rest this is its kinetic energy K after it is lowered a distance d.

(d) Since

K-

LtW

r',

where u is the speed,

tsd u:Vt2K tw:Vz after the block is lowered a distance d. The result found in (c) w-as used. 29

(a) As the body moves along the force is

w:

l_"n'

r

axis from

F*d,r:

I_"n'

-

n6

m the work done by the

4.0

-6rd,n - -3 *1"_',- -3(*'r

- *?)

According to the work-kinetic energy theorem, this is the change in the kinetic energy:

W:A,K:L*@?-r?), where u6 is the initial velocity (at

u1:

t)

and u.s is the final velocity (at r

l2w= I * +'i:

2(-21 J)

2.}ks

(*'r - rl)

: **(r? -

The theorem yields

+ (g.o m/s)2 ,)2 - 6.6mf s.

(b) The velocityof the paftrcle is uy - 5.O^ls when it is at r theorem for n y. The net work done on the particle is W energy theorem yields -3

f).

,?). Thus

- n y. Solve the work-kinetic energy -3(*tr *), so the work-kinetic

- , t-+[(s.0m/s)2 - (8 .O^ls)r] + (3.0m), - 4.7 m. V6N 35

(a) The graph shows F as a function of r tf rs is positive. The work is negative as the object moves from r

F(*) Fs

r

Since the area of a triangle is ] (Uur.)(altitude), the work done from r:.0 to n: n0 is -+(roXFo) and the work done

from

r

The net work is the sum, which is zero.

38

Chapter 7

0

-Fs

(b) The integral for the work is

\

P*o

w: Io'"o ro(X t) dr:Fs(* ")1, -0 43

The power associated with force f is given by P : F . 6, where u* is the velocity of the object on which the force acts. Let d (: 3 7") be the angle between the force and the honzontal. Then P - F.6- Fucos d- 022NX5.O^ls)cos 37":4.9 x I02'W'. 45

(a) The power is given by P

- F'u and the work done by f w'

-

rtz

Jr,

P dt:

rtz

Jr,

from time t1 to time t2 is given by

,, dt.

Since F is the net force, the magnitude of the acceleration is a velocity is us: 0, the velocity as a function of time is given by , "' w: l (tr' l *)t dt - )rr' l *)e3

- F l* and, since the initial : uo * at - (F lm)t Thus

nt^t

t?)

J .'tt

For f1 - 0 and t2

xl2 Yv: iz [ts'o I (t.o s)2 - 0 83 J l5kg

14/

(b) For dr - 1.0 s and t2

L

l

- 2.0 s, oryfl lrz.o w :1 [(s , L lsLs If(zu r), -

(c) For

C1

(1.0 s)r]

- 2.s r.

[(:0,), -(20,),]

- 421

- 2.0 s and tz - 3.0 s,

w:1z fry] lsk L

(d) Substitute u- (Flm)t into P- Fu to obtain P- Ttztl* for the power at any time the end of the third second P - (5.0 N)2(3.0 s) 115 kg - 5.0 'W'.

t.

At

47

The net work Wnet is the sum of the work Wu done by gravrty on the elevator, the work W. done by gravity on the counterweight, and the work W, done by the motor on the system: Wn t - W" + W. + W". Since the elevator moves at constant velocity, its kinetic energy does not change and according to the work-kinetic energy theorem the net work done is zere. This means Wu + W"* W" Chapter

7

39

gravity on it is W"- -TrLegd- -(1200kgX9.8*lr';1S+m): -6.35 x 105J. The counterweight moves downward the same distance, so the work done by gravity on it is W" : rn"gd (950kgX9.8 mlr';1S+m) - 5.03 x 105J. Since Wr: 0, the work done by the motor on the system is W, _ -W" - W. - 6.35 x 105 J - 5.03 x 105 J - 1.32 x 105 J. This work is done in a time interval of At - 3.0 min - 180 s, so the power supplied by the motor to lift the elevator is

p -%-

Lt

1'32

x

180

1o5J s

-7.35x

roz'w'.

63

(a) Take the positive r direction to be in the direction of travel of the cart. In time Lt the cart moves a distance Lr - u Lt, where u is its speed. The work done is W - F* Lr (Fu cos 0) LL where f is the force of the horse and 0 is the angle it makes with the horizontal. Now 6.0 milh 600s, so It - t(40 lbXS.Sftls)cos30ol(600s): 1.8 x 105 ft.lb. (b) The averagepower is P - F*u - Fucos 0 -(401bX8.}ftfs)cos30o:3.0 x 102ft.lb/s. Since 1ft-Ibls - 1.818x 10-'hp, thepoweris P - (3.0x 102ft.lblsxl.818x 10-thplft. lb/s) 0.55 hp. 69

(a) The applied force f is in the direction of the displacement is We : Fd -- (209 NX1.50m) - 314J.

i, so the work done by the force

(b) The crate rises is distance Ly - d stn 0, where 0 is the angle that the incline makes with the horizontal. The work done by the gravitational force of Earth is Ws -(25.0kg)(9.8 mls2xl.50m) sin 25.0"- -155 J. (c) The nonnal force of the incline on the crate is perpendicular to the displacement of the crate, so this force does no work. (d) The net work done on the qate is Wnet: We - 3I4J - 155 J - 159J.

"Ws

7l Let Wr (: 1 10 N) be the first weight hung on the scale and r be the elongation of the spring with this weight on it. Let Wz (- 240 N) be the second weight hung on the scale and n2 be the elongation of the spring when this weight is hung on it. In each case the spring pulls upward with a force that is equal to the weight hung on rt, so according to Hooke's law Wr -- krt and Wz: krz, where k is the spring constant. Now 11 and 12 are not the readings on the scales but nz- nr is the difference of the scale readings. Subtract the two Hooke's law equations to obtain Wz Wr: k(*z - rr). Thus _

,u

Wz /v,/

Wt f v L

n2-rr-

240N 10-3m . greater than mgLl2(. ot Lr, must be greater than Llz(. - 0.57. 85

The force diagram for the ladder is shown on the right. Fs is the force of the ground on the ladder, F* is the force of the wall, and W is the weight of the ladder. The horizontal component of Newton's second law gives F + Fn* - F*- 0 and the vertical component gives Fno W - 0. Fn' Set the net torque about the point where the ladder touches the wall equal to zero. The honzontal component of the force of the ground has a lever arrn of h, the vertical component of the force of the ground has a lever affn of (., where (. is the distance from the foot of the ladder to the wall, the applied force has a lever arrn of (1 dlL)h, where L is the length of the ladder, and the gravitational force has a lever ann of (. 12. Thus Fn*h - Fnrl+ Fh(l - d,lL) + WtlT - 0. The r axis was taken to be horizontal with the positive r direction to the right and the A axis was taken to be vertical with the positive direction upward. The vertical component of the second law gives F ga

-

W

:

200

N. The torque

equation gives

Fn*

Now (. _ J(1om)2 - (8.0m)2 : 6.0m, so ((. lh)Fno - (6.0 m)l(8.0m)(200N) : 150N, I d,lL - 1 (2.0m)110m)-0.80, and (dlzh)W_ (6.0m)(200N/2(8"0m) - 75N. This means D 150N-(0.80)F-75N: 75N-(0.80),F. tgn

If .F' - 50N, Fn*:75N - (0.80)(50N) - 35N and Fn - (35N)i+1ZO0Ni. (b) If F:150N, Fn*:75N- (0.80)(150N;- -45N and Fn - (-45N)i+(200Dj. (a)

(c) When the ladder is on the verge of slipping the frictional force is to the left, in the negative direction. Its magnitude is (0.80)F 75N. If the ladder does not slip this must be less than prFga: (0.38X200N - 76N. The applied force must be less than (75N+ 76N) 1Q.80) 1.9 x 102 N. Thus the applied force that will just start the ladder moving is I.9 x 102 NI.

r

76

Chapter 12

Chapter L3

1

The magnitude of the force of one particle on the other is given by It - Gmflt Lz I ,', where TrLl and Tfi,2 are the masses, T is their separation, and G is the universal gravitational constant. Solve

for r: GmtffLz

F

^

1

rc.67

x 0- 1r N . m2 lke\(s .2ke)e.4 kg) 1

7

At the point where the forces balance GM"*lr?- GMr*|r3, where M" is the mass of Earth, M, is the mass of the Sun, m is the mass of the space probe, rr is the distance from the center of Earth to the probe, and 12 is the distance from the center of the Sun to the probe. Substitute 12 - d - rt, where d is the distance from the center of Earth to the center of the Sun, to find

M"-

M" (d

,2,

- ,t)'

Take the positive square root of both sides, then solve for T1

Values for

_ d\M

\M+tM -

(lso x I .99

x

11

. A little algebra yields

toem) 1030

kg +

A[., Mr, and d can be found in Appendix

5.98

x

1024

kg

- 2.60 x 108 m.

C.

17

The gravitational acceleration is given by as: GMf r2, where M is the mass of Earth and r is the distance from Earth's center. Substitute r- R+ h, where R is the radius of Earth and h is the altitude, to obtain as: GMI(ft+ h)'. Solve for h. You should get hAccording to Appendix C of the text, R- 6.37 x 106m and M - 5.98 x 1024 kg, so

@-R.

h:

(6.67

x 10-ll m3 lr'

.

kgX5.98

x L024 kg)

-6.37 x 106m- 2.6 x

106m.

29

of a uniform sphere is given by p : 3M l4nR3, where M is its mass and R its radius. The ratio of the density of Mars to the density of Earth is (a) The density

Pnr

pn Mp Rtu

j A 5 IX i 0, k* 0 .6 5

I g+ km

is

3

- 0.74. Chapter

13

77

(b) The value of an at the surface of a planet is given by

Mxa

r t'F. R1

0 .6 5

A7I/I- ,r-lVL t' a WHagE:0'11 (c)

r I g+ km

1 75 X

M

ag: GMIR',

0, k*

so the value for Mars is

2

(9.8 mlr2)

- 3.8 ^lt'

If u is the escape speed, then, for a particle of mass m

Gry

**r': and

For Mars

u-V, Fu.67 x 10-

F ,1 :5.0x10'm/s.

37

(a) Use the principle of conservation of energy. Initially the particle is at the surface of the asteroid and has potentral energy [-Li: -GMmlR, where M is the mass of the asteroid, R is its radius, and m is the mass of the particle being fired upward. The initial kinetic energy is i*r'. The parttcle just escapes if its kinetic energy is zeto when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero. Consenration of energy yields -GMmlR+ i*r':0. Replace GMIR with agR, where a,e is the gravitational acceleration at the surface. Then the energy equation becomes -a,sB+ *r'- 0. Solve for u: 2(3.0 m/s ';1soo

2anR

x 103m):1.7 x

103m1

s

(b) Initially the particle is at the surface; the potential energy is [Ii: -GMrnlR and the kinetic energy is Ki comes to rest. The final potential energy is tJy: -GMml(R+ h) and the final kinetic energy is K,; - 0. Conservation of energy yields

GMm -;+;*,2: Replace

GMm

I

R+h

GM with onR2 and cancel m in the energy

equation to obtain onR2

-asR+)*:

(R+ h)

The solution for h is

h-

2anR2

2onR

-

u2

-R

z(g.o m/s2)(soo

- 2(3.0^lrt;1soo :2.5 x 105 m. 78

Chapter 13

x

103

x

m)

to3 m)2

-

(1000 mls)2

(500

x

103

m)

(c) Initially the particle is a distance h above the surface and is at rest. The potenttal energy is U, - -GMml@+ h) and the initial kinetic energy is K,; - 0. Just before it hits the asteroid the potentral energy is U y -GMmlR.Write *.*r'r for the final kinetic energy. Conservation of energy yields

GMm

GMm

R+h

Replace

R

1.

+ - mL)" 2

GM with onR2 and cancel m to obtain onR'

esR*lr'

R+h The solution for u is 2anR2

u-

2(3.0^lrt;1soo x --

1..4

x

103

m/s

103

m)

-

2(3.0^lr';1soo x 500

x 103m+

1000

103 m)2

x

103m

.

39

(a) The momentum of the two-star system is conserved, and since the stars have the same mass, their speeds and kinetic energies are the same. use the principle of conservation of energy. The initial potential energy is Ut,- -GM'lrn where M is the mass of either star and ri is their initial center-to-center separation. The initial kinetic energy is zero since the stars are at rest. The final potential energy is U1: -zGM'lrn since the final separation is ,n12. Write Muz for the final kinetic energy of the system. This is the sum of two terms, each of which is *M r' . Consenration of energy yields

_GMz

__

Ti

_2GIVI2 + Muz T6

The solution for u is (6.67

x

10-rr

m3

lr'.kgX103o kg)

10lo m

-

8.2

x

104

m/s

.

(b) Now the final separation of the centers is ry - 2R: 2 x 105 m, where R is the radius of either of the stars. The final potentral energy is given by LI 1 - -GM'lrf and the energy equation becomes -GM'lrn: -GM'lrt + Mu2. The solution for u is

tlrt

x 10-1rm3 lr' 'keXl030ke) /,u.67 -1.8 x 107m/s.

-

(^iffi

mt*) Chapter 13

79

45

Let l/ be the number of stars in the galaxy, M be the mass of the Sun, and r be the radius of the galaxy. The total mass in the galaxy is IY M and the magnitude of the gravitational force acting on the Sun is F - G M' lr'. The force points toward the galactrc center. The magnitude of the Sun's acceleration is a: 12 lR, where u is its speed. If T is the period of the Sun's motion around the galacttc center then u- 2nRlT and a, -- 4n2RlT'. Newton's second law yields GIY M'I R' : 4r2 M RlT2. The solution for tf is 4r2 R3 iv- GTzM

The period is 2.5

rv

x

10t y, which is 7.88

- (6.67

x

4n2(2'2

101t

r,

so

x 7020 m)3

)(7.88 x I01t r)t (2.0

x 1030 kg)

:

5.1

x 1010

4t (a) The greatest distance between the satellite and E,arth's center (the apogee distance) is Ro: 6.37 x 106m+360 x 103 m - 6.73 x 106m. The least distance (perigee distance) is Rp - 6.37 x 106m+180x 103m - 6.55 x 106m. Here 6.37 x 106m is the radius of Earth. Look at Fig. 13-13 to see thatthe semimajoraxis is o : (Ro+R|)f 2 - (6.73x 106m+6.55x 106 lrri;12 - 6.64x 106m.

(b) The apogee and perigee distances ate related to the eccentricity e by Ro

-

o( I

+ e) and

Rp

Ro

-

Rp

- 2ae. Thus

Rp e: Ro-2aRp RoRo * Rn

6,73x 106m- 6.55 x 106m 6.73

x

106

m * 6.55

x 106 m -0.0136.

6t (a) IJse the law of periods: T2 : (4n'IGM)r3, where M is the mass of the Sun (1.99 x 1030kg) and r is the radius of the orbit. The radius of the orbit is twice the radius of Earth's orbit: r - 2r. - 2(1 50 x 10em):300 x 10em. Thus

T_

4r213

4n2(300 (6.67

x 10- 11 m3 I ,'

x 10e m)3 . kgX r .gg

x

1030

kg)

Divide by (365 dlfiQ{hldX60 minlhx60 s/min) to obtain T

- 2.8y.

(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given by K - G M m l2r, where m is the mass of the asteroid or planet. Notice that it is proportional to m and inversely proportional to r. The ratio of the kinetic energy of the asteroid to the kinetic

80

Chapter 13

energy of Earth is K I K.

KIK.

- (m ln'Le)(r.l i.

- 2.0 x lT-am" and r :2r"

Substitute m

to obtain

7s

(a) Kepler's law of periods gives 4n2(4.20

4r213

(6.67

x

107 m)3

x 10-11N .nP lke'X9.50 x

l02s kg)

-2.15 x

(b) The craft goes a distance2nr tnaperiod, so its speed is us :2r(4.20x 1.23 x 104 mls.

107

104s.

m)l(2.L5x

104

s):

(c) The new speed is ,u:0.98u0

- (0.98XI.23 x 104 mls): l.2I x 104 mls. (d) The kinetic energy of the crafi is K - **r': *tz000kgXr.2\ x 104 mls)z -2.20 x 1011 J. (e) The gravitational potential energy of the planet-craft system is

u--GW r

\_-_

t^_o/

4.20X107m

where the potential energy was taken to be zero when the craft is far from the planet.

(0 The mechanical -2.33 x 10ll J.

energy of the planet-craft system is

(g) The mechanical energy of a satellite is given by E axis. Thus

o"-

GmM

(6.67

2E

E

- K+U - 2.20x 1011 J -4.53 x 1911 1-

:

-GmM l2a, where a is the semimajor

x l0-11N . m2 lke\(3000kgX9.50 x 2(-2.33 x 10rr J)

t02s

kg)

- 4.08 x I07 m.

(h) and (i) The new period is 4r2

a3

(6.67

x

4n2(4.08 x 107 m)3 10-11 N .m2 lkg'X9.50

Thechangeintheperiodis 2.06 x 104s- 2.I5 x 104sorbit is smaller by 9 x 102(4.20 x 107 m)3 s.

x 104s. x l02s kg) -2.06

-9 x 103s. Theperiodforthesecond

79

Use 7;t - GmrTTlrnlr', where ms is the mass of the satellite, mrn is the mass of the meteor, and r is the distance between their centers. The distance between centers is r: R+d - 15m+3m 18 m. Here R is the radius of the satellite and d is the distance from its surface to the center of the meteor. Thus

_ 6.67 x 10-rr N ,b'7;1

.m2 lke')Q0kgX7.0

kg):

^ Z.9

x l0-ll N. Chapter

13

8l

83

(a) The centripetal acceleration of either star is given by o, - u2r, where w is the angular speed and r is the radius of the orbit. Since the distance between the stars is 2r the gravitational force of one on the other is Gmz IQD', where m is the mass of either star. Newton's second law gives Gmz IQD' : murLr. Thus

a-

I

Gm ,t

2

Tt

I

-;l ^Fu,

x 1o-rrN .mlkgl3o. lorkgf

-2'2 x 10-7rcdfs'

(b) As the meteoroid goes from the center of the two-star system to far away the kinetic energy changes by LK- -+mu2 and the potentral energy changes by Ltf - ZGmMlr, where M is the mass of the meteoroid and u is its speed when it is at the center of the two-star system. Since energy is conserved LK + A(J : 0 and 4(6.67

x

10-rr N .m2 lke\(3.0 1.0 x 10ll m

x

1030kg)

:

8.9

x

104

m/s.

87

(a) Since energy is conserved it is the same throughout the motion and there is no variation. (b) The potential energy at the closest distance (perihelion) is

(Je: -GMnMs T'p

-

-5 .40 x

:-(6

.67 x 1o-11 N .m2

(5.98

x lo24 kgX I .99 x 1030 kg) 1.47 x 101r m

(5.98

x

lkg')

1033 J

and at the furthest distance (aphelion) is IJo

ra

The difference is 1 .8

x

T024

kgX | .99

L.52

x

x 1030 kg)

10rr m

1032 J.

(c) Since energy is conserved the vanation in the kinetic energy must be the same as the variation in the potenttal energy, 1.8 x 1032 J. (d) The semimajor axis is a: (ro{ra)12 - (1.47 x 108km+1 .495 x 108 k^)12 - 1.50 x 108km. The kinetic energy at perihelion is

Ke: GMnMs

1l

_l

2")

[;

Now

x

10ll m

- 3 .46 x 10-12 m-1

so

- (6.67 x 1o-rr N .m2 lkg')(s.99 x 1024 kgX r .gg x 1030 kgX3 .46 x l0- m- ) - 2.74 x 1033 J and the speed is u,p 2(2.7 4 x 1033 J)l (5.98 x 1024 kg) - 3 .02 x 104 m/s. 2KIME 82 Chapter 13 Ke

12

I

Since angular momentum is conserved uprp: nara and the speed at aphelion is 't)a: uprolro (3.A2 x 104 mlsxl .47 x 10rr rri;l!.sz x 10rt 2.93 x 104m/s. The varration is 3.02 x 104 mls 2.93 x 104 mls - 9.0 x I02 m/s.

*) -

-

93

Each star is a distance r from the central star and a distance 2r from the other orbiting star, so it is attracted toward the center of its orbit with a force of magnitude

rt-GM: 12

*ry (2r)2

4r2

\r-r

According to Newton's second law this must equal the product of the mass and centripetal acceleration ,2 I,. Each star travels a distance 2rr in a time equal to the period ?, so ?r :2rrf T, and the centripetal acceleration is 4n2 r lT2. Thus

The solution for

? is

4rr3 /2

Chapter 13

83

Chapter 14

1

The air inside pushes outward with a force given by piA, where Pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by PoA, where Po is the pressure outside. The magnitude of the net force is F = (Pi - Po)A. Since 1 atm = 1.013 x 105 Pa,

F = (1.0 atm - 0.96 atm)(1.013 x 105 Pa/atm)(3.4 m)(2.1 m) = 2.9 x 104 N .

J. The change in the pressure is the force applied by the nurse divided by the cross-sectional area of the syringe: F F 42N 5 D.p = A = 7fR2 = 7f(1.1 X 1O- 2 m)2 = 1.1 x 10 Pa.

11 The pressure P at the depth d of the hatch cover is Po + pgd, where p is the density of ocean water and Po is atmospheric pressure. The downward force of the water on the hatch cover is (Po + pgd)A, where A is the area of the cover. If the air in the submarine is at atmospheric pressure then it exerts an upward force of PoA. The minimum force that must be applied by the crew to open the cover has magnitude F = (Po + pgd)A - PoA = pgdA = (l024kg/m3 )(9.8 m/s 2)(100m)(1.2 m)(0.60m) = 7.2 x 105 N. 19

When the levels are the same the height of the liquid is h = (hI + h2)/2, where hl and h2 are the original heights. Suppose hI is greater than h2. The final situation can then be achieved by taking liquid with volume A(h l - h) and mass pA(h l - h), in the first vessel, and lowering it a distance h - h2. The work done by the force of gravity is W = pA(h l - h)g(h - h2)' Substitute h = (hI + h2)/2 to obtain

W

= ~pgA(hl =

~(1.30

=

0.635 J.

X

- h2)2

103 kg/m3 )(9.8 m/s2)(4.00 x 10- 4 m2)(1.56 m - 0.854 m)2

27 (a) Use the expression for the variation of pressure with height in an incompressible fluid: P2 = PI - pg(Y2 - YI). Take YI to be at the surface of Earth, where the pressure is PI = 1.01 X 105 Pa, 84

Chapter 14

and Uz to be at the top of the atmosphere, where the pressure ts 1.3 kg/*'. Then,

Uz-At v

p9

(1.3 kel^'Xq.gr;7s)

pz: 0. Take the density to be

:7'9 x 103m-

7.gkm.

(b) Let h be the height of the atmosphere. Since the density varies with altitude, you must use the integral

Pz: Take

p:

po(l

p:poatA-0

-Alh),

and can be evaluated:

p:

:

Ir^

ps da

.

where po is the density at Earth's surface. This expression predicts that 0 atA-h. Assume g isuniformfrom A-0 to A-h. Nowtheintegral Pz

Since p2

Pt

: pr -

lr^ Poe(r I)

da

: pr

*

pogh

.

0, this means

a\r.vr x 105 2(1.01 r'-r r*/ Pa) I 103m:16km. ' ^v X = - 16 Pog (1.3 kgl^'Xq.8 m/s2) 2pr

h_Lyr: 31

(a) The anchor is completely submerged. It appears to be lighter than its actual weight because the water is pushing up on it with a buoyant force of p*gV , where p* is the density of water and V is the volume of the anchor. Its effective weight (in water) is Weff: W - p*gV, where W is its actual weight (the force of gravity). Thus

v

2o=oN - w"rc-w P*9 = - 2.045 x lo-2 m3 (gg8 kg/m'Xq.8 m/s2)

The density of water was obtained from Table

I4-I

.

of the text.

(b) The mass of the anchor is m - pV, where p is the density of iron. Its weight in arr is W : mg: pgv - (7870kelm'Xg .8^ls\(2.045 x r0-2m3) - 1.58 x 103 N. 35

(a) Let V be the volume of the block. Then, the submerged volume is Vr:2Vf3. According to Archimedes' principle the weight of the displaced water is equal to the weight of the block, so p-V, : pbV , where pu is the density of water, and pa is the density of the block. Substitute V"- 2Vl3 to obtain pa- 2p*13- 2(gg8kg/mt)lZ: 6.7 x \}zkgl^t. The density of water was obtained from Table l4-l of the text. (b) If po is the density of the oil, then Archimedes' principle yields poV, - paV. Substitute V, - 0.90V to obtain po: pbf 0.90: 7.4 x I02kgl^t. 37

(a) The force of gravity mg is balanced by the buoyant force of the liquid pgV": mg Here rra is the mass of the sphere, p is the density of the liquid, and V, is the submerged volume. Chapter

14

85

Thus m sphere, or V"

4nl

lJ?, :

Air in the hollow

sphere,

6

if

(?) (8oo kel^'xo oeom)3 - r zkg

Ol'r'o

ufry, has been neglected.

(b) The density prn of the materral, assumed to be uniform, is given by p,* : m lV , where m is the mass of the sphere and V is its volume. If ri is the inner radius, the volume is

4r

V-

-r?)- T [(0.090 m)3 -

The density is

m rn- v :

L.22kg 9.09

x lo-a m3

(0.080 m)3]

1.3

x

103

- 9.09 x 10-a m3 kgl^t

49

Use the equation of continuity. Let u1 be the speed of the water in the hose and u2 be its speed as it leaves one of the holes . Let A1 be the cross-sectional area of the hose. If there are l/ holes you may think of the water in the hose as l/ tubes of flow, each of which goes through a single hole. The cross-sectional area of each tube of flow is At lN . If A2 is the area of a hole the equation of continuity becomes urAtlN: uzAz. Thus u2: (AtllVAr)ur - (R2 llYr')ut, where R is the radius of the hose and r is the radius of a hole. Thus

u2:

fi

(0.9s cm)2

Tfur,

s3

Suppose that a mass A^m of water is pumped in time Lt The pump increases the potential energy of the water by A^mgh, where h is the vertical distance through which it is lifted, and increases its kinetic energy by i/r*u2, where u is its final speed. The work it does is LW : L*gh+ ).Lmuz and its power is

Lnz ( rP - Lw on* 1.,t) N Ar \'"'," ) Now the rate of mass flow is L* lAf - pAu, where p is the density of water and atea of the hose. The area of the hose is A pAu obtained from Table

I4-I P

-

of the text. Thus

/ \"

pAu I gh+

- (1.s 7 kgls) :66W. 86

Chapter

I4

1rt)

2/

l-. I

(9

L

.B^1s2X3.0 LLL) *' /\J'\t m)

(5'0 m/s)2 2

I I

A is the

-33 Use the equation of continuity: Apr : Azuz. Here A1 is th e cross-sectional arca of the pipe at the top and u1 is the speed of the water there; A2 is the cross-sectional area of the pipe at the bottom and u2 is the speed of the water there. Thus u2 : lg.0 cm2)/(8.0 cm2)] (5.0 mls) z.Sm/s. (b) Use the Bernoulli equation: h + Lpu?+ pghr : pz+ Lpr'r* pghz, where p is the density of water, fu is its initial altitude, and hz is its final altitude. Thus

(t)

pz:

p1 *

)ofr? -

,l) +

ps(hr

- hz) [(s. o mls)2

2

: 2.6 x 10s Pa.

+ (998 ke l^tx9

The density of water was obtained from Table

-

(2.5 m/s)2]

.8^ls2xlo

m)

I4-l of the text.

59

(a) Use the Bernoulli equation: h+ |pu?+ pgfu: pz+ lpu2z* pghz, where fu is the height of the water in the tank, h is the pressure there, and u1 is the speed of the water there; h2 is the altitude of the hole, pz is the pressure there, and u2 is the speed of the water there. p is the density of water. The pressure at the top of the tank and at the hole is atmospheric, So pr : p2. Since the tank is large we may neglect the water speed at the top; it is much smaller than the speed at the hole. The Bernoulli equation then becomes pghl : ipr? + pghz and

U2:

2g(hr

The flow rate is Azuz- (6.5

x

2(g.8*ls2xo.3o m)

10-4 m2)(2.42m1s): 1.6

: 2.42m/s

.

x 10-'*'/r.

(b) Use the equation of continuity: A2u2: Azut, where A3: Azl2 and rh is the water speed where the cross-sectional area of the stream is half its cross-sectional area at the hole. Thus u3 : (Az I At)uz : Zuz - 4.84 ml s. The water is in free fall and we wish to know how far it has fallen when its speed is doubled to 4.84 m/s. Since the pressure is the same throughout the fall, Lputr

+ pghz

h2 h3: 1 : 42s

(4.84mf s)2

-

(2.42mf s)2

2e.8 m/s2)

- 0.90m.

67

(a) The continuity equation yields Au - aV and Bernoulli's equation yields ipr' - Lp + )pvz, where Lp: pz - pr. The first equation gives V- (Ala)u. Use this to substitute for V in the second equation. You should obtain *pr' : Lp + )p(Ala)2rr2 . Solve for u. The result is

2Lp

#) Chapter 14

87

(b) Substitute values to obtain

't):V

-3.06m/s.

The density of water was obtained from Table (64 x 10-4 m2X3.06mls):2.0 x I0-2m3/s.

l4-l of the text.

The flow rate is Au

75

(: 998 kg/-') be the density of water and pt (:

kg/-') be the density of the other liquid. Let d*r, be the length of the water column on the left side, d*n (: 10.9cm) be the length of the water column on the right side, and ds (: 8.0 cm) be the length of the column

Let p

800

of the other liquid. The pressure at the bottom of the tube is given by po + pd,t * p*d*L, where po is atmospheric pressure, and by po * p*d- n These expressions must be equal, so po

* pdt, * p-d*r :

p0

* p*d*n

d-r- P*d-n-Pt'dt': P*

The solution for

d*r is

998

kgl^t

-3.5gcm.

Before the other liquid is poured into the tube the length of the water column on the left side is the same as the water column on the right side, namely 10.0cm. After the liquid is poured it is 3.59cm. The length decreases by 10cm - 3.59cm - 6.41 cm. The volume of water that flows out of the right arrn is T(L 50 cm)2 (6.41 cm) : 45.3 cm3.

88

Chapter 14

Chapter

L5

3 (a) The amplitude is half the range of the displacement, or

trrn:

1

.0 rrlln.

(b) The maximum speedu* is relatedto the amplitude trmby n*: u)frrn, where a is the angular frequency. Since ur - 2nf ,where / is the frequency, 'trrn:2rf rrn:2r(l20Hz)(l .0x 1g-: m): 0.7 5 m/s.

(c) The maximum acceleration is 5.7

x

102

^/r2

arn

.

7 (a) The motion repeats every 0.500 s so the period must be T

* 0.500 s.

(b) The frequency is the reciprocal of the period: f : IIT - IlQ.500s):2.00H2. (c) The angular frequency is a - 2nf :2r(2.00H2) - l2.6radf s.

(d)Theangu1arfrequencyisre1atedtothespringconstantkandthemaSSmbya_\ffi,So

k : mw2 - (0.500 kgX 12.57 radls)' : 79.0 N/m. (e) If trtn is the amplitude, the maximum speed is 't)rn _ urfrrn - 02.57 rud1sx0.350 m) _ 4.40 m/s. (0 The maximum force is exerted when the displacement is a maximum and its magnitude is given by Frn 2

The magnitude of the maximum acceleration is given by arn: tt2trrn, where a is the angular frequency and nnl is the amplifude. The angular frequency for which the maximum acceleration

is9isgivenby,-\MandthecoffeSpondingfrequencyisgivenby Aa

tr2n

: I

l-s

2"y .-,

9.8

^lt'

:498H2.

For frequencies greater than 498zHz the acceleration exceeds g for some part of the motion.

t7 The maximum force that can be exerted by the surface must be less than F'FN or else the block will not follow the surface in its motion. Here, F" is the coefficient of static friction and F71/ is the normal force exerted by the surface on the block. Since the block does not accelerate vertic ally, you know that Flr : mg, where m is the mass of the block. If the block follows the table and moves in simple harmonic motion, the magnitude of the maximum force exerted on it is given by tr - ma,m- mw2rTn acceleration ) u) is the angular frequency, and f is the frequency. The relationship a : 2n f was used to obtain the last form. Chapter 15

89

Substitute It - m(2nf)'r,n and F^r - mg into F < largest amplitude for which the block does not slip is

T/s1) o.o3 rnl-(2nF'9f)2. : (o'soX?'8 (2n x 2.0Ht)2 -

1

m.

A larger amplitude requires a trargw force at the end points of the motion. The surface

cannot

supply the larger force and the block slips.

t9 (a) Let

A

n1 -zcos

/ 2trt\

( , /

be the coordinate as a function of time for particle 1 and

12:

Iror(+.;)

be the coordinate as a function of time for particle 2. Here T is the period. Note that since the range of the motion is A, the amplitudes are both A 12. The arguments of the cosine functions are in radians.

Particle 1 is at one end of its path (*t - Al2) when t ZntlT+116 - 0 or t- -f lI2. That is, particle 1 lags particle 2 by one-twelfth a period. We want the coordinates of the particles at t - 0.50 s. They are

x o'50 s) ntI :4 ro, (2zr l.5s ) 2v\'\)\

and

A lznxo.5os -+=zr\l--0.4334. fr2:^cost 2 \ 1.5s 6/ Their separation at that time is n1 - n2: -0.2504+ 0.433A - 0.1834. (b) The velocities of the particles are given by ?,1

(Znt\ -drt:4 dt T-sm\ t /

and

d"

Lt't-zdt T \T 6/ these expressions for t - 0.50 s. You will find they are both negative,

Evaluate the particles are moving in the same direction.

indicating that

27

When the block is at the end of its path and is momentarily stopped, its displacement is equal to the amplitude and all the energy is potential in nature. If the spring potential energy is taken to be zero when the block is at its equilibrium position, then

E

90

Chapter 15

- lrrl: )r.t x t02N/,o) (0.024m)2 : 3.7 x t0- 2 I

.

29

-

(a) and (b) The total energy is given by E itt*',-, where k is the spring constant and rrn is the amplitude. When r : i*r- the potential energy is f-Jt - *tt*' - ttt*?". The ratio is 1

itt*',"vE itt*',"

1

4

The fraction of the energy that is kinetic is

(c) Since E :L

- :rrnlt/r.

-

K

E-U

E

E

1-1 Y-1 844

UIE:

*tr*h, and fJ: - *tt*',

12

l*',".

Solve

3-

,'l*?*-

Llz for tr. You should get

39

(a) Take the angular displacement of the wheel to be 0 - ?rn cos(2nt lT), where 0,n is the amplitude and T is the period. Differentiate with respect to time to find the angular velocity: O - -(2n lT)0,- sin(2zrt lD. The symbol O is used for the angular velocity of the wheel so it is not confused with the angular frequency. The maximum angular velocity is Qrn

T

39.5 radf s .

0.500 s

(b) When 0 - n12, then 010,"- l12, cos(ZntlT): I12, and

:@:tZlz,

sin(2zrtlT)where the trigonometric identity cosz 4+ sin2 A a_

o

-

-+T o*sin

(r):

- I was used. Thus

(#)

Qrrad)

(f) :

-3 4.2radf

s

.

The negative sign is not significant. During another portion of the cycle its angular speed is +34.2radf s when its angular displacement is n l2rad. (c) The angular acceleration is

0

/

a-

(J*

d2

2n\2

dtr: I t ,)

When

0,n cos(2rt lT)

-

0-n14, (osoo

Again the negative sign is not significant.

\2

'/

(;):

-

1

24

radlr'

Chapter

15 9l

43

(a) A uniform disk pivoted at its center has a rotational inertia of + M R2, where M is its mass and R is its radius. See Table I0-2. The disk of this problem rotates about a point that is displaced from its center by R + L, where L is the length of the rod, so, according to the parallel-axis theorem, its rotational inertia is Lm n2 + M (L + R)2. The rod is pivoted at one end and has a rotational inertia of mL'13, where m is its mass. The total rotational inertia of the disk and rod is I - *mn2 + M(L+ R)2 + |mL2: 1f0.500kgX0.100 m)2 +(0.500kgX0.500m+ 0.100m)z + 1to .2J0kgX0.500 m)2 - 0 .205kg . m2. (b) Put the origin at the pivot. The center of mass of the disk rs,(.a - L+R - 0.500m+0.100m 0.600m away and the center of mass of the rod is (.r- L12: (0.500m)12:0.250m away, on the same line. The distance from the pivot point to the center of mass of the disk-rod system is t _ M la t m(., d_-ffi:

(0.500 kgx0.600 m) + (0.270 kgX0 .250

m) _u.+Til'r. _nA

(c) The period of oscillation is

T

:2r

0.205 kg . m2

+ m)sd

(0.500 kg + 0 .270kgX9 .8^ls2x0 .447 m)

-1.50s.

51

If the torque

exerted by the spring on the rod is proportional to the angle of rotation of the rod and

if the torque tends to pull the rod toward its equilibrium orientation, then the rod will oscillate in simple harmonic motion. If r -- -C0, where T is the torque, 0 is the angle of rotation, and

e

is a constant of proportionality, then the angular frequency of oscillation is a

theperiodisT-2nfa-2rr\re,whereIistherotationa1inertiaoftherod.Thep1anisto find the torque as a function of 0 and identify the constant e in terms of given quantities. This immediately gives the period in terms of given quantities. Let (.s be the distance from the pivot point to the wall. This is also the equilibrium length of the spring. Suppose the rod turns through the angle 0, with the left end moving away from the wall. If L is the length of the rod, this end is now (Llz)sind fuither from the wall and has moved (Ll2)(1 -cos 0)to the right. The spring length is now {ff lDz(I _- cos q2 +Vo+ (LlT)sin 0f', If the angle e is small we may approximate cos I with 1 and sin 0 with 0 in radians. Then the length of the spring is given by (,0+ L0 12 and its elongation is Lr: L0 12. The force it exerts on the rod has magnitude It - k Lr- kL?f 2, where k is the spring constant. Since 0 is small we may approximate the torque exerted by the spring on the rod by r: -FLf 2, where the pivot point was taken as the origin. Thus r - -(kL' lqg. The constant of proportionality C that relates the torque and angle of rotation is e - kL2 f 4. The rotational inertia for a rod pivoted at its center is I - mLz ll2, where m is its mass. See Table L0-2. Thus the period of oscillation is

t--t"l -^ tr,:2r t'r1

92

Chapter 15

mL2 l12 kL2 l4

0.600 kg 3(18s0

N/*) -

0.0653

s

57

(a) You want to solve e-bt/2*

logarithm of both sides to obtain (2* lb)1n3, where the sign was -btlZm reversed when the argument of the logarithm was replaced by its reciprocal. Thus

t_

ffih3-14.3s

(b) The angular frequency is

,- m *

w

4rrr':

V

8.00

N/*

(0.230 kg/s)2

4(1.50 kg)'

1.50 kg

2.31 radf s .

The period is T (14.3 s)/(2.72 s) : 5.27 . 75

(a) The frequency for small amplitude oscillations is the pendulum. This gives

f : (l lzr)\ffi,,

f - (Ilzr)tl (9.80^ls\l(2.0m):0.3

where

L is the length of

5Hz.

(b) The forces acing on the pendulum are the tension force 7 of the rod and the force of gravity mj. Newton's second law yields f + mfi : md, where m is the mass and d is the acceleration of the pendulum. Let d _ d,. + d, where d,. is the acceleration of the elevator and u" is the acceleration of the pendulum relative to the elevator. Newton's second law can then be written m(d - d") + f : md,'. Relative to the elevator the motion is exactly the same as it would be in an inertial frame where the acceleration due to gravity is d - d,". Since j and d," are along the same line and in opposite directions we can find the frequency for small amplitude oscillations byreplacing g with g*a" in the expression f -(Ilzr)\FgIL Thus f^T r2n

ml

9.8

VL

mlrt

+

2.0^lt'

- 0.39H2.

(c) Now the acceleration due to gravity and the acceleration of the elevator are in the same direction and have the same magnitude. That is, d - d," - 0. To find the frequency for small amplitude oscillations, replace g with zero in f - (llzr)\ffi,. The result is zero. The pendulum does not oscillate. 83

[Jse 'uTL stroke, or 0.38 m. Thus

unl:

2n(3.0 Hz)(0.38 m)

:

7

"zm/s.

89

(a) The spring stretches until the magnitude of its upward force on the block equals the magnitude of the downward force of gravity: ky - mg, where A is the elongation of the spring at equilibrium, k is the spring constant, and m is the mass of the block. Thus k

- mg la - (1 .3 kgX9.8 */

sz)l (0.096

m)

:

133

N/-. Chapter

15

93

(b)TheperiodisgivenbyT:Ilf:2nlu):2r1ffik_2rr-0.62S. (c) The frequency is

f : IIT -

Il0.62s

-

L.6Hz.

(d) The block oscillates in simple harmonic motion about the equilibrium point determined by the forces of the spring and gravity. It is started from rest 5.0cm below the equilibrium point so the amplitude is 5.0 cm. (e) The block has maximum speed as it passes the equilibrium point. At the initial position, the block is not moving but it has potenttal energy U,i,

When the block is atthe equilibrium point, the elong atioirof the spring is potential energy is (,1

1,

I

y- -msa. ;ka'^

Write the equation for conservation of energy as [-Li: 2(Lh

- tI r)

A- g.6cm and the

I

,rl **r'and solve for u:

2(-0.44 J + 0.61

m

- 0.51 m f s.

9t (a) The frequency

of oscillation is

- 3.2H2. (b) Because mechanical energy is conserved the maxlmum kinetic energy of the block has the same value as the maximum potential energy stored in the spring, so i*u?^

rrn --

l-m

N Zurn:

l.2kg ,

Affi(5.2mls)

: 0.26 [m.

(c) The position of the block is given by *: nmcos(c,,,t+0\ where n?n - 0.26m and u) - 2nf 2r(3.2H2):2\rudf s. Since r-0 at time t-0, the phase constant O must be either +rl2 or -r, 12. The velocity at t-0 is given by -urfrrn sin@ and this is positive, so 6 must be -TT12, The function is tr : (0.26m) cos[(20 radls)f - Tr l2].

94

Chapter 15

Chapter L5

15

The wave speed u is given by , where r is the tension in the rope and LL is the linear mass density of the rope. The linear mass density is the mass per unit length of rope:

- \mu

m 0.0600 kg o.o3ookg m. p: if ,Jo* Thus 0.0300 kg lm

-

I29 m/s.

t7 (a) In the expression given for y, the quantrty A* is the amplitude and so is 0.1Zmm.

(b) The wave speed is given by u - tFn, where r is the tension in the string and Lr is the linear mass density of the string, so the wavelength is

t-

u

tFrtp

-c

.t

and the angular wave number is

l4lm-r :100H2,

- 2nf :2r(100H2):628rad/s. (d) The positive sign is used since the wave is traveling in the negative r direction. (c) The frequency is

f

so the angular frequency is

a

2t (a) Read the amplitude from the graph. It is the displacement at the peak and is about 5.0 cm. (b) Read the wavelength from the graph. The curye crosses A - 0 at about r : 15 cm and agatn with the same slope at about tr- 55cm, so )- 55cm- 15cm:40cm-0.40m.

(c) The wave speed is

where T is the tension in the string and

u-

pr,

is the linear mass density of the string. Thus

10-3kglm -

L2mf s. Chapter 16

95

(d) The frequency is

f

v 12m/s = - = = 30Hz A 0.40m

and the period is

1

1

T = - = - - = 0.033 s . f 30Hz

(e) The maximum string speed is Um

= WYm = 27i}Ym = 2IT(30Hz)(5.0cm) = 940cm/s = 9.4m/s.

(f) The angular wave number is 27i k = - =

A

27i

0.40m

= 16m- I

(g) The angular frequency is w = 2IT f = 2IT(30 Hz) = 1.9

X

.

102 rad/s.

(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 X 10- 2 m. The formula for the displacement gives y(O,O) = Ym sin ¢. We wish to select ¢ so that 5.0 x 10- 2 sin ¢ = 4.0 X 10- 2. The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select ¢ = 0.93 rad. (i) A positive sign appears in front of w because the wave is moving in the negative ;]; direction. 31 The displacement of the string is given by Y = Ym sin(kx - wt) + Ym sin(kx - wt + ¢) = 2Ym cos(~¢) sin(kx - wt + ~d»,

where ¢

=

IT /2. The amplitude is

A = 2Ym cos(~¢) = 2Ym cos (IT /4) = 1.41Ym·

35 The phasor diagram is shown to the right: Yl m and Y2m represent the original waves and Ym represents the resultant wave. The phasors corresponding to the two constituent waves make an angle of 90° with each other, so the triangle is a right triangle. The Pythagorean theorem gives y~ = + = (3.0 cm)2 + (4.0 cm)2 = 25 cm2 .

Y?m Yim

Thus Ym

=

5.0 cm.

41 Possible wavelengths are given by A = 2L / n, where L is the length of the wire and n is an integer. The corresponding frequencies are given by f = v / A = nv /2L, where v is the wave 96

Chapter 16

tm,llrLlM, where T is the tension in the wire, :M L M is the mass of the

speed. The wave speed is given by u tt is the linear mass density of the wire and obtain the last form. Thus

wire.

T

^TL r2L +

LM

(10.0 mX0.100 kg)

p,

I

was used to

- n(7.glHr)

: 1, f - 7 .9I Hz. (b) For ft:2, f : 15.8 Hz. (c) For fr:3, f :23.7H2. (a) For

rL

43

(a) The wave speed is given by u - \mt, where r is the tension in the string and p is the linear mass density of the string. Since the mass density is the mass per unit length, F : M I L, where M is the mass of the string and L is its length. Thus (96.0 NXS.40 m)

0.l20kg

- 82.0 m/s

.

(b) The longest possible wavelength .\ for a standing wave is related to the length of the string by L - ,\f2, so )- 2L-2(8.40m): 16.8m. (c) The frequency is

f-

ul^

- (82.0^ls)/(16.8m) - 4.88Hz,.

47

(a) Thc resonant wavelengths are given by - zLf n, where L is the length of the string and n ^ given by f - ul^ - nnlzL, where u is the wave is an integer, and the resonant frequencies are speed. Suppose the lower frequency is associated with the integer n,. Then since there are no resonant frequencies betweeo, the higher frequency is associated with n*1. That is, fr: TL't)l2L is the lower frequency and fz: (n * l)u lzL is the higher. The ratio of the frequencies is

fz: n+r fin' The solution for n'

is rL

: -f, fz h

420H2-3I5Hz

r'

f - uf 2L - hl, - (3 l5Hz)13 (b) The longest possible wavelength is ) _ 2L. If f is the lowest possible frequency then u - xf - 2Lf - 2(0.7s mX105 Hr) - 158 m/s. The lowest possible resonant frequency is

53

The waves have the same amplitude, the same angular frequency, and the same angular wave number, but they travel in opposite directions. Chapter 16

97

(a) The amplitude of each of the constituent waves is half the amplitude of the standing wave or 0.50 cm.

(b) Since the standing wave has three loops the string is three half-wavelengths long. If L is the length of the string and is the wavelength, then L -3^12, or ,\ - 2L13:2(3.0m)13 -2.0m. ^ is k - 2nl) - 2nl(2.0m):3.1m-l. The angular wave number (c) If u is the wave speed, then the frequency is

u 3u

I: )- n-

3(100 m/s)

,(3.0"r) The angular frequency is u):2nf :2r(50H2) - 3.1 x I02rudf s. (d) Since the first wave travels in the negative r direction, the second wave must travel in the positive r direction and the sign in front of u must be a negative sign. 6t (a) The phasor dtagram is shown to the right: Ut, Uz, and Us represent the original waves and Urn represents the resultant wave. The honzontal component of the resultant is arnh Ut Az - Ur arl3 - 2yrf 3. The vertical component is U*, : U2 : Ar f 2. The amplitude of the resultant is I

Urn:m:

5 : ,al : o

0.83at

(b) The phase constant for the result ant is 6 _tan-r

u*, _ tan-r ( -1 3 -tan-r:-0.644rad:37" ^M^\ 4 arnh \ 2Yr 13 /

(c) The resultant wave is

:

5

stn(kr - wt + 0 .644 rad) 6At The graph below shows the wave at trme / - 0. As time goes on it moves to the right with speed A

't)

: a lk.

-U'n -Ut 98

Chapter 16

.

69

(a) Take the form of the displacement to be A@,t): arnstn(kr - at). The speed of a point on the cord is u(r, t) - 0y l 0t wave speed, on the other hand, is given by, - alk. The ratio is

^lf

urn : aArn 2narn Ngrn: n u alK 7

(b) The ratio of the speeds depends only on the ratio of the amplitude to the wavelength. Different waves on different cords have the same ratio of speeds if they have the same amplitude and wavelength, regardless of the wave speeds , linear densities of the cords, and the tensions in the cords. 77

(a)

If r is the tension in the wire and

LL

a

is its linear mass density, then the wave speed is mlL, where m is the mass of the wire and L is its

length. Thus

(r20NX1.s0m) 8.70

x 10-3 kg

:

I44mls

.

(b) A one-loop standing wave has two nodes, one a each end, and these are half a wavelength apart, so the wavelength is - 2L - 2(1 .50m) - 3.00m.

^ three nodes, one at each end and one at the midpoint. Since the A two-loop standing wave has nodes are half a wavelength apaft, the wavelength is ,\ - L - 1.50 m. (c) and (d) The frequency is f - ulx. For the one-loop wave f : (I44mls)l(3.00m):48.0H2 and for the two-loop wave f : (I44mls)/(1.50m) :96.6H2. 87

(a) The transverse rope velocity is given by , : aUrn, where a is the angular frequency and is the amplitude. The angular frequency is a - 2n f, where f is the frequency. Thus oa

ant

:

uu

a

m/s 2r(5.0 Hz) 5.0

:

Znf

Arn

- 0.I6m.

(b) The wave speed is u - \trn where r is the tension in the rope and LL is the linear mass density of the rope. The linear mass density is l.L- mlL, where m is the mass of the rope and L is its length. The wave speed is , where is the wavelength. Since the rope is vibrating in ^ its fundamental mode ) - 2L Thus^f

r-puz-irxrf (c) Thc general forrn for the displacement at coordrnate r and time t for a standing wave that has nodes atn -0and r- Iis A:Amstn(2rrlDsrn(2nft. Here Un isthemaximumdisplacement of any of the points along the rope. Since, in this case, the rope is vibrating in its fundamental Chapter

16

99

mode Urn is the maximum displacement of the point at its center and thus has the value calculated in part (a). The displacement at any coordinate r is

a@,t):(0.16m)sin|#")sin|2rr(5.0Hz)t]-(0.I6m)sin[(1.6m-')"]sin[(31s-,)']. 89

(a) The wave speed is given by u- \mu where r is the tension in the rubber band and lr is the linear mass density of the rubber band. According to Hooke's law the tension is r - k L(.. The length of the stretched rubber band is (.+ L(,, so the linear mass density is l.L: ml!+ Lt). The wave speed is k L(,(t + Lt)

(b) The time for a pulse to travel the length of the rubber band is (.+ al t_T_((,+Lt)

m((+ Lt) k L(,((,+

L!)

[f L(. is much less than (, we may neglect the LI in the numerator. Then m[.

KN which is proportional to L l161-. (c) It L(. is much greater than (, we may

u: which is independent of L(..

100

Chapter 16

in the numerator. Then

Chapter L7

5-

Let t y be the time for the stone to fall to the water and t, be the time for the sound of the splash to travel from the water to the top of the well. Then the total time elapsed from dropping the stone to hearing the splash is t_ ty *tr. It d is the depth of the well, then the kinematics of free fall gives d : *gt?, or t y or f" : d/ur.Thus the total time is 2d

t-

d

I

g

This equation is to be solved for d. Rewrite it

us

as

t-

d us

and square both sides to obtain

2d, I

-lL

. L

-z-tasd*\a' u'"

NIow multiply by gu? and realrange to get gd2

-

Zu,(gt

I

rr

s)d +

gu?* :

0.

This is a quadratic equation for d. Its solutions are

d-

2u

r(gt | 'u s) *

ar?@t + ur)2 2s

4g2u?t2

The physical solution must yield d : 0 for t : 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d-40.7mis obtained. 1_

If d, is the distance from the location of the earthquake to the seismograph and u" is the speed of the S waves, then the time for these waves to reach the seismograph is ts: d/ur. Similarly, the time for P waves to reach the seismograph is

Af

tp: d/up.The time delay is

d d - us up --

d(up u

-

u

")

)

sup Chapter

17

101

SO

.l _ u-

lrsup

Lt

(4.5kn'lsx8.0 km/sX3.0 minx60

s/min)

1

^

x

103

km.

Notice that values for the speeds were substituted as given, in km/s, but that the value for the time delay was converted from minutes to seconds. 2 (a) Use

^

- ulf,

where u is the speed of sound

\ _ 343mls 7 '62 x 4.5 x 106 H;-

/t

(b) Now

--

-

\_

lt

Lr

10-5 m '

-

1500

4.5

x

m/s 106

Hr-

3 '33

x 1o-4 m '

be the distance from the closer speaker to the listener. The distance from the other

speaker to the listener is

L2: lW

phase difference at the listener ls

where

Thus

ulf , where u is the speed of sound in tissue. The frequency is the same for air

^ Thus and tissue.

t9 Let

in air and f is the frequency.

d is the distance between the speakers. The

,

2n(Lz

-

Lt)

is the wavelength.

(a) For^ a minimum in intensity at the listener, 0 "\ - 2(Lz L) lQn + 1). The frequency is

u Jt _ r

(2n +

l),

) 2lrry-1,1 2l

-

(2n +

l)n, where n is an integer.

(2n + 1X3 a3 mls)

-37sm]

I

Thus

- (2n+ 1X343 Ht).

To obtain the lowest frequency for which a minimum occurs set n equal to 0. The frequency is f^in,l : 343H2.

(b) To obtain the second lowest frequency set n equal to 1. This means multiply F*i* I by 3. (c) To obtain the third lowest frequency set n equal to 2. This means multiply lr",in, I by 5. For a maximum in intensrty at the listener, d

:

2nn, where n is any positive integer. Thus

)-:lEa-r'] and

:::-u

nu

lW-11

n(343 m/s) (3.7 5 m)2 + (2.00 m)2

-

3.7 5

m

n(686 Hz)

.

(d) To obtain the lowest frequency for which a maximum occurs set n equal to 1. The frequency is f^u*,I : 686 Hz.

102

Chapter 17

(e) To obtain the second lowest frequency set n equal to 2. This means multiply f-in, 1 by 2.

(0 To obtain the third lowest frequency

set

n equal to 3. This means multiply .F*6, t by

3.

25

The intensity is the rate of energy flow per unit area pe{pendicular to the flow. The rate at which energy flows across every sphere centered at the source is the same, regardless of the sphere radius, and is the same as the power output of the source . If P is the power output and I is the intensity adistance r fromthesource,then P- IA:4rr2 I,where A(:4rr2) isthesurface area of a sphere of radius r. Thus P - 4r(2.50 m)2(1.91 x 10-4W/ttr') - 1.50 x I0-2'W'. 29

(a) Let 11 be the original intensity and 12 be the final intensity. The original sound level

h reference intensity. Since Cz

is

: Cr + 30 dB,

(10 dB) Iog(I2l

Iil - (10 dB) log(/t I IO + 30 dB

(10 dB) tog(I2l

Iil -

,

or

Divide by 10dB and use log(I2lIil each side as an exponent

(10 dB) log(/r I

IzlIt:

.

- log(hlIo): Iog(I2lI) to obtain log(I2lIr):3.

Now use

of 10 and recogntze that 1glog(/2

The result is

Iil - 30 dB

/I) :

Izl

I,

103. The intensity is multiplied by a factor

of 1.0 x

103.

(b) The pressure amplitude is proportional to the square root of the intensity so it is multiplied by a factor of .,4000 :32. 43

(a) The string is fixed at both ends and, when vibratittg at its lowest resonant frequency, exactly half a wavelength fits between the ends . If L is the length of the string and l is the wavelength, then - 2L The frequency is f - ulx- ul2L, where u is the speed of waves on the string. Thus ^'u : 2L f - 2(0.220 m)(920 Hz) : 405 m/s.

(b) The wave speed is given by u - \mt where r is the tension in the string and pt is the linear mass density of the string. If M is the mass of the string, then p : M I L since the string is uniform. Thus

r: truz: f*:W(4os (c) The wavelength is

) - 2L - 2(A.220m) :

^ls)z-5e6N.

0.440m. Chopter

17

103

(d) The frequency of the sound wave in arc is the same as the frequency of oscillation of the string. The wavelength is different because the wave speed is different. If u o is the speed of sound in air the wavelength in atr is

,

t\a

ua

l-

^

,l

m/s) 92AHz

343

- 0 .373 m.

45

(a) Since the pipe is open at both ends there are displacement antinodes at both ends and an integer number of half-wavelengths fit into the length of the pipe. If L is the pipe length and A is the wavelength then - 2Lln, where n is an integer. If u is the speed of sound then the resonant ^ f :nf frequencies are givenbV Now L-0.457m, so f :n(344mls)12Q.457m)^-nuf2L. frequencies that lie between 1000H2 and 2000H2, first set 376.4nH2. To find the resonant f - 1000Hz and solve for rL, then set f :2000H2 and agaun solve for rL. You should get 2.66 and 5.32. This means ft:3,4, and 5 are the appropriate values of n There are three resonance frequencies in the given range. (b) For fr:3, f :3(376.4H27 - Il29Hz. (c) For fr: 4, f : 4(376.4H2) - 1506 Hz. 47

The string is fixedatboth ends so the resonantwavelengths are given by,\ - 2Lln,where L is the length of the string and n is an integer. The resonant frequencies are given bV f - n l A: nu f 2L, where t' is the wave speed on the string. Now 'tr : tmt where T is the tension in the string and p is the linear mass density of the string. Thus f : (nlzL),,mr. Suppose the lower frequency is associated with TL : TL1 and the higher frequency is associated with TL : rL1 + I . There are no resonant frequencies between so you know that the integers associated with the given frequencies differ by 1. Thus fi - (nrlzL)tmt and

fz

TLr*l F

TLt tr

1 n

!

1

l-t,

This means

fz h - (llzL)\mt and r : 4L2 ttffz f )t - 4(0.300 m)2(0.650 x 10-t kgl*Xl3 20Hz -

880

Ht)'

:45.3N.

53

Each wire is vibrating in its fundamental mode so the wavelength is twice the length of the wire () 2L) and the frequency is

:

f: )ullT2Ly p) :

(: \mD

l-

is the wave speed for the wire, T is the tension in the wire, and where u linear mass density of the wire.

104

Chapter 17

p is the

in one wire is r and the oscillation frequency of that wire is f r. The tension in the other wire is r * Lr and its frequency is fz. You want to calculate Lr lr for fi: 600 Hz Suppose the tension

and

fz:

606 Hz.

Now

f.-1 E rt n\ LL

and

SO

This means

LrT

fz

Lr

ft

T

G),

1

- (g6H'\t \600 Hrl

I-

o'o2o '

-05 (a) The expression for the Doppler shifted frequency is

.f'-fry * a

us

where f is the unshifted frequency, u is the speed of sound, u p is the speed of the detector (the uncle), and us is the speed of the source (the locomotive). All speeds are relative to the air. The uncle is at rest with respect to the air, so ?/p : 0. The speed of the source is u g the locomotive is moving away from the uncle the frequency decreases and we use the positive sign in the denominator. Thus

f'

U f Ug

343mls \ 343mls + 10.00m/s

)

:

48s.8

Hz.

(b) The girl is now the detector. Relative to the atr she is moving with speed u p toward the source. This tends to increase the frequency and we use the positive sign in the numerator. The source is moving at us: 10.00m/s away from the girl. This tends to decrease the frequency and we use the positive sign in the denominator. Thus (u + u n) - (u + u il and f' - f : 5oo.oHz. (c) Relative to the arr the locomotive is m10v m bo\at u,,,S =20 ).0(0 mtl:s away from the uncle. I-Jse 10.0r the positive sign in the denominator. Relattive irre:t,or thrv