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이 블로그는 할리데이 일반물리학 10판 한글판을 직접 풀어서 올리는 솔루션 블로그 입니다.

정확한 책은 아래의 책에 해당합니다.

1권

일반물리학1 개정10판

저자 JEARL WALKER, DAVID HALIDAY, Robert Resnick

출판 범한서적

출간일 2015년 03월 05일

ISBN 9788971292679

https://book.naver.com/bookdb/book_detail.nhn?bid=8824070

2권

일반물리학 2 개정10판

저자 JEARL WALKER, David Halliday, R. Resnick

출판 범한서적

출간일 2015년 08월 15일

ISBN 9788971292709

https://book.naver.com/bookdb/book_detail.nhn?bid=9473909

직접 풀기에, 오타나 틀린 문제가 있을 수 있습니다.

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목차

1권

1. 측정

2. 직선운동

3. 백터

4. 2차원 운동과 3차원 운동

5. 힘과 운동 – Ⅰ

6. 힘과 운동 – Ⅱ

7. 운동에너지와 일

8. 퍼텐셜에너지와 에너지 보존

9. 질량중심과 선운동량

10. 회전

11. 굴림운동, 토크, 각운동량

12. 평형과 탄성

13. 중력

14. 유체

15. 진동

16. 파동 – Ⅰ

17. 파동 – Ⅱ

18. 온도, 열, 열역학 제1법칙

19. 기체운동론

20. 엔트로피와 열역학 제2법칙

2권

21 Coulomb의 법칙

22 전기장

23 Gauss의 법칙

24 전기퍼텐셜

25 전기용량

26 전류와 저항

27 회로이론

28 자기장

29 전류가 만드는 자기장

30 유도와 유도용량

31 전자기적 진동과 교류

32 Maxwell 방정식, 물질의 자성

33 전자기파

34 영상

35 간섭

36 회절

37 상대론

38 광자와 물질파

39 물질파 더 알아보기

40 원자의 모든 것

41 고체의 전기전도

42 핵물리

43 핵 에너지

44 쿼크, 경입자, 그리고 빅뱅

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ISV SM Ch05 할리데이 10판 5장 솔루션

195

(a) The free-body diagram depicted for the body is

shown to the right. The forces are

F 1 =

r 9 N and F 2 =

r 7 N

The resultant force is

2 2 (9 N) (7 N) 81 N 49 N

130 N 11 N

F= + = +

= =

r

.

null

Using the equation ,

F a m

=

r r we get the magnitude of the acceleration of the body as

11 N 2

1 m/s. 6 kg

F

a m

= = =

r

(b) From the figure, the direction of the acceleration of the body is calculated as follows:

2

1

7 N

tan 0 N 38 , 9 N

F

F

q= = ª fi q= ∞

from which we conclude that the direction of the acceleration of the body is 38 ∞ counter-

clockwise from the positive x direction.

196 CHAPTER 5

We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force

applied on the chopping block is

r r r Fnet=F 1 +F 2 , where the vector addition is done using

unit-vector notation. The acceleration of the block is given by

r r r

a=dF 1 +F 2 i/ .m

(a) In the first case

1 2 ( ) ( ) ( ) ( )

F+F =È3 N iˆ+ 4 N jˆ ̆+È-3 N iˆ+ -4 N jˆ ̆= 0 Î ̊ Î ̊

r r

so

r a= 0.

(b) In the second case, the acceleration

r a equals

(( ) ( ) ) (( ) ( ) )

1 2 2

3 N iˆ 4 N jˆ 3 N iˆ 4 N jˆ ˆ (3 m/s ) j. 2

F F

m

+ + – +

+

= =

r r

(c) In this final situation,

r a is

(( ) ( ) ) (( ) ( ) )

1 2 2

3 N iˆ 4 N jˆ 3 N iˆ 4 N jˆ ˆ (2 m/s )i. 2 kg

F F

m

+ + + –

+

= =

r r

198 CHAPTER 5

As the body is moving with a constant velocity (i.,

r v = constant), the net force acting

on the body is zero and therefore the acceleration of the body is also zero (i.,

r a= 0),

that is,

net 1 2 0

(2 N)ˆ (5 N)ˆ ( N)ˆ ( N)ˆ

F F F ma

i j x i y j

= + = =

= – + +

r r r r

Hence, the other force is

2 1 F = -F= -( 2 N) ˆi+( 5 N) .jˆ

r r

199

The net force applied on the chopping block is

r r r r Fnet=F 1 +F 2 +F 3 ,

where the vector addition is done using unit-vector notation. The acceleration of the

block is given by

Fnet F 1 F 2 F 3 a m m

+ +

= =

r r r r r

(a) The forces exerted by the three astronauts can be expressed in unit-vector notation as

follows:

( )

( )

( ( ) ( ) )

1

2

3

ˆ ˆ ˆ ˆ

(32 N) cos 30 i sin 30 j (27 N) i (16 N ) j

(55 N) cos 0 iˆ sin 0ˆj (55 N) iˆ

(41 N ) cos 60 iˆ sin 60 ˆ (20 N) iˆ ( 35 N ) j.ˆ j

F

F

F

= ∞ + ∞ = +

= ∞ + ∞ =

= – ∞ + – ∞ = –

r

r

r

The resultant acceleration of the asteroid of mass m = 120 kg is therefore

( ) ( ) ( )

2 2

27 .7 i 16 j Nˆ ˆ 55 i Nˆ 20ˆ 35 .5 j Nˆ ˆ ˆ (0 .8 6 m/ s ) i (0 m/ s )j. 1 20 kg

a

+ + + –

= = –

r

(b) The magnitude of the acceleration vector is

( )

2 2 2 2 2 2 2 a= ax +ay = (0 m/s ) + -0 m/s =0 m/s.

r

(c) The vector

r a makes an angle q with the +x axis, where

2 1 1 2

0 m/s tan tan 11. 0 m/s

y

x

a

a

q

– Ê ˆ – Ê- ˆ

= Á ̃= Á ̃= – ∞

Ë ̄ Ë ̄

201

THINK A box is under acceleration by two applied forces. We use Newton’s second

law to solve for the unknown second force.

EXPRESS We denote the two forces as

r r F 1 andF 2. According to Newton’s second law,

F 1 +F 2 =m a,

r r r so the second force is F 2 =ma-F 1.

r r r Note that since the acceleration is in

the third quadrant, we expect F 2

r to be in the third quadrant as well.

ANALYZE

(a) In unit vector notation and

( ) ( ) ( ) ( )

2 ˆ 2 ˆ 2 ˆ 2 ˆ

a= – 12 sin 30 m/s i∞ -12 cos 30 m/s∞ j= – 6 m/s i- 10 .4 m/s j.

r

Therefore, we find the second force to be

( )( ) ( )( ) ( )

( ) ( )

2 1 2 ˆ 2 ˆ ˆ 2 kg 6 m/s i 2 kg 10 m/s j 20 N i

ˆ ˆ 32 N i 20 N j.

F =ma-F

= – + – –

= – –

r r r

(b) The magnitude of

r F 2 is

2 2 2 2 |F 2 |= F 2 x+F 2 y= ( 32 N)- + -( 2 0 N) =38 .2 N.

r

(c) The angle that

r F 2 makes with the positive x-axis is found from

2

2

20 .8 N

t an 0 .6 56 3 2 N

y

x

F

F

f

Ê ˆ –

=Á ̃= =

Ë ̄ –

.

Consequently, the angle is either 33° or 33° + 180° = 213°. Since both the x and y

components are negative, the correct result is f = 213° from the +x-axis. An

alternative answer is 213 ∞ – 360 ∞ = -1 47 .∞

LEARN The result is shown in the figure on the right.

The calculation confirms our expectation that

r F 2 lies in

the third quadrant (same as a

r ). The net force is

( ) ( ) ( )

( ) ( )

net 1 2 20 N iˆ 32 N iˆ 20 N jˆ

ˆ ˆ

12 N i 20 N j

F =F+F = +È- – ̆

Î ̊

= – –

r r r

which points in the same direction as a

r .

202 CHAPTER 5

We note that

2 2 net

ˆ ˆ ˆ ˆ

F =ma=(1 kg) ( 8 m/s )i (6 m/s ) jÈ- + ̆= -( 12 N) i+( 9 N) j. Î ̊

r r

With the other forces as specified in the problem, then Newton’s second law gives the

third force as

F 3

Æ = m a

Æ – F 1

Æ – F 2

Æ = (–30 N) i

^ – (15 N) j

^ .

204 CHAPTER 5

At t=0 s, we have v t( =0 s) ( 6 m/s)= – iˆ+ -( 8 m/s) .ˆj

r Therefore, the angle

v

r makes with +x is

1 1 8 m/s tan tan 50 or 129.. 6 m/s

y v x

v

v

– Ê ˆ -Ê- ˆ

= Á ̃= Á ̃= ∞ – ∞

Ë ̄ Ë- ̄

q

Because v

r is in the third quadrant, we choose the angle of the object’s travel direction as

130°, rounded to three significant digits.

205

To solve the problem, we note that acceleration is the second time derivative of the

position function, and the net force is related to the acceleration via Newton’s second

law. Thus, differentiating

2 3 x t( )= -13 2+ t+4 -3

twice with respect to t, we get

2 2 2

2 8 9 , 8 18.

dx d x t t t dt dt

= + – = –

The net force acting on the particle at t=2 s is
[ ]
2

2

ˆ ˆ ˆ

i (0) 8 18(2) i ( 5 N)i

d x F m dt

= = – = –

r

207

From the slope of the graph we find ax = 3 m/s

2 . Applying Newton’s second law to

the x axis (and taking q to be the angle between F 1 and F 2 ), we have

F 1 + F 2 cosq = m ax fi q = 56∞.

208 CHAPTER 5

For the mass 3m, we have

3 mg- 2 T= 3 ma 3 (1)

For the mass 2m, we have

T- = 0 2 ma 1 (2)

For the mass m, we have

T- = 0 ma 2 (3)

From Eqs. (2) and (3), we get

2 1

a = 2 a

If the string of 2m mass moves by a distance x, the string of mass m moves by distance

2 x. The total distance 3x is shared equally on both sides of the pulley and the mass 3m

comes down by

2 2 0 0

3 1 1 1 from the equation of motion, ( ) 0 2 2 2

x x x x v t at at

È ̆ = – = + = + Í ̇ Î ̊

a 3 =1 1

Adding Eqs. (2) and (3), we get

1 2

1 1

2

2 2

T T ma ma

ma m a

+ = +

= +

That is,

2 T= 4 ma 1 (4)

From Eq. (1), we get

1 1

1

1

3 1

2 3

3 4 3 1.

3 8.

3

8.

1.

3 1 4.

0 5 m/s. 8 8.

mg ma m a

mg ma

a g

a a

a g g g

– = ¥

=

Ê ˆ

=

Á ̃

Ë ̄

=

Ê ¥ ˆ Ê ˆ

=Á ̃ =Á ̃ = =

Ë ̄ Ë ̄

210 CHAPTER 5

THINK We have a piece of salami hung to a spring scale in various ways. The

problem is to explore the concept of weight.

EXPRESS We first note that the reading on the spring scale is proportional to the weight

of the salami. In all three cases (a) – (c) depicted in Fig. 5-34, 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 tension force of the

cord attached to the salami must be the same in magnitude as the weight of the salami

because the salami is not accelerating. Thus the scale reading is mg, where m is the mass

of the salami.

ANALYZE In all three cases (a) – (c), the reading on the scale is

w = mg = (11 kg) (9 m/s

2 ) = 108 N.

LEARN The weight of an object is measured when the object is not accelerating

vertically relative to the ground. If it is, then the weight measured is called the apparent

weight.

211

(a) There are six legs, and the vertical component of the tension force in each leg is

Tsinq where q= 40 ∞. For vertical equilibrium (zero acceleration in the y direction) then

Newton’s second law leads to

6

6

T mg T

mg sin sin

q q

= fi =

which (expressed as a multiple of the bug’s weight mg) gives roughly T mg/ ª0.

(b) The angle q is measured from horizontal, so as the insect “straightens out the legs” q

will increase (getting closer to 90 ∞), which causes sinq to increase (getting closer to

and consequently (since sinq is in the denominator) causes T to decrease.

213

The free-body diagram of the cars is shown on the right. The force exerted by John

Massis is

2 F=2.5mg=2(80 kg)(9 m/s ) 1960 N=.

Since the motion is along the horizontal x-axis, using Newton’s second law, we have

Fx=Fcosq=Max, where M is the total mass of the railroad

cars. Thus, the acceleration of the cars is

2 5 2

cos (1960 N) cos 30 0 m/s. (7 10 N / 9 m/s )

x

F

a M

q ∞ = = = ¥

Using Eq. 2-16, the speed of the car at the end of the pull is

2 vx= 2 a xxD = 2(0 m/s )(1 m)=0 m/s.

214 CHAPTER 5

THINK In this problem we’re interested in the force applied to a rocket sled to

accelerate it from rest to a given speed in a given time interval.

EXPRESS

In terms of magnitudes, Newton’s second law of motion is F = ma, where F =

r Fnet ,

a=| |a

r , and m is the (always positive) mass. The magnitude of the acceleration can be

found using constant acceleration kinematics. Solving v = v 0 + at for the case where it

starts from rest, we have a = v/t (which we interpret in terms of magnitudes, making

specification of coordinate directions unnecessary). The velocity is

v = (1650 km/h) (1000 m/km)/(3600 s/h) = 458 m/s,

so

458 m s 5 (550 kg) 1 10 N. 2 s

v F ma m t

= = = = ¥

LEARN From the expression F=m v t/ , we see that the shorter the time to attain a given

speed, the greater the force required.

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