Tài liệu Physics exercises_solution: Chapter 04 pdf

b The bottle exerts an upward force on the earth, and a downward force on the air.. 4.22: The reaction to the upward normal force on the passenger is the downward normal force, also of

4.1: a) For the magnitude of the sum to be the sum of the magnitudes, the forces must

be parallel, and the angle between them is zero b) The forces form the sides of a right isosceles triangle, and the angle between them is 90 Alternatively, the law of cosines may be used as

 2 2 2 2cos ,2

from which cos 0, and the forces are perpendicular c) For the sum to have 0

magnitude, the forces must be antiparallel, and the angle between them is 180

4.2: In the new coordinates, the 120-N force acts at an angle of 53 from the  -axis, x

or 233 from the x -axis, and the 50-N force acts at an angle of 323 from the x axis

a) The components of the net force are

N32323

cos)N50(233cos)N120

sin)N50(233sin)N120()N250

4.3: The horizontal component of the force is (10N)cos 457.1N to the right and the vertical component is (10N)sin 457.1N down

4.4: a) F x Fcos, where  is the angle that the rope makes with the ramp (θ 30 in this problem), so cos60 030N 69.3N

4.5: Of the many ways to do this problem, two are presented here.

Geometric: From the law of cosines, the magnitude of the resultant is

.N49460

cos)N300)(

N270(2)N300()N270



R

The angle between the resultant and dog A’s rope (the angle opposite the side

corresponding to the 250-N force in a vector diagram) is then

)N300(120sin

cos)N300()N270

sin)N300

4.11: a) During the first 2.00 s, the acceleration of the puck is F/m1.563m/s2

(keeping an extra figure) At t2.00s, the speed is at3.13m/s and the position is

m13.32

)(2.00m/s

3(1/2)(1.56s)

m/s)(2.00(3.13

)m/s1000.3(2

2 14 2

2 6 2

s / m 10 00 3

2 14

t Note that this time is also the distance divided by

the average speed.

c) F  ma(9.1110 31kg)(2.501014m/s2)2.2810 16 N

4.15: F maw(a/g)(2400N)(12m/s2)(9.80m/s2)2.94103N

2.71

160/

F m

F a

4.17: a)mw/g(44.0N)/(9.80m/s2)4.49kg b) The mass is the same, 4.49 kg, and the weight is (4.49kg)(1.81m/s2)8.13N

4.18: a) From Eq (4.9), mw/g (3.20N)/(9.80m/s2)0.327kg.

b) w  mg (14.0kg)(9.80m/s2)137N

4.19: F  ma(55kg)(15m/s2)825N The net forward force on the sprinter is exerted by the blocks (The sprinter exerts a backward force on the blocks.)

4.20: a) the earth (gravity) b) 4 N, the book c) no d) 4 N, the earth, the book, up e) 4 N,

the hand, the book, down f) second g) third h) no i) no j) yes k) yes l) one (gravity) m) no

4.21: a) When air resistance is not neglected, the net force on the bottle is the weight of

the bottle plus the force of air resistance b) The bottle exerts an upward force on the earth, and a downward force on the air

4.22: The reaction to the upward normal force on the passenger is the downward normal

force, also of magnitude 620 N, that the passenger exerts on the floor The reaction to the passenger’s weight is the gravitational force that the passenger exerts on the earth, upward and also of magnitude 650 N 0.452m/s2

m/s /9.80 N 650

N 650 N 620

2 



m F

The passenger’s acceleration is0.452m/s2, downward

)kg100.6(

)s/m80.9)(

kg45

24 2

E E

F

a

4.24: (a) Each crate can be considered a single particle:

AB

F (the force on m due to A m ) and B F (the force on BA m due to B m ) form an A

action-reaction pair

(b) Since there is no horizontal force opposing F, any value of F, no matter how

small, will cause the crates to accelerate to the right The weight of the two crates acts at

a right angle to the horizontal, and is in any case balanced by the upward force of the surface on them

4.25: The ball must accelerate eastward with the same acceleration as the train There

must be an eastward component of the tension to provide this acceleration, so the ball hangs at an angle relative to the vertical The net force on the ball is not zero

4.26: The box can be considered a single particle.

For the truck:

The box’s friction force on the truck bed and the truck bed’s friction force on the box form an action-reaction pair There would also be some small air-resistance force action

to the left, presumably negligible at this speed

4.27: a)

b) For the chair, a y 0 so F y ma y gives

037sin 



mg F

n

N142

.26sin)s/m80.9)(

kg0.65



4.29: tricycle and Frank

T is the force exerted by the rope and f g is the force the ground exerts on the tricycle spot and the wagon

T  is the force exerted by the rope T and T  form a third-law action-reaction pair,

TT

4.30: a) The stopping time is 7.43 10 4 s

s / m 350 ) m 130 0 ( 2 ) 2 / (0

4.31: Take the x -direction to be alongF1

and the y -direction to be along R

Then N

4.33: a) The resultant must have no y-component, and so the child must push with a force

with y-component (140N)sin30(100N)sin6016.6N For the child to exert the

smallest possible force, that force will have no x-component, so the smallest possible

force has magnitude 16.6 N and is at an angle of 270 , or  90 clockwise from the 

x

 -direction

b) 85.6kg (85.6kg)(9.80m/s2) 840N

s / m 0 2

30 cos N 140 60 cos N 100

N100.8(2

)s/m5.1)(

kg106.3(2)/(2

2 7

2 0

2 0

v a

v

so the ship would hit the reef The speed when the tanker hits the reef is also found from

m/s,17.0)

kg106.3(

)m500)(

N100.8(2m/s)5.1()/2

4 2

so the oil should be safe

4.35: a) Motion after he leaves the floor: 2 2 ( 0)

s/m80.9/N890(N

)s/m5.12(kg)850(2

6 2

2 2

4.37: a)

(upward)net F mg

b) When the upward force has its maximum magnitude Fmax (the breaking strength), the net upward force will be Fmaxmg and the upward acceleration will be

.s/m83.5s/m80.9kg80.4

N0

mg F

a

4.38: a) w  mg539N

b)

Downward velocity is decreasing so a

is upward and the net force should be upward

N620s

/m80

mg F

m

F

a

4.39: a) Both crates moves together, so a2.50m/s2

b)

(4.00kg)(2.50m/s2) 10.0N

m a T

4.40: a) The force the astronaut exerts on the rope and the force that the rope exerts on

the astronaut are an action-reaction pair, so the rope exerts a force of 80.0 N on the astronaut b) The cable is under tension c) 2

kg 105.0 N 0

80 0.762m/s



 m F

net force on the massless rope, so the force that the shuttle exerts on the rope must be

80.0 N (this is not an action-reaction pair) Thus, the force that the rope exerts on the

kg 10 9.05 N 0

so,)s/m1040.2()s/m1080.1()

so (i) att 0,a1.8104 m/s2, and (ii) a(0.025s)6.0103m/s2, and the forces are (i) ma2.7104Nand (ii) ma9.0103N

4.42: a) The velocity of the spacecraft is downward When it is slowing down, the

acceleration is upward When it is speeding up, the acceleration is downward

b)

speeding up: wF and the net force is downward

slowing down: wF and the net force is upward

c) Denote the y-component of the acceleration when the thrust is F by 1 a and the y-1

component of the acceleration when the thrust is F by 2 a The forces and accelerations 2

are then related by

1 2

1

a

a w F

1 2 2 1

a a

F a F a w







In this form, it does not matter which thrust and acceleration are denoted by 1 and which

by 2, and the acceleration due to gravity at the surface of Mercury need not be found Substituting the given numbers, with y upward, gives

N

100.16)

s/m80.0(s/m20.1

)N100.25)(

s/m80.0()N100.10)(

s/m20

1

2 2

3 2

3 2

In the above, note that the upward direction is taken to be positive, so that a is negative 2

Also note that although a is known to two places, the sums in both numerator and 2

denominator are known to three places

4.44: a) If the gymnast climbs at a constant rate, there is no net force on the gymnast, so

the tension must equal the weight; T mg

b) No motion is no acceleration, so the tension is again the gymnast’s weight

m ma mg T

w

T     (the acceleration is upward, the same direction as the tension), so  (  a)

g m

m ma mg T

N000,

mg T m

F a

kg2200

N000,

4.46: a) His speed as he touches the ground is

s

/m80.7)m10.3)(

s/m80.9(2

)s/m80.7(2

2 2

F

w

F    soF m(ag)(75.0kg)(50.6m/s2 9.80m/s2)4532N As

a fraction of his weight, this force is  g a 1 6.16

mg F (keeping an extra figure in the

intermediate calculation of a) Note that this result is the same algebraically as 0.603 10mm  1

4.47: a)

b) The acceleration of the hammer head will be the same as the nail,

2 3 2

2

0/2 (3.2m/s) /2(0.45cm)1.13810 m/s

its weight divided by g,4.9N/9.80m/s2 0.50kg, and so the net force on the hammer head is (0.50kg)(1.138103 m/s2)570N This is the sum of the forces on the hammer head; the upward force that the nail exerts, the downward weight and the downward 15-N force The force that the nail exerts is then 590 N, and this must be the magnitude of the force that the hammer head exerts on the nail c) The distance the nail moves is 12 m, so the acceleration will be 4267m/s2, and the net force on the hammer head will be 2133

N The magnitude of the force that the nail exerts on the hammer head, and hence the magnitude of the force that the hammer head exerts on the nail, is 2153 N, or about 2200 N

a) The net force on a point of the cable at the top is zero; the tension in the cable must

be equal to the weight w.

b) The net force on the cable must be zero; the difference between the tensions at the

top and bottom must be equal to the weight w, and with the result of part (a), there is no

tension at the bottom

c) The net force on the bottom half of the cable must be zero, and so the tension in the cable at the middle must be half the weight, w/2 Equivalently, the net force on the

upper half of the cable must be zero From part (a) the tension at the top is w, the weight

of the top half is w/2 and so the tension in the cable at the middle must be

2/

N2.21

The force needed to lift the barbell is given by:

ma w

F

Fnet  lift  barbell The barbell’s mass is (490 N) (9 80 m s ) 50 0 kg   2   , so

N513N23N490

)s/m469.0)(

kg0.50(N

barbell lift

The athlete is not accelerating, so:

Fnet Ffloor Flift wathlete 0

N1395N

882N513athlete lift

F

L

g M

c) Lmgm (g/2), where m is the mass remaining.

L2Mg/3, so m4M/9 Mass 5M /9 must be dropped overboard

4.52: a) m mass of one link

The downward forces of magnitude 2ma and ma for the top and middle links are the

reaction forces to the upward force needed to accelerate the links below

b) (i) The weight of each link is mg(0.300kg)(9.80m/s2)2.94N Using the body diagram for the whole chain:

free-2 2

net 3.53m/s or3.5m/s

kg900.0

N18.3kg

900.0

)N94.2(3N12



m

F a

(ii) The second link also accelerates at 3.53m/s2, so:

and at t  , the acceleration is5 0s

.ˆ)s/m12.0(ˆs/m60.0

N)1075.2

(1.7104 N)iˆ(3.4103 N kˆ

4.54: The velocity as a function of time is v(t)A3Bt2 and the acceleration as a function of time is a(t)6Bt, and so the Force as a function of time is

mBt t

t

0

4 2

4ˆ11

)

4.56: a) The equation of motion, Cv2 m dv dt cannot be integrated with respect to time,

as the unknown function v (t) is part of the integrand The equation must be separated

before integration; that is,

,110

2

v v m Ct

v

dv dt m C

dx v

which may be integrated to obtain

.1

m x x

To obtain x as a function of v, the time t must be eliminated in favor of v; from the

expression obtained after the first integration, 0  v v0 1

m Ctv , so

m x x

b) By the chain rule,

,

v dx

dv dt

dv dx

dv dt

and using the given expression for the net force,

m dx

dv v

v

v x

x m C

m x x

4.57: In this situation, the x-component of force depends explicitly on the y-component of

position As the y-component of force is given as an explicit function of time, v y and y

can be found as functions of time Specifically, a y (k3/m)t, so 2

3/2 )(k m t

246

5 2 3 2 2 1

4 2 3 2 1

3 2 3 2 1

t m

k k t m

k x

t m

k k t m

k v

t m

k k m

k a

x x

ˆ24

ˆ6

ˆ120

2

2 3 4

2 3 2 1

3 3 5

2 3 2 2 1

j i

v

j i

k t

m

k k t m k

t m

k t

m

k k t m k



