CH4

P4.9 Model the fish as a particle under constant acceleration. We use our old standard equations for

constant-acceleration straight-line motion, with x and y subscripts to make them apply to parts of

the whole motion. At t = 0,

( )
vi = 4.00ˆi + 1.00ˆj m/s and rˆi = (10.00ˆi − 4.00ˆj) m

At the first “final” point we consider, 20.0 s later,

( )
v f = 20.0ˆi − 5.00ˆj m/s

υ x 20.0 m/s − 4.00 m/s

(a) ax = = = 0.800 m/s 2
t 20.0s

υ y −5.00 m/s − 1.00 m/s

ay = = = −0.300 m/s 2
t 20.0s

 −0.300 m/s 2 
(b)  = tan −1  2 
= −20.6 = 339 from + x axis
 0.800 m/s 

(c) At t = 25.0 s the fish’s position is specified by its coordinates and the direction of its motion

is specified by the direction angle of its velocity:

1
x f = xi + υ xi t + axt 2
2
1
= 10.0 m + (4.00 m/s)(25.0s) + (0.800 m/s 2 )(25.0s) 2
2
= 360 m
1
y f = yi +  yi t + a y t 2
2
1
= −4.00 m + (1.00 m/s)(25.0s) + ( −0.300 m/s 2 )(25.0s) 2
2
= −72.7 m
υ xf = υ xi + ax t = 4.00 m/s + ( 0.800 m/s 2 ) (25.0s) = 24 m/s
υ yf = υ yi + a y t = 1.00 m/s − ( 0.300 m/s 2 ) (25.0s) = −6.50 m/s
 υy  −1  −6.50 m/s 
 = tan −1   = tan   = −15.2
 υx   24.0 m/s 

P4.25 (a) For the horizontal motion, we have xf = d = 24 m:

 1
x f = xi + υ xi t + axt 2
2
24 m = 0 + i (cos 53)(2.2s) + 0
υi = 18.1m/s

(b) As it passes over the wall, the ball is above the street by

1
y f = yi + υ yi t + a y t 2
2
1
y f = 0 + (18.1m/s)(sin 53)(2.2s) + ( −9.8m/s 2 )(2.2s) 2 = 8.13m
2

So it clears the parapet by 8.13 m – 7 m = 1.13 m.

(c) Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole

flight, we have from the trajectory equation:

 g  2
y f = (tan i ) x f −  2  xf
 2i cos i
2


 9.8m/s2  2
or 6 m = (tan 53) x f −   xf
 2(18.1m/s) cos 53 
2 2

Solving,

(0.041 2 m−1 ) x 2f − 1.33x f + 6 m = 0

and, suppressing units,

1.33  1.332 − 4(0.041 2)(6)

xf =
2(0.0412)

This yields two results:

xf = 26.8 m or 5.44 m

The ball passes twice through the level of the roof.

It hits the roof at distance from the wall

26.8 m – 24 m = 2.79 m

P4.40 From the given magnitude and direction of the acceleration we can find both the centripetal and

the tangential components. From the centripetal acceleration and radius we can find the speed in

part (b). r = 2.50 m, a = 15.0 m/s2.

(a) The acceleration has an inward radial component:

 ac = a cos 30.0 = (15.0 m / s 2 )(cos 30)
= 13.0 m/s 2

(b) The speed at the instant shown can be found by using

2
ac =
r
 = rac = 2.50 m(13.0 m/s)
2

= 32.5 m 2 /s 2
 = 32.5 m/s = 5.70 m/s

(c) a 2 = at2 + ar2

so at = a 2 − ar2 = (15.0 m/s2 )2 − (13.0 m/s2 )2 = 7.50 m/s2

ANS. FIG. P4.40

P4.45 The airplane (AP) travels through the air (W) that can move relative to the ground (G). The

airplane is to make a displacement of 750 km north. Treat north as positive y and west as positive

x.

(a) The wind (W) is blowing at 35.0 km/h, south. The northern component of the airplane’s

velocity relative to the ground is

(AP,G ) y = (υAP,W ) y + (υW ,G ) y = 630 km/h − 35.0 km/h

=595 km/h

We can find the time interval the airplane takes to travel 750 km north:

y = ( υAP,G ) y t →
y 750 km
t = = = 1.26 h
(υ A,P,G ) y
595 km/h

(b) The wind (W) is blowing at 35.0 km/h, north. The northern component of the airplane’s

velocity relative to the ground is

 (υAP,G ) y = (υAP,W ) y + (υW ,G ) y = 630 km/h − 35.0 km/h
=665 km/h

We can find the time interval the airplane takes to travel 750 km north:

y 750 km
t = = = 1.13h
(υ ) AP,G y
665 km/h

(c) Now, the wind (W) is blowing at 35.0 km/h, east. The airplane must travel directly north

to reach its destination, so it must head somewhat west and north so that the east

component of the wind’s velocity is cancelled by the airplane’s west component of

velocity. If the airplane heads at an angle θ measured west of north, then

(υAP,G ) x = (υAP,W ) x + (υW ,G ) x = (630 km/h)sin +( − 35.0 km/h) = 0

sin = 35.0/630 →  = 3.18

The northern component of the airplane’s velocity relative to the ground is

(υAP,G ) y = (υAP,W ) y + (υW ,G ) y = (630 km/h)cos3.18 + 0

= 629 km/h

We can find the time interval the airplane takes to travel 750 km north:

y 750 km
t = = = 1.19 h
(υ AP ,G ) y
629 km/h
CH5

P5.26 (a) The left-hand diagram in ANS.FIG. P5.26(a) shows the geometry of the situation and lets us

find the angle of the string with the horizontal:

cosθ= 28/35.7 = 0.784

or θ = 38.3º

The right-hand diagram in ANS. FIG. P5.26(a) is the free-body diagram. The weight of the bolt is

w = mg = (0.065 kg)(9.80 m/s2) = 0.637 N

(b) To find the tension in the string, we apply Newton’s second law in the x and y directions:

F x = max : −T cos38.3 + Fmagnetic = 0

[1]

F y = may : −T sin 38.3 − 0.637 N = 0

[2]

from equation [2],

0.637 N
T= = 1.03 N
sin 38.3

(c) Now, from equation [1],

Fmagnetic = T cos38.3 = (1.03 N)cos 38.3 = 0.805 N to the right

ANS. FIG. P5.26(a)

*P5.36 (a) First construct a free-body diagram for the 5.00-kg mass as shown in the Figure 5.36a. Since

the mass is in equilibrium, we can require T3–49.0 N = 0 or T3=49.0 N. Next, construct a free-body diagram

for the knot as shown in ANS. FIG. P5.36(a).

Again, since the system is moving at constant velocity, a = 0, and applying Newton’s second law in

component form gives

F x = T2 cos50.0 − T1 cos 40.0 = 0

F y = T2 sin 50.0 + T1 sin 40.0 − 49.0N = 0
 Solving the above equations simultaneously for T1 and T2 gives T1 = 31.5 N and T2 = 37.5 N and above we

found T3 = 49.0 N.

(b) Proceed as in part (a) and construct a free-body diagram for the mass and for the knot as shown in

ANS. FIG. P5.36(b). Applying Newton’s second law in each case (for a constant-velocity system), we find:

T3 − 98.0 N = 0

T2 − T1 cos 60.0° = 0

T1 sin 60.0° − T3 = 0

Solving this set of equations we find:

T1 = 113 N, T2 = 56.6 N, and T3 = 98.0 N

ANS. FIG. 5.36(a)

ANS. FIG. 5.36(b)

P5.42 m1 = 2.00 kg, m2 = 6.00 kg, θ = 55.0°

(a) The forces on the objects are shown in ANS. FIG. P5.42.

(b) ∑Fx = m2g sin θ − T = m2a and

T − m1 g = m1a
m2 g sin  − m1 g
a=
m1 + m2
(6.00 kg)(9.80 m/s 2 ) sin 55.0 − (2.00 kg)(9.80 m/s 2 )
=
2.00 kg + 6.00 kg
= 3.57 m/s 2

(c) T = m1 (a + g) = (2.00 kg)(3.57 m/s2 + 9.80 m/s2) = 26.7 N

(d) Since υi = 0, υf = at = (3.57 m/s2)(2.00 s) = 7.14 m s.

 ANS. FIG. P5.42

*P5.47 We use the particle under constant acceleration and particle under a net force models. Newton’s

law applies for each axis. After it leaves your hand, the block’s speed changes only because of

one component of its weight:

F x = max
−mg sin 20.0 = ma
υ2f = υi2 + 2a( x f − xi )

ANS. FIG. P5.47

Taking υf = 0, υi = 5.00 m/s, and a = − g sin (20.0°) gives, suppressing units,

0 = (5.00)2 − 2 (9.80) sin (20.0°)(xf − 0)

or

25.0
xf = = 3.73m
2(9.80)sin(20.0)
CH6

P6.6 (a) The car’s speed around the curve is found from

235m
= = 6.53 m/s
36.0s

1
(2 r ) = 235 m,
This is the answer to part (b) of this problem. We calculate the radius of the curve from 4

which gives r = 150 m.

The car’s acceleration at point B is then

2 
a r =   toward thecenter
 r 
( 6.53m/s )
2

= at 35.0 north of west

150 m

( ( )
= ( 0.285 m/s 2 ) cos35.0 −ˆi + sin 35.0ˆj )
(
= −0.23ˆi + 0.163ˆj m/s 2 )
(b) From part (a), υ = 6.53 m/s

(c) We find the average acceleration from

a avg =
( v − vi )
t

=
( 6.53ˆj − 6.53ˆi ) m/s
36.0 s

(
= −0.181ˆi + 0.181ˆj m/s 2 )
P6.16 (a) We apply Newton’s second law at point A, with υ = 20.0 m/s, n = force of track on roller

coaster, and R = 10.0 m:

Mυ2
F = = n − Mg
R

ANS. FIG. P6.16

From this we find

 Mυ2 ( 500 kg ) ( 20.0 m/s 2
)
n = Mg + = ( 500 kg ) ( 9.80 m/s 2 ) +
R 10.0 m
n = 4900 N + 20000 N = 2.49  104 N

(b) At point B, the centripetal acceleration is now downward, and Newton’s second law now gives

Mυ2
 F = n − Mg = −
R

The maximum speed at B corresponds to the case where the rollercoaster begins to fly off the track, or when

n = 0. Then,

Mυ2max
− Mg = −
R

which gives

υmax = Rg = (15.0m) ( 9.80m/s 2 ) = 12.1 m/s

P6.25 The water moves at speed

2 r 2 (0.120 m)
υ= = = 0.104 m/s
T 7.25 s

The top layer of water feels a downward force of gravity mg and an outward fictitious force in the turntable

frame of reference,

mυ2 m ( 0.104 m/s )

2

= = m 9.01 10−2 m/s 2

r 0.12 m

It behaves as if it were stationary in a gravity field pointing downward and outward at

 0.0901 m/s 2 
tan −1  2  = 0.527
 9.8 m/s 

Its surface slopes upward toward the outside, making this angle with the horizontal.

P6.59 (a) The wall’s normal force pushes inward:

 Finward = mainward

becomes

mυ2 m  2 R  4 2 Rm
2

n= =   =
R R T  T2

The friction and weight balance:

 Fupward = maupward
becomes

ANS. FIG. P6.59

+ f − mg = 0

so with the person just ready to start sliding down,

f s = s n = mg

Substituting,

4 2 Rm
s n = s = mg
T2

Solving,

4 2 Rs
T2 =
g

gives

4 2 R s
T=
g

(b) The gravitational and friction forces remain constant. (Static friction adjusts to support the weight.)

The normal force increases. The person remains in motion with the wall.

(c) The gravitational force remains constant. The normal and friction forces decrease. The person slides

relative to the wall and downward into the pit.

CH7

P7.4 Assuming the mass is lifted at constant velocity, the total upward force exerted by the two men

equals the weight of the mass: Ftotal = mg = (653.2 kg)(9.80 m/s2) = 6.40 × 103 N. They exert this upward

force through a total upward displacement of 96 inches (4 inches per lift for each of 24 lifts). The total work

would then be

Wtotal = (6.40 103 N)[(96in)(0.0254 m/1in)] = 1.56 104 J

P7.13 Let θ represent the angle between A and B . Turning by 25.0º makes the dot product larger, so the

angle between C and B must be smaller. We call it θ – 25.0º. Then we have

5A cos θ = 30 and 5A cos (θ – 25.0º) = 35

Then

A cos θ = 6 and A (cos θ cos 25.0º + sin θ sin 25.0º) = 7

Dividing,

cos 25.0º + tan θ sin 25.0º = 7/6

or tan θ = (7/6 – cos 25.0º)/sin 25.0º = 0.616

Which gives θ = 31.6º. Then the direction angle of A is

60.0º – 31.6º = 28.4º

Substituting back,

A cos 31.6 = 6 so A = 7.05 m at 28.4

P7.21 (a) The force mg is the tension in each of the springs. The bottom of the upper (first) spring

moves down by distance x1 = |F|/k1 = mg/k1. The top of the second spring moves down by this distance, and

the second spring also stretches by x2 = mg/k2. The bottom of the lower spring then moves down by distance

mg mg 1 1
xtotal = x1 + x2 = + = mg  + 
k1 k2  k1 k2 

(b) From the last equation we have

x1 + x2
mg =
1 1
+
k1 k2
This is of the form

 1 
F =  ( x1 + x2 )
 1/ k1 + 1/ k2 

The downward displacement is opposite in direction to the upward force the springs exert on the load, so we

may write F = –keff xtotal, with the effective spring constant for the pair of springs given by

1
keff =
1 / k1 + 1 / k2

P7.42 (a) We take the zero configuration of system potential energy with the child at the lowest point

of the arc. When the swing is held horizontal initially, the initial position is 2.00 m above the zero level.

Thus,

u g = mgy
= (400 N)(2.00 m)
= 800 J

(b) From the sketch, we see that at an angle of 30.0° the child is at a vertical height of (2.00 m) (1 – cos

30.0º) above the lowest point of the arc. Thus,

ug = mgy = (400 N)(2.00 m)(1 − cos 30.0) = 107 J

(c) The zero level has been selected at the lowest point of the arc. Therefore, ug = 0 at this location.

ANS. FIG. P7.42

CH8

P8.6 (a) Define the system as the block and the Earth.

∆K + ∆U = 0

1 2 
 mυ B − 0  + ( mghB − mghA ) = 0
2 

1
m B2 = mg ( hA − hB )
2

υB = 2 g ( h4 − hB )

υ B = 2 ( 9.80 m/s 2 ) ( 5.00 m − 3.20 m ) = 5.94 m/s

ANS. FIG. P8.6

Similarly,

υC = 2 g ( hA − hC )
υC = 2 g ( 5.00 − 2.00 ) = 7.67 m/s

(b) Treating the block as the system,

1 2 1
= K = mυC − 0 = ( 5.00 kg )( 7.67 m/s ) = 147 J
2
Wg
A →C 2 2

P8.15 (a) The spring does positive work on the block:

1 2 1 2
Ws = kxi − kx f
2 2

Ws = (500 N/m) ( 5.00 10 −2 m ) − 0

1 2

2
= 0.625 J

Applying ∆K = Ws:

1 2 1 2
mυ f − mυi
2 2
1
= Ws → mυ2f − 0 − Ws
2
 ANS. FIG. P8.15

so

2 (Ws )
f =
m
2(0.625)
= m/s = 0.791 m/s
2.00

(b) Now friction results in an increase in internal energy fkd of the block-surface system. From

conservation of energy for a nonisolated system,

Ws = K + Eint
K = Ws − f k d

1 2 1 2
mυ f − mυi = Ws − f k d = Ws − s mgd
2 2
mυ f = 0.625 J − (0.350) ( 2.00 kg ) ( 9.80 m/s 2 ) ( 0.050 0 m )
1 2
2
1
( 2.00 kg ) υ2f = 0.625 J − 0.343 J = 0.282 J
2
2(0.282)
υf = m/s = 0.531 m/s
2.00

P8.22 For the Earth plus objects 1 (block) and 2 (ball), we write the energy model equation as

( K1 + K 2 + U1 + U 2 ) f
− ( K1 + K 2 + U1 + U 2 )i
=  Wother forces − f k d

ANS. FIG. P8.22

Choose the initial point before release and the final point after each block has moved 1.50 m. Choose U = 0

with the 3.00-kg block on the tabletop and the 5.00-kg block in its final position.

So K1i = K2i = U1i = U1f = U2f = 0

We have chosen to include the Earth in our system, so gravitation is an internal force. Because the only

external forces are friction and normal forces exerted by the table and the pulley at right angles to the motion,

Wother forces = 0

We now have

1 1
m1 2f + m2 2f + 0 + 0 − 0 − 0 − 0 − m2 gy2i = 0 − f k d
2 2

where the friction force is

f k = k n = k mA g

The friction force causes a negative change in mechanical energy because the force opposes the motion.

Since all of the variables are known except for f, we can substitute and solve for the final speed.

1 1
m1υ2f + m2 2f − m2 gy2i = − f k d
2 2
2 gh ( m2 − k m1 )
2 =
m1 + m2
2 ( 9.80 m/s 2 ) (1.50 m) 5.00 kg − 0.400 ( 3.00 kg ) 
=
8.00 kg
= 3.74 m/s

*P8.37 (a) The fuel economy for walking is

1 h  3 mi  1 kcal   1.30 108 J 

    = 423 mi/gal
220 kcal  h  4 186 J   1 gal 

(b) For bicycline:

1 h  10 mi  1 kcal   1.30 108 J 

    = 776 mi/gal
400 kcal  h  4 186 J   1 gal 