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Openstax College Physics
Instructor Solutions Manual
Chapter 4
CHAPTER 4: DYNAMICS: FORCE AND
NEWTON’S LAWS OF MOTION
4.3 NEWTON’S SECOND LAW OF MOTION: CONCEPT OF A SYSTEM
1.
A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s 2 . What is the net
external force on him?
Solution
net F = ma = (63.0 kg)(4.20 m/s²) = 265 N
2.
If the sprinter from the previous problem accelerates at that rate for 20 m, and then
maintains that velocity for the remainder of the 100-m dash, what will be his time for
the race?
Solution
Running from rest, runner attains a velocity of
v 2 = v02 + 2ax = 0 + 2(4.20 m/s 2 )(20 m)
or v  12.96 m/s at end of acceleration.
Find the time for this first part using x = 20 m  0  0.5at12 , or t 1  3.09 s . For the
second part, with no acceleration, x2  vt2 , or t 2  x2 /v  80 m/12.96 m/s  6.17 s .
Total time  3.09 s  6.17 s  9.26 s (a record!)
3.
A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it
is 60.0 N. Calculate its acceleration.
Solution
a = net F/m = 60.0 N/4.50 kg = 13.3 m/s 2 of the cart.
Openstax College Physics
4.
Solution
Instructor Solutions Manual
Chapter 4
Since astronauts in orbit are apparently weightless, a clever method of measuring
their masses is needed to monitor their mass gains or losses to adjust diets. One way
to do this is to exert a known force on an astronaut and measure the acceleration
produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s
acceleration is measured to be 0.893 m/s 2 . (a) Calculate her mass. (b) By exerting a
force on the astronaut, the vehicle in which they orbit experiences an equal and
opposite force. Discuss how this would affect the measurement of the astronaut’s
acceleration. Propose a method in which recoil of the vehicle is avoided.
(a) m =
net F
50.0 N
=
 56.0 kg
a
0.893 m/s 2
(b) a meas  aastro  aship , where : aship 
mastro aastro
.
mship
If the force could be exerted on the astronaut by another source (other than the
spaceship), then the spaceship would not experience a recoil.
5.
Solution
In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the
force of friction opposing the motion is 24 N, what force F (in newtons) is the person
exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is
removed. How far will the mower go before stopping?


net F  F  f  F  net F  f  51 N  24 N  75 N
f  ma  a 
f 24 N

 1.0 m/s 2 so using v 2  v02  2ax
m 24 kg
0  1.5 m/s   2  1.0 m/s 2  x  x  1.1 m
2
6.
The same rocket sled drawn in Figure 4.31 is decelerated at a rate of 196 m/s 2 . What
force is necessary to produce this deceleration? Assume that the rockets are off. The
mass of the system is 2100 kg.
Solution
net F  ma  (2100 kg)(196 m/ s 2 )  4.12  105 N
Openstax College Physics
Instructor Solutions Manual
Chapter 4
7.
(a) If the rocket sled shown in Figure 4.32 starts with only one rocket burning, what is
its acceleration? Assume that the mass of the system is 2100 kg, and the force of
friction opposing the motion is known to be 650 N. (b) Why is the acceleration not
one-fourth of what it is with all rockets burning?
Solution
(a) Use the thrust given for the rocket sled in Figure 4.8, T  2.59 10 4 N . With only
one rocket burning, net F  T  f or
net F T  f 2.59  10 4 N  650 N
a


 12.0 m/s 2
m
m
2100 kg
(b) The acceleration is not one-fourth of what it was with all rockets burning because
the frictional force is still as large as it was with all rockets burning.
8.
Solution
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of
1000 km/h? (Such deceleration caused one test subject to black out and have
temporary blindness.)
a=
v
1000 km/h
=
= 253 m/s 2
t
1.1 s
9.
Suppose two children push horizontally, but in exactly opposite directions, on a third
child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N,
friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the
system of interest if the acceleration of the child in the wagon is to be calculated? (b)
Draw a free-body diagram, including all forces acting on the system. (c) Calculate the
acceleration. (d) What would the acceleration be if friction were 15.0 N?
Solution
(a) The system is the child in the wagon plus the wagon.
(b)
N
Fl
m
f
w
Fr
Openstax College Physics
Instructor Solutions Manual
Chapter 4
(c) net F  Fr  F1  f  ma so that
Fr  F1  f 90.0 N  75.0 N  12.0 N

 0.130 m/s 2
m
23.0 kg
in the direction of the second child' s push
a
(d) net F  90.0 N  75.0 N  15.0 N  0  a  0 m/s 2
10.
A powerful motorcycle can produce an acceleration of 3.50 m/s 2 while traveling at
90.0 km/h. At that speed the forces resisting motion, including friction and air
resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes
the motion of an object.) What force does the motorcycle exert backward on the
ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?
Solution
net F  F  f  ma  F  ma  f  (245 kg)(3.50 m/s 2 )  400 N  1.26  10 3 N
11.
The rocket sled shown in Figure 4.33 accelerates at a rate of 49.0 m/s 2 . Its passenger
has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat
exerts against his body. Compare this with his weight by using a ratio. (b) Calculate
the direction and magnitude of the total force the seat exerts against his body.
Solution
2
3
(a) Fh  ma  (75.0 kg)(49.0 m/s )  3.68  10 N and
w  mg  (75.0 kg)(9.80 m/s 2 )  7.35  10 2 N
Fh ma a 49.0 m/s 2

 
 5.00 times greater th an weight
w mg g 9.80 m/s 2

Fh
(b)

Fne

N

net F  [(Fh )2  (N)2 ]1/2
net F  [(3675 N) 2  (735 N) 2 ]1/2  3750 N
N
θ  tan 1 
 Fh

  11.3 from horizontal

Openstax College Physics
12.
Solution
Instructor Solutions Manual
Chapter 4
Repeat the previous problem for the situation in which the rocket sled decelerates at
a rate of 201 m/s 2 . In this problem, the forces are exerted by the seat and restraining
belts.


2
4
4
(a) Fh  ma  75.0 kg   201 m/s  1.507  10 N   1.51  10 N
1.507  10 4
2
w  mg  7.35  10 N; Ratio =
 20.5 times greater th an weight
7.35  10 2
(b)

Fne

N


Fh


net F   1.507  10 4 N  735 N 
2
N
 Fh
  tan 1 
2

1/ 2
 1.51  10 4 N

  2.79 from horizontal

13.
The weight of an astronaut plus his space suit on the Moon is only 250 N. How much
do they weigh on Earth? What is the mass on the Moon? On Earth?
Solution
wMoon  mg Moon
m
wMoon
250 N

 150 kg
g Moon 1.67 m/s 2
wEarth  mg Earth  150 kg 9.8 m/s 2   1470 N  1.5  10 3 N
Mass does not change, so the astronaut's mass on both Earth and the Moon will be
150 kg.
14.
Suppose the mass of a fully loaded module in which astronauts take off from the
Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its acceleration
in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If
it could, calculate its acceleration.
Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 4
(a) ma  FT  w
FT  w FT
3.00  10 4 N
a

 g Moon 
 1.67 m/s 2
4
m
m
1.00  10 kg
 (3.00  1.67) m/s 2  1.33 m/s 2
(b) The module cannot take off from the earth, since FT  w of the earth.
4.4 NEWTON’S THIRD LAW OF MOTION: SYMMETRY IN FORCES
15.
What net external force is exerted on a 1100-kg artillery shell fired from a battleship if
the shell is accelerated at 2.40 10 4 m/s 2 ? What force is exerted on the ship by the
artillery shell?
Solution
net F  ma  (1100 kg)( 2.40 10 4 m/s 2 )  2.64 107 N
Force exerted on ship = 2.64 107 N opposite to shell’s direction of motion, by
Newton’s 3rd law.
16.
A brave but inadequate rugby player is being pushed backward by an opposing player
who is exerting a force of 800 N on him. The mass of the losing player plus equipment
is 90.0 kg, and he is accelerating at 1.20 m/s 2 backward. (a) What is the force of
friction between the losing player’s feet and the grass? (b) What force does the
winning player exert on the ground to move forward if his mass plus equipment is 110
kg? (c) Draw a sketch of the situation showing the system of interest used to solve
each part. For this situation, draw a free-body diagram and write the net force
equation.
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution (a) net F  F  f  ma  f  F  ma  800 N  (90.0 kg)(1.20 m/s 2 )  692 N
(b) net F  F  f  ma 
F  ma  f  (110 kg  90.0 kg)(1.20 m/s 2 )  692 N  932 N
(c)
F = 800 N
F = 932 N
f = 692 N
System for (b)
f = 692 N
System for (a)
4.5 NORMAL, TENSION, AND OTHER EXAMPLES OF FORCES
17.
Two teams of nine members each engage in a tug of war. Each of the first team’s
members has an average mass of 68 kg and exerts an average force of 1350 N
horizontally. Each of the second team’s members has an average mass of 73 kg and
exerts an average force of 1365 N horizontally. (a) What is the acceleration of the two
teams? (b) What is the tension in the section of rope between the teams?
Openstax College Physics
Solution (a) f1 . . .
m1 . . .
Instructor Solutions Manual
. . . f1
f1
m1 m1
F2
f2
f2
m2 m2
Chapter 4
...
f2
...
. . . m2
F1
net F  Ma ; f1  1350 N ; f 2  1365 N
9( f 2  f1 )  9(m1  m2 )a ; m1  68 kg ; m2  73 kg
a
f 2  f1 1365 N  1350 N

 0.1064 m/s 2  0.11 m/s 2 ;
m1  m2
68 kg  73 kg
Thus, the heavy team wins. Note that the difference 1365 N  1350 N  15 N
limits the answer to two significant figures.
(b )
T  9 f 1  9m1 a  T  9m1 a  9 f1
T  9(68 kg )(0.1064m/s 2 )  9(1350 N)  1.2  10 4 N
18.
What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her
straight up at 7.50 m/s 2 ? Note that the answer is independent of the velocity of the
gymnast—she can be moving either up or down, or be stationary.
Solution
net F   Ft  mg  ma  Ft  m(a  g )  (45.0 kg)(7.50 m/s 2  9.80 m/s 2 )  779 N
19.
(a) Calculate the tension in a vertical strand of spider web if a spider of mass
8.00 10 5 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand
of spider web if the same spider sits motionless in the middle of it much like the
tightrope walker in Figure 4.17. The strand sags at an angle of 12 below the
horizontal. Compare this with the tension in the vertical strand (find their ratio).
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution (a) T  mg  (8.00 10-5 kg)(9.80 ms 2 )  7.84 10-4 N
(b) T  
F1
mg
8.00 10 -5 (9.80 m/s 2 )


 1.89 10 -3 N
2 sin θ 2 sin θ
2 sin 12.0
T
 2.41 times the tension in the vertical strand
T
20.
Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he
climbs at a constant speed? (b) What is the tension in the rope if he accelerates
upward at a rate of 1.50 m/s 2 ?
Solution (a) net F  T  mg  ma  0 (a  0 if constant speed)
T  mg  (60.0 kg)(9.80 m/s 2 )  588 N
(b) net F  T  mg  ma
T  m(g  a)  (60.0 kg)(9.80 m/s 2  1.50 m/s 2 )  678 N
21.
Show that, as stated in the text, a force F exerted on a flexible medium at its center
and perpendicular to its length (such as on the tightrope wire in Figure 4.17) gives rise
F^
to a tension of magnitude T =
.
2sin(q )
Solution
y
T


F

T

x
Net Fy  F  2T sin   0
F  2T sin 
T
F
2 sin 
Openstax College Physics
22.
Instructor Solutions Manual
Chapter 4
Consider the baby being weighed in Figure 4.34. (a) What is the mass of the child and
basket if a scale reading of 55 N is observed? (b) What is the tension T1 in the cord
attaching the baby to the scale? (c) What is the tension T2 in the cord attaching the
scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the
situation indicating the system of interest used to solve each part. The masses of the
cords are negligible.
Solution
(a)
55 N
 5.6 kg
9.80 m/s 2
(b) 55 N
(c) T2  T1  mg  55 N  (0.500 kg) 9.80 m/s 2   60 N
(d)
4.6 PROBLEM-SOLVING STRATEGIES
23.
A 5.00 105 - kg rocket is accelerating straight up. Its engines produce 1.250  107 N
of thrust, and air resistance is 4.50  106 N . What is the rocket’s acceleration?
Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s
laws of motion.
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution
Using the free body diagram: net F  T  f  mg  ma, so that
a
24.



T  f  mg 1.250 107 N  4.50 106 N  5.00 105 kg 9.80 m/s 2

 6.20 m/s 2
5
m
5.00 10 kg
The wheels of a midsize car exert a force of 2100 N backward on the road to
accelerate the car in the forward direction. If the force of friction including air
resistance is 250 N and the acceleration of the car is 1.80 m/s 2 , what is the mass of
the car plus its occupants? Explicitly show how you follow the steps in the ProblemSolving Strategy for Newton’s laws of motion. For this situation, draw a free-body
diagram and write the net force equation.
Solution Using the free body diagram: net F  F  f  ma, so that
m
25.
F  f 2100 N  250 N

 1.03  10 3 kg
a
1.80 m/s 2
Calculate the force a 70.0-kg high jumper must exert on the ground to produce an
upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how
you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution Step 1. Use Newton’s Laws of Motion.
Step 2. Given : a  4.00 g  (4.00)(9.80 m/s 2 )  39.2 m/s 2 ; m  70.0 kg
Find F .
Step 3.
 F  F  w  ma, so that F  ma  w  ma  mg  m(a  g )
F  (70.0 kg)[(39.2 m/s 2 )  (9.80 m/s 2 )]  3.43  10 3 N
The force exerted by the high-jumper is actually down on the ground, but F is up
from the ground to help him jump.
Step 4. This result is reasonable, since it is quite possible for a person to exert a force
of the magnitude of 10 3 N .
26.
When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by
pushing straight down on the mat. Calculate the force she must exert if her
deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you
follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.
Solution
Using the free body diagram:
net F  F  W  F  mg  ma , where a  7.00 g and m  40.0kg.
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solving for the force gives F  ma  mg  m(a  g )  m(7.00 g  g) and substituting
in the values gives F  (40.0 kg)(8.00  9.80 m / s 2 )  3.14 103 N
27.
A freight train consists of two 8.00 104  kg engines and 45 cars with average
masses of 5.50 10 4 kg . (a) What force must each engine exert backward on the track
to accelerate the train at a rate of 5.00  10 –2 m/s 2 if the force of friction is
7.50  105 N , assuming the engines exert identical forces? This is not a large frictional
force for such a massive system. Rolling friction for trains is small, and consequently
trains are very energy-efficient transportation systems. (b) What is the force in the
coupling between the 37th and 38th cars (this is the force each exerts on the other),
assuming all cars have the same mass and that friction is evenly distributed among all
of the cars and engines?
Solution
(a) net F  Ma  2 F  f 
Ma  f 2(80,000kg)  45(55,000 kg)(5.00  10  2 m / s 2 )  7.50  10 5 N
F

2
2
5
 4.41  10 N
(b) The 37th car has 8 cars (45-38) behind it net F   M a  F   f   8ma, and since
friction is evenly distributed: f  
f (8)
.
47
Substituting gives:
F   8ma 
28.
8f
8(7.50  10 5 N)
 8(55,000 kg)(5.00  10 2 m / s 2 ) 
 1.50  10 5 N
47
47
Commercial airplanes are sometimes pushed out of the passenger loading area by a
tractor. (a) An 1800-kg tractor exerts a force of 1.75  104 N backward on the
pavement, and the system experiences forces resisting motion that total 2400 N. If the
acceleration is 0.150 m/s 2 , what is the mass of the airplane? (b) Calculate the force
exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced
by the airplane. (c) Draw two sketches showing the systems of interest used to solve
each part, including the free-body diagrams for each.
Openstax College Physics
Solution
Instructor Solutions Manual
(a) net F  Ma  (ma  mt )a  F  f , so that : ma 
ma 
Chapter 4
F f
 mt ,
a
1.75 10 4 N  2400 N
 1800 kg  9.89 10 4 kg
2
0.150 m/s
(b) net F  F   f   ma a
 F   ma a  f   (9.89 104 kg)(0.150 m/s 2 )  2200 N  1.70 104 N
(c)
F'
F
System for (a)
29.
f'
f
System for (b)
A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the
car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an
acceleration of 0.550 m/s 2 ? The mass of the boat plus trailer is 700 kg. (b) What is the
force in the hitch between the car and the trailer if 80% of the resisting forces are
experienced by the boat and trailer?
Solution (a) net F  F  f  Ma  (mb  mt  mc )a
f  F  (mb  mt  mc )a  1900 N  (1800 kg)(0.550 m/s 2 )  910 N
(b) net F   (mb  mt )a  F   0.8 f
 F   (700 kg)(0.550 m/s 2 )  0.8  910 N  1110 N  1.11103 N
30.
(a) Find the magnitudes of the forces F1 and F2 that add to give the total force Ftot
shown in Figure 4.35. This may be done either graphically or by using trigonometry.
(b) Show graphically that the same total force is obtained independent of the order of
addition of F1 and F2 . (c) Find the direction and magnitude of some other pair of
vectors that add to give Ftot . Draw these to scale on the same drawing used in part (b)
or a similar picture.
Solution (a) Since F2 is the y -component of the total force:
Openstax College Physics
Instructor Solutions Manual
Chapter 4
F2  Ftot sin 35  (20 N)sin35   11.47 N  11 N .
And F1 is the x -component of the total force:
F1  Ftot cos35  (20 N)cos35  16.38 N  16 N .
(b)
(c) For example, use vectors as shown in the figure.

F2'

Ftot
15°
20°

F1'
F1 is at an angle of 20 from the horizontal, with a magnitude of F1cos20  F1
F1 
F1
16.38 N

 17.4 N  17 N
cos20 cos20
F2 is at an angle of 90 from the horizontal, with a magnitude of
F2  F2  F1sin 20  5.2 N
31.
Two children pull a third child on a snow saucer sled exerting forces F1 and F2 as
shown from above in Figure 4.36. Find the acceleration of the 49.00-kg sled and child
system. Note that the direction of the frictional force is unspecified; it will be in the
opposite direction of the sum of F1 and F2 .
Solution Each force vector needs to be broken into its components:
Openstax College Physics
Instructor Solutions Manual
Chapter 4
F1x  10 cos 45 N and F1 y  10 sin 45 N
F2 x  8 cos 30 N and F2 y  8 sin 30 N
Ftotx  F1x  F2 x  14 N
Ftoty  F1 y  F2 y  3.07 N
F  14 2  3.07 2 N   14.33 N
Because there is a frictional force of 7.5N,
net F  F  f  14.33 N  7.5 N  6.83 N
F  ma  a 
tan  
Ftoty
Ftotx
F 6.83 N

 0.14 m/s 2
m 49 kg
   12.37
The child-sled combo are accelerating at 0.14 m/s 2 at 12.37 north of east.
32.
Suppose your car was mired deeply in the mud and you wanted to use the method
illustrated in Figure 4.37 to pull it out. (a) What force would you have to exert
perpendicular to the center of the rope to produce a force of 12,000 N on the car if the
angle is 2.00°? In this part, explicitly show how you follow the steps in the ProblemSolving Strategy for Newton’s laws of motion. (b) Real ropes stretch under such forces.
What force would be exerted on the car if the angle increases to 7.00° and you still
apply the force found in part (a) to its center?
Solution (a) Use Figure 4.37 as the free body diagram.
net Fx  T cos   T cos   0
net Fy  F  T sin   T sin   0
F  2T sin   2(12,000 N)(sin 2.00)  837.6 N  838 N
(b) T 
F
837.6 N

 3.44  103 N
2 sin  2 sin 7.00
Openstax College Physics
33.
Instructor Solutions Manual
Chapter 4
What force is exerted on the tooth in Figure 4.38 if the tension in the wire is 25.0 N?
Note that the force applied to the tooth is smaller than the tension in the wire, but
this is necessitated by practical considerations of how force can be applied in the
mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for
Newton’s laws of motion.
Solution Step 1: Use Newton’s laws since we are looking for forces.
Step 2: Draw a force diagram:
Step 3: Given T  25.0 N , find Fapp . Using Newton’s laws gives Σ Fx  0, so that
applied force is due to the y -components of the two tensions:
Fapp  2T sin   225.0 N  sin 15  12.9 N
The x -components of the tension cancel.  Fx  0
Step 4: This seems reasonable, since the applied tensions should be greater than the
force applied to the tooth.
34.
Figure 4.39 shows Superhero and Trusty Sidekick hanging motionless from a rope.
Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is
negligible. (a) Draw a free-body diagram of the situation showing all forces acting on
Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above
Superhero. (c) Find the tension in the rope between Superhero and Trusty Sidekick.
Indicate on your free-body diagram the system of interest used to solve each part.
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution (a)
(b) Using the upper circle of the diagram,
F
Using the lower circle of the diagram,
F
y
y
 0 , so that T '  T  wB  0 .
 0 , giving T  wR  0 .
Next, write the weights in terms of masses: wB  mB g , wR  mR g .
Solving for the tension in the upper rope gives:
T '  T  wB  wR  wB  mR g  mB g  g (mR  mB )
Plugging in the numbers gives: T '  9.80 m/s 2 55.0 kg  90.0 kg   1.42  10 3 N
Using the lower circle of the diagram, net
F
y
 0 , so that T  wR  0 . Again,
write the weight in terms of mass: wR  mR g. Solving for the tension in the
lower rope gives: T  mR g  (55.0 kg) 9.80 m/s 2   539 N
35.
A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0
below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is
60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must
the nurse exert to move at a constant velocity?
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution (a)
(b) To move at a constant velocity, net Fx  0  F cos 35.0  f . Since f  60.0 N ,
F  60N / cos 35.0  73N .
38.
Unreasonable Results (a) Repeat Exercise 4.29, but assume an acceleration of
1.20 m/s 2 is produced. (b) What is unreasonable about the result? (c) Which premise
is unreasonable, and why is it unreasonable?
Solution (a) net F  F  f  Ma  (mb  mt  mc ) a
f  1900 N  (1800 kg)(1.20 m/s 2 )   260 N
(b) It is unreasonable that the force of friction be in the direction of motion.
(c) The unreasonable premise is that an acceleration of 1.20 m/s 2 could be produced.
39.
Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass
of 1.50 106 kg at takeoff, the engines of which produce a thrust of 2.00  10 6 N ? Do
not neglect gravity. (b) What is unreasonable about the result? (This result has been
unintentionally achieved by several real rockets.) (c) Which premise is unreasonable,
or which premises are inconsistent? (You may find it useful to compare this problem to
the rocket problem earlier in this section.)
Solution (a) net F  ma  T  mg 
T  mg 2.00  10 6 N  1.50  10 6 kg 9.80 m/s 2
a

  8.47 m/s 2
6
m
1.50  10 kg



(b) There is not enough thrust to take off a  0 .
(c) The thrust is not large enough for the given rocket mass to leave the ground.
4.7 FURTHER APPLICATIONS OF NEWTON’S LAWS OF MOTION
Openstax College Physics
40.
Instructor Solutions Manual
Chapter 4
A flea jumps by exerting a force of 1.20  10 5 N straight down on the ground. A
breeze blowing on the flea parallel to the ground exerts a force of 0.500  10 6 N on
the flea. Find the direction and magnitude of the acceleration of the flea if its mass is
6.00 10 7 kg . Do not neglect the gravitational force.

Fne x
Solution
N

Fne y
f


Fne
w



net Fy  N  w  N  mg  1.20  10 5 N  6.00  10 7 kg 9.80 m/s 2  6.12  10 6 N
net Fx  f  0.500  10 6 N
 net Fx 
  4.67

net
F
y 

  tan 1 
net F  (net Fx ) 2  (net Fy ) 2

 
2
 0.500  10 6 N  6.12  10 6 N

2 1/ 2
 6.14  10 6 N, so that
net F 6.14  10 6 N
a

 10.2 m/s 2
7
m
6.00  10 kg
 10.2 m/s 2 at 4.67 from vertical
41.
Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in
Figure 4.40. (These muscles are called the medial and lateral heads of the
gastrocnemius muscle.) Find the magnitude and direction of the total force on the
Achilles tendon. What type of movement could be caused by this force?
Solution Force on Achilles tendon F  2(200 N)cos20  376 N . Force used to raise heel.
42.
A 76.0-kg person is being pulled away from a burning building as shown in Figure
4.41. Calculate the tension in the two ropes if the person is momentarily motionless.
Include a free-body diagram in your solution.
Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 4
T1
15°
m
10°
T2
w
T1sin15   T2 cos 10
T2  0.263T1
T1cos15  T2sin10   mg

T1cos15  0.263T1sin10   (76.0 kg) 9.80 m/s 2

1.012T1  744.8 N
T1  735.9 N  736 N
T2  0.263T1  193.6 N  194 N
43.
Integrated Concepts A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to
join another dolphin in play. What average force was exerted to slow him if he was
moving horizontally? (The gravitational force is balanced by the buoyant force of the
water.)
Solution
net F  ma ; a 
44.
Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average
force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b)
How far does he travel?
v  v0
, so that
t
m(v  v 0 )
 7.50 m/s  12.0 m/s 
net F 
 (35.0 kg) 
   68.5 N
t
2.30 s


F
Ft 650 N  0.800 s
Solution (a)
v  at ; a 
v 

 7.43 m/s
m
(b) x 
m
70.0 kg
1 2 1 F 2 1 650 N
at    t  
 (0.800 s) 2  2.97 m
2
2 m
2 70.0 kg
Openstax College Physics
45.
Instructor Solutions Manual
Chapter 4
Integrated Concepts A large rocket has a mass of 2.00 106 kg at takeoff, and its
engines produce a thrust of 3.50  107 N . (a) Find its initial acceleration if it takes off
vertically. (b) How long does it take to reach a velocity of 120 km/h straight up,
assuming constant mass and thrust? (c) In reality, the mass of a rocket decreases
significantly as its fuel is consumed. Describe qualitatively how this affects the
acceleration and time for this motion.
Solution (a) net F  ma  T  mg so that
T  mg 3.50  10 7 N - (2.00  10 6 kg)(9.80 m/s 2 )
a

 7.70 m/s 2
m
2.00  10 6 kg
(b) 120 km/h  33.33 m/s. Since v  at , we know that t 
v 33.33 m/s

 4.33 s
a 7.70 m/s
(c) As the mass of the rocket decreases, the acceleration increases for the same
thrust, and therefore it takes much less than 4.33 s to reach a velocity of
120 km/h.
46.
Integrated Concepts A basketball player jumps straight up for a ball. To do this, he
lowers his body 0.300 m and then accelerates through this distance by forcefully
straightening his legs. This player leaves the floor with a vertical velocity sufficient to
carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor.
(b) Calculate his acceleration while he is straightening his legs. He goes from zero to
the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts
on the floor to do this, given that his mass is 110 kg.
Solution (a) v 2  v02  2 g ( y  y0 ), where y  y0  0.900 m, and v  0 m/s.
v0  [2 g  y  y0 ]1/2  [2(9.80 m/s 2 )(0.900 m)]1/2  4.20 m/s
2
2
(b) v  v0  2a y  y0 , where y  y0  0.300 m, and v0  0 m/s
v2
(4.20 m/s) 2
a

 29.4 m/s 2
2( y  y0 ) (2)(0.300 m)
Openstax College Physics
Instructor Solutions Manual
Chapter 4
(c)
net F  ma  F  w  F  mg
F  ma  mg  ma  g   110 kg(29.4 m/s 2  9.80 m/s 2 )  4.31103 N
47.
Integrated Concepts A 2.50-kg fireworks shell is fired straight up from a mortar and
reaches a height of 110 m. (a) Neglecting air resistance (a poor assumption, but we
will make it for this example), calculate the shell’s velocity when it leaves the mortar.
(b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the
shell in the tube as it goes from zero to the velocity found in (a). (c) What is the
average force on the shell in the mortar? Express your answer in newtons and as a
ratio to the weight of the shell.
Solution (a) v 2  v02  2 g(y  y0 ), y0  v  0
v0  (2 gy)1/2  [2(9.80 m/s 2 )(110 m)]1/2  46.43 m/s  46.4 m/s
(b) v 2  v02  2a( y  y0 ), y0  v0  0
v 2 (46.43 m/s) 2
a

 2.395 103 m/s 2  2.40 103 m/s 2
2 y (2  0.450 m)
(c) net F  ma  F ( F includes weight of shell)
F  (2.50 kg)(2.395  10 3 m/s 2 )  5.99  10 3 and
w  mg  (2.50 kg)(9.80 m/s 2 )  24.5 N
F 5990 N

 245
w 24.5 N
48.
Integrated Concepts Repeat Exercise 4.47 for a shell fired at an angle 10.0 from the
vertical.
Openstax College Physics
Instructor Solutions Manual
Chapter 4
Solution (a) v y2  v02y  2 g ( y  y0 ); y0  v  0
v0 y  2 gy   [2(9.80 m/s 2 )(110 m)] 1/2  46.43 m/s
1/2
v0 y  v0 sin  0  v0 
v 0y
sin  0

46.43 m/s
 47.2 m/s
sin 800
(b) v 2  v02  2a( y  y0 ); y0  v0  0, so that a 
v 2 (47.15 m/s) 2

 2.47 103 m/s 2
2 y 2  0.450 m
(c) net F  ma  F (weight of shell included in F )
F  (2.50 kg)(2.47 103 m/s 2 )  6175  6.18 103 N ; w  mg  24.5 N
F 6175 N

 252
w 24.5 N
49.
Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a)
The elevator accelerates upward from rest at a rate of 1.20 m/s 2 for 1.50 s.
Calculate the tension in the cable supporting the elevator. (b) The elevator
continues upward at constant velocity for 8.50 s. What is the tension in the cable
during this time? (c) The elevator decelerates at a rate of 0.600 m/s 2 for 3.00 s.
What is the tension in the cable during deceleration? (d) How high has the
elevator moved above its original starting point, and what is its final velocity?
Solution (a)
T
m
w
net F  ma  T  w  T  mg , m  1700 kg .
a  1.20 m/s 2 , so the tension is :
T  ma  g   (1700 kg)(1.20 m/s 2  9.80 m/s 2 )  1.87  10 4 N
(b) a  0 m/s 2 , so the tension is: T  w  mg(1700 kg)(9.80 m/s 2 )  1.67  10 4 N
Openstax College Physics
Instructor Solutions Manual
Chapter 4
(c) a  0.600 m/s 2 , but down :
T  mg  a   (1700 kg)(9.80 m/s 2  0.600 m/s 2 )  1.56  10 4 N
(d) v3
t3
v2
t2
v1
t1
y3
y2
y1
1 2 1
a1t1  (1.20 m/s 2 )(1.50 s) 2  1.35 m and
2
2
v1  a1t1  (1.20 m/s 2 )(1.50)  1.80 m/s
y1 
y 2  v1t 2  (1.80 m/s )(8.50 s)  15.3 m
y 3  v 2 t  a3t 32  (1.80 m/s )(3.00 s)  0.5(0.600 m/s )(3.00 s) 2  2.70 m
v3  v 2  a3t 3  1.80 m/s  (-0.600 m/s 2 )(3.00 s)  0 m/s
y1  y 2  y3  1.35 m  15.3 m  2.70 m  19.35 m  19.4 m and vfinal  0 m/s
50.
Unreasonable Results (a) What is the final velocity of a car originally traveling at 50.0
km/h that decelerates at a rate of 0.400 m/s 2 for 50.0 s? (b) What is unreasonable
about the result? (c) Which premise is unreasonable, or which premises are
inconsistent?
Solution (a) 50.0 km/h  13.89 m/s; v  v0  at  13.89 m/s  (0.400 m/s 2 )(50.0 s)   6.11 m/s
(b) It is unreasonable that a car could end up going in the opposite direction by the
brakes.
(c) f  s mg , a 
f
a 0.400 m/s 2
 s g, so that s  
 0.0400. This is a reasonable
m
g 9.80 m/s 2
coefficient of friction, so what is an unreasonable premise is that the car could
decelerate at 0.400 m/s 2 for 50 s. This inconsistent with the car having an initial
speed of 50 km/h.
Openstax College Physics
51.
Instructor Solutions Manual
Chapter 4
Unreasonable Results A 75.0-kg man stands on a bathroom scale in an elevator that
accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons
and compare it with his weight. (The scale exerts an upward force on him equal to its
reading.) (b) What is unreasonable about the result? (c) Which premise is
unreasonable, or which premises are inconsistent?
Solution (a)
Using a 
v  v0
gives:
t
v  v0 30.0 m/s  0 m/s

 15.0 m/s 2 net F  F  w  ma ,
t
2.00 s
F  ma  mg  m(a  g )  75.0 kg 15.0 m/s 2  9.80 m/s 2  1860 N.
a


F m(a  g ) 15.0 m/s  9.80 m/s


 2.53
w
mg
9.80 m/s 2
2
2
(b) The value (1860 N) is more force than you expect to experience on an elevator.
(c) The acceleration a  15.0 m/s 2  1.53g is much higher than any standard elevator.
The final speed is too large (30.0 m/s is VERY fast)! The time of 2.00 s is not
unreasonable for an elevator.
4.8 EXTENDED TOPIC: THE FOUR BASIC FORCES—AN INTRODUCTION
52.
(a) What is the strength of the weak nuclear force relative to the strong nuclear force?
(b) What is the strength of the weak nuclear force relative to the electromagnetic
force? Since the weak nuclear force acts at only very short distances, such as inside
nuclei, where the strong and electromagnetic forces also act, it might seem surprising
that we have any knowledge of it at all. We have such knowledge because the weak
nuclear force is responsible for beta decay, a type of nuclear decay not explained by
other forces.
Openstax College Physics
Solution
53.
Solution
(a)
Fw 1013

 1013
Fs
1
(b)
Fw 1013
 2  1011
Fem 10
Instructor Solutions Manual
Chapter 4
(a) What is the ratio of the strength of the gravitational force to that of the strong
nuclear force? (b) What is the ratio of the strength of the gravitational force to that of
the weak nuclear force? (c) What is the ratio of the strength of the gravitational force
to that of the electromagnetic force? What do your answers imply about the influence
of the gravitational force on atomic nuclei?
(a)
(b)
(c)
Fg
Fs
Fg
Fw
Fg
Fem

10 38
 10 38
1

10 38
 10  25
13
10

10 38
 10 36 . The gravitational force has a very small influence on atomic
2
10
nuclei.
54.
What is the ratio of the strength of the strong nuclear force to that of the
electromagnetic force? Based on this ratio, you might expect that the strong force
dominates the nucleus, which is true for small nuclei. Large nuclei, however, have sizes
greater than the range of the strong nuclear force. At these sizes, the electromagnetic
force begins to affect nuclear stability. These facts will be used to explain nuclear
fusion and fission later in this text.
Solution
Fs
1
 2  10 2
Fem 10
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