= ∆
=
=
=
=
2. The height could be determined by analyzing the vertical component of velocity.
2 2
f i
2 2
f i
2 2
2
2
2
(0 m/s) (5.2 m/s) 2( 9.8 m/s ) 1.4 m
v v a d v v
d a
d
= + ∆
∆ = −
= −
−
∆ =
3. The three forms of friction would use up some of the elastic potential energy intended to launch the spring. To compensate, the spring would need to be stretched an extra amount.
4. The energy to stretch the string would come from food you ate. The energy stored in the spring would be converted to kinetic energy as it left. The kinetic energy would be partially converted into gravitational potential as it rises, and than reconverted back to kinetic as it lands. As it lands on the floor and slides to a stop, the kinetic energy is converted into sound and thermal energy.
5. The spring would cause a mass to move up and down with decreasing amplitude until it came to rest.
6. The concepts and equations can be used to design and manufacture vehicles, beds, and a wide variety of other products.
CHAPTER 4 SELF QUIZ
(Page 225)
True/False
1. T
2. F The work done by gravity is zero.
3. F The work you do on the backpack is negative.
4. FThe gravitational potential energy decreases in proportion to the distance fallen.
5. F This does not refute the law of conservation of energy because some energy is converted into other forms, such as heat (thermal energy) in the ball and the floor.
6. T 7. T
8. FMaximum speed occurs at the equilibrium position, but elastic potential energy is at a minimum at the point of minimum extension of the spring.
9. FA long damping time would not be appropriate for a bathroom scale. It would be appropriate for a “jolly-jumper” toy.
Multiple Choice
10. (c) 11. (c) 12. (e) 13. (d) 14. (e) 15. (a)
16. (d) W
=
( cos )F θ∆
d(mg sin )(cos180 )θ L
= °
CHAPTER 4 REVIEW
(Pages 226–229)
Understanding Concepts
1. (a) No work is done because the force is perpendicular to the displacement.
(b) The work is negative because the student is exerting a force opposite the direction of motion.
(c) The work is negative because gravity is exerting a force opposite the direction of motion.
(d) Assuming a level roadway, the work done is zero because the force is perpendicular to the displacement.
(e) No work is done because the electrical force is perpendicular to the displacement.
(f) No work is done because the tension is perpendicular to the displacement.
2. The force must be applied perpendicular to the object for it to do no work on the object.
3. The normal force can do work on an object. For example, when you jump, you push down on the ground and the normal force pushes up on you and accelerates you up, giving you kinetic energy.
4. (a) No work is being done on the swimmer because the balancing forces forward and backward produce no motion.
(b) Work is done on the student to speed him up, but after that, there is no work being done on the student. Once the student reaches the speed of the current, the only force is the upward buoyant force, which is perpendicular to the waters surface. Technically, a small amount of work is being done by gravity as they whole river/student system is pulled closer to the earth.
5. (a) The velocity will be changing because Newton’s second law states that if a net force acts on an object, it will accelerate (i.e., change its velocity).
(b) It is possible the speed is constant if the particle is travelling in a circle.
(c) The kinetic energy is proportional to the square of the speed and the mass. Since both can remain constant under these conditions, the kinetic energy may be constant.
6. Agree. In the absence of friction (including air resistance), all of the gravitational potential energy will be converted into kinetic energy. The mass will cancel in each term.
7. (a) Damped vibrations are useful in the suspension of a vehicle.
(b) Damped vibrations are not useful in a pendulum clock.
8. It is not possible to have a motion that is not damped. Such a device would be a perpetual motion machine that cannot exist. The force of friction within the system cannot be avoided.
9. m = 0.425 kg
∆ y = 11.8 m (a) W = ?
2
( cos ) ( cos )
(0.425 kg)(9.80 m/s )(cos180 )(11.8 m) 49.1 J
W F d
mg y
W
θ θ
= ∆
= ∆
= °
= −
The work gravity does on the ball on the way up is –49.1 J.
(b) W = ?
2
( cos ) ( cos )
(0.425 kg)(9.80 m/s )(cos 0 )(11.8 m) 49.1 J
W F d
mg y
W
θ θ
= ∆
= ∆
= °
=
The work gravity does on the ball on the way down is 49.1 J.
10. F = 9.3 N W = 87 J
∆d = 11 m
θ = ?
1
The angle between the applied force and the horizontal is 32
°
.11. Let the subscript C represent the child, and TO represent the toboggan.
mC = 25.6 kg mTO = 4.81 kg
∆ y = 27.3 m (a) W C = ?
First, find the actual distance:
sin 27.3
The applied force, F A, will be equal to the component of gravity down the hill, mg sin φ
C
(4.81 kg)(9.80 N/kg)(cos 0 )(27.3) 1.29 10 J
(c) The total work on the child and toboggan during the slide will be equal to the work done to take them up the hill. Using the equation derived in part (a),
T
3 T
( cos )(27.3)
(25.6 kg 4.81 kg)(9.80 N/kg)(cos0 )(27.3) 8.14 10 J
The total work on the child and the toboggan is 8.14 103 J.
12. m = 73 kg
θ = 9.3
°
vi = 4.2 m/s
∆d = ?
The relation between the distance along the slope and the vertical height is:
To calculate the change in distance:
T1 T2 2(9.8 m/s)(sin9.3 ) 5.6 m
The skier would travel 5.6 m along the hill before stopping.
13. An increase of 50% is equivalent of multiplying by 1.5:
2
Since the negative value is not admissible, the original speed of the object was 8.90 m/s.
14. m = 7.0 109 kg
The gravitational potential energy is 2.5 1012 J.
(b) The work done by on the pyramid by one person in 20 years is:
( ) ( )
To calculate the total number of people:
12
There were 3 103 workers involved.
15. m = 45 kg
(45 kg)(9.80 N/kg)(cos180 )(0.66 m) 2.9 10 J
(45 kg)(9.80 N/kg)(cos0 )(0.66 m) 2.9 10 J
The work done by the weightlifter on the mass is 2.9 102 J.
(c) ∆ E g = ?
g
2 g
(45 kg)(9.80 N/kg)(0.66 m) 2.9 10 J
The change in gravitational potential energy is 2.9 102 J.
16. m = 47 g = 0.047 kg
The speed of the stick just before it hits the ground is 9.2 m/s.
(b) If air resistance was included, the answer in part (a) would be slightly smaller. The energy from gravity would be shared between the thermal energy and the kinetic energy.
17. y1 – 27 m
The speed of the stick just before it hits the ground is 29 m/s.
18. F = 1.5 102 N [22
°
below the horizontal]m = 18 kg
∆d = 1.6 m µk = 0.55
(a) F N = ? F K = ?
To calculate the normal force:
N A
N A
2
2 N
0
sin 22 0
sin 22
(1.5 10 N)(si n 22 ) (18 kg )(9.8 N/ kg ) 232.6 N
2.3 10 N F y
F F mg
F F mg
F
Σ =
− ° − =
= ° +
= × ° +
=
= ×
To calculate the force of friction:
K K N
2 K
(0.55)(232.6 N ) 127.9
1.3 10 N
F F
F
=
µ=
=
= ×
The normal force on the box is 2.3 102 N. The force of friction on the box is 1.3 102 N.
(b) v = ?
If the box is starting from rest, there is no initial kinetic energy.
th K
2 K
2
K
K
2
( cos ) 1
2
2( cos ) 2
2( cos ) 2
2(1.5 10 N)(cos22 )(cos0 )(1.6 m) 2(127.9 N)(1.6 m) 18 kg
1.4 m/s W E E
F d F d mv
mv F d F d
F d F d
v m
v θ
θ θ
= +
∆ = ∆ +
= ∆ − ∆
∆ − ∆
=
× ° ° −
=
=
The final speed of the box is 1.4 m/s.
(c) E th = ?
th K
2 th
(127.9 N)(1.6 m) 2.0 10 J E F d
E
= ∆
=
= ×
The amount of thermal energy produced is 2.0 102 J.
19. m = 1.2 103 kg
The speed with which the gymnast leaves the trampoline is 8.40 m/s.
21. The work in the area under the graph. Consider the area in three parts:
triangle 1 rectangle triangle 2
1 1 2 2
The person does 42 J of work.
22. x = 0.418 m
The force constant of the spring is 239 N/m.
(b) x = 0.150 m F x = ?
(239 N/m)(0.150 m)
The force required to stretch the spring is 35.9 N.
(c) To stretch it 0.150 m:
To compress it 0.300 m:
2
The work required is 2.69 J to stretch it, and 10.8 J to compress it.
23. k = 22 N/m
(22 N/m)(0.035 m) 2(4.2 10 N)
The eraser will slide 0.32 m along the desk.
24. k = 75 N/m A = 0.15 m v = 1.7 m/s x = 0.12 m
m = ?
For SHM, all of the energy is E e when at the maximum amplitude, A:
( ) ( )
The mass of the block is 0.21 kg.
25. m = 0.42 kg k = 38 N/m
A = 5.3 cm = 0.053 m (a) E e = ?
Maximum energy occurs at full amplitude.
2
The maximum energy of the mass-spring system is 0.053 J.
(b) v = ?
Maximum speed occurs when there is no elastic potential energy:
T1 T2
38 N/m(0.053 m) 0.42 kg
The maximum speed of the mass is 0.50 m/s.
(c) x = 4.0 cm = 0.040 m
The speed of the mass is 0.33 m/s.
(d) x = 4.0 cm = 0.040 m
(38 N/m)(0.040 m) (0.42 kg)(0.33 m/s)
2 2
The total energy is 0.053 J. The results are the same.
Applying Inquiry Skills
26. (a) One text weighs 2.0 kg and has a thickness of 3.6 cm. The first one doesn’t need to be raised at all.
g1 g2 g3 g4 g5
1 2 3 4 5
1 2 3 4 5
( )
( 2.0 kg )(9.8 N/ kg )(0 m 0.036 m 0.072 m 0.108 m 0.144 m) 7.1 J
W E E E E E
mgy mgy mgy mgy mgy mg y y y y y
W
= + + + +
= + + + +
= + + + +
= + + + +
=
You would have to do 7.1 J of work.
(b) Some errors would be determining the mass of the text. Also, the thickness may compress the books on the bottom.
Not all texts may have the same mass.
27. The calculator would need to know the distance the box was pushed.
28.
29. The tractor seat would need to have a strong spring to absorb large bumps, and a shock absorber to prevent “launching”
the driver. There would have to be smaller springs on top of that to absorb small vibrations. Damping would be important to prevent resonance.
Making Connections
30. Roller coasters are often shut down in high winds because of the loss of energy that may occur due to increased air resistance. Cold can cause parts to shrink and increase frictional forces beyond safe limits.
31. (a) The ball on track B will arrive first. Shortly after the start, the vertical drop of the ball on track B causes an increase in the speed, which it will enjoy for the majority of the race. At the end, it will slow down to the same speed that the ball on track A has just accelerated to.
(b) Racing cyclists use this in a variety of ways. The sprinters stay high on the track until ready to make a break for it, converting all of their stored gravitational energy into kinetic. Team racers use the change to minimize the work of the rider. When the leader is ready to give up his spot, he rides up the hill a bit, converting some of his kinetic energy into gravitational potential energy, reducing his speed. Once the last team mate has passed, he can drop down again, gaining the gravitational potential energy back as kinetic energy without needing to supply it from his own body power.
32. (a) v1 = 0 m/s v2 = ?
T1 T2
2 2
1 1 2 2
2 2
1 1 2 2
2
2 1 2
2 1 2
2
2
1 1
2 2
2 2
2 2
2 ( )
2(9.80 m/s )(37.8 m 17.8 m) 19.8 m/s
E E
mv mgy mv mgy v gy v gy
v gy gy v g y y
v
=
+ = +
+ = +
= −
= −
= −
=
The speed of the coaster at position C is 19.8 m/s.
(b) v1 = 5.00 m/s v2 = ?
T1 T2
2 2
1 1 2 2
2 2
1 1 2 2
2 2
2 1 1 2
2
2 1 1 2
2 2
2
1 1
2 2
2 2
2 2
2 ( )
(5.00 m/s) 2(9.80 m/ s )(37.8 m 17.8 m) 20.4 m/s
E E
mv mgy mv mgy v gy v gy
v v gy gy v v g y y
v
=
+ = +
+ = +
= + −
= + −
= + −
=
The speed of the coaster at position C is 20.4 m/s.
33. (a) The mass is not necessary because it cancels out of the mathematical equations.
(b) You might expect the speed would be 5.0 m/s more at the second point, but the reality is that the small kinetic energy supplied by having a speed going over the first hill is negligible compared with the large amount of gravitational potential energy.
34. a = 12 g [upward]
(a) F A = ?
A
A
A
( )
(0.12 )
1.12 F y ma F mg ma
F ma mg m a g m g g
F mg
Σ =
− =
= +
= +
= +
=
The force required is 1.12mg . (b) W = ?
( cos )
(1.12 )(cos 0 )( ) 1.12
W F d
mg y
W mg y
=
θ∆
= ° ∆
= ∆
The work done is 1.12mg ∆ y.
35. m = 1.5 kg
k = 2.1 103 N/m
∆ y = 0.37 m x = ?
T1 T 2
5.439 14.7 1050 1050 14.7 5.439 0
E E
Using the quadratic equation:
( 14.7) ( 14.7)2 4(1050)( 5.439) 2(1050)
14.7 151.9 2100
0.079 m or 0.065 m (negative value inadmissable) 0.079 m
The maximum distance the spring is compressed is 0.079 m.
36. m = 0.55 g = 5.5 10 –4 kg
∆d = 95 cm = 0.95 m
∆d x = 3.7 m E e = ?
First we must calculate the time required for the vertical drop (vi = 0):
2
To calculate the horizontal speed:
3.7 m
We know that initial kinetic energy came from elastic potential energy, therefore:
T1 T 2
The elastic potential energy stored was 0.019 J.
37. y1 = 16 m y2 = 9.0 m H = ?
First we must solve for the launch speed if v1 = 0:
T1 T2
2 2
1 1 2 2
2
1 2 2
2
2 1 2
2 1 2
2
2
1 1
2 2
2 2
2 2
2 ( )
2(9.8 m/s )(16 m 9.0 m) 11.71 m/s
E E
mv mgy mv mgy gy v gy
v gy gy v g y y
v
=
+ = +
= +
= −
= −
= −
=
Resolve vector components:
As just clearing the wall, the horizontal component will be the same, and the vertical component must produce a 30º angle to the vertical.
To calculate the vertical speed: 14.35 m/s [down]
y
Determine how far the skier has dropped from the launch point:
( )( )
The speed of the ball when it “swishes” through the hoop is 8.4 m/s.
39. g = 2.0 v1 = 6.0 y2 = 5.0
S = ?
Solve for the launch speed at the top of the ramp where y1 = 0:
T1 T2
Analyze projectile motion to find the change in time:
2 i
2
2
2
1 ( ) 2
5.0 (4.0sin 30 ) 1( 2.0)( ) 2
5 2 ( )
( ) 2 5 0
y y
d v t a t
t t
t t
t t
∆ = ∆ + ∆
− = ° ∆ + − ∆
− = ∆ − ∆
∆ − ∆ − =
Using the quadratic equation:
( 2) ( 2)2 4(1)( 5) 2(1)
2 4.9 2
3.45 units or 1.45 units (dismiss negative answer) 3.45 units
t
t
− − ± − − −
∆ =
= ±
= −
∆ =
To calculate horizontal range (S ):
(4.0cos30 )(3.45) 12 units
S v t x
S
= ∆
= °
=
The shuttle lands a distance of 12 units from the ramp.
CHAPTER 5 MOMENTUM AND COLLISIONS Reflect on Your Learning
(Page 230)
1. (a) The expression refers to once you start playing well, it is easier to keep playing well (or scoring, or winning).
(b) The physics meaning refers to a specific quantity. The everyday use of the word momentum can mean continuing to do well and inertia.
2. (a) The momentum of car A is less than the momentum of car B.
(b) The momentum of bicycle and rider A is less than the momentum of bicycle and rider B.
(c) The momentum of the large truck A is greater than the momentum of car B.
3. (a) The phrase “follow-through” refers to continue to swing the racket/club even after contact with the ball is made.
This allows the racket/club to be in contact with the ball for a longer period of time.
(b) “Follow-through” affects the momentum (or change in it).
4. (a) One technique that could be used is to break the total area up into eight different rectangles. The height of each would be an estimate. Each rectangle’s area could be determined and the sum would give the total area.
(b) kg m2 kg m
N s s
s s
F t ⋅ = ⋅ = ⋅ × = ⋅
5. The car should be designed with crunch zones to absorb the energy. If elastic bumpers were used, the car would rebound and be able to impact other objects.