This document discusses simple harmonic motion and elasticity. It begins by defining simple harmonic motion as back-and-forth motion caused by a restoring force proportional to displacement, with displacement centered around an equilibrium position. It then discusses Hooke's law, where the restoring force of an ideal spring is directly proportional to displacement. Several equations for simple harmonic motion are presented, including those relating displacement, velocity, acceleration, period, frequency, and amplitude. Examples are provided to illustrate these concepts for springs and pendulums undergoing simple harmonic motion.
Introduction to oscillations and simple harmonic motionMichael Marty
Physics presentation about Simple Harmonic Motion of Hooke's Law springs and pendulums with derivation of formulas and connections to Uniform Circular Motion.
References include links to illustrative youtube clips and other powerpoints that contributed to this peresentation.
This Unit is rely on introduction to Simple Harmonic Motion. the contents was prepared using the Curriculum of NTA level 4 at Mineral Resources Institute- Dodoma.
Introduction to oscillations and simple harmonic motionMichael Marty
Physics presentation about Simple Harmonic Motion of Hooke's Law springs and pendulums with derivation of formulas and connections to Uniform Circular Motion.
References include links to illustrative youtube clips and other powerpoints that contributed to this peresentation.
This Unit is rely on introduction to Simple Harmonic Motion. the contents was prepared using the Curriculum of NTA level 4 at Mineral Resources Institute- Dodoma.
Oscillation is the repetitive variation, typically in time, of some measure about a central value (often a point of equilibrium) or between two or more different states. The term vibration is precisely used to describe mechanical oscillation. Familiar examples of oscillation include a swinging pendulum and alternating current.
Oscillations occur not only in mechanical systems but also in dynamic systems in virtually every area of science: for example the beating of the human heart (for circulation), business cycles in economics, predator–prey population cycles in ecology, geothermal geysers in geology, vibration of strings in guitar and other string instruments, periodic firing of nerve cells in the brain, and the periodic swelling of Cepheid variable stars in astronomy. Contents
Oscillation is the repetitive variation, typically in time, of some measure about a central value (often a point of equilibrium) or between two or more different states. The term vibration is precisely used to describe mechanical oscillation. Familiar examples of oscillation include a swinging pendulum and alternating current.
Oscillations occur not only in mechanical systems but also in dynamic systems in virtually every area of science: for example the beating of the human heart (for circulation), business cycles in economics, predator–prey population cycles in ecology, geothermal geysers in geology, vibration of strings in guitar and other string instruments, periodic firing of nerve cells in the brain, and the periodic swelling of Cepheid variable stars in astronomy. Contents
Phyisics explaination on simple harmonic motion for first and second year university students , includes practical and theory about waves and some practical applications
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
CHAPTER 18
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections 18.1-18.4)
Objectives:
a) Define the various ways a force and couple do work.
b) Apply the principle of work and energy to a rigid body.
APPLICATIONS
The work of the torque developed by the driving gears on the two motors on the mixer is transformed into the rotational kinetic energy of the mixing drum.
The work done by the compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of
its roller and wheels
2. Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy:
T = 0.5m(vG)2 + 0.5IGw2
Since
vG = rGw, T = 0.5(IG + m(rG)2)w2 = 0.5IOw2
Similar to Simple harmonic motion and elasticity (20)
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
3. Simple Harmonic Motion
Back and forth motion that is caused by a force that is directly
proportional to the displacement. The displacement centers
around an equilibrium position.
xFs
4. Springs – Hooke’s Law
One of the simplest type
of simple harmonic
motion is called
Hooke's Law. This is
primarily in reference to
SPRINGS.
kxorkxF
k
k
xF
s
s
N/m):nitConstant(USpring
alityProportionofConstant
The negative sign only
tells us that “F” is what is
called a RESTORING
FORCE, in that it works in
the OPPOSITE direction
of the displacement.
5. 10.1 The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
The restoring force on an ideal spring is xkFx
6. 10.2 Simple Harmonic Motion and the Reference Circle
tAAx coscos
DISPLACEMENT
8. 10.2 Simple Harmonic Motion and the Reference Circle
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
T
f
1
T
f
2
2
amplitude A: the maximum displacement
9. 10.2 Simple Harmonic Motion and the Reference Circle
VELOCITY
tAvv
v
Tx sinsin
max
10. 10.2 Simple Harmonic Motion and the Reference Circle
Example 3 The Maximum Speed of a Loudspeaker Diaphragm
The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm.
(a)What is the maximum speed of the diaphragm?
(b)Where in the motion does this maximum speed occur?
11. 10.2 Simple Harmonic Motion and the Reference Circle
tAvv
v
Tx sinsin
max
(a)
sm3.1
Hz100.12m1020.02 33
max
fAAv
(b)The maximum speed
occurs midway between
the ends of its motion.
12. 10.2 Simple Harmonic Motion and the Reference Circle
ACCELERATION
tAaa
a
cx coscos
max
2
13. 10.2 Simple Harmonic Motion and the Reference Circle
FREQUENCY OF VIBRATION
m
k
tAax cos2
tAx cos
xmakxF
2
mAkA
14. 10.2 Simple Harmonic Motion and the Reference Circle
Example 6 A Body Mass Measurement Device
The device consists of a spring-mounted chair in which the astronaut
sits. The spring has a spring constant of 606 N/m and the mass of
the chair is 12.0 kg. The measured
period is 2.41 s. Find the mass of the
astronaut.
15. 10.2 Simple Harmonic Motion and the Reference Circle
totalm
k
2
total km
T
f
2
2
astrochair2total
2
mm
T
k
m
kg77.2kg0.12
4
s41.2mN606
2
2
2
chair2astro
m
T
k
m
16. 10.3 Energy and Simple Harmonic Motion
A compressed spring can do work.
18. 10.3 Energy and Simple Harmonic Motion
DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy is the energy that a spring
has by virtue of being stretched or compressed. For an
ideal spring, the elastic potential energy is
2
2
1
elasticPE kx
SI Unit of Elastic Potential Energy: joule (J)
19. 10.3 Energy and Simple Harmonic Motion
Conceptual Example 8 Changing the Mass of a Simple
Harmonic Oscilator
The box rests on a horizontal, frictionless
surface. The spring is stretched to x=A
and released. When the box is passing
through x=0, a second box of the same
mass is attached to it. Discuss what
happens to the (a) maximum speed
(b) amplitude (c) angular frequency.
20. 10.3 Energy and Simple Harmonic Motion
Example 8 Changing the Mass of a Simple Harmonic Oscilator
A 0.20-kg ball is attached to a vertical spring. The spring constant
is 28 N/m. When released from rest, how far does the ball fall
before being brought to a momentary stop by the spring?
21. 10.3 Energy and Simple Harmonic Motion
of EE
2
2
12
2
12
2
12
2
12
2
12
2
1
ooooffff kymghImvkymghImv
oo mghkh 2
2
1
m14.0
mN28
sm8.9kg20.02
2
2
k
mg
ho
22. 10.4 The Pendulum
A simple pendulum consists of a particle attached to a frictionless
pivot by a cable of negligible mass.
only)angles(small
L
g
only)angles(small
I
mgL
23. A simple pendulum consists of a particle attached
to a frictionless pivot by a cable of negligible mass
When the particle is pulled away from its
equilibrium position by an angle and
released, it swings back and forth
Gravity causes the back-and-forth rotation about the axis. The rotation speeds
up as the particle approaches the lowest point and slows down on the upward
part of the swing.
We denote the position of the pendulum along the circle of arc by, s, where we
choose s=0 to correspond to the equilibrium. In terms of the angle, which the
cord of the pendulum makes with the vertical direction we have
There is no acceleration along the direction of the cord so we have a tangential net
force:
But for small angles
From Hook’s law, the restoring force is
Equating (4) and (6) substituting (5)
But the a frequency which gives
Previously we found that
𝜔 = 2𝜋𝑓 =
2𝜋
𝑇
then the period (T) of the pendulum given by
𝑇 =
2𝜋
𝜔
= 2𝜋
𝑙
𝑔
24. 13. REASONING AND SOLUTION From the drawing given with the problem statement, we see that
the kinetic frictional force on the bottom block (#1) is given by
fk1 = µk(m1 + m2)g (1)
and the maximum static frictional force on the top block (#2) is
MAX
s2 s 2f m g
(2)
Newton’s second law horizontally applied to the bottom block gives
F – fk1 – kx = 0 (3)
Newton’s second law applied to the top block gives MAX
s2 0f kx (4)
a. To find the compression x, we have from Equation (4) that
x = MAX
s2f /k = µsm2g/k = (0.900)(15.0 kg)(9.80 m/s
2
)/(325 N/m) = 0.407 m
b. Solving Equation (3) for F and then using Equation (1) to substitute for fk1, we find
that
F = kx + fk1 = kx + µk(m1 + m2)g
F = (325 N/m)(0.407 m) + (0.600)(45.0 kg)(9.80 m/s
2
) = 397 N
25. 26. REASONING The work done in stretching or compressing a spring is given
directly by Equation 10.12 as 2 21
0 f2
W k x x , where k is the spring constant and x0
and xf are, respectively, the initial and final displacements of the spring from its
equilibrium position. The work is positive if the restoring force and the displacement
have the same direction and negative if they have opposite directions.
SOLUTION
a. The work done in stretching the spring from +1.00 to +3.00 m is
2 22 2 21 1
0 f2 2
46.0 N/m 1.00 m 3.00 m 1.84 10 JW k x x
b. The work done in stretching the spring from –3.00 m to +1.00 m is
2 22 2 21 1
0 f2 2
46.0 N/m 3.00 m 1.00 m 1.84 10 JW k x x
c. The work done in stretching the spring from –3.00 to +3.00 m is
2 22 21 1
0 f2 2
46.0 N/m 3.00 m 3.00 m 0 JW k x x
26. 27. REASONING As the block falls, only two forces act on it: its weight and the
elastic force of the spring. Both of these forces are conservative forces, so the falling
block obeys the principle of conservation of mechanical energy. We will use this
conservation principle to determine the spring constant of the spring. Once the spring
constant is known, Equation 10.11, /k m , may be used to find the angular
frequency of the block’s vibrations.
SOLUTION
a. The conservation of mechanical energy states that the final total
mechanical energy Ef is equal to the initial total mechanical energy E0, or Ef = E0
(Equation 6.9a). The expression for the total mechanical energy of an object
oscillating on a spring is given by Equation 10.14. Thus, the conservation of total
mechanical energy can be written as
2 2 2 2 2 21 1 1 1 1 1
f f f f 0 0 0 02 2 2 2 2 2
f 0
mv I m g h k y mv I m g h k y
E E
Before going any further, let’s simplify this equation by noting which variables are
zero. Since the block starts and ends at rest, vf = v0 = 0 m/s. The block does not rotate,
so its angular speed is zero, f = 0 = 0 rad/s. Initially, the spring is unstretched, so
that y0 = 0 m. Setting these terms equal to zero in the equation above gives 21
f f 02
m g h k y m g h
Solving this equation for the spring constant k, we have that
2
0 f
2 21 1
f2 2
0.510 kg 9.80 m/s 0.120 m
83.3 N/m
0.120 m
mg h h
k
y
b. The angular frequency of the block’s vibrations depends on the spring
constant k and the mass m of the block:
27. 43. REASONING As the ball swings down, it reaches it greatest speed at the lowest
point in the motion. One complete cycle of the pendulum has four parts: the
downward motion in which the ball attains its greatest speed at the lowest point, the
subsequent upward motion in which the ball slows down and then momentarily comes
to rest. The ball then retraces its motion, finally ending up where it originally began.
The time it takes to reach the lowest point is one-quarter of the period of the
pendulum, or t = (1/4)T. The period is related to the angular frequency of the
pendulum by Equation 10.4, T = 2/. Thus, the time for the ball to reach its lowest
point is
1
4
1 2
4
t T
The angular frequency of the pendulum depends on its length L and the acceleration
g due to gravity through the relation /g L (Equation 10.16). Thus, the time is
1 2 1 2
4 4 2
L
t
gg
L
2
0.85 m
0.46 s
2 2 9.80 m/s
L
t
g
28. 45. REASONING
a. The angular frequency of a simple pendulum can be found directly from
Equation 10.16 as /g L , where g is the magnitude of the acceleration due
to gravity and L is the length of the pendulum.
b. The total mechanical energy of the pendulum as it swings back and forth is the
gravitational potential energy it has just before it is released, since the pendulum
is released from rest and has no initial kinetic energy. The reason is that friction
is being neglected, and the tension in the cable is always perpendicular to the
motion of the bob, so the tension does no work. Thus, the work done by
nonconservative forces, such as friction and tension, is zero. This means that the
total mechanical energy is conserved (see Equation 6.9b) and is the same at all
points along the motion, including the initial point where the bob is released.
c. To find the speed of the bob as it passes through the lowest point of the swing, we
will use the conservation of energy, which relates the total mechanical energy at
the lowest point to that at the highest point.
a. The angular frequency of the pendulum is
2
9.80 m/s
3.5 rad/s
0.79 m
g
L
(10.16)
b. At the moment the pendulum is released, the only type of energy it has is its
gravitational potential energy. Thus, its potential energy PE is equal to its initial
total mechanical energy E0, so PE = E0. According to Equation 6.5, the potential
energy of the pendulum is PE = mgh, where m is the mass of the bob and h is its
height above its equilibrium position (i.e., its position when the pendulum hangs
straight down). The drawing shows that this height is related to the length L of
the pendulum by 1 cos5.50h L . Thus, the total mechanical energy of the
pendulum is
0
2
1 cos5.50
0.24 kg 9.80 m/s 0.79 m 1 cos5.50 10 J
E mgh mgL
c. As the bob passes through the lowest point of the swing, it has only kinetic
energy, so its total mechanical energy is 21
f f2
E mv . Since the total mechanical
energy is conserved f 0 ,E E we have that
21
f 02
mv E
Solving for the final speed gives
0
f
2 2 10 J
0.27 m/s
0.24 kg
E
v
m
29. 10.4 The Pendulum
Example 10 Keeping Time
Determine the length of a simple pendulum that will
swing back and forth in simple harmonic motion with
a period of 1.00 s.
2
2
L
g
T
f
m248.0
4
sm80.9s00.1
4 2
22
2
2
gT
L
2
2
4
gT
L
30. 10.5 Damped Harmonic Motion
In simple harmonic motion, an object oscillated
with a constant amplitude.
In reality, friction or some other energy
dissipating mechanism is always present
and the amplitude decreases as time
passes.
This is referred to as damped harmonic
motion.
32. 10.6 Driven Harmonic Motion and Resonance
When a force is applied to an oscillating system at all times,
the result is driven harmonic motion.
Here, the driving force has the same frequency as the
spring system and always points in the direction of the
object’s velocity.
33. 10.6 Driven Harmonic Motion and Resonance
RESONANCE
Resonance is the condition in which a time-dependent force can transmit
large amounts of energy to an oscillating object, leading to a large amplitude
motion.
Resonance occurs when the frequency of the force matches a natural
frequency at which the object will oscillate.
34. 10.7 Elastic Deformation
Because of these atomic-level “springs”, a material tends to
return to its initial shape once forces have been removed.
ATOMS
FORCES
35. 10.7 Elastic Deformation
STRETCHING, COMPRESSION, AND YOUNG’S MODULUS
A
L
L
YF
o
Young’s modulus has the units of pressure: N/m2
37. 10.7 Elastic Deformation
Example 12 Bone Compression
In a circus act, a performer supports the combined weight (1080 N) of
a number of colleagues. Each thighbone of this performer has a length
of 0.55 m and an effective cross sectional area of 7.7×10-4 m2. Determine
the amount that each thighbone compresses under the extra weight.
41. 10.7 Elastic Deformation
Example 14 J-E-L-L-O
You push tangentially across the top
surface with a force of 0.45 N. The
top surface moves a distance of 6.0 mm
relative to the bottom surface. What is
the shear modulus of Jell-O?
A
L
x
SF
o
xA
FL
S o
2
32
mN460
m100.6m070.0
m030.0N45.0
S
42. 10.7 Elastic Deformation
VOLUME DEFORMATION AND THE BULK MODULUS
oV
V
BP
The Bulk modulus has the units of pressure: N/m2
44. 10.8 Stress, Strain, and Hooke’s Law
HOOKE’S LAW FOR STRESS AND STRAIN
Stress is directly proportional to strain.
Strain is a unit less quantity.
SI Unit of Stress: N/m2
In general the quantity F/A is called the stress.
The change in the quantity divided by that quantity is called the
strain:
ooo LxLLVV
47. The change in length of the wire is, L FL0 /YA
, where the force F is equal to the tension T in the wire. The tension in the wire can be found by
applying Newton's second law to the two crates.
SOLUTION The drawing shows the free-body diagrams for
the two crates. Taking up as the positive direction, Newton's
second law for each of the two crates gives
(1)
(2)
T m1g m1a
T m2g –m2a
TT
m1g
m2g
Solving Equation (2) for a, we find a
T m2g
m2
. Substituting into Equation (1) gives
T 2m1g
m1
m2
T 0
Solving for T we find T
2m1m2g
m1 m2
2(3.0 kg)(5.0 kg)(9.80 m/s2
)
3.0 kg + 5.0 kg
37 N
Using the value given in Table 10.1 for Young’s modulus Y of steel, we find,
therefore, that the change in length of the wire is given by Equation 10.17 as
L
(37 N)(1.5 m)
(2.0 1011
N/m2
)(1.3 10–5
m2
)
2.1 10–5
m
48. sin12F f mg m a
11 2 5 2 4
0 2 2
2.0 10 N/m 7.8 10 m 2.0 10 m
m
sin12 61 kg 1.1 m/s 68 N 61 kg 9.80 m/s sin12
Y A L
L
m a f mg
0
Y A L
L
F
To determine F, we examine the following free-body diagram of the skier. For
convenience, the +x direction is taken to be parallel to the slope and to point
upward (see the drawing).
Three forces act on the skier in the x direction: (1) the towing force (magnitude = F),
(2) the frictional force (magnitude = f ) exerted on the skis by the snow, and (3) the
component of the skier’s weight that is parallel to the x axis (magnitude =
W sin12 = mg sin 12). This component is shown to the right of the free-body
The net force acting on the skier has a magnitude of sin12F f mg
. According to Newton’s second law (see Section 4.3), this net force is equal to the skier’s mass times the
magnitude of her acceleration, or
49. 58. REASONING AND SOLUTION F = S(X/L0)A for the shearing force. The shear modulus S for copper is given in Table 10.2.
From the figure we also see that tan = (X/L0) so that
6
1 1
10 2 2
6.0 10 N
tan tan 0.091
4.2 10 N/m 0.090 m
F
SA
59. REASONING AND SOLUTION The shearing stress is equal to the force per
unit area applied to the rivet. Thus, when a shearing stress of 5.0 10
8
Pa is applied
to each rivet, the force experienced by each rivet is
Nm)(5.0Pa) 42–38
109.310105.0())(Stress()Stress( 2
rAF
Therefore, the maximum tension T that can be applied to each beam, assuming
that each rivet carries one-fourth of the total load, is 5
4 1.6 10 NF .
60. REASONING Both cylinders experience the same force F. The magnitude of
this force is related to the change in length of each cylinder according to Equation
10.17: F Y(L/ L0)A. See Table 10.1 for values of Young’s modulus Y. Each
cylinder decreases in length; the total decrease being the sum of the decreases for each
cylinder.
SOLUTION The length of the copper cylinder decreases by
Lcopper
FL0
YA
FL0
Y(r
2
)
(6500 N)(3.0 10–2
m)
(1.110
11
N/m
2
)(0.25 10
–2
m)
2 9.0 10
–5
m
Similarly, the length of the brass decreases by
Lbrass
(6500 N)(5.0 10–2
m)
(9.01010
N/m2
) (0.2510–2
m)2
1.810–4
m
Therefore, the amount by which the length of the stack decreases is 2.7 10–4
m
50. 61. REASONING AND SOLUTION Equation 10.20 gives the desired result.
Solving for V/V0 and taking the value for the bulk modulus B of aluminum from
Table 10.3, we obtain
V
V0
P
B
1.01105
Pa
7.1 10
10
N/m
2 1.4 10
–6
63. REASONING AND SOLUTION From the drawing we have x = 3.0 × 10−3
m
and
A = 2 rx = 2(1.00 10
–2
m)(3.0 10
–3
m)
We now have Stress = F/A. Therefore,
F = (Stress)A = (3.5 10
8
Pa)[21.00 10
–2
m)(3.0 10
–3
m)] = 6.6 104
N
51. 81. REASONING Since the
surface is frictionless, we
can apply the principle of
conservation of mechanical
energy, which indicates
that the total mechanical
energy of the spring/mass
system is the same at the
instant the block contacts the bottle (the final state of the system) and at
the instant shown in the drawing (the initial state). Kinetic energy 21
2
mv
is one part of the total mechanical energy, and depends on the mass m and
the speed v of the block. The dependence of the kinetic energy on speed
is critical to our solution. In order for the block to knock over the bottle,
it must at least reach the bottle. When launched with the minimum speed
v0
shown in the drawing, the block will reach the bottle with a final speed
of vf
= 0 m/s. We will obtain the desired initial speed v0
by solving the
energy-conservation equation for this variable.
0.080 m
v0
x = 0 m
0.050 m
52. SOLUTION The conservation of mechanical energy states that the final
total mechanical energy Ef
is equal to the initial total mechanical energy
E0
. The expression for the total mechanical energy for a spring/mass
system is given by Equation 10.14, so that we have
2 2 2 2 2 21 1 1 1 1 1
2 2 2 2 2 2f f f f 0 0 0 0
f 0
mv I mgh kx mv I mgh kx
E E
Since the block does not rotate, the angular speeds f
and 0
are zero.
Moreover, the block reaches the bottle with a final speed of vf
= 0 m/s
when the block is launched with the minimum initial speed v0
. In
addition, the surface is horizontal, so that the final and initial heights, hf
and h0
, are the same. Thus, the above expression can be simplified as
follows: 2 2 21 1 1
2 2 2f 0 0
kx kx mv
In this result, we are given no values for the spring constant k and the mass m.
However, we are given a value for the angular frequency . This frequency is given
by Equation 10.11
k
m
, which involves only the ratio k/m. Therefore, in
solving the simplified energy-conservation expression for the speed v0
, we will divide
both sides by m, so that the ratio k/m can be expressed using Equation 10.11.
2 22 1 11
2 22 2 20 0f
0 f 0
or
kx mvkx k
v x x
m m m
Substituting
k
m
from Equation 10.11, we find
2 22 2
0 f 0
7.0 rad/s 0.080 m 0.050 m 0.44 m/sv x x
53. Can we apply work energy theorem?
Work done by non-conservative forces
is equal to the change in the
energy of the system
𝑊𝑁𝐶 = 𝐸𝑓 − 𝐸𝑖
𝑊ℎ𝑎𝑡 𝑖𝑠 𝑜𝑢𝑟 𝐸?