Download Chapter 5 Energy

Chapter 5
Energy
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
work
lever systems
energy
theoretical mechanical advantage
potential energy
actual mechanical advantage
kinetic energy
efficiency
power
Conservative and Nonconservative Systems
Establish the difference between a conservative and a nonconservative system.
Conservation of Energy
Explain the principle of conservation of energy.
Energy Problems
Apply the principle of conservation of energy to solve mechanics problems.
Lever Systems
Determine the theoretical mechanical advantage of human body lever systems.
Efficiency
Calculate the efficiency of a machine or human in action.
Power
Determine the power required for a given process or activity.
PREREQUISITES
Before beginning this chapter you should have achieved the goals of Chapter 3,
Kinematics, and Chapter 4, Forces and Newton's Laws.
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Chapter 5
Energy
OVERVIEW
Newton's three Laws of Motion provide a useful model for viewing the motion of
objects we encounter in our daily experience. However, the model becomes difficult
to apply in situations where the forces involved vary significant with time and
displacement. To provide a parallel assessment procedure, the concept of work and
energy was introduced. This alternate model for viewing our environment allows
for a simple solution to many problems.
SUGGESTED STUDY PROCEDURE
To begin your study of this chapter, become familiar with the following Chapter
Goals: Definitions, Conservation of Energy, Energy Problems, Lever Systems,
Efficiency and Power. A discussion of each of the terms listed under Definitions can
be found in the first section of this Study Guide chapter. Next, read text sections 5.15.13 and study the examples provided at the end of each section. Please remember
that the answer to questions asked in the text are given in this Study Guide chapter,
section two.
At the end of the chapter, read the Chapter Summary and complete Summary
Exercises 1-7 and 11-15. Check your answers against those provided. Now, complete
Algorithmic Problems 1-6 and check each answer. Then do Exercises and Problems
1, 2, 3, 5, 7, 12, 15, 18, 19, and 22. For additional work on the goals of this chapter, see
the third section of this Study Guide chapter. Finally, attempt the Practice Test. If
you have difficulty with any of the concepts, refer to the appropriate chapter section
for additional work. This study procedure is outlined below.
--------------------------------------------------------------------------------------------------------------------Chapter Goals
Suggested
Summary
Algorithmic
Exercises
Text Readings Exercises
Problems
& Problems
--------------------------------------------------------------------------------------------------------------------Definition
5.1, 5.2, 5.3
1-7
1,2
Conservation of
5.4, 5.5, 5.6,
11-14
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Energy
5.7, 5.8, 5.11
Energy Problems 5.4, 5.5, 5.6,
1, 3, 6
3, 5, 12,
5.8, 5.9, 5.13
18, 19
Lever Systems
5.12
5, 6, 13
4
7
Efficiency
5.10
7, 14
22
Power
5.9
15
2, 5
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DEFINITIONS
WORK
Applied force produces a displacement; magnitude of work done is given by product of
force component parallel to the displacement and the magnitude of the displacement.
Note that work as used in physics requires a force component acting in the direction
of motion as well as motion. So, for example, the force that a string exerts on a whirling
object connected to it is always perpendicular to the direction of motion so the work
done by the force is zero.
ENERGY
Property of a system which causes changes in its own state or state of its surroundings;
measure of ability to do work (physical).
Energy is an intrinsic property of a system. We always compute energy from other
measured properties of a system. Energy is a scalar quantity. There are many forms of
energy; mechanical, electrical, chemical, thermal, etc. There are two kinds of mechanical
energy; potential and kinetic.
POTENTIAL ENERGY
Ability to do work as a result of position or configuration.
A most common form of potential energy is the energy which results from
changing the height of objects above the earth. Have you ever wondered why it is more
tiring to climb up stairs than to come down them? The changes in potential energy that
occur may be an answer.
KINETIC ENERGY
Energy of motion of a body.
The idea that the kinetic energy of a moving object is proportional to the square of
the speed is well-illustrated by data about automobiles. The distance required to stop a
moving automobile increases with the square of the speed, as does the fatality rate in
accidents.
POWER
Time rate of doing work or using energy.
The increased power of large automobile engines is most manifest in the ability of an
automobile to have its speed rapidly increased.
LEVER
A simple machine. Essentially a rigid body with a fixed pivot.
A lever is only one kind of simple machine. Others, such as the wheel and axle, the
inclined plane, and the screw, are only briefly mentioned in the textbook.
THEORETICAL MECHANICAL ADVANTAGE
Ratio of distance through which the applied force acts to the distance through which the
load moves.
ACTUAL MECHANICAL ADVANTAGE
Ratio of the load to the applied force.
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EFFICIENCY
Ratio of work accomplished to work supplied.
The efficiency of a simple machine is also equal to the ratio of the actual mechanical
advantage to the theoretical mechanical advantage.
ANSWERS TO QUESTIONS FOUND IN THE TEXT
SECTION 5.2 Work
Any force you apply to an object which has a component which is not in the
direction of motion is not doing work. When you lift an object you do work, but as
you carry the object to a new location, the vertical force you exert to hold up the box
does not do any work as long as the motion is in a horizontal direction.
SECTION 5.3 Energy
The chemical energy stored in a battery is converted to electrical energy to operate a
portable radio and then the electrical energy is converted to the sound energy which
you hear coming from the radio.
SECTION5.5 Potential Energy
Whenever you lift or lower an object near the surface of the earth you change its
gravitational potential energy. The potential energy of water that can flow down
from high elevations is converted into electrical energy through the use of
hydroelectric stations.
SECTION 5.7 Conservative and Nonconservative Forces
1. Each time the boy slides up a hill he will rise a smaller amount. Each trip he loses
some of his energy to frictional, non-conservative forces working on his sled.
Finally he will come to rest in the valley between the hills.
2. The total energy of a non-conservative system will always decrease as the nonconservative forces dissipate the energy of the system.
3. All real systems are non-conservative. It is only in our imaginations that we have
a system which never suffers a loss of energy to dissipative forces.
4. The final equilibrium state of a system will be its state of lowest energy.
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EXAMPLES
ENERGY PROBLEMS
1. A professional skier of 60 kg mass is on a ski jump ramp like the one shown See
Fig. 5-1. (not recommended for beginners!). At point C the ramp is pointed up at
an angle of 45ø. Point A is 40 meters above the lowest point of the ramp and point
C is 10 meters above the lowest point. The distance along the ramp from point A
to point C is 81 meters. Assume the potential energy of the skier at point B is
zero.
(a) What is the potential energy of the skier at point A?
(b) Assuming no friction, what is his speed when he gets to point B if he starts from
rest at A?
(c) Suppose for artificial snow the force of friction is a constant equal to 5 newtons.
How much energy does the skier lose in going from A to C?
(d) What is the maximum height to which the skier can rise above B if he starts from
rest at A on artificial snow, and jumps off the ramp at point C?
What Data Are Given
The mass of the skier = 60 kg; distance travelled on the ramp = 81 m; starting
elevation = 40 m; friction force = 0 for part (b) and 5 N for parts (c) and (d). The
angle of incline of the ramp is 45ø.
What Data Are Implied?
It is assumed that the skier does no work on himself through the use of his muscles.
In addition, the only dissipative force considered is that artificial snow frictional
force. The gravitational acceleration constant is assumed to be 9.80 m/s2.
What Physics Principles Are Involved?
This problem involves the use of the work-energy theorem, the conservation of
energy, and the expressions for gravitational potential energy, and kinetic energy.
What Equations Are to be Used?
Kinetic energy = KE = 1/2 mv2
(5.2)
Potential energy = PE = mgh
(5.4)
Work done = F • d
(5.1)
Work-energy theorem; net work on a system = change in energy
W = ΔKE + ΔPE
(5.9)
Projectile motion concepts from Chapter 4.
Algebraic Solutions
Let m = mass of skier; h = starting height of point A.
(a) Potential energy at A = mghA
(b) Speed at B; conservation of energy gives
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PEB + KEB = PEA + KEA
(5.11)
since friction is neglected for this part of the problem. Both the kinetic energy at A
and the potential energy at B are zero.
Can you explain why?
KEB = PEA
(½) mvB2 = mgh
The Speed at B = vB = SQR RT[2gh]
(1)
(c) Energy loss = work done by friction = friction force times the distance
(d) Maximum height is found by applying the conservation of energy concepts to his
motion through the air. At point C he has a kinetic energy,
KEc = mghA - mghC - work done by friction
where hA and hC are the heights of points A and C respectively. His speed at C is
given by SQR RT[2KEC]/m, so his horizontal velocity,
vhorz = SQR RT[2KEC)/m] cos θ
(2)
where θ is the angle of the ramp. At the peak of the jump path his vertical
component of velocity will equal zero so his total energy will be equal to his
potential energy plus his kinetic energy because of his constant velocity in the
horizontal direction. This total energy must be equal to the energy he had when he
left the ramp at point C.
Energy at peak of jump = kinetic energy at point C = KEC
mghpeak + (½) mvhorz2 = KEC = mghA - mghC - friction work
(3)
Since everything in this equation is known expect the peak height, hpeak it can be
found by combining Equations (2) and (3)
hpeak = (KEC/mg) - (v horz2/2g) = (KEC/mg) - (2KECcos2θ/ 2mg)
hpeak = (KEC mg)(1 - cos2θ)
(4)
Numerical Solutions
(a) Potential energy at A = (60 kg)(9.8 m/s2)(40 m) = 2.4 x 104 m2kg/s2
PEA = 2.4 x 104 joules
(b) vB = SQR RT[2(9.8)(40) m2/s2] = 1.2 x 102 m/s
(c) Energy loss = (5 N)(81 m) = 405 joules
(d) Kinetic energy at C = (60)(9.8)(40 - 10) J - 405 joules
KEC = 1.72 x 104 J
hpeak = [1.72 x 104 / (60)(9.8)] (1 - cos2 45ø)
hpeak = 1.46 x 101 m ≈15 meters above point C
Thinking About the Answers
You should check to see that all of the units came out correctly. You can see in this
problem that the friction force made little difference in the energy at point C and the
height of the skier peak. In equation (4) we can see that if the angle θ is zero, the
height of the peak is zero and if the angle θ is 90ø the peak height will be a
maximum. In this particular case, since the angle is 45ø the launch velocity is equally
divided between its horizontal and vertical components and the kinetic energy at the
peak is one-half the total kinetic energy at point C. Did you notice that the zero point
for potential energy was shifted to point C for answering part (d)? Can you rework
the problem without making that shift?
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LEVER SYSTEMS
2. Lever systems are only one of several kinds of simple machines. Shown below, in
addition to the lever, are the crank and axle, inclined plane, pulley system, and
screw jack. Using the symbols shown on each machine, calculate the actual
mechanical advantage, the theoretical mechanical advantage, and the efficiency of
each machine.
(i)
(iii) Levers
Type I
Type II
Type III
Standing on tiptoe
Nutcracker
Jawbone
(iv) Pulley Systems
(v) Screw Jack
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What Data Are Given?
In each case a load of W newtons is moved by the application of a force of F
newtons. The distances for both the load W and the applied force F are given.
What Data Are Implied?
Any explicit calculation of frictional forces is not done. Instead all of the frictional
forces are included in the value of W that can be moved by the force F.
What Physics Principles Are Involved?
The definitions of theoretical mechanical advantage, actual mechanical
advantages, and efficiency are needed.
What Equations Are to be Used?
Theoretical mechanical advantage (TMA) =
distance applied force moves/distance move load
(5.15)
Actual mechanical advantage (AMA) = load W/force F
(5.16)
Efficiency (e) = work done by the machine/energy supplied to the machine (5.14)
Algebraic Solutions
(i) AMA = W/F;
TMA = 2πl/2πr = l/r;
e =Wr/Fl
(ii) AMA = W/F; TMA = d/h;
e = W • h/F • d
(iii) AMA = W/F; TMA = l/d; e = Wd/Fl; same for all three types of lever
(iv) AMA = W/F; TMA = l/d;
e = Wd/Fl
Note for the case shown for every 1 m the force moves, the W goes up only 1/4
m; so TMA = 4. This is determined by the geometry of the pulley system. The
system shown has 4 ropes supporting the bottom pulleys so the TMA is 4. Can
you use the geometry of a pulley system to derive this algorithm for computing
the TMA of a pulley system?
(v) AMA = W/F; TMA = 2πl/p;
e = Wp/2πFl
EFFICIENCY
3. A mile runner felt that he needed to eat 1.3 ounces (37 gm) of a candy bar every
time he ran a mile in 4 1/2 minutes. Estimate his energy efficiency. Use Table 5.1
and Table 5.2 for the constants you need.
What Data Are Given?
The time of running and the amount of candy eaten, 4.5 minutes and 37 gm
respectively.
What Data Are Implied?
To calculate efficiency of energy use you need to calculate the energy input and
the energy used in a useful way. You can use Table 5.1 and your knowledge that
sugar is a carbohydrate to convert 37 gm of sugar to its approximate energy
value since its energy conversion value is 4 kcal/gam. Likewise, you can use
Table 5.2 to estimate the rate at which the runner uses energy, 11.4 kcal/minute.
What Physics Principles Are Involved?
The definition of energy efficiency must be used in conjunction with some
concepts implied by the data in Table 5.1 and Table 5.2 . You can take the
product of the amount of food eaten in grams times the energy conversion factor
( Table 5.1) to compute the total energy input. You can take the product of the
running times the energy use rate (Table 5.2 ) to calculate the use of energy.
What Equations Are to be Used?
Efficiency = Useful work done/energy input
Energy Input = (mass of food) (energy conversion factor)
Energy Consumption = useful work = (time of activity) (energy use rate)
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Algebraic Solution
E = U.W.D. / E.I. = (time)(kcal/min) / (mass)(kcal/gm)
Numerical Solution
E = (4.5 min)(11.4 kcal/min) / (37 gm)(4 kcal/gm) = 51.3 kcal/148 kcal = 35%
Thinking About the Answer
This runner seems to be 35% efficient. That means that 65% of the energy input is
not used in running. Where does it go?
POWER
4. What are the power ratings in watts of a person engaged in various activities? You
may use the values in Table 5.2 for this problem.
What Data Are Given?
One person engaged in various human activities.
What Data Are Implied?
The energy use rate for human sitting, resting, etc., are given in Table 5.2 The
watt is an energy use rate of 1 joule per second.
What Physics Principles Are Involved?
You need to use the definition of power as given in Equation 5.12.
Power = Work Done/Time to Do it (5.12)
What Equations Are to be Used?
Power = Work/Time = Energy Used/Time (5.12)
The data in Table 5.2 are already given in units of energy (kcal) per unit time
(minutes) and only need to be converted to joules and seconds respectively.
Algebraic Solution
Power (Watts) = Energy Used (joules) / Time (seconds)
joules = (4187 J/kcal)(no. of kcal)
seconds = 60 sec./min. (time in minutes)
Power (watts) = (4187/60) (kcal/min.) = 69.8 (kcal/min.)
Numerical Solutions - Data from Table 5.2
Power (watts) sitting = 69.8 (2.0 kcal/min.) = 140 W
Power (watts) resting = 69.8 (1.2 kcal/min.) = 84 W
Power (watts) sleeping = 69.8 (1.1 kcal/min.) = 77 W
Power (watts) walking = 69.8 (3.8 kcal/min.) = 270 W
Power (watts) bicycling = 69.8 (6.9 kcal/min.) = 470 W
Power (watts) swimming = 69.8 (8.0 kcal/min.) = 560 W
Power (watts) skiing = 69.8 (9.0 kcal/min.) = 630 W
Power (watts) running = 69.8 (11.4 kcal/min.) = 796 W
Power (watts) climbing up stairs = 69.8 (12.0 kcal/min.) = 838 W
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PRACTICE TEST
1. A professional skier of 100 kg mass is on a ski ramp See Fig. (not recommended
for beginners). At point C the ramp is pointed straight up. Point A is 40 meters
above the ground. Point C is 10 meters above the ground. The distance from point A
to point C along the ramp is 150 meters. The potential energy at B is zero.
________a. What is the potential energy of the skier at point A?
________b. If the friction force along the path is a constant 50 N, what is the work
done by friction in going from point A to point C?
________c. What is the kinetic energy of the skier at point C?
________d. What is the maximum height h, above the end of the ramp (point C) to
which the skier can rise in his trajectory?
2. A force of 6 N is required to raise a weight of 30 N by means of a pulley system. If
the weight is raised 1 meter in one second while the applied force moves through a
distance of 8 m _______a. What is the actual mechanical advantage of the pulley system?
_______b. What is the theoretical mechanical advantage of the pulley system?
_______c. What is the efficiency of this system?
_______d. What is the power dissipated in the system?
3. The human arm See Fig. is being tested for mechanical advantage. In this test, the
biceps arm action is utilized with a weight of 10 Newtons. In the configuration
shown,
the force of the muscle is found to be 85
Newtons.
_______a. Find the TMA of the arm.
_______b. Find the AMA of the arm.
_______c. Calculate the arm's efficiency.
ANSWERS:
1. 40,000 J,
2. 5,
3. 0.125,
7500 J,
8,
0.12,
22,500 J,
62.5%,
0.94
22.5 meters above C.
18 watts
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