of 30
8/3/2019 m416
1/30
Project PHYSNET Physics Bldg. Michigan State University East Lansing, MI
MISN-0-416
ENERGY OF A SYSTEM OF PARTICLES
F`
w
Lw
Lm
Pm
E
Pw
F`m
F`m
1
ENERGY OF A SYSTEM OF PARTICLES
by
F. Reif, G. Brackett and J. Larkin
CONTENTS
A. Kinetic Energy and Work
B. Potential Energy
C. Conservation of Energy
D. Energy Transformations
E. Macroscopic and Internal Energy
F. Direction of Energy-Transformation ProcessesG. Macroscopic Equilibrium
H. Summary
I. Problems
2
8/3/2019 m416
2/30
8/3/2019 m416
3/30
MISN-0-416
ENERGY OF A SYSTEM OF PARTICLES
A. Kinetic Energy and Work
B. Potential Energy
C. Conservation of Energy
D. Energy Transformations
E. Macroscopic and Internal Energy
F. Direction of Energy-Transformation ProcessesG. Macroscopic Equilibrium
H. Summary
I. Problems
Abstract:
In the preceding units we discussed energy and the conservation of energyin the simple case of a system consisting of a single particle. We shall nowgeneralize these ideas to systems consisting of many particles. Thus weshall be able to apply energy arguments to many practical situations andto deal with systems as complex as biological organisms.
5
MISN-0-416 A-1
SECT.
A KINETIC ENERGY AND WORK
Consider a system consisting of several interacting particles labeled1, 2,. 3, . . . . (For example, the system might consist of the sun andthe planets.) Then the change dK of the kinetic energy of each particleduring a small enough displacement is related to the work W done onthis particle by all forces so that
dK1 = W1, dK2 = W2, . . . (A-1)
By adding corresponding sides of these equations for all particles in thesystem we obtain
dK1 + dK2 + . . . = W1 + W2 + . . . (A-2)
But since the sum of the changes on the left side is equal to the changein their sum, we can write
d(K1 + K2 + . . .) = W1 + W2 + . . . (A-3)
Let us introduce the convenient abbreviations
K = K1 + K2 + . . . (A-4)
andW = W1 + W2 + . . . (A-5)
Then we can write the relation (A-3) in the simple form
dK = W (by all forces) (A-6)
Note that W is the sum of the works done by all forces acting on eachparticle in the system.
The quantity K in Eq. (A-4) is called the kinetic energy of the sys-tem, i.e.,
Def. Kinetic energy of a system: The sum of thekinetic energies of all the particles in the system.
(A-7)
The quantity W in Eq. (A-5) is called the work done on the system ina small displacement. Here we have used this general definition:
Def.
Work done on a system: The work done on a
system by specified forces is the sum of the worksdone by these forces on all particles in the system.
(A-8)
6
8/3/2019 m416
4/30
8/3/2019 m416
5/30
MISN-0-416 A-4
(a) (b)
(c) (d)
v`
rest v`
v`
v`
v`
v`
v`
Fig. A-2.
K+ dr
`dr`
Br-
2 1
1=10-
meter10
Fig. A-3.
* In the case of a string of constant length, or in the case of anincompressible liquid, the average distance between adjacent
atoms interacting with each other remains constant. Hencethe work done by the mutual forces between atomic particlesis then also zero.
Finding the Kinetic Energy of a System (Cap. 2a)
A-1Figure A-2 shows four possible states of an isolated system con-sisting of two helium atoms, each of mass m. The system has the
same momentum P = mv in each state. Write an expression (in termsof m and v) for the systems kinetic energies Ka, Kb, Kc, and Kd inthese states. Does the system have the same kinetic energy in each state?(Answer: 106) (Suggestion: [s-1])
Knowing About Work and Kinetic Energy for Systems
A-2A positively-charged potassium ion (K+) and a negatively-charged bromide ion (Br) form an isolated system as they ap-
proach each other to form a potassium bromide (KB r) molecule (Fig. A-3). (a) Suppose that the potassium ion, of mass 7.0 1026 kg, has aninitial velocity of 800 m/s toward the right, while the bromide ion, of mass14 1026 kg, has an initial velocity of 400 m/s toward the left. Whatis the initial kinetic energy of this system? (b) These ions then movethrough the small displacements illustrated in Fig. A-3. (These displace-ments are exaggerated for clarity.) If each ion exerts on the other anattractive electric force of magnitude 1 1011 N, what is the small workW done on the system by all forces during this motion? (c) What is the
final kinetic energy of the system after the ions have moved through those
9
MISN-0-416 A-5
displacements? (Answer: 102) (Suggestion: [s-7])
10
8/3/2019 m416
6/30
8/3/2019 m416
7/30
MISN-0-416 B-3
(a) (b) (c)
initial state intermediate state standard state
1 1
1
2 2
2
3 3 3
Fig. B-1: Work done in bringing an isolated system of threeparticles from some initial state to the standard state.
The preceding conclusion can be stated in these general terms:
The potential energy of an isolated system of particlesis the sum of the potential energies between all inter-acting pairs of particles.
(B-6)
Since the potential energy of a pair of particles depends only on the dis-tance between them, the potential energy U of a system depends thusonly on the distances between all the interacting particles in the system.
Finding the Potential Energy of a System (Cap. 2b)
B-1(a) What is the Coulomb potential energy of a system consistingof two protons separated by a distance of 1.0 1015 meter, a
typical separation within a nucleus? Use the values 1.6 1019 C for thecharge of a proton, where C is the abbreviation for a coulomb of chargeand ke = 9 10
9 N m2/C2. (b) In a beryllium nucleus, four protons arearranged in space so that each proton is separated from the others bythe distance described in part (a). (See Fig. B-2.) What is the Coulomb
potential energy of the system consisting of these four protons? (Answer:104) (Suggestion: [s-8])
Knowing About Work and Potential Energy for Systems
B-2A lithium nucleus consists of three protons, each separated fromthe others by the distance specified in part (a) of problem B-1.
(a) In the initial state a of a fusion reaction, a fourth proton is very farfrom the lithium nucleus. What is the initial Coulomb potential energyUa of the system consisting of all four protons? (b) The distant proton
13
MISN-0-416 B-4
Fig. B-2.
then approaches the lithium nucleus, and the strong nuclear force bindsthe four protons together to form a beryllium nucleus. In this final stateb, the system consisting of the four protons is arranged as described inpart (b) of problem B-1. What is the final Coulomb potential energy Ubof this system? (c) What is the work Wab done on this system by theCoulomb electric forces due to the protons as the system goes from the
state a to the state b? (Answer: 101)
14
8/3/2019 m416
8/30
8/3/2019 m416
9/30
8/3/2019 m416
10/30
8/3/2019 m416
11/30
8/3/2019 m416
12/30
8/3/2019 m416
13/30
8/3/2019 m416
14/30
8/3/2019 m416
15/30
8/3/2019 m416
16/30
8/3/2019 m416
17/30
8/3/2019 m416
18/30
MISN-0-416 G-5
S
BA
F
`
A
F`
S
F`
B
;
LA LB
Fig. G-3.
20cm
40cm
CM
F`
d
F`
j F`
g
0
15
Fig. G-4.
(i.e., 84 pound) applied to his wrist. Hence Eq. (G-8) shows that the max-imum magnitude of the force which can be supplied by the biceps muscle
is (375 newton)(27.5 cm/5.0 cm) = 2060 newton (i.e., about 460pound).This is an impressively large force.
Now: Go to tutorial section G.
APPLYING THE TORQUE CONDITION FOR MACROSCOPICEQUILIBRIUM
G-1Figure G-3 shows a simple scale used throughout the world tomeasure the mass M of an object (such as the fish shown). The
scale consists of a bar of negligible mass, which is free to rotate about asupporting ring passing through the bar at a point S, and a metal blockof mass m which can be moved along the bar. The object is hung froma hook at the point A on one end of the bar, a distance LA from thepoint S, and the block is moved along the bar until the bar is horizontaland in macroscopic equilibrium (i.e., at rest). In this situation, the blockis located at a point B, a distance LB from the point S. The blockthen exerts on the bar at the point B a downward force FB equal inmagnitude to the blocks weight, and the object exerts on the bar at the
point A a downward force FA equal in magnitude to the objects weight.(a) To relate the objects mass M to the known mass m of the block,apply the equilibrium torque condition with an axis passing through thepoint S at which the supporting force FS is applied to the bar. Write anequation expressing M in terms of m and the known distances LA andLB. (b) Suppose that m = 1 kg and LA = 0.1 meter. What is the objectsmass M if LB = (1/2)LA = 0.05 meter, if LB = LA = 0.1 meter, and ifLB = 2LA = 0.2 meter? (Answer: 122) (Suggestion: [s-3])
G-2
Because of the force Fd exerted on the humerus (the bone in theupper arm) by the deltoid muscles attached to the shoulder, a
35
MISN-0-416 G-6
18cm
F`
t
F`
a
F`
n
O
15
37
3.0cm
Fig. G-5.
man can keep his 15 kg arm outstretched horizontally (see Fig. G-4). In
this situation, the shoulder joint exerts on the arm a force Fj , and the
earth exerts on the arm a gravitational force Fg applied at the arms centerof mass (CM). What is the magnitude Fd of the force the deltoid muscles
must exert to maintain the arm in macroscopic equilibrium? (Since theforce Fj is unknown, apply the equilibrium torque condition with an axis
passing through the point O at which Fj is applied.) (Answer: 115)
G-3Figure G-5 shows the forces acting on the foot of an 80 kg man ina crouching position. The normal force Fn exerted by the floor
thus has a magnitude equal to half the mans weight. In this situation,what is the magnitude of the force Fa exerted by the Achilles tendonand calf muscles? (Since the force Ft exerted on the foot by the tibia is
unknown, choose an axis passing through the point O at which this forceis applied. You may neglect the gravitational force on the foot, since itis negligible compared with the other forces on the foot.) (Answer: 120)More practice for this Capability: [p-4], [p-5]
36
8/3/2019 m416
19/30
8/3/2019 m416
20/30
MISN-0-416 H-3
into macroscopic energy or random internal energy. For our purposes, thesystem S0 consisting of the car, the earth, and the air is isolated, and wecan regard the earth as fixed relative to an inertial reference frame. Sincethe earths kinetic energy is thus zero, the macroscopic energy Emac of
the system S0 is just the cars energy E = 1/2mv2 + mgy (i.e., the sumof the cars macroscopic kinetic and gravitational potential energies). (a)Suppose that the frictional forces on the car due to the air and the roadare negligible, so that the system S0 is non-dissipative. What happens tothe random internal energy Eran, the internal energy Eint, and the macro-scopic energy Emac of the system S0? Are we correct in saying, as we didin Unit 415, that the cars energy E is conserved? (b) Suppose that thesefrictional forces are not negligible, so that the system S0 is dissipative.What happens to the random internal energies Eran, Eint, and Emac in
this case? Are we correct in saying, as we did in Unit 415, that the carsenergy E is not conserved in this case? (Answer: 126)
39
MISN-0-416 I-1
SECT.
I PROBLEMS
I-1Conservation of momentum and energy in a ballistics experiment:A bullet of mass 2.0 gram is fired horizontally with an initial speed
of 500 m/s toward a 1.0 kg wood block at rest on a flat, smooth surface(Fig. I-1). The total external force on the system consisting of bullet andblock is thus negligible. (a) What is the blocks speed just after the bullethits and becomes embedded in it? (b) If we choose a standard positionat the level of the blocks center, the macroscopic gravitational potentialenergy of this system is zero during the collision. What is the systemsmacroscopic energy Emac just before and just after the collision? (c)
Is Emac conserved during the collision? If not, find the changes Emacand Eint in the systems macroscopic and internal energies during thecollision. (Answer: 121) (Suggestion: [s-6])
I-2Distance of closest approach for two protons: In a scattering ex-periment, a proton of mass m and charge q collides head-on with
another proton initially at rest in a target. When it is far from thetarget, the incident proton has a speed v, and the two protons form anisolated system during their collision. If the protons interact only by the
Coulomb electric force, what is the distance R between them at the timeduring their collision when both protons have the same speed v/2? (Thisdistance R is the distance of closest approach, or the smallest distancebetween the protons during their collision.) (Answer: 117) (Suggestion:[s-4])
I-3Scattering angle for colliding protons: A glancing collision of twoprotons is illustrated in Fig. I-2. When the protons are far apart
before the collision, proton 1 has a velocity V and proton 2 is at rest(Fig. I-1a). When the protons are far apart after the collision, the two
Fig. I-1.
V`
v`
1
v`
2
1
2
12
(a) (b)
q
Fig. I-2.
40
8/3/2019 m416
21/30
MISN-0-416 I-2
protons have velocities v1 and v2 (Fig. I-2b). By applying the principlesof conservation of momentum and conservation of energy to the isolatedsystem consisting of the two protons, show that the angle between v1and v2 is equal to 90. (Hint: Use conservation of momentum to draw a
vector diagram showing the triangle formed by V, v1, and v2. Then useconservation of energy to show that the sides of this triangle are relatedby the Pythagorean Theorem.) (Answer: 123)
I-4Force exerted at the ankle joint: The bones in a joint suffer verylittle wear even though they exert very large forces on each other,
presumably because the region between the bones is very well lubricated.To illustrate the magnitudes of such forces, use your results in problemG-3 and the equation of motion to find the magnitude Ft of the force
exerted by the tibia on the ankle joint in a crouching man. (Answer:119) (Suggestion: [s-2])
41
MISN-0-416 I-3
42
8/3/2019 m416
22/30
8/3/2019 m416
23/30
MISN-0-416 Tutorial Supplement gtu-3
Let us now find the torques exerted by the remaining forces FA and FB.On the following diagrams, construct the component vectors of FA andFB parallel and perpendicular to the line joining the point O to thepoints A and B, and indicate with a small arrow the counterclockwise
direction perpendicular to this line. Using the diagrams, express thecomponent F for each force, and the torque exerted by each force, interms of symbols for known and desired information.
-
F
`
B
30
30
BB
AA
OO
F`
A
LA
LB
FA: Component =
Torque =
FB: Component =
Torque =
Use your results to write the torque equation for macroscopic equilib-rium in terms of symbols for known and desired information.
-
(Answer: 2)
IMPLEMENTATION:
(1) Solve algebraically for desired quantities.Write an equation for the magnitude FB in terms of symbols for knownquantities.
- FB =
Write an equation for the magnitude FC in terms of FB and knownquantities.
- FC =
45
MISN-0-416 Tutorial Supplement gtu-4
(2) Substitute known values and find the desired quantities.
- FB =
FC =
CHECKING:
Check that the work is correct, that the results have the correct signsand units, and that the magnitudes are not unreasonably large or small.
The method we have illustrated is useful in applying the torque conditionfor macroscopic equilibrium to any problem. In particular, it should help
you solve systematically the problems in text section G.
(Answer: 6) Now: Go to text problem G-1.
46
8/3/2019 m416
24/30
8/3/2019 m416
25/30
8/3/2019 m416
26/30
MISN 0 416 A S l t 1 MISN 0 416 A S l t 2
8/3/2019 m416
27/30
MISN-0-416 Answers Supplement an-1
ANSWERS TO PROBLEMS
1.Motion Macroscopic energy Internal energy
a becomes larger becomes smallerb remains same remains samec becomes larger becomes smallerd becomes smaller becomes largere becomes smaller becomes larger
2.
F
`
B
3030
B B
A A
O O
F`
A
ccw = counterclockwise
ccwccw
~~ ~~
FA: Component = FA sin30, Torque = FA sin30
LA.
FB: Component = FB sin30, Torque = FB sin30
LB.
FA sin30LA FB sin30
LB = 0, or FALA FBLB = 0.
3. a. FL + FR = 800 N
b. FL = 200 N, FR = 600 N
4. Yes, FO. Yes, FC
5. E = (1/2)mv2 kee2/r1 kee
2/r2 + kee2/R
6. FB = FALA/LB, FC = 800 N FB, FB = 300 N, FC = 500N
7. a. Eran and Emac remain the same. Kmac and Umac are the same.Height is the same.
b. Eran becomes larger, Emac becomes smaller. Kmac is the same,Umac is smaller. Height is smaller. Satellite at rest on the earthssurface (or all parts of it at rest on the earths surface).
8. Fe = 2.5W, two and a half times the mans weight!
53
MISN-0-416 Answers Supplement an-2
101. a. 6.9 1013 J
b. 1.4 1012 J
c. 7 1013 J
102. a. 3.3 or 3.4 1020 J (either is acceptable)
b. 3 1021 J
c. 3.6 or 3.7 1020 J
103. a. 2 1021 J
b. the same as in part a.
104. a. 2.3 1013 J
b. 6(2.3 1013 J), or 1.4 1012 J
105. E = (1/2)mnv2n + (1/2)mev
21 + (1/2)mev
22 + U1,n + U2,n + U1,2
106. Ka = (1/2)mv2, Kb = (5/16)mv2, Kc = (5/18)mv2, Kd = (1/4)mv2.No.
107. a. 0.1 or 10 percent
b. 3.4 106 J
c. 8.5 hours! I wouldnt.
108. a. E = mv20
b. E = mv2 + keq2/R
c. E = keq2/R
109. Emac becomes smaller, Eint becomes larger
110. Yes, because it is an isolated system of atomic particles.
111. (a), (b), and (e)
112. a. It becomes larger.
b. Yes.
113. a. Speed remains the same. Emac and Eint remain the same.
b. Speed becomes smaller. Emac becomes smaller, Eint becomeslarger.
114. a. Kmac: same, Umac: larger, Emac: larger, Eint: smaller
b. Kmac : same, Umac: smaller, Emac: smaller, Eint: larger
54
MISN 0 416 Answers Supplement an 3 MISN 0 416 me 1
8/3/2019 m416
28/30
MISN-0-416 Answers Supplement an-3
115. Fd = 1.1 103 N
116. It increases.
117. R = 4keq2/mv2
118. a. Eran and Emac both remain the same.
b. Kmac is the same, Umac is the same.
c. equal to
119. Ft = 1.4 103 N
120. Fa = 1.0 103 N
121. a. 1.0m/s
b. Before: Emac = 2.5 102 J. After: Emac = 0.5 J.
c. No! Emac = 250J, Eint = +250 J
122. a. M = m(LB/LA)
b. If LB = (1/2)LA, M = (1/2)m = 0.5kg. IfLB = LA, M = m =1kg. IfLB = 2LA, M = 2m = 2 kg.
123. From momentum conservation, V = v1 + v2. From energy conserva-tion, V2 = v21 + v
22 . Since the triangle formed by the velocities has
sides related by the Pythagorean theorem, it is a right triangle and = 90
qv`
1v`
2
V`
124. (a) and (e) are true.125. a. Eran becomes larger, Emac becomes smaller.
b. Kmac is the same, Umac is smaller.
c. smaller than
d. The bob hangs vertically at rest.
126. a. Eran, Eint, and Emac remain the same. Yes.
b. Eran and Eint become larger, Emac becomes smaller. Yes.
55
MISN-0-416 me-1
MODEL EXAM
GIVEN INFORMATION:
Potential energies due to the interaction of two particles:
Coulomb potential energy: U = keq1q2/R
Gravitational potential energy: U = Gm1m2/R
1. Motion of a diver. Consider an isolated system S0 consisting of aswimming pool, the air, the earth, and a diver who dives from the highdiving board into the water. The diver strides out along the board,bounces once, dives into the water, swims to the side of the pool, andrests there. At the beginning and end of this motion, all macroscopicparticles in the system S0 can be considered to be at rest.
For the motion described, state whether the final values of the macro-scopic and internal energies of the system S0 are larger than, smallerthan, or the same as the initial values of those energies. (Take themacroscopic potential energy of this system to be the gravitationalpotential energy due to the interaction of macroscopic particles with
the earth.)
2. Energy of a positronium atom. A positron is an atomic particlehaving the same mass me as an electron but a positive charge +e equalin magnitude to the negative charge e of the electron. Suppose thatan electron and a positron interact by the Coulomb electric force, andbriefly form an isolated system called a positronium atom. In thissystem, the electron and positron are separated by a distance D andmove with the same speed v in a common orbit around the systemscenter of mass.
a. Using the symbols provided, write an expression for the energy ofthis system due to all interactions.
b. After a brief interval, the electron and positron in this isolated sys-tem begin to move away from each other with decreasing speed.Does the energy E of this system increase, decrease, or remain con-stant?
56
MISN 0 416 me 2
8/3/2019 m416
29/30
MISN-0-416 me-2
Brief Answers:
1. Macroscopic energy: smallerInternal energy: larger
2. a. E = mev2 kee
2/D
b. remain constant
57 58
8/3/2019 m416
30/30
59 60
