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PHY 113 C General Physics I
11 AM-12:15 PM MWF Olin 101
Plan for Lecture 9:
1. Review (Chapters 1-8)
2. Exam preparation advice –
3. Example problems
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PHY 113 C Fall 2013-- Lecture 9
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PHY 113 C Fall 2013-- Lecture 9
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iclicker question
What is the best way to prepare for Thursday’s exam?
A. Read Lecture Notes and also reread Chapters 18 in Serway and Jewett.
B. Prepare equation sheet.
C. Solve problems from previous exams.
D. Solve homework assignments (both graded and
ungraded) from Webassign Assignments 1-8
E. All the above
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PHY 113 C Fall 2013-- Lecture 9
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iclicker question
Have you (yet) accessed the online class
lecture notes from previous classes?
A. yes
B. no
iclicker question
Have you (yet) accessed the passed Webassign
Assignments (with or without the answer key)?
A. yes
B. no
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PHY 113 C Fall 2013-- Lecture 9
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Access to previous exams
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PHY 113 C Fall 2013-- Lecture 9
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Previous exam access -- continued
Overlap with
2013 schedule
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PHY 113 C Fall 2013-- Lecture 9
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Comments on preparation for next Thursday’s exam –
continued
What you should bring to the exam (in addition to your
well-rested brain):
 A pencil or pen
 Your calculator
 An 8.5”x11” sheet of paper with your favorite
equations (to be turned in together with the exam)
What you should NOT use during the exam
 Electronic devices (cell phone, laptop, etc.)
 Your textbook
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PHY 113 C Fall 2013-- Lecture 9
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F  ma
dv
a
dt
dr
v
dt
Problem solving skills
Math skills
Equation Sheet
Advice:
1. Keep basic concepts and equations at the top of your head.
2. Practice problem solving and math skills
3. Develop an equation sheet that you can consult.
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iclicker exercise
Does the previous slide annoy you?
A. yes
B. no
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Problem solving steps
1. Visualize problem – labeling variables
2. Determine which basic physical principle(s) apply
3. Write down the appropriate equations using the variables
defined in step 1.
4. Check whether you have the correct amount of
information to solve the problem (same number of
knowns and unknowns).
5. Solve the equations.
6. Check whether your answer makes sense (units, order of
magnitude, etc.).
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Likely exam format (example from previous exam)
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Likely exam format (example from previous exam)
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Review of slides from previous lectures
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Mathematics Review -- Appendix B Serwey & Jewett
Quadratic equation :
ax 2  bx  c  0
c
 b  b 2  4ac
x
2a
a
b
Differential calculus :
d n
at  ant n 1
dt
d t
e   e t
dt
d
sin( t )   cos( t )
dt
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q
Integral calculus :



Trigonometry :
b
cos q 
c
a
sin q 
c
a
tan q 
b
n 1
at
at n dt 
n 1
1
et dt  et

sin( t )dt  
PHY 113 C Fall 2013-- Lecture 9
1

cos( t )
14
One dimensional motion -Summary of relationships
dx
v(t ) 
dt
dv
a (t ) 
dt
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
t
x(t )   v(t ' )dt '
t0
t

v(t )   a (t ' )dt '
t0
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Special relationships between t,x,v,a for constant a:
General relationship :
dx
v(t ) 
dt
dv
a (t ) 
dt
t

x(t )   v(t ' )dt '
t0
t

v(t )   a (t ' )dt '
t0
Special relationship :
v(t )  v0   at  v0  at
1 2
1 2
x(t )  x0   v0 t  at  x0  v0t  at
2
2
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Introduction
of vectors
Vector addition:
a+b
b
a
Vector subtraction:
a
a–b
-b
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Treatment of vectors in
component form
by
b  bx xˆ  by yˆ
ab  c
bx
ay
a  ax xˆ  ay yˆ
ax
For a  a x xˆ  a y yˆ and b  bx xˆ  by yˆ
a  b  a x  bx xˆ  a y  by yˆ  c
c  c x2  c y2
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Vectors relevant to motion in two dimenstions
Displacement: r(t) = x(t) i + y(t) j
Velocity: v(t) = vx(t) i + vy(t) j
vx 
Acceleration: a(t) = ax(t) i + ay(t) j
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PHY 113 C Fall 2013-- Lecture 9
dx
dt
dv
ax  x
dt
vy 
dy
dt
ay 
dv y
dt
19
Visualization of the position vector r(t) of a particle
r(t1)
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r(t2)
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Visualization of the velocity vector v(t) of a particle
dr
r (t 2 )  r (t1 )
vt  
 lim
dt t2 t1 0 t 2  t1
v(t)
r(t2)
r(t1)
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Visualization of the acceleration vector a(t) of a particle
dv
v (t 2 )  v (t1 )
at  
 lim
dt t2 t1 0 t 2  t1
v(t2)
v(t1)
r(t2)
r(t1)
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a(t1)
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Projectile motion (near earth’s surface)
j
vertical direction (up)
r (t )  x(t )ˆi  y (t )ˆj
v (t )  v (t )ˆi  v (t )ˆj
x
y
a(t )   gˆj
g = 9.8 m/s2
i
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PHY 113 C Fall 2013-- Lecture 9
horizontal
direction
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Projectile motion (near earth’s surface)
r (t )  x(t )ˆi  y (t )ˆj
dr
v (t ) 
 v x (t )ˆi  v y (t )ˆj
dt
dv
a(t ) 
  gˆj
dt
 vt   v  gtˆj note that vt  0   v
i
i
1 2ˆ
 r t   ri  v i t  gt j note that r t  0   ri
2
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Projectile motion (near earth’s surface)
Trajectory equation in vector form:
r t   ri  v i t  gt ˆj
1
2
2
v (t )  v i  gtˆj
Trajectory equation in component form:
xt   xi  v xi t
v x (t )  v xi
2
1


y t  yi  v yi t  2 gt
v y (t )  v yi  gt
Aside: The equations for position and velocity written
in this way are call “parametric” equations. They are
related to each other through the time parameter.
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Diagram of various trajectories reaching the same
height h=1 m:
y
q
x
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Projectile motion (near earth’s surface)
Trajectory equation in component form:
xt   xi  v xi t  xi  vi cos q i t
y t   yi  v yi t  gt  yi  vi sin q i t  gt
v x (t )  v xi  vi cos q i
1
2
2
1
2
2
v y (t )  v yi  gt  vi sin q i  gt
Trajectory path y(x); eliminating t from the equations:
x  xi
t
vi cos q i


x  xi
1  x  xi 
y  x   yi  vi sin q i
 2 g

vi cos q i
v
cos
q
i 
 i
 x  xi 

y  x   yi  tan q i  x  xi   g 
vi cos q i 

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1
2
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2
Isaac Newton, English physicist and mathematician
(1642—1727)
1. In the absence of a
net force, an object
remains at constant
velocity or at rest.
2. In the presence of a
net force F, the
motion of an object of
mass m is described
by the form F=ma.
3. F12 =– F21.
http://www.newton.ac.uk/newton.html
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Newton’s second law
F=ma
Types of forces:
Fundamental
Approximate
Gravitational
F=-mg j
Empirical
Friction
Electrical
Support
Magnetic
Elastic
Elementary
particles
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Example of two dimensional motion on a frictionless
horizontal surface
F1  F2  ma
F1  F2
a
m
m
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Example – support forces
Fg  mgyˆ
Fsupport  Fapplied
Fsupport acts in direction  to surface
(in direction of surface “normal”)
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Example of forces in equilibrium
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Example: 2-dimensional forces
A car of mass m is on an icy (frictionless)
driveway, inclined at an angle t as shown.
Determine its acceleration.
Conveniently tilted
coordinate system:
Along y : n  mg cos q  0
Along x : mg sin q  ma x
a x  g sin q
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Another example:
Note: we are using a tilted coordinate frame
i
vf=0
vi
mg
Along incline : x(t )  xi  vi t  12 g sin q t 2
v(t )  vi  g sin q t
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Friction forces
The term “friction” is used to describe the category of forces
that oppose motion. One example is surface friction which
acts on two touching solid objects. Another example is air
friction. There are several reasonable models to quantify
these phenomena.
 Fapplied
Surface friction: f  
Normal force between
 N
surfaces
Material-dependent
coefficient
at low speed
 Kv
D
Air friction:
2
at high speed
 K v
K and K’ are materials and
shape dependent constants
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surface friction force
Models of surface friction forces
fs,max=sn
(applied force)
Coefficients s , k depend on the surfaces; usually, s > k
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Consider a stationary block on an incline:
n  mg cos q  0  n  mg cos q
n
f  mg sin q  0  f  mg sin q
If f  f S ,max   S n   S mg cos q
Then  S mg cos q  mg sin q
 S  tan q
q mg cos q
mg
q
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mg sin q
PHY 113 C Fall 2013-- Lecture 9
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Consider a block sliding down an inclined surface;
constant velocity case
n  mg cos q  0  n  mg cos q
n
f  mg sin q  0  f  mg sin q
If f   K n   K mg cos q
Then  K mg cos q  mg sin q
f=kn
 K  tan q
mg
q
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Summary
 K  tan q when block moves at constant velocity
 S  tan q when block is just about to slip
n
q
mg cos q
mg
q
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mg sin q
PHY 113 C Fall 2013-- Lecture 9
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Uniform circular motion and Newton’s second law
r
F  ma
v2
a c   rˆ
r
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F
Definition of work:
dr
ri
rj
rf
Wi f   F  dr
ri
Units of work :
Work  Newtons meters  Joules
1 J  0.239 cal
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Example:
rf
xf
ri
xi
Wi  f   F  dr   Fx dx  (5 N )(4m)  12 5 N 2m   25 J
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Work and potential energy
rf
Definit ionof work :
Wi  f   F  dr
ri
Definition of potential energy :
r
U r   Wref r    F  dr
rref
Note: It is assumed that F is conservative
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Review of energy concepts:
rf
Definition of work :
Wi  f   F  dr
ri
Definition of kinetic energy :
1 2
K  mv
2
Work - kinetic energy theorem :
f
total
i f
W
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1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
PHY 113 C Fall 2013-- Lecture 9
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Summary of work, potential energy, kinetic energy
relationships
Work - kinetic energy theorem :
f
total
i f
1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
total
i f
W
W
W
 U
total
i f
W
conservative
i f
f
W
 U  W
i
dissipative
i f
dissipative
i f
 U r f   U ri   W
dissipative
i f
 K f  Ki
Rearranging : K f  U f  K i  U i  W
dissipative
i f
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Example problem from Webassign #8
A baseball outfielder throws a 0.150-kg baseball at a speed of
37.2 m/s and an initial angle of 31.0°. What is the kinetic energy
of the baseball at the highest point of its trajectory?
vf
yf
vi
qi
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Example problem from Webassign #8
f
h
h
The coefficient of friction between the
block of mass m1 = 3.00 kg and the surface
in the figure below is μk = 0.440. The system
starts from rest. What is the speed of the
ball of mass m2 = 5.00 kg when it has fallen
a distance h = 1.85 m?
dissipative
K f  U f  K i  U i  Wi 
f
dissipative
Wi 
  fh    k m1 gh
f
dissipative
K f  K i  U i  U f   Wi 
f
1
m1  m2 v 2f  0  m2 gh   k m1 gh
2
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Example problem from Webassign #8
A block of mass m = 3.40 kg is
released from rest from point A and
slides on the frictionless track
shown in the figure below. (Let ha =
6.70 m.)
.
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Example problem
from 2012 Exam #2:
conservative
Wi 
f
 U f  U i 
 U x f   U  xi 
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Example problem from Webassign #5
 T1 cos q1  T2 cos q 2  0
T1 sin q1  T2 sin q 2  T3  0
T3  m1 g  0
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