PHY 113 C General Physics I
11 AM – 12:15 PM MWF Olin 101
Plan for Lecture 7:
Chapter 7 -- The notion of work and energy
1. Definition of work
2. Examples of work
3. Kinetic energy; Work-kinetic energy
theorem
4. Potential energy and work; conservative
forces
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PHY 113 C Fall 2013 -- Lecture 7
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7.3,7.15,7.31,7.34
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PHY 113 C Fall 2013 -- Lecture 7
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Webassign questions for Assignment 6 -- #1
52. Consider a large truck carrying a heavy load, such as
steel beams. … Assume that a 10,000-kg load sits on the
flatbed of a 20,000-kg truck initially moving at vi=12 m/s.
Assume that the load on the truck bed has a coefficient of
static friction of mS=0.5. When the truck is braked at
constant force, it comes to rest in a distance d. What is
the minimum stopping distance d such that the load
remains stationary relative to the truck bed throughout the
breaking?
f  ma if
m
a
f  m S n  m S mg
vi
f
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PHY 113 C Fall 2013 -- Lecture 7
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Webassign questions for Assignment 6 -- #1 -- continued
m
a
vi
f
f  m a if
f  m S n  m S mg
 m a  m S mg
or
a  mS g
iclicker exercise -Do we have enough information to calculate a?
A. Yes
B. No
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PHY 113 C Fall 2013 -- Lecture 7
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Webassign questions for Assignment 6 -- #1 -- continued
m
a
vi
f
For constant acceleration a :
vt   vi  at
1 2
xt   xi  vi t  at
2
After some algebra :
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2a xt   xi   vt   v
PHY 113 C Fall 2013 -- Lecture 7
2
2
i
5
Webassign questions for Assignment 6 -- #4
A block of mass 3 kg is pushed
up against a wall by a force P
that makes an angle of q=50o
with the horizontal. ms=0.25.
Determine the possible values
for the magnitude of P that allow
the block to remain stationary.
f
N
Vertical forces :  ms N  P sin q  mg  0
Horizontal forces : P cos q  N  0
f
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mg
PHY 113 C Fall 2013 -- Lecture 7
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f
N
Vertical forces :  ms N  P sin q  mg  0
Horizontal forces : P cos q  N  0
f
mg
N  P cos q
 ms P cos q  P sin q  mg  0
mg
Solving for P : P 
sin q  ms cos q
3  9.8 N
P
sin 50o  0.25 cos 50o
31.72 N

48.57 N
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Webassign questions for Assignment 6 -- #6
F
F  ma
v2
In this case : a   rˆ
r
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Preparation for the introduction of work:
Digression on the definition of vector “dot” product
q
A  B  AB cos q
Note that if q  90 , then A  B  0
o
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PHY 113 C Fall 2013 -- Lecture 7
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Digression: definition of vector “dot” product -- continued
q
A  B  AB cos q
Example :
A  5, B  15, q  120
o
A  B  (5)(15) cos120  37.5 (scalar)
o
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PHY 113 C Fall 2013 -- Lecture 7
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Digression: definition of vector “dot” product –
component form
q
Suppose A  Ax ˆi  Ay ˆj and B  Bx ˆi  B y ˆj
A  B  Ax Bx  Ay B y
Note that the result of a vector dot product is a scalar.
Example : A  2ˆi  4ˆj and B  1ˆi  3ˆj
A  B  2 1   4 3  10
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PHY 113 C Fall 2013 -- Lecture 7
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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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Units of work:
work = force · displacement = (N · m) = (joule)
•Only the component of force in the direction of the
displacement contributes to work.
•Work is a scalar quantity.
•If the force is not constant, the integral form must be used.
•Work can be defined for a specific force or for a combination
of forces
rf
W1   F1  dr
ri
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rf
W2   F2  dr
rf
W1 2   (F1  F2 )  dr  W1  W2
ri
PHY 113 C Fall 2013 -- Lecture 7
ri
13
iclicker question: A ball with a weight of 5 N follows the
trajectory shown. What is the work done by gravity from the
initial ri to final displacement rf?
2.5m
rf
ri
1m
1m
(A) 0 J
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10 m
(B) 7.5 J
(C) 12.5 J
(D) 50 J
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Gravity does
negative work:
rf
Gravity does
positive work:
ri
r  0
r  0
mg
mg
ri
W=mg(rfri)<0
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rf
W=mg(rfri)>0
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rf
Wi  f   F  dr
Work done by a variable force:
ri
In this case :
rf
xf
ri
xi
Wi  f   F  dr   Fx dx
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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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Example – spring force:
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Fx = - kx
PHY 113 C Fall 2013 -- Lecture 7
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F
Positive work
x
Negative work
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Detail:
rf
xf
xf
ri
xi
xi
Wi  f   F  dr   Fx dx 
  kx
1
2
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2 xf
xi
PHY 113 C Fall 2013 -- Lecture 7
  kx dx

 k x x
1
2
2
f
2
i

20
More examples:
Suppose a rope lifts a weight of 1000N by 0.5m at a constant
upward velocity of 4.9m/s. How much work is done by the
rope?
(A) 500 J (B) 750 J
(C) 4900 J (D) None of these
Suppose a rope lifts a weight of 1000N by 0.5m at a constant
upward acceleration of 4.9m/s2. How much work is done by
the rope?
(A) 500 J (B) 750 J
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(C) 4900 J (D) None of these
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Another example
n
q
FP
fk
mg
xi
xf
Assume FP sinq <<mg
Work of gravity?
Work of FP?
Work of fk?
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0
FP cos q (xf-xi)
mkn (xf-xi)=mk(mg- FP sin q (xf-xi)
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iclicker exercise:
Why should we define work?
A. Because professor like to torture
students.
B. Because it is always good to do work
C. Because it will help us understand
motion.
D. Because it will help us solve the energy
crisis.
 Work-Kinetic energy theorem.
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Back to work:
F
dr
ri
rj
rf
Wi  f   F  dr
ri
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Why is work a useful concept?
Consider Newton’s second law:
 Ftotal · dr= m a · dr
Ftotal = m a
rf
rf
rf
rf
rf
dv
dv dr
dv
 Ftotal  dr   ma  dr   m  dr   m  dt   m  vdt
dt
dt dt
dt
ri
ri
ri
ri
ri
Wtotal = ½ m vf2 - ½ m vi2
Kinetic energy (joules)
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Introduction of the notion of Kinetic energy
Some more details:
Consider Newton’s second law:
 Ftotal · dr= m a · dr
Ftotal = m a
rf
rf
rf
tf
tf
dv
dv dr
dv
F

d
r

m
a

d
r

m

d
r

m

dt

m
 vdt
 total




dt
dt dt
dt
ri
ri
ri
ti
ti
tf
vf
f
dv
1
1 2 1 2


 m  vdt   mdv  v   d  mv  v   mv f  mvi
dt
2
 2
ti
vi
i 2
Wtotal = ½ m vf2 - ½ m vi2
Kinetic energy (joules)
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PHY 113 C Fall 2013 -- Lecture 7
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Kinetic energy:
K = ½ m v2
units: (kg) (m/s)2 = (kg m/s2) m
N
m
= joules
Work – kinetic energy relation:
Wtotal = Kf – Ki
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Kinetic Energy-Work theorem
f
Wi  f
1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
iclicker exercise:
Does this remind you of something you’ve
seen recently?
A. Yes
B. No
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Kinetic Energy-Work theorem
f
Wi  f
1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
Note : if Ftotal is constant :
Wi  f   Ftotal  dr  Ftotal  r f  ri   ma  r f  ri 
f
i
 2a  r
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f
r  v
i
2
f
v
2
i
PHY 113 C Fall 2013 -- Lecture 7
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Kinetic Energy-Work theorem
f
Wi  f
1 2 1 2
  Ftotal  dr  mv f  mvi
2
2
i
Example: A ball of mass 10 kg, initially at rest falls a
height of 5m. What is its final velocity?
0
Wi  f
1 2 1 2
 mgh  mv f  mvi
2
2
v f  2 gh  (2)(9.8)(5m)  9.899m / s
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vi  0
PHY 113 C Fall 2013 -- Lecture 7
i
h
v f  ??
f
30
Example
A block, initially at rest at a
height h, slides down a
frictionless incline. What is
its final velocity?
h=0.5m
h
Wi  f
1 2 1 2
 mgh  mv f  mvi
2
2
0
v f  2 gh  (2)(9.8)(0.5m)  3.13m / s
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Example
A block of mass m slides on a horizontal surface with
initial velocity vi, coming to rest in a distance d.
vi
vf=0
d
1. Determine the work done during this process.
2. Analyze the work in terms of the kinetic friction
force.
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Example -- continued
vi
vf=0
f
d
Wi  f
1 2 1 2
1 2
 mv f  mvi   mvi
2
2
2
rf
xf
ri
xi
Wi  f   F  dr    fdx   fd   m k mgd
1 2
 mvi  m k mgd
2
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PHY 113 C Fall 2013 -- Lecture 7
or
vi2
mk 
2 gd
33
Example
k
A mass m initially at rest and
attached to a spring compressed a
distance x=-|xi|, slides on a
frictionless surface. What is the
velocity of the mass when x=0 ?
1 2 1 2 1 2 0
Wi  f  kxi  mv f  mvi
2
2
2
k
vf 
xi
m
For m  0.5kg k  5 N / m xi  0.2m
5
0.2  0.63m / s
vf 
0.5
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Special case of “conservative” forces
conservative  non-dissipative
For non - dissipative forces F, it is possible to write
Wi  f   F  dr  U r f   U ri 
f
i
Example of gravity near surface of Earth :
Wi  f   F  dr  mg  y f  yi   mgy f  mgyi 
f
i
 U (r f )  mgy f
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and
U (ri )  mgyi
PHY 113 C Fall 2013 -- Lecture 7
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k
Example of spring force :
f
xf
i
xi

Wi  f   F  dr    k x dx   12 kx 2f  12 kxi2
 U (r f )  kx
1
2
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2
f
and

U (ri )  kx
PHY 113 C Fall 2013 -- Lecture 7
1
2
2
i
36
iclicker exercise:
Why would you want to write the work as the
difference between two “potential” energies?
A. Normal people wouldn’t.
B. It shows a lack of imagination.
C. It shows that the work depends only on the
initial and final displacements, not on the
details of the path.
Define potential energy function :
r
U r     F  dr
rref
Note that
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dU
Fx  
dx
PHY 113 C Fall 2013 -- Lecture 7
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Work-Kinetic Energy Theorem for
conservative forces:
Wi  f
1 2 1 2
 U r f   U ri   mv f  mvi
2
2
Or :
1 2
1 2
mv f  U r f   mvi  U ri   E (constant)
2
2
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Energy diagrams
1 2 1 2
E  mv  kx
2
2
Note : when x  xmax :
v  0, E 
1
2
2
kxmax
Note : when x  0 :
U (0)  0,
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PHY 113 C Fall 2013 -- Lecture 7
1
2
mv 2 
1
2
2
kxmax
39
Example: Model potential energy function U(x)
representing the attraction of two atoms
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