Transcript Document

Physics 115 2013
Final Review
Calculus is about “rates of change”.
A TIME RATE is anything divided by time.
CHANGE is expressed by using the Greek letter,
Delta, D.
For example: Average SPEED is simply the “RATE at
which DISTANCE changes”.
The MEANING?
d (kt 3 )
2
 3kt
dt
For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8
meters.
The derivative, however, tell us how our DISPLACEMENT (x)
changes as a function of TIME (t). The rate at which
Displacement changes is also called VELOCITY. Thus if we
use our derivative we can find out how fast the object is
traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12
m/s
Derivative of a power function
Unit Vector Notation
iˆ - unit vector  1 in the x direction
ˆj - unit vector  1 in the y direction
kˆ - unit vector  1 in the z direction
The proper terminology is to use the
“hat” instead of the arrow. So we
have i-hat, j-hat, and k-hat which are
used to describe any type of motion
in 3D space.
How would you write vectors J and K in unit
vector notation?
J  2iˆ  4 ˆj
K  2iˆ  5 ˆj
Example
A boat moves with a velocity of 15 m/s, N in a river which flows with a
velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with
respect to due north.
Rv  82  152  17 m / s
8.0 m/s, W
15 m/s, N
Rv

8
Tan   0.5333
15
  Tan1 (0.5333)  28.1
17 m / s @ 28.1 WofN
The Final Answer :
17 m / s  118.1
 8iˆ m / s  15 ˆj m / s

Dot Products in Physics
Consider this situation: A force F is applied to a
moving object as it transverses over a
frictionless surface for a displacement, d.
As F is applied to the object it will increase the
object's speed!
But which part of F really
causes the object to increase
in speed?
It is |F|Cos θ ! Because it is parallel to the displacement d
In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be
defined as "WORK" (check your equation sheet!)
A  B  A B cos
W  F  x  F x cos
Work is a type of energy and energy DOES NOT
have a direction, that is why WORK is a scalar or in
this case a SCALAR PRODUCT
(AKA DOT PRODUCT).
Example
Suppose a person moves in a straight line from the lockers( at a position x = 1.0 m)
toward the physics lab(at a position x = 9.0 m). “To the right” is taken as positive,
as shown below
The answer is positive so the person must have been traveling horizontally in the
positive direction.
Example
Suppose the person turns around!
The answer is negative so the person must have been traveling horizontally in the
negative direction
What is the DISPLACEMENT for the entire trip?
Dx  x final  xinitial  1.0 1.0  0m
What is the total DISTANCE for the entire trip?
8  8  16 m
Instantaneous Velocity
Instantaneous velocity is a
measure of an object’s
displacement per unit time at a
particular point in time.
Example: A body’s position is defined as:
4ˆ
ˆ
d (7t i  j )
dx
t
v

dt
dt
v  lim Dt 0
Dx
Dt
dx
v
dt
x(t )  7t 3iˆ 
4ˆ
j , v(t )  ?
t
3
4 ˆ
ˆ
v(t )  [21t i  2 j ] m / s
t
2
Instantaneous Acceleration
Instantaneous velocity is a measure
of an object’s velocity per unit
time at a particular point in time.
If the velocity of an object is defined as:
v(t )  [21t 2iˆ 
4 ˆ
j ] m / s, a(t )  ?
2
t
4 ˆ
ˆ
d (21t i  2 j )
dv
t
a


dt
dt
2
a  lim Dt 0
dv
a
dt
8
[ 42tiˆ  3 ˆj ] m / s / s
t
Dv
Dt
What do the “signs”( + or -) mean?
Quantity
Positive
Negative
Displacement
Your position has
changed toward the
positive.
Your position has
changed toward the
negative.
Velocity
You are traveling in
the +x or +y
direction.
You are traveling in
the –x or –y
direction.
Acceleration
If moving in the
positive direction,
you are speeding
up. If moving in the
negative direction,
you are slowing
down.
If moving in the
positive direction,
you are slowing
down. . If moving in
the negative
direction, you are
speeding up.
The 3 Kinematic equations
There are 3 major kinematic
v

v

at
o
equations than can be
used to describe the
2
1
motion in DETAIL. All are x  xo  vo t  2 at
used when the
2
2
acceleration is CONSTANT. v  v  2a( x  x )
o
o
Kinematics for the VERTICAL Direction
All 3 kinematics can be used to analyze one dimensional
motion in either the X direction OR the y direction.
v  vo  at  v y  voy  gt
2
2
1
1
x  xo  vox t 
at  y  yo  voy t 
gt
2
2
2
2
2
2
v  vox  2a ( x  xo )  v y  voy  2 g ( y  yo )
Examples
A stone is dropped at rest from the top of a cliff. It is
observed to hit the ground 5.78 s later. How high is the
cliff?
What do I
know?
v = 0 m/s
oy
g = -9.8 m/s2
yo=0 m
t = 5.78 s
What do I
want?
y=?
Which variable is NOT given and
NOT asked for?
Final Velocity!
y  yo  voy t  1 gt 2
2
y  (0)(5.78)  4.9(5.78) 2
y  -163.7 m
H =163.7m
Examples
A pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that
during the windup and delivery the ball covers a displacement of 2.5
meters. This is from the point behind the body when the ball is at rest to
the point of release. Calculate the acceleration during his throwing
motion.
What do I
know?
vo= 0 m/s
x = 2.5 m
v = 43.5 m/s
What do I
want?
a=?
Which variable is NOT given and
NOT asked for?
TIME
v  v  2a( x  xo )
2
2
o
43.5  0  2a(2.5  0)
2
a
2
378.5 m/s/s
Examples
How long does it take a car at rest to cross a 35.0 m intersection after
the light turns green, if the acceleration of the car is a constant 2.00
m/s/s?
What do I
know?
vo= 0 m/s
x = 35 m
a = 2.00 m/s/s
What do I
want?
Which variable is NOT given and
NOT asked for?
Final Velocity
t=?
x  xo  vox t  1 at 2
2
35  0  (0)  1 (2)t 2
2
t  5.92 s
Examples
A car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds.
What was the acceleration?
What do I
know?
vo= 12.5 m/s
v = 25 m/s
t = 6s
What do I
want?
Which variable is NOT given and
NOT asked for?
DISPLACEMENT
a=?
v  vo  at
25  12.5  a(6)
a
2.08 m/s/s
Summary
There are 3 types of MOTION graphs
• Displacement(position) vs. Time
• Velocity vs. Time
• Acceleration vs. Time
There are 2 basic graph models
• Slope
• Area
Summary
v (m/s)
x (m)
a (m/s/s)
area = x
t (s)
t (s)
area = v
t (s)
Comparing and Sketching graphs
One of the more difficult applications of graphs in physics is when given a certain
type of graph and asked to draw a different type of graph
List 2 adjectives to describe the SLOPE or VELOCITY
1. The slope is CONSTANT
2. The slope is POSITIVE
x (m)
t (s)
How could you translate what the
SLOPE is doing on the graph
ABOVE to the Y axis on the graph
to the right?
v (m/s)
t (s)
Example
v (m/s)
x (m)
t (s)
t (s)
1st line
• The slope is constant
• The slope is “-”
2nd line
• The slope is “0”
3rd line
• The slope is “+”
• The slope is constant
Example – Graph Matching
What is the SLOPE(a) doing?
a (m/s/s)
The slope is increasing
v (m/s)
t (s)
a (m/s/s)
t (s)
t (s)
a (m/s/s)
t (s)
A pictorial representation of forces complete
with labels.
FN
Ff
T
T
W1,Fg1
or m1g
m2g
•Weight(mg) – Always
drawn from the center,
straight down
•Force Normal(FN) – A
surface force always drawn
perpendicular to a surface.
•Tension(T or FT) – force in
ropes and always drawn
AWAY from object.
•Friction(Ff)- Always drawn
opposing the motion.
Ff
FN
mg
Since the Fnet = 0, a system moving at a constant
speed or at rest MUST be at EQUILIBRIUM.
TIPS for solving problems
• Draw a FBD
• Resolve anything into COMPONENTS
• Write equations of equilibrium
• Solve for unknowns
10-kg box is being pulled across the table to the right at a
constant speed with a force of 50N at an angle of 30
degrees above the horizontal.
a) Calculate the Force of Friction
Fax  Fa cos  50cos30  43.3N
a) Calculate the Normal Force
Ff  Fax  43.3N
FN
Ff
Fa
Fay
30
Fax
mg
FN  m g!
FN  Fay  m g
FN  m g  Fay  (10)(9.8)  50 sin 30
FN  73N
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.
Fs  x
k  Constantof Proportion
ality
k  Spring Constant(Unit : N/m)
Fs  kx or  kx
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.
Hooke’s Law from a Graphical Point of View
Fs  kx
Suppose we had the following data:
x(m)
Force(N)
0
0
0.1
12
0.2
24
0.3
36
0.4
48
Fs
x
k  Slope of a F vs. x graph
k
Force vs. Displacement
y = 120x + 1E-14
R2 = 1
80
70
0.5
60
0.6
72
Force(Newtons)
60
50
k =120 N/m
40
30
20
10
0
0
0.1
0.2
0.3
0.4
Displacement(Meters)
0.5
0.6
0.7
Example
A load of 50 N attached to a spring hanging vertically stretches the spring
5.0 cm. The spring is now placed horizontally on a table and stretched
11.0 cm. What force is required to stretch the spring this amount?
Fs  kx
50  k (0.05)
k
1000 N/m
Fs  kx
Fs  (1000)(0.11)
Fs 
110 N
The acceleration of an object is directly
proportional to the NET FORCE and inversely
proportional to the mass.
a  FNET
a
1
a
m
FNET
 FNET  m a
m
FNET   F
Tips:
•Draw an FBD
•Resolve vectors into components
•Write equations of motion by adding and
subtracting vectors to find the NET FORCE.
Always write larger force – smaller force.
•Solve for any unknowns
A 10-kg box is being pulled across the table to the
right by a rope with an applied force of 50N.
Calculate the acceleration of the box if a 12 N
frictional force acts upon it.
FN
Ff
mg
Fa
In which direction,
is this object
accelerating?
The X direction!
So N.S.L. is worked
out using the forces
in the “x” direction
only
FNet  m a
Fa  F f  m a
50  12  10a
a  3 .8 m / s 2
A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected
to a cable that passes over a pulley and then is fastened to a hanging mass,
m2 = 11.0 kg as shown below. Find the acceleration of each mass and the
tension in the cable.
FN
FNet  m a
m2 g  T  m2 a
T
T
m1g
T  m1a
m2 g  m1a  m2 a
m2 g  m2 a  m1a
m2 g  a(m2  m1 )
m2g
a
m2 g
(11)(9.8)

 7.7 m / s 2
m1  m2
14
FNet  m a
m2 g  T  m2 a
T  m1a
T  (3)(7.7)  23.1 N
Where does the calculus fit in?


dv
d 2x
F  ma  m  m
dt
dt
There could be situations where you are given
a displacement function or velocity function.
The derivative will need to be taken once or
twice in order to get the acceleration. Here is
an example.
You are standing on a bathroom scale in an elevator in a tall
building. Your mass is 72-kg. The elevator starts from rest and
travels upward with a speed that varies with time according to:
v(t )  3t  0.20t 2
When t = 4.0s , what is the reading on the bathroom scale (a.k.a.
Force Normal)?
Fnet  m a
dv d (3t  0.20t )
a

 3  0.40t FN  m g  m a  FN  m a  m g
dt
dt
FN  (72)(9.8)  (72)(4.6) 
a(4)  3  0.40(4)  4.6 m/s/s
1036.8 N
2
TWO types of Friction
• Static – Friction that keeps an object at rest and
prevents it from moving (no sliding)
• Kinetic – Friction that acts during motion (two surfaces
sliding)
Force of Friction
• The Force of Friction is
directly related to the
Normal Force.
F f  FN
  constantof proportionality
  coefficient of friction
Fsf   s FN
The coefficient of
Fkf   k FN
friction is a unitless
constant that is
specific to the
material type and
usually less than one.
Example
A 1500 N crate is being pushed
across a level floor at a constant
speed by a force F of 600 N at an
angle of 20° below the horizontal
as shown in the figure.
Fa
a) What is the coefficient of kinetic
friction between the crate and the
floor?
F f   k FN
FN
Fay
F f  Fax  Fa cos  600(cos20)  563.82N
FN  Fay  m g  Fa sin   1500
20
FN  600(sin 20)  1500 1705.21N
Fax
563.82   k 1705.21
 k  0.331
Ff
mg
Example
If the 600 N force is instead pulling the block
FN
at an angle of 20° above the horizontal
as shown in the figure, what will be the
acceleration of the crate. Assume that
the coefficient of friction is the same as
found in (a)
FNet  m a
20
Fax
Ff
Fax  F f  m a
Fa cos  FN  m a
Fa cos   (m g  Fa sin  )  m a
600cos 20  0.331(1500 600sin 20)  153.1a
563.8  428.57  153.1a
a  0.883 m / s 2
Fa
mg
Fay
Inclines

Ff
FN
m g cos


mg
m g sin 



Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or
equilibrium
•Solve
Example
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a
frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed
incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the
incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic
friction and (c) the tension in the string.
T
FN
Ff
m2gcos40
m2g
m1
40
m2gsin40
m1g
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( Ff  T )  m2a
40
T
FNET  m a
Example
FNET  m a
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( Ff  T )  m2a
x  vox t  1 at 2
2
1  0  1 a ( 4) 2
2
a  0.125 m / s 2
T  4(.125)  4(9.8)  39.7 N
m2 g sin   F f  T  m2 a
m2 g sin   F f  (m1a  m1 g )  m2 a
m2 g sin    k FN  m1a  m1 g  m2 a
m2 g sin    k m2 g cos  m1a  m1 g  m2 a
m2 g sin   m1a  m1 g  m2 a   k m2 g cos
k 
m2 g sin   m1a  m1 g  m2 a
m2 g cos
k 
56.7  0.5  39.2  1.125
 0.235
67.57
Horizontally Launched Projectiles
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for the
“y” direction. And for this we use kinematic #2.
2
1
x  vox t 
at
2
x  voxt
Remember, the velocity is
CONSTANT horizontally, so
that means the acceleration is
ZERO!
y  1 gt 2
2
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is equal to
ZERO.
Horizontally Launched Projectiles
Example: A plane traveling with a
horizontal velocity of 100 m/s is
500 m above the ground. At
some point the pilot decides to
drop some supplies to
designated target below. (a) How
long is the drop in the air? (b)
How far away from point where
it was launched will it land?
What do I
know?
What I want to
know?
vox=100 m/s
t=?
x=?
y = 500 m
voy= 0 m/s
g = -9.8 m/s/s
y  1 gt 2  500  1 (9.8)t 2
2
2
102.04  t 2  t  10.1 seconds
x  voxt  (100)(10.1) 
1010 m
Vertically Launched Projectiles
There are several
things you must
consider when doing
these types of
projectiles besides
using components. If
it begins and ends at
ground level, the “y”
displacement is
ZERO: y = 0
Vertically Launched Projectiles
You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
vo

vox
voy
x  voxt
y  voy t  1 gt 2
2
vox  vo cos 
voy  vo sin 
Example
A place kicker kicks a football with a velocity of 20.0 m/s and at
an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
vox  vo cos 
vox  20 cos 53  12.04 m / s
  53
voy  vo sin 
voy  20sin 53  15.97 m / s
Example
A place kicker kicks a
football with a velocity of
20.0 m/s and at an angle
of 53 degrees.
(a) How long is the ball in
the air?
What I know
vox=12.04 m/s
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
y  voy t  1 gt 2  0  (15.97)t  4.9t 2
2
2
15.97t  4.9t  15.97  4.9t
t  3.26 s
What I want
to know
t=?
x=?
ymax=?
Example
A place kicker kicks a
football with a velocity of
20.0 m/s and at an angle
of 53 degrees.
(b) How far away does it
land?
What I know
vox=12.04 m/s
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
x  voxt  (12.04)(3.26) 
What I want
to know
t = 3.26 s
x=?
ymax=?
39.24 m
Example
A place kicker kicks a football
with a velocity of 20.0 m/s
and at an angle of 53
degrees.
(c) How high does it travel?
CUT YOUR TIME IN HALF!
What I know
vox=12.04 m/s
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
What I want
to know
t = 3.26 s
x = 39.24 m
ymax=?
y  voy t  1 gt 2
2
y  (15.97)(1.63)  4.9(1.63) 2
y  13.01 m
A special case…
What if the projectile was launched from the ground at an angle and did
not land at the same level height from where it started? In other
words, what if you have a situation where the “y-displacement”
DOES NOT equal zero?
Let's look at the second kinematic closely!
Assuming it is shot from the ground. We see we have one squared term
variable, one regular term variable, and a constant number with no
variable. What is this?
A QUADRATIC EQUATION!
Circular Motion and New’s 2nd Law
2
Recall that according to
Newton’s Second Law, the
acceleration is directly
proportional to the Force.
If this is true:
v
FNET  m a ac 
r
2
mv
FNET  Fc 
r
Fc  Centripetal Force
Since the acceleration and the force are directly
related, the force must ALSO point towards the center.
This is called CENTRIPETAL FORCE.
NOTE: The centripetal force is a NET FORCE. It could be
represented by one or more forces. So NEVER draw it in an
F.B.D.
Examples
Top view
What is the minimum coefficient of static friction
necessary to allow a penny to rotate along a 33 1/3 rpm
record (diameter= 0.300 m), when
the penny is placed at the outer edge of the record?
F f  Fc
FN
mg
Side view
Ff
m v2
FN 
r
m v2
m g 
r
v2

rg
rev 1 min
33.3
*
 0.555rev
sec
min 60 sec
1sec
 1.80 sec
T
rev
0.555 rev
2r 2 (0.15)
vc 

 0.524 m / s
T
1.80
v2
(0.524) 2


 0.187
rg (0.15)(9.8)
Examples
The maximum tension that a 0.50 m
string can tolerate is 14 N. A 0.25-kg
ball attached to this string is being
whirled in a vertical circle. What is
the maximum speed the ball can
have (a) the top of the circle, (b)at
the bottom of the circle?
m v2
FNET  Fc  m ac 
r
m v2
T  mg
 r (T  m g)  m v2
r
r (T  m g)
0.5(14  (0.25)(9.8))
v

m
0.25
v  5.74 m / s
T
mg
Examples
At the bottom?
T
mg
m v2
FNET  Fc  m ac 
r
m v2
T  mg 
 r (T  m g)  m v2
r
r (T  m g)
0.5(14  (0.25)(9.8))
v

m
0.25
v  4.81 m / s
N.L.o.G – Putting it all together
m1m2
r2
G  constantof proportion
ality
G  UniversalGravitational Constant
Fg 
 27
G  6.67x10
Fg  G
Nm 2
m1m2
r2
Fg  m g  Use this when you are on theearth
Fg  G
m1m2
 Use this when you are LEAVING the earth
r2
kg 2
Kepler’s 3rd Law – The Law of Periods
"The square of the period of any planet is proportional to
the cube of the semi major axis of its orbit."
Gravitational forces are centripetal, thus we
can set them equal to each other!
Since we are moving in a circle we can
substitute the appropriate velocity formula!
The expression in the RED circle derived by setting the
centripetal force equal to the gravitational force is called
ORBITAL SPEED.
Using algebra, you can see that everything in
the parenthesis is CONSTANT. Thus the
proportionality holds true!
Example
A 2-kg sliding puck whose initial velocity magnitude is v1 = 10
m/s strikes a wall at a 30 degree angle and bounces off. If it
leaves the wall with a velocity magnitude of v2 = 10 m/s, and
if the collision takes a total of 0.02 seconds to complete,
what was the average force applied to the puck by the
wall?
There is something you need to consider:
Momentum is a VECTOR!!!
Let’s look at this problem using a X-Y axis for reference
Example cont’
If we did the same thing for the Y direction we would discover
that the Force Net is equal to ZERO!
The temptation is to treat momentum as a SCALAR...DO
NOT DO THIS! SIGNS COUNT!
Momentum is conserved!
The Law of Conservation of Momentum: “In the absence
of an unbalanced external force, the total momentum
before the collision is equal to the total momentum
after the collision.”
po (truck )  m vo  (500)(5)  2500kg * m / s
po ( car )  (400)(2)  800kg * m / s
po (total )  3300kg * m / s
ptruck  500* 3  1500kg * m / s
pcar  400* 4.5  1800kg * m / s
ptotal  3300kg * m / s
Several Types of collisions
Sometimes objects stick together or blow apart. In this
case, momentum is ALWAYS conserved.
p
before
  p after
m1v01  m2 v02  m1v1  m2 v2
When 2 objects collide and DON’T stick
m1v01  m2 v02  mtotal vtotal
When 2 objects collide and stick together
mtotal vo (total )  m1v1  m2 v2
When 1 object breaks into 2 objects
Elastic Collision = Kinetic Energy is Conserved
Inelastic Collision = Kinetic Energy is NOT Conserved
Work
The VERTICAL component of the force DOES NOT
cause the block to move the right. The energy imparted to
the box is evident by its motion to the right. Therefore
ONLY the HORIZONTAL COMPONENT of the force
actually does WORK.
When the FORCE and DISPLACEMENT are in the SAME
DIRECTION you get a POSITIVE WORK VALUE. The
ANGLE between the force and displacement is ZERO
degrees. What happens when you put this in for the
COSINE?
When the FORCE and DISPLACEMENT are in the
OPPOSITE direction, yet still on the same axis, you get a
NEGATIVE WORK VALUE. This negative doesn't mean
the direction!!!! IT simply means that the force and
displacement oppose each other. The ANGLE between the
force and displacement in this case is 180 degrees. What
happens when you put this in for the COSINE?
When the FORCE and DISPLACEMENT are
PERPENDICULAR, you get NO WORK!!! The ANGLE
between the force and displacement in this case is 90
degrees. What happens when you put this in for the
COSINE?
Example
W  F r
W  F r cos
A box of mass m = 2.0 kg is moving over a
frictional floor ( uk = 0.3) has a force whose
magnitude is F = 25 N applied to it at an angle of
30 degrees, as shown to the left. The box is
observed to move 16 meters in the horizontal
direction before falling off the table.
W  F r
W  F r cos
W  25 16 cos30
W  346.4 Nm
W  346.4 J
a) How much work does F do before taking the
plunge?
Example cont’
What if we had done this in UNIT VECTOR notation?
F  21.65iˆ  12.5 ˆj
W  ( Fx  rx )  ( Fy  ry )
W  (21.65  16)  (12.5  0)
W  346.4 Nm
W  346.4 J
much work does the FORCE
Example cont’ How
NORMAL do and Why?
Fn
W  F r
W  F r cos
There is NO WORK since “F”
and “r” are perpendicular.
W  FN 16 cos90
Ff
W 0J
How much does the internal energy of the system increase?
34.08 J
Elastic Potential Energy
The graph of F vs.x for a
spring that is IDEAL in
nature will always
produce a line with a
positive linear slope.
Thus the area under the
line will always be
represented as a
triangle.
NOTE: Keep in mind that this can be applied to WORK or can be conserved with any
other type of energy.
Elastic potential energy
W   F ( x)dx   (kx)dx
x
x
x 0
x 0
W   (kx)dx  k  xdx
x2 x
W  k | | x 0  W  U spring  1 kx2
2
2
Elastic “potential” energy is a fitting term as springs STORE energy when there
are elongated or compressed.
Energy is CONSERVED!
W  DK   D U
K  K o  (U  U o )
K  K o  U  U o
Ko  U o  K  U
Energybefore  Energyafter
Example
A 2.0 m pendulum is released from rest when the support
string is at an angle of 25 degrees with the vertical. What
is the speed of the bob at the bottom of the string?

Lcos
L
h
EB = EA
UO
= K
mgho
= 1/2mv2
gho
= 1/2v2
1.83
= v2
1.35 m/s = v
h = L – Lcos
h = 2-2cos
h = 0.187 m
Ballistic Pendulum
• Event includes a collision during which
momentum conservation is the best model
and post-collision motion during which energy
conservation is the best model.
• Examples: last three problems on HW,
Workshop 14,