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”.

d ( kt
3
)

3 kt
2 dt
For example, if t = 2 seconds, using x(t) = kt 3 =(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 = 3kt 2 =3(1)(2) 2 = 12 m/s

i
ˆ k
ˆ j
ˆ
unit vecto
unit vecto
unit vecto
r

1 in the
 x direction The proper terminology is to use the
“hat” instead of the arrow. So we r

1 in the
 y direction have i-hat , j-hat , and k-hat which are used to describe any type of motion r

1 in the
 z direction in 3D space.
How would you write vectors J and K in unit vector notation?
J

2 i
ˆ 
4
ˆ j
K

2 i
ˆ 
5
ˆ j

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.
R v

8 2 
15 2 
17 m / s
8.0 m/s, W
15 m/s, N
Tan
 
 
8

0 .
5333
15
Tan

1
( 0 .
5333 )

28 .
1

R v

The Final Answer :
17 m / s @ 28 .
1

WofN
17 m / s

118 .
1


8 i
ˆ m / s

15
ˆ j m / s

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).

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.

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?
D x
 x final
 x initial

1 .
0

1 .
0

0 m
What is the total DISTANCE for the entire trip?
8

8

16 m

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: v
 dx dt
 d ( 7 t
3 i
ˆ 
4 t dt
ˆ j ) v
 lim v
 dx dt x ( t )

7 t
3 i
ˆ 
4 t
D t

0
D
D x t j
ˆ
, v ( t )

?
v ( t )

[ 21 t
2 i
ˆ 
4 t
2
ˆ j ] m / s

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 )

[ 21 t
2 i
ˆ 
4 t
2 j
ˆ
] m / s , a ( t )

?
a
 lim
D t

0
D v
D t a
 dv dt a
 dv dt
 d ( 21 t
2 i
ˆ 
4 t
2 dt
ˆ j )
 [ 42 t i
ˆ 
8 t
3
ˆ j ] m / s / s

Quantity
Displacement
Velocity
Acceleration
Positive
Your position has changed toward the positive.
You are traveling in the +x or +y direction.
If moving in the positive direction, you are speeding up. If moving in the negative direction, you are slowing down.
Negative
Your position has changed toward the negative.
You are traveling in the –x or –y direction.
If moving in the positive direction, you are slowing down. . If moving in the negative direction, you are speeding up.

There are 3 major kinematic equations than can be used to describe the motion in DETAIL.
All are used when the acceleration is CONSTANT.
v x v
2


 v o x o v
2 o


 at v o t
2 a

( x
1

2 x at
2 o
)

All 3 kinematics can be used to analyze one dimensional motion in either the X direction OR the y direction.
v
 v o
 at
 v y
 v oy
 gt x v
2

 x o
2 v ox
 v ox t


2 a ( x
1
2 at
2
 x o
)

 y v
2 y

 y o
2 v oy

 v oy t

2 g ( y
1
2
 gt y o
)
2

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 oy
= 0 m/s g = -9.8 m/s 2 y o
=0 m t = 5.78 s
What do I want?
y = ?
Which variable is NOT given and
NOT asked for?
Final Velocity!
y
 y o
 v oy t
 1
2 gt
2 y

( 0 )( 5 .
78 )

4 .
9 ( 5 .
78 )
2 y

-163.7 m
H =163.7m

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?
What do I want?
Which variable is NOT given and
NOT asked for?
TIME v o
= 0 m/s x = 2.5 m v = 43.5 m/s a = ?
v
2  v o
2 
2 a ( x
 x o
)
43 .
5
2 a


0
2
378.5 m/s/s

2 a ( 2 .
5

0 )

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?
v o
= 0 m/s x = 35 m a = 2.00 m/s/s
What do I want?
t = ?
Which variable is NOT given and
NOT asked for?
Final Velocity x
 x o
 v ox t
 1
2 at
2
35

0

( 0 )
 1
2
( 2 ) t
2 t

5.92 s

A car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds.
What was the acceleration?
What do I know?
v o
= 12.5 m/s v = 25 m/s t = 6s
What do I want?
a = ?
Which variable is NOT given and
NOT asked for?
DISPLACEMENT v
 v o
 at
25

12 .
5
 a ( 6 ) a

2.08 m/s/s

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

x (m) t (s) v (m/s)
a (m/s/s) area = x t (s) area = v t (s)

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 x (m)
List 2 adjectives to describe the SLOPE or VELOCITY
1.
2.
The slope is CONSTANT
The slope is POSITIVE 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)

v (m/s) x (m) t (s)
•
•
1 st line
The slope is constant
The slope is “-”
•
•
3 rd line
The slope is “+”
The slope is constant
•
2 nd line
The slope is “0” t (s)

What is the SLOPE(a) doing?
The slope is increasing a (m/s/s) v (m/s) a (m/s/s) t (s) t (s) a (m/s/s) t (s) t (s)

A pictorial representation of forces complete with labels.
F
N
F f
W
1
,Fg
1 or m
1 g
T
T m
2 g
•
• Weight(mg) – Always drawn from the center, straight down
Force Normal(F
N
) – A surface force always drawn perpendicular to a surface.
• Tension(T or F
T
) – force in ropes and always drawn
AWAY from object.
• Friction(Ff)Always drawn opposing the motion.

F f mg
F
N

Since the F net
= 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 a) Calculate the Normal Force F ax
F f


F a
F ax cos



50 cos 30

43 .
3 N
43 .
3 N
F f
F
N mg
30
F a
F ax
F ay
F
N
F
N
F
N
F
N
 mg !

F ay
 mg
 mg

F ay

73 N

( 10 )( 9 .
8 )

50 sin 30

One of the simplest type of simple harmonic motion is called Hooke's Law.
This is primarily in reference to SPRINGS.
F s k

 x
Constant of Proportion ality k
F s

Spring Constant(U nit
 kx or
 kx
: N/m)
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
Suppose we had the following data: x(m) Force(N)
0 0
0.1
0.2
0.3
0.4
12
24
36
48
0.5
0.6
60
72 40
30
20
10
0
0
80
70
60
50
0.1
F s
 kx k k

F s x

Slope of a F vs.
x graph
Force vs. Displacement y = 120x + 1E-14
R
2
= 1
0.6
0.7
k =120 N/m
0.2
0.3
0.4
Displacement(Meters)
0.5

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?
F s

50
 kx k ( 0 .
05 ) k

1000 N/m
F s
F s
F s
 kx

( 1000 )( 0 .
11 )

110 N

The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass.
a

F
NET a

1 m
F
NET
 
F a

F
NET m

F
NET
 ma
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.
F f
F
N mg
F a
In which direction, is this object accelerating?
The X direction!
F
Net
F a


F f ma
 ma
50

12

10 a
So N.S.L. is worked out using the forces in the “x” direction only a

3 .
8 m / s
2

A mass, m
1
= 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, m
2
= 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable.
F
N
T
F m
Net
2 g

 ma
T
 m
2 a

T
T m
1 a m
1 g m
2 g m
2 g m
2 g


 m
1 a
 m
2 a m
2 a a ( m
2

 m
1 a m
1
) m
2 g a
 m
1 m
2
 g m
2

( 11 )( 9 .
8 )
14

7 .
7 m / s
2

F
Net
 ma m
2 g
T


T m
1 a
 m
2 a
T

( 3 )( 7 .
7 )

23 .
1 N


F

There could be situations where you are given
 m a
 m dv dt
 m d
2 x dt 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 )

3 t

0 .
20 t
2
When t = 4.0s , what is the reading on the bathroom scale (a.k.a.
Force Normal)?
F net
 ma a
 dv a ( 4 ) dt


3
 d ( 3 t

0 dt
0 .
40 ( 4 )
.
20
 t
2
)

3
4.6 m/s/s

0 .
40 t F
N
F
N
 mg

 ma

( 72 )( 9 .
8 )

1036.8 N
F
N
 ma
( 72 )( 4 .
6 )
 mg


• Static – Friction that keeps an object at rest and prevents it from moving (no sliding)
• Kinetic – Friction that acts during motion (two surfaces sliding)

• The Force of Friction is directly related to the
Normal Force.
F f
 

F
N constant of proportion ality
  coefficien t of friction
F sf
F kf


 s
F
N
 k
F
N
The coefficient of friction is a unitless constant that is specific to the material type and usually less than one.

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.
a) What is the coefficient of kinetic friction between the crate and the floor?
F ay
F ax
F f
F a
20
F
N
F f
F f
  k
F
N


F ax
F ay

F a cos

 mg


600 (cos 20 )
F a sin
 
1500

563 .
82 N
F
N
F
N

600 (sin 20 )

1500
563 .
82
 k
  k
1705 .
21

0 .
331

1705 .
21 N mg

If the 600 N force is instead pulling the block 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)
F
Net
F ax

 ma
F f
 ma
F a
F a cos cos





F
N

( mg

 ma
F a sin
600 cos 20

0 .
331 ( 1500


)
 ma
600 sin
F f
20 )

153 .
1 a
563 .
8

428 .
57

153 .
1 a a

0 .
883 m / s
2 mg
F
N
F a
20
F ax
F ay

F f mg cos

 mg
 mg sin

F
N




Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or equilibrium
•Solve

Masses m
1
= 4.00 kg and m
2
= 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m
1 is held at rest on the floor and m
2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m
2 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.
m
2 gcos40
T
T
F f
40
F
N
F
NET
T

 m
1 g ma
 m
1 a

T
 m
1 a
 m
1 g m
2 g sin
 
( F f

T )
 m
2 a m
1 m
2 g
40 m
2 gsin40 m
1 g

F
NET
T

 m
1 g ma
 m
1 a m
2 g sin
 

T
( F f


T ) m
1 a

 m
1 g m
2 a
T

4 (.
125 )

4 ( 9 .
8 )

39 .
7 N x
 v ox t
 1
2 at
2
1

0
 1
2 a ( 4 )
2 a

0 .
125 m / s
2 m
2 g sin
 
F f

T
 m
2 a m
2 g sin
 
F f

( m
1 a
 m
1 g )
 m
2 a m
2 g m
2 g m
2 g
 k
 sin sin sin


 m
2 g


 k
 k
F
N m
2 g
 m
1 a cos


 m
1 g m
1 a
 m
1 a sin
 m
1 g
 m
2 a
  m
2 m
1 g a
 cos
 m
1 g




 k m
2 a m
1 g g
 m
2 a cos
 m
2 a m
2
 k

56 .
7

0 .
5

39 .
2

1 .
125
67 .
57

0 .
235

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.
x
 v t ox
 1
2 at
2 x
 v t ox
Remember, the velocity is
CONSTANT horizontally, so that means the acceleration is
ZERO!
y
 1
2 gt
2
Remember that since the projectile is launched horizontally, the INITIAL
VERTICAL VELOCITY is equal to
ZERO.

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?
v ox
=100 m/s y = 500 m v oy
= 0 m/s g = -9.8 m/s/s y
 1
2
102.04
gt
2  
500
 1
2
( 9.8) t
2
  
10.1 seconds x
 v t ox

What I want to know?
t = ?
x = ?
(100)(10.1)

1010 m

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

You will still use kinematic #2, but YOU MUST use
COMPONENTS in the equation.
v o
 v ox v oy
 ox
 oy
 1
2 gt
2 v ox
 v o v oy
 v o cos
 sin


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?
  53 v ox
 v o v ox
 v oy
 v o cos sin

 v oy


/

/

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 What I want to know v ox
=12.04 m/s t = ?
v oy
=15.97 m/s x = ?
y max
=?
y = 0 g = - 9.8 m/s/s y
 v t oy
 1
2 gt
2  
(15.97) t

15.97
t
 
4.9
t
2 
15.97

4.9
t

4.9
t
2 t

3.26 s

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 What I want to know v ox
=12.04 m/s t = 3.26 s v oy
=15.97 m/s x = ?
y max
=?
y = 0 g = - 9.8 m/s/s x
 v t ox

(12.04)(3.26)

39.24 m

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 What I want to know v ox
=12.04 m/s t = 3.26 s v oy
=15.97 m/s x = 39.24 m y max
=?
y = 0 g = - 9.8 m/s/s y
 v t oy
 1
2 gt
2 y
 y

13.01 m
2

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!

nd
Recall that according to
Newton’s Second Law, the acceleration is directly proportional to the Force.
If this is true:
F
NET
 ma a c
 v
2 r
F
NET
F c

F c
 mv r

Centripeta l
2
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.

Top view
F
N mg
Side view
F f
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

F c

F
N
 mv
2 r
 mg
 mv
2 r
  v
2 rg rev
33 .
3 min
1 min
*
60 sec

0 .
555 rev sec
1 sec
0 .
555 rev v c

2
 r
T


1 .
80 sec rev
2

( 0 .
15 )
1 .
80


T
0 .
524 m / s
  v
2 rg
( 0 .
524 )
2

( 0 .
15 )( 9 .
8 )

0 .
187

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?
F
NET

F c
 ma c
 mv
2 r mv
2
T
 mg
 v v r
 r ( T
 mg m

5 .
74 m / s
)
 r ( T

 mg )

0 .
mv
25
2
0 .
5 ( 14

( 0 .
25 )( 9 .
8 ))
T mg

At the bottom?
F
NET

F c
 ma c
 mv
2 r
T v
 mg

 mv r ( T r
 mg ) m
2
 r ( T

 mg )
 mv
2
0 .
5 ( 14

( 0 .
25 )( 9 .
8 ))
0 .
25 v

4 .
81 m / s
T mg

F g
G
 m
1 m
2 r
2
 constant of proportion ality
G

Universal Gravitatio nal Constant
G

6 .
67 x 10

27 Nm
2 kg
2
F g

G m
1 m
2 r 2
F g
 mg

Use this when you are on the earth
F g

G m
1 m
2 r
2

Use this when you are LEAVING th e earth

rd
"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!

A 2-kg sliding puck whose initial velocity magnitude is v
1
= 10 m/s strikes a wall at a 30 degree angle and bounces off. If it leaves the wall with a velocity magnitude of v
2
= 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

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!

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.” p o ( truck ) p o ( car )
 mv o


( 400 )( 2 )
( 500 )( 5 )

2500

800 kg * m / s kg * m / s p o ( total )

3300 kg * m / s p truck p car


500
400 *
* 3
4 .
5


1500 kg
1800 kg
* m
* m
/
/ s s p total

3300 kg * m / s

Sometimes objects stick together or blow apart. In this case, momentum is ALWAYS conserved.
 p before
  p after m
1 v
01 m
1 v
01

 m
2 v
02 m
2 v
02

 m
1 v
1 m
 total m
2 v total v
2 m total v o ( total )
 m
1 v
1
 m
2 v
2
When 2 objects collide and DON’T stick
When 2 objects collide and stick together
When 1 object breaks into 2 objects
Elastic Collision = Kinetic Energy is Conserved
Inelastic Collision = Kinetic Energy is NOT Conserved

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?

W
W

F
 r

F r cos

W

25 16 cos 30
W
W

346 .
4 Nm

346 .
4 J
W
W

F
 r

F r cos

A box of mass m = 2.0 kg is moving over a frictional floor ( u k
= 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.
a) How much work does F do before taking the plunge?

What if we had done this in UNIT VECTOR notation?
F
W

21 .
65 i
ˆ 
12 .
5
ˆ j

( F x
 r x
)

( F y
 r y
)
W
W
W

( 21 .
65

16 )

( 12 .
5

0 )

346 .
4 Nm

346 .
4 J

F f
How much work does the FORCE
NORMAL do and Why?
F n
W
W

F
 r

F r cos

W
W


0
F
N
J
16 cos 90
There is NO WORK since “F” and “r” are perpendicular.
How much does the internal energy of the system increase?
34.08 J

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.

W
W
 
F ( x ) dx
 
( kx ) dx
  x x

0
( kx ) dx
 k
 x x

0 x dx
W
 k | x
2
| x x

0

W
2

U spring
 1
2 kx
2
Elastic “potential” energy is a fitting term as springs STORE energy when there are elongated or compressed.

W
 D
K
  D
U
K

K o
 
( U

U o
)
K
K o

K o

U o



U
K

U

U o
Energy before

Energy after

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 h = L – Lcos
 h = 2-2cos
 h = 0.187 m
E
B
= E
A
U
O
= K mgh o
= 1/2mv 2 gh o
= 1/2v
1.83 = v 2
2
1.35 m/s = v

• 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,