# Physics 115 2012

Document Sample

```					Physics 115 2012

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?

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
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?
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.

8.0 m/s, W

15 m/s, N

Rv    q

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!)
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?

What is the total DISTANCE for the entire trip?
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:
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:
What do the “signs”( + or -) mean?
Quantity       Positive               Negative
changed toward the     changed toward the
positive.              negative.
Velocity       You are traveling in   You are traveling in
the +x or +y           the –x or –y
direction.             direction.
Acceleration   If moving in the       If moving in the
positive direction,    positive direction,
you are speeding up.   you are slowing
If moving in the       down. . If moving in
negative direction,    the negative
you are slowing        direction, you are
down.                  speeding up.
The 3 Kinematic equations
There are 3 major kinematic
equations than can be
used to describe the
motion in DETAIL. All are
used when the
acceleration is CONSTANT.
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.
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        What do I      Which variable is NOT given and
Final Velocity!
v = 0 m/s
oy                y=?
g = -9.8 m/s2
yo=0 m
t = 5.78 s
-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           What do I             Which variable is NOT given and
know?               want?                                     TIME
vo= 0 m/s             a=?
x = 2.5 m
v = 43.5 m/s

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         What do I             Which variable is NOT given and
Final Velocity
vo= 0 m/s            t=?
x = 35 m
a = 2.00 m/s/s

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       What do I      Which variable is NOT given and

vo= 12.5 m/s       a=?                    DISPLACEMENT

v = 25 m/s
t = 6s

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

a
=
pe

pe
slo

slo
area = v
area = x

t (s)                   t (s)                 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
v
=

x (m)                      2. The slope is POSITIVE
pe
slo

t (s)
v (m/s)
How could you translate what the
SLOPE is doing on the graph
ABOVE to the Y axis on the graph
to the right?
t (s)
Example
v (m/s)
x (m)

t (s)
t (s)

1st line                      2nd line
• The slope is constant       • The slope is “0”
• The slope is “-”

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)

a (m/s/s)               t (s)

t (s)

a (m/s/s)                      t (s)

t (s)
Free Body Diagrams
A pictorial representation of forces complete
with labels.
FN                   •Weight(mg) – Always
T               drawn from the center,
straight down
Ff                             •Force Normal(FN) – A
T      surface force always drawn
perpendicular to a surface.
W1,Fg1                    •Tension(T or FT) – force in
or m1g                    ropes and always drawn
AWAY from object.
m2g   •Friction(Ff)- Always drawn
opposing the motion.
Free Body Diagrams

Ff        FN

mg
New’s 1st Law and Equilibrium
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
Example
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

b) Calculate the Normal Force

FN    Fa
Fay
Ff                30
Fax
mg
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.
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           12
0.2           24
0.3           36
0.4           48
0.5           60
0.6           72                      k =120 N/m
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?

110 N
1000 N/m
Newton’s Second Law
The acceleration of an object is directly
proportional to the NET FORCE and inversely
proportional to the mass.

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
Newton’s 2nd Law
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           In which direction,
Fa   is this object
accelerating?
Ff
The X direction!
mg
So N.S.L. is worked
out using the forces
in the “x” direction
only
Example
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
T

T

m1g

m2g
Example (cont.)
Where does the calculus fit in?
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:

When t = 4.0s , what is the reading on the bathroom scale (a.k.a.
Force Normal)?

4.6 m/s/s                      1036.8 N
TWO types of Friction
• Static – Friction that keeps an object at rest
and prevents it from moving
• Kinetic – Friction that acts during motion
Force of Friction
• The Force of Friction is
directly related to the
Normal Force.
The coefficient of
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
a) What is the coefficient of kinetic
across a level floor at a constant   friction between the crate and the
speed by a force F of 600 N at an    floor?
angle of 20° below the horizontal
as shown in the figure.

Fa        FN
Fay
20
Fax

Ff
mg
Example
If the 600 N force is instead pulling the block
FN   Fa
at an angle of 20° above the horizontal                       Fay
as shown in the figure, what will be the                20
acceleration of the crate. Assume that                  Fax
the coefficient of friction is the same as
found in (a)
Ff
mg
Inclines
q
q
Ff                FN                 q
q

q

Tips
mg
q          •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
m
2
m2gcos40          Ff
40
T
m 2g
m1                             40
m2gsin40

m 1g
Example
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.

Remember, the velocity is        Remember that since the
CONSTANT horizontally, so        projectile is launched
that means the acceleration is   horizontally, the INITIAL
ZERO!                            VERTICAL VELOCITY is equal to
ZERO.
Horizontally Launched Projectiles
Example: A plane traveling with a       What do I        What I want to
horizontal velocity of 100 m/s is    know?            know?
500 m above the ground. At
some point the pilot decides to      vox=100 m/s      t=?
drop some supplies to
designated target below. (a)         y = 500 m        x=?
How long is the drop in the air?
(b) How far away from point          voy= 0 m/s
where it was launched will it
land?                                g = -9.8 m/s/s

1010 m
10.1 seconds
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     voy

q
vox
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?

/s
0   m
0.
=2
vo

q = 53
Example
A place kicker kicks a            What I know     What I want
football with a velocity of                    to know
20.0 m/s and at an angle       vox=12.04 m/s   t=?
of 53 degrees.                 voy=15.97 m/s   x=?
(a) How long is the ball in
y=0             ymax=?
the air?
g = - 9.8
m/s/s

3.26 s
Example
A place kicker kicks a          What I know     What I want
football with a velocity of                  to know
20.0 m/s and at an angle  vox=12.04 m/s      t = 3.26 s
of 53 degrees.               voy=15.97 m/s   x=?
(b) How far away does it        y=0             ymax=?
land?                        g = - 9.8
m/s/s

39.24 m
Example                            What I know     What I want
to know
A place kicker kicks a football    vox=12.04 m/s   t = 3.26 s
with a velocity of 20.0 m/s     voy=15.97 m/s   x = 39.24 m
and at an angle of 53
degrees.                        y=0             ymax=?
g = - 9.8
(c) How high does it travel?       m/s/s

CUT YOUR TIME IN HALF!

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?
Circular Motion and New’s 2nd Law
Recall that according to
Newton’s Second Law,
the acceleration is
directly proportional to
the Force. If this is true:

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          What is the minimum coefficient of static friction
necessary to allow a penny to rotate along a 33 1/3 rpm
Top view      record (diameter= 0.300 m), when
the penny is placed at the outer edge of the record?

FN
Ff

mg

Side view
Examples
The maximum tension that a 0.50 m
string can tolerate is 14 N. A 0.25-kg    T   mg
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?
Examples
At the bottom?

T

mg
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.”
Several Types of collisions
Sometimes objects stick together or blow apart. In this
case, momentum is ALWAYS conserved.

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
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
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.

a) How much work does F do before taking the
plunge?
Example cont’
What if we had done this in UNIT VECTOR notation?
Example cont’ How much work does the FORCE
NORMAL do and Why?
Fn

There is NO WORK since “F”
and “r” are perpendicular.

Ff

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

Elastic “potential” energy is a fitting term as springs STORE energy when there
are elongated or compressed.
Energy is CONSERVED!
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?

q   Lcosq        h = L – Lcosq
L   h = 2-2cosq
h = 0.187 m
h

EB = E A

UO      = K
mgho    = 1/2mv2
gho     = 1/2v2
1.83    = v2
1.35 m/s = v

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 0 posted: 6/21/2013 language: English pages: 67