Download Topic 2 Mechanics Part 2 2015-17

IB PHYSICS
NOTES and STUDY GUIDE
TOPIC 2 MECHANICS – Part 2
TOPIC 6(New)
Newton's laws
Linear Momentum
Work, Energy and Power
Topic 6 (New)Circular Motion
Topic 6 ( New) Gravitation
1
1. Concept of Force : Free Body Diagrams
2
Tsokos text Read p. 67 see examples : fig. 3.9 , p. 65 : fig. 3.4 ,
p. 67 – 68 : 2,5,8,10
Tsokos pp. 67 – 68 : 2,5,8,10
3
IB Example This question is about power and efficiency.
A bus is travelling at a constant speed of 6.2 m s–1 along a section of road that is inclined at an
angle of 6.0° to the horizontal.
The bus is represented by the black dot shown below. Draw a labeled sketch to
represent the forces acting on the bus.
●
4
2. Newton's Laws of Motion
Dynamics: while kinematics dealt with only motion, dynamics looks at why things move.
Force is a push or a pull upon an object resulting from the object's interaction with another
object. It is a vector quantity (magnitude and direction) and is measured in Newtons [N].
Forces are divided into 2 categories:
Contact forces (friction, air resistance or drag, tension, compression, normal force, lift,
thrust, buoyancy, etc.)
Non-contact forces (weight/gravity, magnetic, and electrostatic).
Newton’s First Law of Motion – Inertia – At Rest or Constant Velocity
Aristotle (384 -322 B.C.) believed that the natural state of a body was at rest and a force
was required to keep an objection in motion. Furthermore, the greater the force the
greater the speed. Some 2000 years later Galileo found that to push an object across a
surface that was smoother requires less force to achieve the same speed. Applying a
lubricant further reduced friction and thus reduced the force required. He went on to
theorize that if there was no friction an object would continue to move with constant
velocity without further application of forces. Galileo interpreted friction as a force that is
the same as a push or a pull. His theory explained a lot more phenomena and allowed for
more verifiable predictions. It was on this foundation that Newton built his famous ‘three
laws of motion’ in 1687.
NEWTON’S FIRST LAW OF MOTION
An object at rest tends to stay at rest and an object in motion tends to remain in motion
with constant velocity, unless acted on by an unbalanced force (a non-zero net force).
OR
If the net force acting on a body is equal to zero, the body will move with constant
velocity and zero acceleration.
^ CP DVD: Demo - Inertia of a cylinder
As an equation it can be written as:
F 0
http://video.yahoo.com/watch/3826981/10465439
or
http://video.yahoo.com/watch/3826981/10465439
F 0 F 0
x
y
(Newton's First Law of Motion; no net force)
Note:
No net force and constant velocity.
Translating means moving from one place to another. Translational equilibrium occurs
when a body is not accelerating. That is, the net force acting on the body is zero.
Examples of Newton's 1st Law (and hence translational equilibrium) include any object that
i) is stationary and ii) is moving with constant velocity. Because of friction it is difficult
to observe constant velocity on Earth.
A book on a table has 2 forces: i) gravity pulling down and ii) table pushing up. But forces
are equal; hence net force is zero, so book remains at rest.
An example of constant velocity on Earth would be an ice hockey puck sliding on ice which
is approximately frictionless. The is no air resistance in space so an object flying off
into space is also an example of Newton's 1st Law.
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The tendency of a body to maintain its state of rest or in uniform motion in a straight line is
called inertia. Hence Newton’s First Law is often called the law of inertia.
^ CP DVD: Demo - Newton’s Law of Inertia; Toilet Paper roll; Demo – Table Cloth Trick
http://video.yahoo.com/watch/3826981/10465439
Example 4-1 (Giancoli 2005; 74)
Newton’s Second Law - F = ma ( Mass and Acceleration)
Think of a shopping cart: if you double the force you push with then the acceleration will
double, hence a
Fnet. But the acceleration also depends on the mass: the greater the
mass the less the acceleration, thus a 1/m. Think of the force required to start
(accelerate) an empty shopping cart compared with a full one.
NEWTON’S SECOND LAW
The acceleration of an object is directly proportional to the net external force
acting on it and is inversely proportional to its mass. The direction of
acceleration is in the same direction as the net force acting on the object
As an equation it can be written:
(Newton’s Second Law of Motion)
F
a
m
i.e.
 F  ma
 F  ma
 F  ma
x
x
y
y
or
6
Note:
External forces only (i.e. can’t pull yourself along by your belt);
Mass must be constant (i.e. not a rocket losing fuel);
1 Newton = amount of net force required to accelerate a body of 1kg to 1m/s 2
i.e. [1 N = 1 kg.m/s2]
Alternate units: 1 lb = 1 slug.ft/s2 1 dyn = 1 g.cm/s2
[1 N = 105 dyn = 0.2248 lb]
^ C.P. DVD: Newton's 2nd Law; Free-Fall Acceleration Explained
7
Examples 4-2 and 4-3 (Giancoli 2005, 76-77)
8
Newton’s Third Law – Action – Reaction
The force acting on an object is always applied by another object (e.g. a horse pulls a cart,
a person pushes a shopping trolley, a hammer pushes on a nail, a magnet attracts a paper
clip, etc). But Newton realized that things are not so one-sided (as a hammer hits a nail the
nail accelerates and likewise the hammer quickly decelerates to zero).
NEWTON’S THIRD LAW
Whenever one object exerts a force on a second object, the second object exerts an equal
and opposite force on the first.
OR
If body A exerts a force on body B (an “action”) then body B exerts an equal and opposite
force on body A (a “reaction”). These two forces have the same magnitude but opposite
direction.
Often paraphrased as “to every action there is an equal and opposite reaction” but it is
very important to remember that the ‘action’ force and the ‘reaction’ force are acting on
different objects. Forces are equal and opposite and are referred to as ‘action-reaction’
pairs.
As an equation it can be written:
FA on B = - FB on
A
(Newton’s Third Law of Motion)
^ CP DVD: Demo – Action & Reaction; A & R of Rifle + Bullet
Note:
Action-reaction pairs never act on the same object. The easiest way to ensure you have
an action-reaction pair is to reverse the subscripts.
Law also applies to non-contact forces (e.g. gravitational force).
Reaction forces from inanimate objects occur because no matter how hard a material is it
is elastic to some degree.
Any body, such as a rope, that has forces applied at its ends is said to be in tension (FT =
T).
FAB in Giancoli (2005, 79) represents the force on A by B (which is opposite to subscripts
used above).
* Brainpop – Newton’s Laws
9
Examples 4-4 and 4-5 (Giancoli 2005, 79-80)
NOTE:
Newton's 1st and 2nd laws of motion are used to determine an objects state of motion (at
rest, moving with constant velocity or accelerating). To determine an objects state of
motion you consider the net force acting on one object only!
Newton's 3rd law of motion is used to determine action-reaction force pairs. It deals with
forces acting two different objects. As such, it cannot be used to determine an objects
state of motion.
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Mass and Weight
Mass is the amount of matter in an object or more precisely is a measure of inertia (the
more mass a body has, the harder it is to change its state of motion). Mass is a scalar
quantity which is measured on a beam balance. The standard SI unit is the kilogram [kg].
Mass is constant throughout the universe.
Weight varies as a body travels through the universe depending on the value of g. Weight
is a vector quantity and is measured by a spring scale. The SI unit is the Newton (N).
Fg  w  mg
(Weight force acting on an object; Newton’s 2nd Law)
Note:
Weight changes depending on where an object is in the universe due to variation in the
acceleration due to gravity e.g. an object on the Moon will weigh about 1/6 as much as it
did on Earth, because gravity on the Moon is 1/6 weaker than on Earth (gmoon=1.67m/s2) but
mass (and hence inertia) is constant anywhere in the universe.
When an object is at rest on the Earth’s surface the gravitational force doesn’t disappear
(can still be measured on a spring scale). Why then, doesn’t the object move? From
Newton’s 1st and 2nd Law: if no net force acts on an object, it remains at rest. Therefore
the ground/surface must exert this upward force. This contact force occurs because the
table is compressed slightly and due to its elasticity, it pushes back on the object [Fig 4-15
Giancoli 2002, 87]. The contact force that acts perpendicular to a common surface of
contact is usually referred to as the Normal Force (‘normal’ means perpendicular) and
labeled
F or FN in diagrams.
Magnitude of any vector is always positive (as it does not include direction).
^ CP DVD: Definition of Newton; Demo - Weight-Mass Distribution]
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Wilson Buff Conceptual Questions pp. 132 – 133: 1-7, 12, 17, 41, 42
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IB Examples:
1. A skydiver of mass 80 kg falls vertically with a constant speed of 50 m s–1. The upward
force acting on the skydiver is approximately
A.
0 N.
B.
80 N.
C.
800 N.
D.
4000 N.
2. A car of mass 1000 kg accelerates on a straight, flat, horizontal road with an acceleration
a = 0.3 m s–2.
The driving force F on the car is opposed by a resistive force of 500 N.
The net (resultant) force on the car is
A.
200 N.
B.
300 N.
C.
500 N.
D.
800 N.
3. This question is about forces.
An athlete trains by dragging a heavy load across a rough horizontal surface.
The athlete exerts a force of magnitude F on the load at an angle of 25° to the
horizontal.
(a)
Once the load is moving at a steady speed, the average horizontal frictional force
acting on the load is 470 N.
Calculate the average value of F that will enable the load to move at constant
speed.
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3. Newton's Laws : Vector Component Review
Review Wilson Buffa 34, 37, 45 below
14
Text Tsokos Read p. 71 and study fig. 4.3 and 4.4 , p. 72 Q1
When the object is in equilibrium the following rules must apply :
The forces going up must equal the forces going down : Σ Fy = 0
The forces going right must equal the forces going left : Σ Fx = 0
15
Tsokos pp. 74: 2,3, 10
16
Inclined Planes
17
Tsokos p. 82 fig. 5.12
Wilson Buffa p. 97 : 45 Review
p. 137 : 72
18
Text Tsokos p. 75 : 14
Do AP 2005
Free Response # 2 a and b only
A simple pendulum consists of a bob of mass 1.8 kg attached to a string of length 2.3 m. The pendulum is
held at an angle of 30" from the vertical by a light horizontal string attached to a wall, as shown above.
(a) On the figure below, draw a free-body diagram showing and labeling the forces on the bob in the
position shown above.
(b) Calculate the tension in the horizontal string.
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IB Exam Nov. 2009 Paper 2 # 5
Note b) (i): treat as free fall object v2 = u2 + 2 as
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4. Linear Momentum : formulas , Units and Newton's 2nd and 3rd law
Formulas - Units
p  mv
Linear momentum ( p) is defined as the product of an object’s mass and its
velocity.
(Linear momentum defined)
The standard SI units: kg ms-1 = N s]
Explain unit the kg m s-1 and how it is derived from the formula for momentum. Also
explain how kg m s-1 can be converted to N s.
1N = 1 kg ms-2
1 Ns = ( kgms-2 ) (s) = kgms-1
Momentum depends on mass and velocity and is the reason a small guy from Asia can run
and knock over a big guy from Texas. It is the reason why a bullet has a big force of
impact and why a person can karate chop a stack of bricks. Momentum is the reason for
collisions , air bags , brakes and many things involving inertia and motion.
Newton's 2nd law
In order for an object to have a change in velocity there must be a net force acting . So a
change in momentum ( mass and velocity) also requires a force.
F 
p
t

F = mv2 – mv1
∆t
Allows analysis of problems not possible with F  ma . For example, collisions
and problems where mass is not constant (e.g. motion of a rocket).
If mass is constant then:
F 
p mv

 ma
t
t
(Newton’s 2nd Law).
21
Tsokos read p. 87 , p. 88: Q1
Note: Q1
22
Newton's 2nd and 3rd laws
23
TSOKOS , p. 95-96 1, 7
24
5. Linear CHANGE IN Momentum: Impulse
Changes in Momentum = ∆P = Impulse
∆P = mv2 - mv1
∆P = Impulse = Force x time
I = F ∆t = m ∆v
∆P = F ∆t
F = ∆P
∆t
F = ∆P
∆t
=
mv2 - mv1
∆t
Changes in momentum occur with changes in mass and velocity. When mass is constant
, the velocity changes causing acceleration. And what causes this acceleration is force.
Time is an important factor in changing momentum. If you apply a force over a small time
you get a small change in momentum. If you apply a force over a large time you get a
large change in momentum.
This change in momentum based on force and time is called impulse. Impulse changes
momentum in much the same way that force changes velocity causing acceleration.
CONSTANT VELOCITY = NO FORCE = NO IMPULSE = NO COLLISION
Case 1:
To increase the momentum apply the greatest force possible over the longest period of
time possible.
Ex. Long range cannons have long barrels. The long barrels allow more time for the force
to act resulting in more change in velocity = change in momentum. Result is the cannons
go further due to higher impulse.
Ex. Golf ball – club
25
Golf balls go further with more impact time (impulse time) between the club and ball.
Case 2: Decreasing Momentum Over a Long Time
I = ∆P =
m
V

T
F
Force is reduced because hay extends time of impact ( increasing contact time).
I = ∆P =
m
V

Ft
Time of impact is reduced by the wall causing large force.
26
In both cases the momentum of the truck is the same.
Other examples: landing with legs bent, safety net for acrobats, catching a baseball.
Case 3 : Decreasing Momentum Over a Short Time
The girl causes a large impulse to the bricks in a short time and produces a large force.
I = ∆P = F
t
27
28
Text Tsokos p. 95 : 2
29
6. Linear Momentum : Law of Conservation of Momentum – Collisions
Important principle in Physics and used to analyze collisions of objects ranging from
subatomic particles to automobiles.
ELASTIC COLLISIONS
Net momentum before collision = Net momentum after collision
Total Kinetic Energy is conserved.
Drawing: Momentum is transferred from ball1 to ball 2 .
Ball 2 at rest , initial velocity ( v0 ) = 0 .
1
2
1
2
CORRECT
1
2
Law of Conservation of Momentum and Collisions
The total momentum of an isolated system of bodies remains constant (i.e. when no
external forces act on a system, the total momentum of the system stays the same).
pbefore  p after
mAu A  mBuB  mAvA  mB vB
(Conservation of momentum for two objects colliding)
NOTE: FORMULA NOT IN DATA BOOKLET
u denotes initial velocity (or velocity before collision)
v denotes velocity after collision
Isolated systems only, where the only forces present are those between the objects of the
system (no external forces).
Internal forces are forces that particles exert on each other.
External forces are forces from some object outside of the system e.g. friction.
30
Tsokos read pp. 91-92 : Q 10 - 12
31
INELASTIC COLLISIONS
Inelastic Collisions
In an inelastic collision KE is conserved. The law of conservation of energy still holds true.
Completely Inelastic Collision ( v
 v A  vB )
A completely inelastic collision is a collision where the two bodies stick and move together
as one after the collision. e.g. a bullet imbedding itself in wood.
mAu A  mBuB  mA  mB v
(Completely inelastic collision)
NOTE: FORMULA NOT IN DATA BOOKLET
In any isolated system where external forces are neglected momentum is conserved.
Total Kinetic energy is conserved with inelastic collisions.
32
Text Tsokos pp. 96-97 : 8, 14, 15.
33
HANDOUT Review Formula : Momentum , Impulse, Force and Collisions
a. P = mv
∆P = mv2 - mv1
p. 88 Q1 basic definition : Ns = kgms-1
b. Changes in Momentum = ∆P = Impulse
c. F = ∆P
∆t
=
mv2 - mv1
see Tsokos p. 95-96 : 1, 2b, 7
∆t
d. F = ∆P = m∆v =
∆t
∆t
ma
( newton’s 2nd Law F = ma)
e. ∆P = Impulse = Force x time
I = F ∆t
f.
mAu A  mBuB  mAvA  mB vB
law of cons. Of momentum elastic
collisions
g.
mAu A  mBuB  mA  mB v
inelastic collisions
34
Impulse and Force -Time Graph Tsokos pp. 89 – 92
∆P = Impulse = Force x time
I = F ∆t = area under curve ( usually a triangle) = ½ base x height = ½ time x ForceMAX
∆P = mv2 - mv1
The magnitude of the impulse delivered to the ball ( by the raquet) is equal to the product
of the average force acting on the ball ( by the raquet) times the time interval.
Impulse is the area under the curve of a force – time graph and equal the total
momentum change of the mass.
I = F∆t = area under curve ( usually a triangle) = ½ base x height = ½ time x ForceMAX
35
The thick curve ( bold print) represents the force that brings the body to rest over a short
period of time. The force is large and the time is short . Similar to a truck being stopped
by a wall:
The thin curve represents the force that brings the body to rest over a long period of time.
The force is small and the time is long. Similar to a truck being stopped by a hay stack:
The areas of the curve are the same since they both represent the same change in
momentum.
36
Tsokos pp. 90 – 91 Q6 , Q7
NOTE : average force Favg = total change in momentum / total change in time. Do not
confuse with maximum force ( Fmax) as seen in next problem.
37
Average Force vs. Maximum Force ( Fmax) :
IB questions will want you to calculate the maximum force ( Fmax) from graphs. Use
Impulse (I) = change in momentum (∆P ) = area under curve = ½ base x height
= ½ time x Fmax
NOTE : Fmax = height of triangle
From Q6 above change in momentum (∆P ) = 2.25 Ns
Q7 part 2 :
38
IB Example
Assume the ∆P (change in momentum) of the ball is 43 Ns
39
IB EXAM May 2009 Paper 2
( b) The ball rebounds with a speed of 1.97 ms-1. Show that the impulse given to the ball is
0.313Ns.
40
7. Work : Formulas , Work at angle θ , Work due to gravity
Important to DRAW ALL PROBLEMS REGARDING FORCES AND WORK
Formulas
Work in physics is given a very specific meaning to describe what is accomplished by the
action of a force when it acts on an object as the object moves through a distance.
Work is defined to be the product of displacement and the component of the force parallel
to the displacement.
When a constant force acts in the same direction as displacement the work done by the
force is:
Work = Force x Displacement
W  Fs
(Work defined; F is parallel to s)
Note:
Work is a scalar quantity.
The standard SI Unit of Work is the Joule [1J = 1Nm].
W = work and w = weight.
If a force is pushed at an angle to the displacement then only the component of the force
parallel to the displacement does work. When F and s have different directions, take the
component of F in the direction of s: W  ( F cos  ) s or:
W  Fs cos 
(Work; θ is the angle between F and s)
Note:
When calculating work F is always positive. Direction of F is not important because work
is a scalar quantity.
When θ between 0o and 90o then work is positive.
When θ
o
the W = 0.
When θ between 90o and 180o then work is negative.
In general when a body does positive work on another body, the 2 nd body does an equal
and opposite amount of work on the 1st body (Newtons 3rd Law).
Text Tsokos read pp. 99-100 : Q1
p. 112: 1, 3
41
42
Work at an Angle vs. Horizontal work ( no angle)
When pulling a crate at an angle you have an upward force ( Fy) that reduces the
downward force of gravity and subsequently the weight. The normal force is also reduced
along with friction. All this proves that it is easier to pull something at an angle compared
to pulling it horizontally. See and review examples 4.10 and 4.11 below.
43
+ Work and - Work
- work
+ work
44
Work due to gravity = 0 : θ = 900 force perpendicular to displacement
Tsokos Text read p. 101 and study fig. 7. 5
Work done BY gravity on an object that is moving horizontally ( by another force), is zero
because the angle at which gravity acts is completely vertical with no horizontal
component i e perpendicular to the displacement, ( 900 ) .
Work due to gravity on an object that is falling or thrown upward does not = zero: θ
= 1800 force parallel to displacement
Tsokos text read p. 101 and study fig. 7.6
45
Work is done by gravity on a falling object or a projectile that is thrown up ward ( review
free fall) . In this case the force of gravity is acting parallel to the direction of displacement
, θ = 1800. Work falling = + mgh Work upward = - mgh ( gravity acting opposite
direction of displacement).
Work due to gravity is independent of path followed = mgh
Tsokos text read p. 102 and study fig. 7.7
The work done by gravity is independent of the path followed and depends only on the
vertical distance separating the initial and final positions.
Work done in holding something still = 0 no displacement
Tsokos text read top of p. 102
46
Wilson / Buffa pp. 143-144 ex. 5.1, 5.2
47
General Questions about work:
1. What are the units for work?
2. If you push against the wall of a building are you doing any work? Explain
3. You are carrying a backpack across the room. What is the work done by your vertical
carrying force?
Tsokos text p. 113 : 5
Classic Inclined Plane see below:
A 3.00 kg block slides down a frictionless plane inclined 20.0 0 to the horizontal. If the
length of the plane is 1.50 m, how much work is done? What was the force?
DRAW:
48
8. Network
Net (Total) Work
Since work is a scalar quantity you can simply calculating the W done by each individual
force and then add then together:
Wnet   Wi
OR find net force by vector addition and:
W  F s cos
`
49
50
9. Hooke's Law – Spring Constant T = kx : Network by a variable
force
NET WORK = AREA UNDER CURVE ( LINE) OF F vs. X (displacement) GRAPH
Tsokos text read pp. 100-101 and study fig. 7.4
Hooke's law is about springs and states that the further you stretch a spring the more force
is required. The tension ( T) or force in the spring is directly proportional to the
displacement (x) or how far the spring is stretched.
T = kx or F=kx
k is a spring constant and is based on the material or stiffness of the spring. The stiffer the
spring the greater value of k.
Network = area under F x graph
The network done by a variable force ( such as the spring) is equal to the area under the
Force displacement graph. Since the area represents a triangle work = ½ base x height:
W=½bh
base = displacement (x))
height = Tension in spring ( T) = kx
W = ½ ( x) ( kx)
51
W = ½ k x2
When the spring is being stretched from x1 to x2
W = ½ k ( x22 - x12 )
52
10.Work Related to Energy Part 1 :Kinetic and Potential Energy :
Energy is one of the most important concepts in science. By definition , energy is the
ability to do work. Kinetic energy is the energy of motion. The work – energy theorem
relates the work done on an object to the kinetic energy ( K) of that object:
The net work done on an object by all forces acting on it is equal to the
kinetic energy of the object.
W = ΔEk CHANGE in Ek
change in
( unit: Joules J = N m)
For example you apply a force to a box. Work is being done on the box and the box
moves. Since the box is moving it now has kinetic energy. So work is the change in
kinetic energy of the box.
If net force = 0 there is no work done and no change in kinetic energy.
Kinetic Energy
* Brainpop: Kinetic Energy
Kinetic energy is the energy associated with a body in motion.
Kinetic Energy is:
E
1 2 p2
mv 
2
2m
( E k Kinetic Energy defined)
Note:
The IB data booklet uses the symbol EK for kinetic energy.
Kinetic energy is a scalar quantity (depends only on mass and velocity not direction).
The standard SI units is the Joule [1kg.m2/s2 = 1(kg.m/s2)m = 1Nm = 1J].
Kinetic energy can be positive or zero but it can never be negative (see equation).
Work and Kinetic Energy - VEOCITY
NO CHANGE IN V = NO WORK
The net work done on an object is related to its velocity: velocity increases when ∑W
decreases when
; and is constant when ∑W = 0.
;
Work-Energy Theorem: The net work done on an object is equal to the change in its
kinetic energy.
W = E 2 - E1
(Work-Energy Theorem)
Note:
Is valid when F is not constant and with curved (non-linear) motion.
When ∑W is positive then E2 > E1 (i.e. Ek is increasing and thus the speed is increasing).
53
When ∑W = 0 then E2 = E1 and thus speed is constant.
When ∑W is negative then E2 < E1 (i.e. Ek is decreasing and thus the speed is
decreasing).
Must use an inertial frame of reference.
W = Ek
= ½ mv2
W = F s = Ek
= ½ mv2
54
Formulas
W = Ek =
W=Fx
→ F= W
x
Remember Newton's 2nd Law F = ma
from
Ek = ½ m v2 :
v = √ 2Ek
m
55
Text Tsokos p. 104 : Q5 , Q6
p. 113 : 6, 11
56
Wilson / Buffa Ex. 5.5 p. 151 Finding speed of puck and work needed to stop puck,
mass given
Potential Energy (Stored Energy)
*
Potential Energy is energy associated with the position of a system (not its motion).
There are three types of potential energy:
Gravitational potential energy (PE of a diver is converted into KE as she falls).
Elastic potential energy (energy stored in the diving board as the diver jumps on it or
energy stored in a compressed spring).
Electrical potential energy (covered later).
Gravitational Potential Energy
Gravitational potential energy is defined as the energy a body has because of its height
relative to a given point.
EP = mgh ( mgy )
(Gravitational potential energy defined)
Gravitational Potential Energy Difference
Experimentally we can only quantify changes in potential energy. Gravitational potential
energy difference is defined as the work that must be done by an external force to move
an object through a vertical displacement Δh. The work done by the external force is
stored as potential energy.
Δ EP = mg Δh (Gravitational potential energy difference)
57
Note:
IB data booklet uses EP for gravitational potential energy.
Gravitational potential energy is always measured in relation to a zero level.
Choice of origin (h = 0) is a matter of convenience. It is usually easiest to define h = 0 as
the lowest point in the problem.
PEg is gravitational potential energy of the system not gravitational potential energy of a
body (because PEg includes both the body and the Earth).
58
10.
Work Related to Energy PART 2 : Conservation of Energy
Conservation of Energy is one of the most important laws in Physics.:
Energy can not be created or destroyed; it may be transformed from one form into
another, but the total amount of energy of a system never changes.
The work- energy theorem applies to changes in kinetic and potential energy that can
also be used to explain changes in thermal, mechanical , electrical and nuclear
energy as well. It is important to understand how energy changes or transforms form
one form to another or from one location to another.
Drawing:
Ep  10 000 J
Ek  0
Ep  7500 J
Ek  2500 J
Ep  5000 J
Ek  5000 J
Ep  2500 J
Ek  7500 J
Ep  0 J
Ek  10 000 J
The circus diver at the top of the pole has a potential energy ( U) of 10 000 J and a
kinetic energy ( K) of 0. As he dives, his potential energy is converted into kinetic
energy. At each successive interval the total energy ( E = U + K) ) is conserved and
stays a constant 10 000 J. At the bottom , just before he hits, all his energy is kinetic
( 10 000J) and his potential energy is 0.
Mechanical Energy (E) is defined as the sum of the kinetic and the potential energy.
Ek = ½ m v2
Ep = m g y
Etot = Ep + Ek
Ek1  Ep1  Ek2  Ep2
(Conservation of mechanical energy; no frictional forces)
59
Note:
Einitial  E final i.e. the total mechanical energy is conserved.
Equations work for varying forces and curved motion (only the initial and final heights are
relevant not the path taken).
Again it is usually easiest to define the origin (h = 0) as the lowest point in the problem.
Conservation of Mechanical Energy is:
E  Ek  Ep = Constant
(Mechanical energy is conserved when only gravity does work)
For example, when a ball is thrown vertically Ek is converted to Ep; and on the way
down Ep is converted back to Ek. But E is always constant (provided only gravity does
work i.e. no air resistance)
Ex. A painter on a scaffold drops a 1.5 kg can of paint from a height of 6.0 m. Calculate
the kinetic , potential and total mechanical energy at the following heights : 6 m , 4 m ,
0 m.
E total = Ep + Ek
6 m : Ek = 0 J
Ep = m g y = (1.5) ( 9.8) ( 6) = 88.2J
E total = all potential Ep = 88.2 J
4m: Etotal = 88.2 J
Ep= m g y = 58.8 J
Since Energy is conserved E total = Ep + Ek →
Ek = 29.4 J
0m: Etotal = 88.2 J
Ep = 0 J
E total = all kinetic Ek = 88.2 J
60
Wilson/ Buffa p. 175 : 66 a, b , 67 a, b
66. A person standing on a bridge at a height of 115m above a river drops a 0.25 kg
rock. A) What is the rock's mechanical energy at the time of release relative to the
surface of the river? b) What are the rock's kinetic, potential, and mechanical
energies after it has fallen 75.0 m?
67. A 0.30 kg ball is thrown vertically upward with an initial speed of 10.0 m s-1. If the
potential energy is taken as zero at its initial position, find the kinetic, potential and
mechanical energies at :
a) its initial position (assuming ball is already in motion )
b) 2.50 m above the initial position.
61
11. Conservation of Energy Problems and Manipulating Formulas
Ek1  Ep1  Ek2  Ep2
½ mv21 + mgy1 = ½ mv22 + mgy2
Although this formula is somewhat long, many times some of the terms go to zero and
you can remove them form the equation, to help solve easier.
For example when the object is at rest ½ mv21 = 0
For example when the object is on the ground mgy2 = 0
Thus many time you can simplify formula to : ½ mv2 = mgy
Also you can take the entire equation , divide by m and reduce it to:
½ mv21 + mgy1 = ½ mv22 + mgy2
= ½ v21 + gy1 = ½ v22 + gy2
m
.
Again simplifying the formula to ½ v2 = gy
From this we get the very important formula v = √ 2gy
62
Review Free Fall Formulas
v = v0 + gt →
at rest
v = gt
v2 = v02 + 2ax → at rest
v2 = 2ay → v = √ 2gy ( note g replaces a and y replaces x)
Ex. In the drawing a 0.50 kg ball is thrown up vertically with an initial velocity of 10 m/s.
a) What is the change in kinetic energy between the starting point and the ball´s
maximum height?. b) What is the change between the ball´s potential energy between
the starting point and the ball´s maximum height?
63
At top
v at top = 0 → Ek = 0
Ep = mgy ( height)
Etotal = Ep + Ek
At top all Ek has been converted to Ep = mgy
a) ΔEk = Ek2 – Ek1
Ek = ½ m v2
At maximum height v = 0 so Ek = 0
At starting point Ek = ½ (0.5 kg) ( 10m/s)2 = 25 J
ΔEK = 0 – 25 J = - 25 J
Negative because losing kinetic energy as ball goes upward.
b) A change in kinetic energy is equal to the change in potential energy:
Δ Ek = Δ Ep
In this case the loss in Ek = gain in Ep
25J = + 25 J
Δ Ep= + 25 J.
64
Wilson / Buffa p. 175 : 66 c, 67c
66. A person standing on a bridge at a height of 115m above a river drops a 0.25 kg rock.
c) What are the rock's kinetic, potential, and mechanical energies just before it hits the
water?
67. A 0.30 kg ball is thrown vertically upward with an initial speed of 10.0 m s-1. If the
potential energy is taken as zero, find the kinetic, potential and mechanical energies at :
c) its maximum height?
65
12. Three ways to calculate speed at which object hits the ground
knowing height (y) or mass (m)
1. Knowing height ( y) :
From ΔEp = ΔEk
m g y = ½ m v2
v=√2gy
Ex. A painter on a scaffold drops a 1.5 kg can of paint from a height of 6.0 m. At what
speed does the can hit the ground?
v=√2gy
v = 10.8 m/s
2. Knowing height and mass. Similarly the problem can be solved knowing total potential
energy ( m g y) is converted to kinetic energy and using this formula to calculate v:
From Ek = ½ m v2
v = √ 2K
m
Ek = Ep = m g y = ( 1.5 kg) ( 9.8) ( 6.0m) = 88.2 J
v = √ 2 ( 88.2) =
10.8 m/s
1.5
3. Knowing height only. Similarly the problem can be solved from the free fall formulas :
y = v0 t + ½ g t2
and
from rest v0 = 0 thus
v = gt
t = √ 2y
g
= √ 2 (6.0M) = 1.11 s
9.8
since v = g t = (9.8) ( 1.11) = 10.8 ms-1
66
Calculating maximum height knowing initial velocity:
A 0.50 kg ball is thrown up vertically with an initial velocity of 10 m/s. What is its maximum
height the ball reaches?
v at top = 0
From v2 = v02 – 2 g y
y = v02
2g
OR
NOTE: all Ek converted to Ep
½ mv2 = mgy
y = v02
2g
y = 5.1 m
NOTE : y is independent of mass and depends on v0 . This agrees with the fact that all
objects fall at the same rate regardless of their mass.
Wilson/ Buffa p. 175 : 68
68. What is the maximum height of a 0.30 kg ball is thrown vertically upward with an initial
speed of 10.0 m s-1.
Roller Coaster Problems pp 57 ap text
Tsokos Text p. 106: Q10, Q11 p. 113 : 9a only
READ p. 112 fig. 7.22
67
READ p. 112 fig. 7.22
68
69
13. Power and Efficiency
Power
Brainpop: Power
*
Definition of work has no reference to time but it is often necessary to know how quickly
work can be done. Power is defined as the rate at which work is performed.
Power = work (or energy)  time
P=W
t
Note:
Power is a scalar quantity.
The standard SI Unit – Watt [1W = 1J/s] (be careful not to confuse it with the symbol for
work)
100W light bulb converts 100 J of electrical energy into light and heat energy every
second.
1 hp = 746 W (i.e. 1hp = ¾ kW).
Power can also be calculated force and velocity:
P
W Fs
s

and v  therefore:
t
t
t
P = Fv
(Power, force and velocity related)
The Commercial Unit of Electricity – The Kilowatt-hour
kWh = Power x time = Work = Energy = J
E=Pt
1 kWh is the amount of energy used in 1hr (3600s) when the power consumed is 1kW
(103 J/s).
Hence, the kWh is a unit for energy not power [1 kWh = 103 J/s x 3600s = 3.6 x 106 J = 3.6
MJ].
Note: This conversion is in the data booklet.
70
Text Tsokos p. 109 : Q 16
p. 114 : 15,16
E = Pt
71
EFFICIENCY
Efficiency
Efficiency is defined as the amount of useful work performed per amount of available
energy.
Efficiency 
Usefuloutput
Efficiency = Power output
Totalinput
Total Power input ( of object)
Efficiency = Work output
Total Work input ( of object)
Input vs. output:
Input is the power or work rating on how much the motor - machine can theoretically
produce. Since all motors - machines are not 100% efficient the out put is always a
percentage of the total possible input.
% Efficiency : sometimes the efficiency rating is expressed as a %. For example if a
machine has an efficiency rating of 0.75 it is 75% efficient.
Note:
Efficiency is always less than or equal to one, it is never greater than one i.e a machine
can not be more than 100 % efficient.
Equation can be used for work or power.
Text Tsokos p. 110 Q 18 , p. 114 : 18
72
14.
Uniform Circular Motion
Drawing : Centripetal Acceleration = Change in direction without change in
speed.

v  Tangential speed
Fc = centripetal force towards center
ac = centripetal acceleration follows Fc - towards center
Fc
Fc
v
The centripetal acceleration ( ac ) of an object in uniform circular motion is NOT in the
same direction as the tangential velocity vector. If it were, the object would accelerate and
motion would not be uniform. Therefore, centripetal acceleration is a good example of
how you can cause acceleration just by changing direction without changing speed.
As a matter of fact, the centripetal acceleration is directed towards the center. This
centripetal force ( Fc causing the acceleration) causes the velocity vector to continuously
change direction thereby maintaining uniform circular motion. The centripetal force is
always perpendicular to the direction of motion and does no work. Therefore, there is no
change in kinetic energy and no change in speed (energy-work theorem) but a change
in velocity.
If there were no centripetal force the object would move in a straight line.
73
EXAMPLES
Wilson/ Buffa p. Ex. 7.7 Tension in string causes Centripetal force
74
Car on a curve - Friction
The friction between the tires and the road provides the centripetal force towards the
center of circular motion. No friction and the car skids off the road. Roadways are
designed to bank and provide centripetal force to prevent cars from skidding out of control.
The coefficient of friction between tire and road is also taken into account.
Sattelites, Moons , Planets – Gravity
The force of gravity causes the centripetal force needed to keep planets in orbit ( orbital
speed = tangential velocity).
Water in bucket fig. 7.27 p. 255 ; Roller coaster loop fig. 7.28 p. 256
75
FORMULAS
UCM is motion where an object moves in a circle with constant speed. Acceleration is not
constant and its magnitude is given by:
mv 2
 F  mac 
r
v 2 4 2 r
ac   2
r
T
(UCM, magnitude only)
Where: v = speed [m/s].
r = radius of the circle [m].
T = period, the time for one complete cycle (time for one revolution of a circle) [s (or
s/rev)].
Note:
The direction of v (and hence s) a tangent to the curve at
any point.
ac (aka centripetal acceleration) has direction at
perpendicular to v (and s) inward along the radius
[Figure 5-3 (Giancoli 2005; 108)].
ac is not constant. It is the magnitude only as direction
constantly changes.
Note:
Learn to derive the two equations above.
The period-frequency relationship is very important.
A commonly used unit for frequency is [rpm = rev/min]
76
EXAMPLE
Two masses, independently , are suspended from light strings and are in uniform motion
as illustrated below. m1 = 2.5 kg and m2 = 3.5kg The tension in the strings are T1= 4.5 N
and T2 = 2.9 N. Find a) the centripetal accelerations and b) the magnitude of the
tangential velocities.
77
Dynamics of UCM
Only by Newton’s 2nd Law is relevant to UCM. The Magnitude of a is constant, therefore
magnitude of F is constant The direction of F is the same as a, towards the center of the
circle.
mv 2
 F  mac 
r
ac = v2
r
A net force must be present in UCM otherwise the body would move in a straight line with
constant velocity. The net force is directed to the center of the circle and is
sometimes referred to as a centripetal force but be careful because this centripetal
force is not a new force, it is a net force that is created by other forces such as gravity,
friction, tension or the normal force.
Positive direction for FBDs is towards the center of the circle (same direction as a). Be
careful with vertical circles (down is positive at the top, up is positive at the bottom).
As with ma,
mv 2
r
is not an individual force, it is the result of the net force. It should not be
included in FBDs nor in determining the net force.
F is perpendicular to s at any given point, therefore does no work.
Example 5-22 Rounding a flat curve (Young and Freedman 2000; 142)
The BMW Z3 roadster, of mass 1000kg, is rounding a flat, unbanked curve with radius
230m. If the friction between the tires and the road is 8500N, find the maximum speed at
which the driver can take the curve without sliding out.
Tsokos p. 124 : 1 b) only, 2
78
79
LAB CENTRIPETAL FORCE
80
15.
Law of Gravitation Tsokos pp. 127 – 130
Why are the planets, moon and sun all nearly spherical? Why do some satellites circle the
Earth in 90 minutes, while the moon takes 27 days for the trip? And why don’t satellites fall
back to Earth? The study of gravitation provides answers for these and many related
questions.
Newton’s Universal Law of Gravitation
Every point mass of matter in the universe attracts
every other point mass with a force that is proportional
tothe product of the masses of the two particles and
inversely proportional to the square of the distance
between them.
Translating this into an equation, we have:
Fg 
GMm
r2
(Newton’s Universal Law of Gravitation)
Where: G = universal gravitational constant = 6.67 x 10-11 N.m2kg-2
M = mass of the larger object (IB uses m1 and m2)
Note:

Point mass assumption: If the separation between
the two objects is large compared to their radii we
can treat spherical objects as point masses particles with all their mass concentrated at the
center (Figure 12-2 Young and Freedman 2000 p.359) and r =
distance between the two centers of the
spheres.

Gravitational forces always act along a line joining
the two particles (Figure 12-1 Young and freedman 2000; 359).

Even when the masses of the two particles are
different, the two interacting forces have equal
magnitude (and form an action-reaction pair
Newton’s 3rd Law).
Don’t confuse g with G. ^ CP DVD - Jolly’s Method of measuring

of
G

The equation is found under Topic 6 (page 8) in the
IB data booklet.

Planets, moons and stars are spherical because all
particles are attracted to each other and tend to
move to minimize the distance between them, just like a lump of clay forms a
sphere if you squeeze it equally from all sides.

At points inside the Earth gravity does not decrease as 1/r2 indicates because some
of the mass of the Earth is on the opposite side of the body to the center and pulls
in the opposite direction. At the center of the Earth the gravitational force on the
body is zero.
81

Example 5-10 (Giancoli 2005; 119)

Mass of earth = 5.98 x 1024 kg

Mass of sun = 1.99 x 1030 kg
Tsokos p. 128 : Q1, Q2 p. 130 : 1 a) mass of earth = 5.98 x 1024 kg ,
mass of moon = 7.35 x 1022 kg , distance between moon and earth = 3.83 x 108 m
distance between sun and earth = 1.5 x 1011 m
82
How Force varies with m, M or R2
Many questions will ask multiplicative changes regarding the formula
1)
2)
3)
4)
F =GMm
R2
For example if you double the distance what happens to the Force ?
If you double M what happens to the force?
If you double M and m what happens to the force?
If you double M and m and reduce the distance between the objects by half , what
happens to the force?
The Force of Gravity on objects on the earth’s surface.
There is no separation between the object and the planet. Since the gravity of the earth
acts at its center the distance between the mass of the object and the mass of the earth
used in the formula is actually the radius of the earth ( 6.38 x106 m)
F =GMm
RE2
This Force of gravity on an object on the earth is also equal to the weight of the object.
Remember the weight of the object is the measurement of the gravitational force = mg so
we have:
F =GMm
RE2
=
weight = mg
Gravitational field strength ( g) = acceleration due to gravity of a given planet of
mass
g = GM
R2
83
This is derived from above :
F =GMm
RE2
GMm
R2
=
= mg
weight = mg

solving for g :
g = GM
R2
Do not confuse finding (g) gravity of a planet with finding F which is the total
gravitational force between 2 planets or 2 objects ( M and m). Remember g is also
called the gravitational field strength and the acceleration due to gravity of a particular
planet.
radius of the earth ( 6.38 x106 m)
Tsokos : p. 128 : Q3 , Q4, ; p. 131 : 3,4,
84
Gravitational field strength ( g) and gravitational force (F) definition :
Physicists wondered how a mass knows the presence of another mass nearby that will
attract it. They developed the idea of a gravitational field. A mass M is said to create a
gravitational field in the space around it. This means that when another mass ( m) is
placed at some point near M it feels the gravitational field.
The gravitational field strength ( g) at a certain point is the force per unit mass ( F/m)
experienced by a small point mass m…. at that point :
g= F
m
=
g = GM
R2
If M is the mass of the earth then the gravitational field strength (g) is the acceleration
due to gravity (g) at a distance R from the earth. Be careful with the wording:
Example 1:
Gravitational field strength at a point may be defined as
A.
the force on a small mass placed at the point.
B.
the force per unit mass on a small mass placed at the point.
C.
the work done to move unit mass from infinity to the point.
D.
the work done per unit mass to move a small mass from infinity to the point.
Note:

Gravitational field strength is a vector quantity.

SI units: [Nkg-1 or ms-2].

g = 9.81m/s2 only applies to situations very close to the Earth's surface (< 1000m).

M = mass of larger object e.g. the mass of the planet.

For points on the Earth’s surface r = rE = radius of the Earth (6.38 x 106m).

For points above the Earth’s surface the additional distance must be added to rE.

Equations work equally well for all other bodies e.g. the Moon.
Example 5-13 (Giancoli 2005; 121)
85
Example 2:
Why does an astronaut weigh less on the moon than on Earth?
a. The astronaut has less mass on the moon.
b. The astronaut is farther from Earth’s center when he or she is on the moon.
c. The gravitational field strength is less on the moon’s surface than on Earth’s
surface.
d. The astronaut is continually in free fall because the moon orbits Earth.
Example 3
Example
The acceleration of free fall of a small sphere of mass 5.0 × 10
–3
kg when close to
–2
the surface of Jupiter is 25 ms . The gravitational field strength at the surface of
Jupiter is
–4
–1
A.
2.0 × 10
B.
1.3 × 10
C.
25 N kg .
D.
5.0 × 10 N kg .
–1
N kg .
–1
N kg .
–1
3
–1
Example 4
Planet X has radius R and mass M. Planet Y has radius 2R and mass 8M.
Which one of the following is the correct value of the ratio
gravitatio nal field strength at surface of planet X
?
gravitatio nal field strength at surface of planet Y
A.
4
B.
2
C.
1
2
D.
1
4
Example
86
Probl
Problems
1. A 61.5 kg student sits at a desk 1.25 m away from a 70.0 kg student.
What is the magnitude of the gravitational force between the two
students? (G = 6.673  10 Nm
2. Two trucks with equal mass are attracted to each other with a
gravitational force of 6.7  10 N. The trucks are separated by a
distance of 3.0 m. What is the mass of one of the trucks? (G = 6.673 
10 Nm /kg )
The
87
REVIEW CIRCULAR MOTION
A satellite orbits the Earth at constant speed as shown below.
satellite
Earth
(a)
(b)
Draw on the diagram
(i)
an arrow labelled F to show the direction of the gravitational force of the
Earth on the satellite.
(ii)
an arrow labelled V to show the direction of the velocity of the satellite.
Although the speed of the satellite is constant, it is accelerating. Explain why it is
accelerating.
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
(c)
Discuss whether or not the gravitational force does work on the satellite.
.................................................................................................................................
.................................................................................................................................
...............................................................................................................................
88
NEW FORMULAS – Circular Motion
Only watch first 3 min. of video below:
https://www.youtube.com/watch?v=9QDKnQm_poI
Angular Speed (𝜔 ) and Linear ( or tangential) Speed ( v)
As you increase your distance from the center ( r increases) your tangential – linear
speed (v) increases. An object on the outer part of the circle will cover a greater
circumference or circular path ( 2 π r ) in the same amount of time ( t for one revolution)
and there fore have a greater speed. v is proportional to r :
v = distance
time
=
2πr
T
However your angular speed (𝜔 ) does not change. Angular speed 𝜔 , just like the word
says ( angular), is based on the angle (Θ) which does not change and so your angular
speed does not change anywhere on the circle ( as long as Θ is the same) :
𝜔= Θ
t
89
Sometimes you need to calculate tangential or linear speed ( v) from angular speed
(𝜔 ) or vice versa :
v = 𝜔r
v = linear speed or tangential speed
𝜔 = angular speed
𝜔= Θ
t
( NOT IN DATA BOOKLET)
Rads and degrees: use rads but if given degrees remember:
3600 = 2 π = 6.28 rad
1 rad = 57.30
Be able to derive :
Since Circumference = 2 π r and T = time for one revolution so
v = distance
time
T=
2π
𝜔
=
2πr
T
(NOT IN DATA BOOKLET)
From Topic 9 Simple Harmonic Motion
a= 𝜔2r
90
Some Problems…… see solutions on next page 
1. Convert the following:
a) 3600 = ______ rad
b) 1800 = ______ rad
c) 900 = ______ rad
d) 0.79 rad = _______ degrees
e) 2.1 rad = _______ degrees
f) π rad = _______ degrees
2. Do all points on a wheel rotating about a fixed axis through its center have the same
angular speed? Same tangential speed? Explain
3. A wheel is spinning at a constant angular speed about an axis through its center.
A) Which points on the wheel will have the greatest and smallest tangential speeds?
B) Which points on the wheel will have the greatest and smallest angular speeds?
4. If a particle is rotating with an angular speed of 3.5 rads per second
a) what is the speed in degrees s-1 ?
b) how long will it take for the particle to go through one revolution?
5. A merry go round makes 24 revolutions in 3.0 minutes.
a) What is the averages angular speed in degrees s-1
b) rads s-1
c) what is the tangential speed of a person 4.0m from the center?
d) what is the tangential speed of a person 5.0m from the center?
6. A wheel of radius 1.5 m rotates at a uniform speed . If a point on the rim of the wheel
has a centripetal acceleration of 1.2ms-2 , what is the point’s tangential speed?
91
Some Problems
1.
a) 3600 = ___2 π or 6.28 rad
b) 1800 = __π
or 3.14 ____ rad
0
c) 90 = __ π / 2
___ rad
d) 0.79 rad = __45_____ degrees
e) 2.1 rad = ____120___ degrees
f) π rad = __180_____ degrees
2. Same angular speed but different tangential speeds :
An object on the outer part of the circle will cover a greater circumference or circular path (
2 π r ) in the same amount of time ( t for one revolution) and there fore have a greater
speed. v is proportional to r :
v = distance =
2πr
time
T
However your angular speed (𝜔 ) does not change. Angular speed 𝜔 , just like the word
says ( angular), is based on the angle (Θ) which does not change and so your angular
speed does not change anywhere on the circle ( as long as Θ is the same) :
𝜔= Θ
t
3. a) Greatest tangential speeds at points farthest from the center; lowest tangential
speeds at points closest to the center.
b) All points have the same angular speed.
4. a) 2000/ sec
b) One rev. = 360 0
𝜔= Θ
t
t = Θ = 360
𝜔
2000/ sec
5. a) 24 rev x 360
3min x 60 s
= 1.8 s
= 480 s-1
b) 0.84 rads s-1
c) v = 𝜔 r
= (0.84 rads s-1 )( 4.0m) = 3.5 ms-1
d) v = 𝜔 r
= (0.84 rads s-1 )( 5.0m) = 4.2 ms-1
6.
v = √ar = √(1.2)(1.5) = 1.3ms-2
92