Download VU2 Movement 2008

VCE Physics
Unit 2
Topic 1
Movement
View physics as a system of thinking about the world rather than information
that can be dumped into your brain without integrating it into your own belief
systems.
Unit Outline
To achieve this outcome the student should use scientific methods, data,
theories and knowledge to:
•
•
•
•
•
•
•
•
•
Describe non-uniform and uniform motion along a straight line graphically;
Analyse motion along a straight line graphically, numerically and algebraically;
Describe how changes in movement are caused by the actions of forces;
Model forces as external actions through the centre of mass point of each body;
Explain movement in terms of the Newtonian model and some of its assumptions,
including Newton’s 3 laws of motion, forces act on point particles, and the ideal,
frictionless world.
Compare the accounts of the action of forces by Aristotle, Galileo and Newton.
Apply the vector model of forces including vector addition, vector subtraction and
components to readily observable forces including weight, friction and reaction
forces;
Model mathematically work as force multiplied by distance for a constant force and as
area under the force versus distance graph.
Interpret energy transfers and transformations using an energy conservation model
applied to ideas of work, energy and power, including transfers between
– kinetic energy and gravitational potential energy close to the Earth’s surface;
– potential energy and kinetic energy in springs;
Chapter 1
Introduction
1.0 An Ideal World
To make life easier for
Physics students
situations or events
which require
mathematical analysis
are often described as
occuring in an ideal,
frictionless world.
In the ideal world an object
under the influence of Earth’s
gravity will accelerate at 9.8 ms-2
throughout its journey never
reaching a terminal velocity.
In the ideal world the
laws of motion apply
exactly, eg. objects
which are moving will
continue to move
with the same speed
unless or until
something occurs to
change this.
In the ideal world energy
transformations are always
100% efficient, so that the
potential energy of a
pendulum at the top of its
swing is all converted to
Kinetic Energy (motion
energy) at the bottom.
In the ideal world perpetual motion
machines are commonplace.
1.1 The S.I. System
In 1960, the “General Conference of Weights and Measures” , a Paris based
international organisation, agreed that one set of units would be adopted
world wide for the measurement of physical quantities.
This system is called the Systeme Internationale d’Units, or more simply the
S.I. System.
The system is used and recognised worldwide and defines 7 fundamental
units.
Physical Quantity
S.I. Unit
Symbol
Length
metre
m
Mass
kilogram
kg
Time
second
s
Electric Current
ampere
A
Temperature
kelvin
K
Luminous Intensity
candela
cd
Amount of Substance
mole
mol
All other units are derived from these 7 fundamentals.
A derived unit is the force unit, the Newton,
which is found from mass x length x 1/(time)2
Thus the Newton has
dimensions kg x m x s-2
1.2
Length; metre [m]
S.I. Definitions
Current; ampere [A]
It is the distance light
It is that current which produces
travels, in a vacuum,
a force of 2 x 10-7 N between two
in 1/299,792,458th of
parallel wires which are 1 metre
a second.
apart in a vacuum.
Temperature; kelvin [K]
It is 1/273.16th of the
thermodynamic temperature of
the triple point of water.
Amount of Substance; mole [mol]
Mass; kilogram [kg] It is the mass of
It is the amount of substance that
a platinum-iridium cylinder kept at
contains as many elementary
Sevres in France. It is now the only
units as there are atoms in 0.012
basic unit still defined in terms of a
kg of 12C
material object.
Luminous Intensity; candela [cd]
It is the intensity of a source
Time; second [s]
of light of a specified
It is the length of time
frequency, which gives a
taken for 9,192,631,770
specified amount of power in
periods of vibration of
a given direction.
the caesium-133 atom
Questions
to occur.
Fundamentals
QUESTIONS
1. Which of the following quantities have fundamental units and which
have derived ?
Quantity (Unit)
Fundamental
Power (Watts)
Derived
√
Distance (metre)
√
Time (second)
√
Force (Newton)
√
Energy (Joule)
√
Mass (kilogram)
√
√
Electrical Resistance (ohms)
Temperature (kelvin)
√
Electric Current (amperes)
√
Fundamentals
2. From which of the fundamental units do the following derive their units ?
Quantity (Unit)
e.g Force (Newton)
Acceleration (ms-2)
Momentum (kgms-1)
Impulse (Newton.second)
Velocity (ms-1)
Work (Joule)
Note: W = F.d
Fundamental Units
Mass (kg), length (m), time (s)
Length (m), time (s)
Mass (kg), length (m), time (s)
Mass (kg), length (m), time (s)
Length (m), time (s)
Mass (kg), length (m), time (s)
Fundamentals
3. Show that 1 ms-1 = 3.6 kmh-1
Two relevant conversion factors are: 1 km = 1000 m, 1 h = 3600 s
These can be written as
1km
or
1000m
1000m
and
1km
1h
or
3600 s
3600s
1h
Which ones to use ?
Easy, you want to end up with km on the top line and h on the bottom
1m x 1km x
s
1000m
so 1 ms-1 = 3.6 kmh-1
3.6
3600s
1h
1.3
Position
In order to specify the position of an object we first need to define an ORIGIN
or starting point from which measurements can be taken.
For example, on the number line, the point 0 is taken as the origin and all
measurements are related to that point.
-40 -35 -30 -25 -20 -15 -10 -5
0
5
10
15 20 25 30 35 40
Numbers to the right of zero are labelled positive
Numbers to the left of zero are labelled negative
A number 40 is 40 units to the right of 0
A number -25 is 25 units to the left of 0
Questions
Position
4. What needs to be defined before the position of any object can be
specified ?
A zero point needs to be defined before the position of an object can be
defined
5. (a) What distance has been covered when an object moves from
position +150 m to position + 275 m ?
Change in position = final position – initial position
= +275 – (+150) = + 125 m. Just writing 125 m is OK
(b) What distance has been covered when an object moves from
position + 10 m to position -133.5 m ?
Change in position = final position – initial position
= -133.5 – (+10) = - 143.5 m. Negative sign IS required
Chapter 2
Vectors & Scalars
2.0 Scalars and Vectors
Before proceeding further we need to define two new quantities:
SCALAR QUANTITIES
These are completely defined by
•A Number and
•A Unit
Examples of scalars are:
Temperature 170, Mass 1.5 kg
VECTOR QUANTITIES
These are completely defined by
•A Number
•A Unit and
•A Direction
Examples of vectors are:
Displacement 25 km West,
Force 14 Newtons South
Vectors are usually represented by an ARROW, with the length of the arrow
indicating the size of the quantity and the direction of the arrow the direction
of the quantity.
N
This vector represents a
Force of 4 N, acting North West
Questions
Vectors and Scalars
6. Which of the following quantities are scalars and which vectors ?
Quantity
Unit
Scalar
Vector
Distance
metre
Momentum
kgms-1 East
Kinetic Energy
joule
Acceleration
ms-2 N45oE
√
Gravitational Field Strength
Nkg-1 downwards
√
Displacement
metre sideways
√
Age
years
Velocity
ms-1 West
Temperature
oC
√
√
√
√
√
√
2.1
Vector Addition &
Subtraction
Vectors can be at any angle to one another and still be added.
This can be done in two ways:
Draw accurate, scale vectors on graph paper and measure the size
and direction of the result of the addition, called the “resultant vector”
Draw sketch vectors and use trig and algebraic methods to calculate
the size and direction of the resultant.
The tail of the second adds to
the head of the first
ADDITION
=
+
5.0 units
SE
5.0 units
NE
7.1 units East
The resultant is
drawn from the tail
of the first to the
head of the second
SUBTRACTION
5.0 units
NE
_
=
5.0 units
SE
To subtract, reverse the direction
of the negative vector then add.
5.0 units
NE
+
5.0 units
NW
=
7.1 units
North
2.2
Vector Components
A single vector can be broken up into two or more parts called COMPONENTS.
This process is useful when, for example, trying to find the vertical and
horizontal parts of a force which is accelerating a mass through the Earth’s
atmosphere.
FH and FV are the COMPONENTS of
the force F.
F = 5 x 106 N
At present, the total
force is directed at 30o
above the horizontal
FV
300
30o
FH
The Horizontal component of the force (FH) can be found using trig methods:
FH = F cos 30o
= (5 x 106) ( 0.866)
= 4.3 x 106 N
Similarly for the Vertical component (FV),
FV = F sin 30o
= (5 x106)(0.5)
= 2.5 x 106 N
Questions
Vector Addition
7. What is the resultant force when 2 forces (6.0 N west and 4.0 N south) act on
an object at the same time ?
6 N west
θ
θ = tan-1 4/6
= 33.7o
4 N south
Resultant Force = √(6)2 + (4)2 = 7.2 N
Vector Subtraction
8. Calculate the change in velocity of an object initially travelling at 8.5 ms-1
East whose final velocity was 8.5 ms-1 West. (remember Change in Velocity =
Final Velocity – Initial Velocity)
8.5 ms-1 West
8.5 ms-1 West
+
8.5 ms-1 East
8.5 ms-1 West
17 ms-1 West
=
=
Vector Components
9. An boy fires a stone from slingshot. The stone leaves with a velocity of 27
ms-1 at an angle 320 above the horizontal. Calculate the vertical and horizontal
components of the stone’s velocity.
Vv
θ = 32o
VH
VH = 27 Cos 32o
= 22.9 ms-1
Vv = 27 sin 32o
= 14.3 ms-1
Vector addition/subtraction
10. Calculate the acceleration of a car whose velocity changes from 16 ms-1 west
to 21 ms-1 north in 1.5 seconds (acceleration = change in velocity/change in time)
Acceleration is a vector quantity so a vector calculation is required to calculate it.
Initial Velocity
16 ms-1 West
Final Velocity
21 ms-1 North
Resultant Velocity
= √(21)2 + (16)2
= 26.4 ms-1
Change in Velocity = VF – Vi
-
=
+
=
θ
Acceleration = change in velocity/change in time
= 26.4/1.5
= 17.6 ms-2 at N37.3oW
θ = tan-1 16/21
= 37.3o
Chapter 3
Kinematics
3.0 Distance & Displacement
Distance is a SCALAR quantity. It
has a Unit (metres) but no
Direction.
Distance is best defined as “How far
you have travelled in your journey”
Displacement is a VECTOR quantity
Having both a Unit (metres) and a
Direction.
Displacement is best defined as
“How far from your starting point you
are at the end of your journey”
The difference between these two quantities is easily illustrated with a
simple example. You are sent on a message from home to tell the butcher
his meat is off.
Positive Direction
2 km
At this point in the journey , Distance travelled = 2 km and Displacement = + 2 km
At the end of the journey, Distance travelled = 2 + 2 = 4 km
while Displacement = +2 + (-2) = 0 km
3.1 Speed & Velocity
These two terms are used interchangeably in the community but strictly
speaking they are different:
Velocity is the time rate of change
of displacement, i.e.,
Velocity = Displacement
Time
Speed is the time rate of change of
distance, i.e.,
Speed = Distance
Time
Speed is a SCALAR QUANTITY,
having a unit (ms-1), but no
direction.
Thus a speed would be:
100 kmh-1 or,
27 ms-1
Velocity is a VECTOR QUANTITY,
having a unit (ms-1) AND a direction.
Thus a velocity would be:
100 kmh-1 South or
- 27 ms-1
3.2 Acceleration
Acceleration is defined as the time
rate of change of velocity, i.e.,
Acceleration = Velocity
Time
There is no scalar measurement
of acceleration, so acceleration
MUST always be quoted with a
direction.
Acceleration is a VECTOR QUANTITY
having both a unit (ms-2) and a
direction.
Typically, Acceleration means
an increase in velocity over
time, while Deceleration means
a decrease in velocity over time.
v
a
When v and a are in the same direction,
the car accelerates and its velocity will
increase over time.
a
v
When v and a are in the opposite
direction, the car decelerates and
its velocity will decrease over time.
3.3 Instantaneous & Average
Velocity
The term velocity can be
misleading, depending upon
whether you are concerned with an
Instantaneous or an Average value.
The best way to illustrate the
difference between the two is with You take a car journey out of a city to
an example.
your gran’s place in a country town 90
km away. The journey takes you a total of
2 hours.
The average velocity for this journey,
vAV = Total Displacement = 90 = 45 kmh-1
Total Time
2
However, your instantaneous velocity measured at a particular time during
the journey would have varied between 0 kmh-1 when stopped at traffic
lights, to, say 120 kmh-1 when speeding along the freeway.
Average and Instantaneous velocities are rarely the same.
Unless otherwise stated, all the problems you do in this section of the
course require you to use Instantaneous Velocities.
Questions
Kinematics
11. A runner completes a 400 m race (once around the track) in 21 seconds what
is:
(a) her distance travelled (in km), (b) her displacement (in km), (c) her speed (in
ms-1) and (d) her velocity (in ms-1) ?
(a) Distance = 0.4 km
(b) Displacement = 0 km
(c) Speed = distance/time = 400/21 = 19ms-1
(d) Velocity = displacement/time = 0/21 = 0 ms-1
Acceleration
12. A roller coaster, at the end of its journey, changes it’s velocity from 36 ms-1 to
0 ms-1 in 2.5 sec. Calculate the roller coaster’s acceleration.
a = change in velocity/change in time = (0 – 36)/2.5 = - 14.4 ms-2
Chapter 4
Motion by Graphs
4.0
Graphical Relationships
It is often useful and convenient
to represent information about
things like position, velocity,
acceleration etc., using graphs.
Graphs “tell you a story”.
You need to develop the skills andThere are two basic types of graphs used
abilities to “read the story”.
in Physics:
(a) Sketch Graphs – give a “broad brush”
picture of the general relationship
between the two quantities graphed.
(b) Numerical Graphs – give the exact
mathematical relationship between the
two quantities graphed and may be
used to calculate or deduce numerical
values.
4.1
Sketch Graphs
Sketch graphs have labelled axes
but no numerical values, they give a
general broad brush relation
between the quantities.
Velocity
Distance
Displacement
Time
The Story:
The
Story:
The
object
begins its journey
The
The
Story:
Story:
As
time
passes
itsthe
at
the
origin
at t = 0. the
As
time
As
As
time
time
passes
passes,
passes
its displacement
displacement
gets
velocity
distance
remains
of
the
increases at a constant rate
larger
atfrom
an increasing
constant.
object
itsSo time
(slope
is constant).
rate.
rate
ofis
change
of of
This
starting
a graph
point
does
an
displacement
which
equals
This
the graphatof an
object
not is
change.
travelling
velocity
is
constant.with
object
moving
constant
This
the
velocity
graph
of a
This
is ais
graph
of
an object
constant
acceleration
travelling
at constant
stationary
object
velocity
Questions
Sketch Graphs
Distance
13 (a)
Distance versus time graph. As time
passes displacement remains the
same. This is the graph of a
stationary object
Time
Displacement
(b)
Displacement versus time graph. As
time passes its displacement is
increasing in a uniform manner. This
is a graph of an object travelling at
constant velocity.
Time
Sketch Graphs
(c)
Velocity
Velocity versus time graph. As time
passes the velocity of the object
remains the same. This is a graph
of an object travelling at constant
velocity.
Time
(d)
Displacement
Time
Displacement versus time graph. As
time passes its displacement gets
larger at an increasing rate. This is a
graph of an accelerating object. (if the
shape is parabolic the object is
increasing its speed in a uniform
fashion i.e. it has a constant
acceleration)
4.2 Exact Graphical Relationships
The graphs you are required to interpret mathematically are those where
distance or displacement, speed or velocity or acceleration are plotted
against time.
The information available from these graphs are summarised in the table
given below.
Graph Type
Distance or
Displacement
versus
Time
Speed or Velocity
versus
Time
Acceleration
versus
Time
Obtain from Slope of
the Graph
Obtain from Area
Under the Graph
Distance or
Displacement
Speed or
Velocity
No Useful
Information
Speed or
Velocity
Acceleration
Distance or
Displacement
Acceleration
No Useful
Information
Velocity
Read directly from
the Graph
Learn this table off by heart.
Put it on any cheat sheet you are allowed to use.
Questions
Graphical Interpretation
14. Given below is the Distance vs Time graph for a cyclist riding along a
straight path.
(a) In which section (A,B,C or D) is
Distance
the cyclist stationary ?
A
B
C
D
(b) In which section is the cyclist
travelling at her slowest (but not zero)
20
speed ?
(c) What is her speed in part (b)
above ?
10
(d) What distance did she cover in
the first 40 seconds of her journey ?
(e) In which section(s) of the graph is
Time (s)
her speed the greatest ?
0
(f) What is her displacement from
20 30 40 50 60
10
her starting point at t = 50 sec ?
(a) Stationary in section C
(b) Section B
(c) Travels 10 m in 20 s  speed = 10/20 = 0.5 ms-1
(d) 20 m (read directly from graph)
(e) Section D (travels 20 m in 10 s) speed = 2 ms-1
(f) Displacement at t = 50 s is 0 m (i.e., back at starting point)
Graphical Interpretation
15. Shown below is the Velocity vs Time
graph for a motorist travelling along a
straight section of road.
(a) What is the motorist's
Velocity
displacement after 4.0 sec ?
(b) What is the motorists
acceleration during this 4.0
sec period ?
(c) What distance has the
Time(s)
motorist covered in the 20.0
sec of his journey ?
(d) What is the motorist's
(a) Displacement = area under velocity time graph. displacement at t = 20.0 sec
Between t = 0 and t = 4 s. Area = ½ (10 x 4) = 20 (e) What happens to the
motorists velocity at t = 20.0
m
sec? Is this realistic ?
(b) Acceleration = slope of velocity time graph
(f) Sketch an acceleration vs
= (10 – 0)/(4 – 0) = 2.5 ms-2
time graph for this journey.
(c) Distance = area under graph (disregarding
(ms-1)
10
8
6
4
2
0
1
-2
-4
-6
-8
-10
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
signs) Total area = ½(10 x 4) + (6 x 10) + ½(10 x
2) + ½(9 x 2) + (6 x 9) = 20 + 60 + 10 + 9 + 54 =
153 m
(d) Displacement = area under graph (taking signs
into account) = ½(10 x 4) + (6 x 10) + ½(10 x 2) ½(9 x 2) - (6 x 9) = 20 + 60 + 10 - 9 – 54 = 27 m
(e) Velocity falls from 9 ms-1
to zero in no time – no
realistic, as it would require
an infinite deceleration to
achieve this.
Graphical Interpretation
15, continued
a
(e)
To Infinity
2.5
4
-4.5
-5
10 12 14
t
20
Graphical Interpretation
16. An object is fired vertically upward on a DISTANT PLANET. Shown below is
the Velocity vs Time graph for the object. The time commences the instant the
object leaves the launcher
(a) What is the acceleration of
Velocity (ms-1)
the object ?
30
(b) What is the maximum height
attained by the object ?
(c) How long does the object
take to stop ?
Time (s) (d) How far above the ground is
0
2
4
6
8
10
12
the object at time t = 10.0 sec ?
-30
(a) Acceleration = slope of velocity time graph.
Slope = (30 – 0)/(0 – 6) = -5.0 ms-2
(b) Displacement = area under velocity time graph = ½ (6 x 30) = 90 m
(c) Stops at t = 6.0 sec
(d) The rocket has risen to a height of 90 m in 6 sec. It then falls a distance of
½ (4 x 20) = 40 m, so it will be 90 – 40 = 50 m above the ground at t = 10 s
Chapter 5
The Equations of
Motion
5.0 The Equations of Motion
The Equations of Motion are a set of equations
linking displacement, velocity, acceleration and
time.
They allow calculations of these quantities
without the need for graphical representations.
The 3 main equations are:
v = u + at
Where,
v2 = u2 + 2as
u = initial velocity (ms-1)
s = ut + ½at2
v = final velocity (ms-1)
a = acceleration (ms-2)
s = displacement (m)
t = time (s)
THESE EQUATION CAN ONLY BE USED IF
THE ACCELERATION IS CONSTANT
When using the equations, always list out the
information given and note what you need to find,
then choose the most appropriate equation.
+ve
u=
v=
a=
s=
t=
In some cases you also need to define a positive direction, up or down for
vertical motion, left or right for horizontal motion questions
5.1 Motion Under Gravity
Objects (close to the surface)
The acceleration in this case is
falling through the Earth’s
ALWAYS directed downward.
gravitational field are subject
Objects thrown or fired directly
to a constant acceleration of
upwards would thus have their
-2
9.8 ms .
velocity and acceleration in
Since the acceleration is
opposite directions.
constant this motion can be
analysed by the equations of
motion.
The calculations using the equations of
v = u + at
v2 = u2 + 2as
s = ut + ½at2
motion always ignore the effects of
friction and air resistance
You need to go through the
same process of listing
information
and deciding on a positive direction
+ve
u=
v=
a=
s=
t=
Questions
Equations of Motion
17. A truck travels from rest for 10.0 sec with an acceleration of 3.0 ms-2.
Calculate the truck's final velocity and total distance travelled.
List information: u = 0, v = ?, a = 3.0 ms-2, s = ?, t = 10 s
firstly find v, use v = u + at  v = 0 + (3.0)(10) = 30 ms-1
then find x use x = ut + ½at2  (0)(10) +½(3.0)(10)2 = 150 m.
18. A ball rolling down an inclined plane from rest travels a distance of 20.0 m
in 4.00 sec. Calculate its acceleration and its final speed
List information: u = 0, v = ?, a = ?, x = 20.0 m, t = 4.0 s
Firstly find a, use x = ut + ½at2  20.0 = (0)(4.0) + ½a(4.0)2  a = 1.25 ms-2
The find v, use v = u + at  v = 0 + (1.25)(4.0)  5.0 ms-1
Equations of Motion
19. The speed of a freewheeling skateboard travelling on a level surface falls
from 10.0 ms-1 to 5.00 ms-1 in moving a distance of 30.0 m. If the rate of
slowdown is constant, how much further will the skateboard travel before
coming to rest ?
List information u = 10 ms-1, v = 5.0 ms-1 ,a = ?, x = 30 m, t = ?
Cannot get to answer in 1 step. First find acceleration
Use v2 = u2 + 2ax  a = (v2 – u2)/2x  a = - 1.25 ms-2
Now new information u = 5.0 ms-1, v = 0, a = -1.25ms-2, x = ?, t = ?
Use v2 = u2 + 2ax  x = (v2 – u2)/2a  x = 10 m
20. A bullet leaves the barrel of a gun aimed vertically upwards at 140 ms-1.
How long will it take to reach its maximum height ? (Ignore air resistance
and use g = 10 ms-2) .
List information (up is +ve) u = 140 ms-1 v = 0, a = -10 ms-2, x = ? t = ?
Use v = u + at  t = (v – u)/a = (0 – 140)/-10 = 14 s
Chapter 6
Forced Change
6.0
What
is
a
Force
?
"A force is an interaction between
two material objects involving a
push or a pull."
How is this different from the usual
textbook definition of a Force
Forces are like conversations in
simply being a “push or a pull” ? that:
First, a force is an "interaction".
To have a force, you have to
You can compare a force to another
have 2 objects - one object
common interaction - a conversation.
pushes, the other gets pushed.
A conversation is an interaction between 2
In the definition, "(material)
people involving the exchange of words
objects" means that both objects
(and ideas).
have to be made out of matter Some things to notice about a
atoms and molecules. They both
conversation (or any interaction) are:
have to be "things", in the sense
To have a conversation, you need two
that a chair is a "thing".
people. One person can't have a
A force is something that
conversation
happens between 2 objects. It is
A conversation is something that happens
not an independently existing
between two people.
"thing" (object) in the sense that
It is not an independently existing "thing"
a chair is an independently
(object), in the sense that a chair is an
existing "thing".
Questions
independently existing "thing".
Force
21. A force is an interaction between 2 objects. Therefore a force can be likened to
A: Loving chocolate
B: Fear of flying
C: Hatred of cigarettes
D: Having an argument with your partner
22. Between which pair can a force NOT exist ?
A: A book and a table
B: A person and a ghost
C: A bicycle and a footpath
D: A bug and a windscreen
6.1 What Kinds of Forces Exist ?
For simplicity sake, all forces (interactions) between
objects can be placed into two broad categories:
1. Contact forces are types
of forces in which the two
interacting objects are
physically contacting each
other.
Examples of contact forces
Force is a quantity which is
include frictional forces,
measured using the derived
tensional forces, normal
metric unit known as the
forces, air resistance forces,
Newton.
and applied forces.
One Newton (N) is the amount of
force required to give a 1 kg
2. Field Forces are forces in which
mass an acceleration of 1 ms-2.
the two interacting objects are not
So 1N = 1 kgms-2
in contact with each other, yet are
able to exert a push or pull despite
Force is a vector quantity,
a physical separation.
you must describe both the
Examples of field forces include
magnitude (size) and the
Gravitational Forces, Electrostatic
direction.
Forces and Magnetic Forces
Questions
Contact or Field Forces
23. Classify the following as examples of either Contact or Field forces in action
(or maybe both acting at the same time).
EXAMPLE
CONTACT
FORCE
(a) A punch in the nose
√
(b) A parachutist free falling
√
(c) Bouncing a ball on the ground
√
FIELD
FORCE
√
(d) A magnet attracting a nail
√
(e) Two positive charges repelling
each other
√
(f) Friction when dragging a
refrigerator across the floor
√
(g) A shotput after leaving the
thrower’s hand
√
√
6.2
What
Do
Forces
Do
?
Forces affect motion. They can:
BEGINNING MOTION:
A constant force (in the same
direction as the motion) produces
an ever increasing velocity.
• Begin motion
• Change motion
• Stop motion
• Have no effect
FR
NO EFFECT:
A total applied
force smaller
than friction
will not move
the mass
CHANGING MOTION:
A constant force (at right angles to the
motion) produces an ever changing
direction of velocity.
STOPPING MOTION:
A constant force (in the opposite
direction to the motion) produces
an ever decreasing velocity.
6.3
Where Forces Act
Forces acting on objects
must have a point of
application, a place
where the force acts.
For Contact Forces the
point of application is
simply the point at
which the force initiator
contacts the object.
Force of carton on finger
Force of finger on carton
C of M
For Field Forces, the
only one applicable in
movement being gravity,
will act through the
centre of mass of the
object
Gravitational Force
Questions
Net Force
24. A body is at rest. Does this necessarily mean that it has no force acting
on it ? Justify your answer.
NO – A body will remain at rest if the NET FORCE acting is zero – it could
have any number of forces acting on it. So long as these forces add to zero it
will remain at rest.
25. Calculate the net force acting on the object in each of the situations
shown. (a)
(b)
900 N
1200 N N
300 N Left
(c)
0N
75 N
95 N
20 N Left
250 N
250 N
(d)
300 N Down
150 N
450 N
6.4
Forces in 2 Dimensions
Forces can act in any direction and the total or resultant force is the
vector sum of all the forces acting.
Tim, Tom and Tam, the triplets,
are fighting over a teddy bear.
Each exerts a different force.
What will be the net force on the
bear ?
Tom
Tim
Tam
A force diagram
shows each boy’s
contribution
Add the vectors head to tail.
FTOM = 25 N
FTIM = 26 N
Tim
FTAM = 18 N
FRESULTANT
The resultant force is the vector joining the
starting point to the finishing point
The bear will then accelerate in the direction of the resultant force
Questions
Net Force
26. Tim, Tom and Tam, the triplets, are fighting over a teddy bear. Each exerts a
different force. A force diagram shows each boy’s contribution. What will be the
net force on the bear ?
FTIM = 35 N
Resolve FTIM and FTOM
to give F = 9 N left
Then resolve this force with FTAM
9N
θ
X
12 N
FTOM = 26 N
FTAM = 12 N
X = √(92 + 122)
= 15 N
θ = sin-1 (9/12)
= 48.60
Net force is 15 N directed at S48.6oW
6.5
Weight
Weight is the outcome of a gravitational field acting on a mass
Weight is a FORCE and is measured in Newtons.
Its direction is along the line joining the centres of the two bodies which,
between them, generate the Gravitational Field.
1 kg
9.8 N
Near the surface of the Earth, each kilogram of mass is
attracted toward the centre of the earth by a force of 9.8 N.
(Of course each kilogram of Earth is also attracted to the
mass by the same force, Newton 3)
So, the Gravitational Field Strength near the Earth’s
surface = 9.8 Nkg-1
Weight and mass are NOT the same, but they are related
through the formula:
W = mg
Where:
W = Weight (N)
m = mass (kg)
g = Grav. Field
Strength (Nkg-1)
Questions
1 kg
Mass & Weight
27. Fill in the blank spaces in the table based on a person whose mass on earth
is 56 kg
Planet
Mass on planet
(kg)
Grav Field Strength
(Nkg-1)
Earth
56
9.81
Mercury
56
0.36
20.2
Venus
56
0.88
20.2
Jupiter
56
26.04
1458.2
Saturn
56
11.19
626.6
Uranus
56
10.49
Weight on planet
(N)
549.4
587.4
6.6
Reaction Force
All objects on and near the Earth’s
surface are subject to the
gravitational force.
Any object subject to a net or
resultant force will accelerate in the
direction of that force (Newton 2).
Why then do objects placed on a
table on the Earth’s surface remain
stationary ?
There must be a force equal in size
and opposite in direction to cancel
out the gravitational force.
There is such a force. It is called the
REACTION or NORMAL FORCE.
R
W
W
Because there is no net or resultant
force on the vase, it remains
stationary on the table
Remove the table, the reaction force
disappears and the vase accelerates
under the action of W, until it
encounters the floor and probably
smashes.
The Reaction Force only exists as a
result of the action of the weight of
the vase acting on the table top and
as such the reaction force does not
exist as an isolated force in its own
right.
Note: W and R are NOT an action reaction pair. Why?
Because when R disappears W does not.
Chapter 7
Centre of Mass
7.0 Centre of Mass
In order to deal with large objects it is useful to think of all the object’s mass
being concentrated at one point, this point being the Centre of Mass of the
object.
For regularly shaped
objects eg. squares or
rectangles, cubes or
spheres the Centre of Mass
of the object is in the
C of M
geometric centre of the
object
For odd shaped
objects such as a
boomerang, the
Centre of Mass may
Centre of Mass
fall outside the
perimeter of the
object.
The C of M is the point around which the object it
will spin if a torque or turning force is applied to
the object.
7.1 Translation and Rotation
When a Force acts C through
the Centre of Mass (of M) of an
object or structure, it causes
Translational Motion, ie. The
object moves in the direction of
the applied force according to
Newton’s 2nd Law. (see slide
6.3)
A Force acting through the Centre of Mass
causes Translational Motion only
When the force is applied to
another part of the object or
structure, a TORQUE or TWISTING
FORCE or TURNING MOMENT is
applied and Rotational as well as
Translational motion occurs
A Force acting at a point other than the C of M will
cause BOTH Translational AND Rotational Motion.
Chapter 8
Newton’s Laws
8.0
Aristotle
to
Newton
It was Galileo (1564 – 1642) who
Attempts to explain the “causes
of motion” (a field of study
called dynamics), were first
recorded in the time of the
ancient Greek philosopher
Aristotle (384 – 322 BC).
Aristotle
It was believed that constant
speed required a constant force.
This seemed logical as
everyone could see that a
horse needed to apply a
constant pull to haul a cart at
constant speed.
Galileo
However there were problems
with the theories which could
not, for example, explain why
falling objects tended to
increase their speed in the
absence of any visible force or
why heavenly bodies behaved
differently than those on earth.
Newton
was the first to define the
property of matter we call
INERTIA, (matter’s tendency to
resist changes in its motion), with
his law which said “when no
force exists a body will stay at
rest or move with constant speed
Philosophers prior to Newton
believed a set of laws covering
motion on earth could be
developed, but they needed to be
modified to explain the motions of
heavenly bodies.
Isaac Newton (1642 – 1727) was
the first to realise there WAS a
universal set of laws which could
describe the motion of ALL
bodies, BUT these laws had to be
modified for use within the
friction riddled confines of the
Earth and its atmosphere.
Questions
Aristotle to Newton
Scientist
Statement
Newton
Constant speed requires constant force
Aristotle
Defined the property of matter called inertia
Galileo
A universal set of laws applicable everywhere but must
be modified for use on earth
8.1 Newton’s Laws
Newton developed 3 laws which cover all aspects of
motion (provided objects travel at speeds are well
below the speed of light).
Law 1
(The Law of Inertia)
A body will remain at rest, or in a state of uniform
motion, unless acted upon by a net external force.
Law 2
The acceleration of a body is directly proportional to
net force applied and inversely proportional to its
mass. Mathematically, a = F/m more commonly
written as F = ma
Law 3
Newton, at age 26
(Action Reaction Law)
For every action there is an equal and
opposite reaction.
Motion at or near the speed of light
is explained by Albert Einstein’s
Theory of Special Relativity.
8.2
Objects want
to keep on
doing what
they are
doing
Newton’s
Newton’s 1st Law states:
A body will remain at rest, or
in a state of uniform motion,
unless acted upon by a net
external force.
Another way of saying this is:
st
1
Law
If NO net external force
exists
No Net Force
means No
Acceleration
There is no experiment that
can be performed in an
isolated windowless room
which can show whether the
room is stationary or moving
at constant velocity.
Newton 1 deals with non
accelerated motion.
It does not distinguish
between the states of
“rest” and “uniform
Most importantly:
It requires an
motion” (constant
unbalanced force
Force is NOT needed to
velocity).
to change the
keep
an
object
in
motion
velocity of an
As far as the law is
object
concerned these are the
same thing (state).
Is this how you understand the world works ?
8.3
Newton’s
nd
2
Law
Using the
Newton’s 2nd Law states:
formula
The acceleration of an object as
FNET = ma is
produced by a net force is
only valid for
directly proportional to the
situations
magnitude of the net force FNET,
where the
in the same direction as the net
mass remains
force, and inversely proportional
constant
to the mass of the object.
Mathematically, a = FNET/m more Newton actually expressed his 2nd law
in terms of momentum.
commonly written as FNET = ma
The Net Force on
Newton 2 deals with accelerated
an object equals
motion.
the rate of
change of its
FNET is the VECTOR SUM
momentum
of all the forces acting
on an object.
Momentum (p) = mass x velocity
The acceleration and
FNET are ALWAYS in the
So, FNET = change in momentum = Δp = mΔv = ma
same direction.
change in time
Δt
Δt
8.4
Newton’s 3rd Law
Newton's 1st and 2nd Laws tell you
what forces do.
Newton's 3rd Law tells you what
forces are.
2. People associate
action/reaction with "first
an action, then a reaction”
For example, first Suzie
annoys Johnnie (action) then
For every action
Johnny says "Mommy!
there is an equal
and opposite
Suzie’s annoying me!"
reaction
(reaction).
This statement is correct,
This is NOT an example what
but terse and confusing.
is going on here!
You need to understand
The action and reaction
that it means:
forces exist at the same time.
"action...reaction" means that
"equal" means :
forces always occur in pairs.
Both forces are equal in magnitude.
Single, isolated forces never
Both forces exist at exactly the same time.
happen.
They both start at exactly the same instant,
"action " and "reaction "
and they both stop at exactly the same
are unfortunate names for a
instant.
couple of reasons :
They are equal in time.
1. Either force in an interaction
"opposite" means that the two forces
can be the "action" force or the
always act in opposite directions "reaction" force.
exactly 180o apart.
Questions
Newton’s Laws
29. At what speeds are Newton’s Laws applicable ?
At speeds way below the speed of light
30. Newton’s First Law:
A: Does not distinguish between accelerated motion and constant velocity motion
B: Does not distinguish between stationary objects and those moving with constant
acceleration
C: Does not distinguish between stationary objects and those moving with constant
velocity
D: None of the above
31. Newton’s Second Law:
A: Implies that for a given force, large masses will accelerate faster than small masses
B: Implies that for a given force, larger masses will accelerate slower than smaller
masses
C: Implies that for a given force, the acceleration produced is independent of mass
D: Implies that for a given force, no acceleration is produced irrespective of the mass.
Newton’s laws
32. Newton’s Third Law:
A: Does not distinguish which force of a pair is the “action” force and which is the
“reaction” force.
B: Implies that both action and reaction forces begin and end at the same instant
C: Implies that forces always exist in pairs
D: All of the above.
33. Which of Newton’s Laws require that the vector sum of all the forces acting is
needed before a calculation of acceleration can be made ?
A: Newton’s 1st Law
B: Newton’s 2nd Law
C: Newton’s 3rd Law
D: Newtons 1st and 2nd Laws
Newton’s 2nd Law
34. A car of mass 1250 kg is travelling at a constant speed of 78 kmh-1 (21.7 ms-1). It
suffers a constant retarding force (from air resistance, friction etc) of 12,000 N
(a) What is the net force on the car when travelling at its constant speed of 78 kmh-1 ?
At constant velocity, acc = 0 thus ΣF = 0
(b) What driving force is supplied by the car’s engine when travelling at 78 kmh-1 ?
At constant velocity ΣF = 0, so driving force = retarding force = 12,000 N
(c) If the car took 14.6 sec to reach 78 kmh-1 from rest , what was its acceleration
(assumed constant) ?
Use eqns of motion
u = 0 ms-1 , v = 21.7 ms-1, a = ?, x = ?, t = 14.6 s
use v = u + at -> 21.7 = 0 + 14.6(a) -> a = 1.49 ms-2
8.5 The Horse and Cart Problem
If the horse and cart exert equal and opposite forces on
each other, how come the combination can move ?
An explanation hinges on a couple of
simple points: (Lets assume no friction)
FHC
FCH
1. An object accelerates (or not) because of
the forces that push or pull on it. (Newton 2)
2. Only the forces that act on an object can
cancel. Forces that act on different objects
FRH
FHR
don't cancel - after all, they affect the
Why does the cart accelerate?
motion of different objects!
Looking at the cart alone, just one
Why does the horse accelerate?
force is exerted on it, (FHC) - the
There are 2 forces acting on the horse.
force that the horse exerts on it.
The cart pulls the horse backwards (FCH),
The cart accelerates because the
and the road pushes the horse forward (FRH).
horse pulls on it!
The net force is the vector sum of these two
The cart’s acceleration equals the
forces.
net force on it divided by its mass
The horse’s acceleration equals the net force
There are 2 pairs of Newton 3
on it divided by the its mass.
forces in this situation: If F
NET on the horse is zero, what happens ?
FHC and FCH
The obvious answer is the horse and cart are at rest.
FHR and FRH
BUT, they could also be moving at constant speed !
Questions
Newton 1
Newton’s 3rd Law
35. Explain why, if a cart exerts an equal an opposite force on a horse as the
horse exerts on the cart, the combination is able to move forward.
It is the forces that act on the individual components (i.e., on the horse or cart
individually) that decide whether each will move.
If the friction of the road on the horse’s feet is larger than the force of the cart on
the horse then the horse will accelerate.
If the force of the horse on the cart is greater than the frictional force acting on
the cart then the cart will accelerate.
36. A car mass 1500 kg is towing a trailer of mass 750 kg. The car/trailer
combination accelerate at 3.4 ms-2. The trailer suffers a constant retarding
force of 500 N, while the car suffers a constant retarding force of 1000 N.
Calculate the net force acting on the trailer.
The net force is that force that provides the acceleration. From Newton 2, ΣF = ma
ΣF = (750)(3.4) = 2550 N (2.55 x 103 N)
Calculate the driving force supplied by the car’s engine.
The driving force must (i) overcome friction and (ii) provide extra force to accelerate
the combination
ΣF = (1000 + 500) + ma = 1500 + (1500 + 750)(3.4) = 1500 + 7650
= 9150 N (9.15 x 103 N)
8.6 Momentum and Impulse
Newton described Momentum as the “quality of motion”, a measure of the
ease or difficulty of changing the motion of an object.
Momentum is a vector quantity having both magnitude and direction.
Mathematically,
p = mv
Where,
p = momentum (kgms-1)
m = mass (kg)
v = velocity (ms-1)
In order to change the momentum of an object a mechanism for that change is
required.
This mechanism of change is called Impulse.
Where,
Mathematically,
I = Impulse (N.s)
I = Ft
F = Force (N)
t = Time (s)
The relationship between momentum and impulse can be derived from
Newton’s 2nd Law:
F = ma and a = v/t, so F = mv/t
Rearranging we get:
Ft = mv
ie. Impulse = Momentum
8.7 Conservation of Momentum
The concept of Momentum is particularly useful in analysing collisions.
This is because of the Law of Conservation of Momentum which states:
IN AN ISOLATED SYSTEM, TOTAL MOMENTUM IS CONSERVED.
The term “isolated system” means no external forces are acting in the situation
under investigation.
In a crash situation, where the vehicle comes to a
halt after, say, hitting a tree, both its velocity and momentum fall to zero.
The apparently “lost” momentum, has, in fact, been
transferred via the tree to the Earth.
Since the Earth has a huge mass (6 x 1024 kg). The
change in its velocity is so small as to be negligible.
In the crash mentioned, the momentum change is a
fixed quantity so the Impulse (the product of F and t) is
also a fixed quantity.
However the individual values of F and t can vary as long as the multiply to
give that fixed value.
If t, the time during which the crash occurs, can be lengthened, then the force
which needs to be absorbed by the car and its occupants is reduced.
Modern vehicles use this concept in crumple zones and air bags as both are
designed to extend the time and so reduce the force.
Questions
Momentum
37. A car (and its occupants) is of total mass of 2250 kg and is travelling at 50
kmh-1 . Approaching, head on, is a motorcycle (and rider) of total mass
350kg travelling at 180 kmh-1
(a) Which vehicle (car or bike) has the greater momentum ?
Firstly need to convert speeds to ms-1
50 kmh-1 = 13.9 ms-1 : 180 kmh-1 = 50.0 ms-1
p4wd = mv = (2250)(13.9) = 31275 kgms-1
pcycle = mv = (350)(50) = 17500 kgms-1
So 4WD has the greater momentum.
(b) They collide head on and stick together. What velocity will the “wreck”
have immediately after collision ?
Assume 4WD is travelling to the right and motorcycle to the left.
So Σp = p4wd - pcycle = 31275 – 17500 = 13775 kgms-1 to the right.
If the vehicles stick together total mass = 2250 + 350 = 2600 kg
So Σp = mtotalv  v = Σp/mtotal = 13775/2600 = 5.3 ms-1 to the right
Momentum and Impulse
38. While talking on a mobile phone a truck driver loses concentration and runs
off the road and hits a tree. His speed goes from 20 ms-1 to 0 ms-1 in 0.7 sec.
his truck has a mass of 42 tonnes (1 tonne = 1000 kg)
(a) Calculate his change in momentum
Change in momentum = final mom – initial mom = mvfinal - mvinitial = m (vfinal – vinitial)
= 4.2 x 104(0 – 20)
= - 8.4 x 105 kgms-1 (negative sign can be omitted from answer as it only indicates
direction of momentum change)
(b) Calculate the Impulse during the collision
Since change in momentum = impulse. Impulse = 8.4 x 105 Ns
(c) Calculate the force he will experience during the collision
I = Ft  F = I/t = (8.4 x 105)/0.7 = 1.2 x 106 N
Airbags/Crumple Zones
39. Explain why, in a modern car equipped with seat belts and an air bag , he
would likely survive the collision whereas in the past, with no such safety
devices, he would most likely have been killed.
The change in momentum in any collision is a fixed value thus impulse is also
fixed, but the individual values of F and t can vary as long as their product is the
that fixed value.
In modern vehicles seat belts and crumple zones are designed to increase to
time it takes to stop thus necessarily reducing the force needed to be absorbed
by the driver because Impulse = Ft.
This reduced force will lead to reduced injuries.
In the old days the driver would have been “stopped” be some hard object like a
metal dashboard and his time to stop would have been much shorter and thus
the force experienced would have been larger leading to more severe injury
and likely death.
Chapter 9
Work, Energy & Power
9.0
Work
In Physics, the term WORK is very
strictly defined.
When a force moves an object through
a distance, work has been done.
Mathematically:
W=Fxd
Work is a SCALAR
quantity, meaning it
has a magnitude but
no direction.
Where,
W = Work (Joules)
F = Force (N)
d = distance (m)
If a force is applied and the object does not
move, NO WORK has been done.
If the force applied is constant,
the work done can be calculated
from the formula, W = F x d
But, if the force varies during the
course of doing the work, as in
compressing a spring, the work
must be calculated from the area
under the force versus distance
graph
Force
Area = Work done
Distance
Questions
Work
40. Calculate the work done on a refrigerator when a net force of 125 N acts
over a distance of 4.5 m
Work = Fxd = (125) x (4.5) = 562.5 J (5.63 x 102 J)
41. The graph shows the force required to
compress a spring
Force
6000 (kN)
4500
(a) Calculate the work done in compressing
3000
the spring by 3.0 cm.
1500
Work done = area under graph
= ½ base x height
Distance (cm)
-2
6
1.0 2.0 3.0 4.0
= ½ (3.0 x 10 )(4.5 x 10 )
= 1.35 x 105 J
(b) Calculate the further work required to compress the spring from 3.0 cm to 4.0 cm
Further work to compress from 3.0 cm to 4.0 cm
= area under graph between these two distances
= area of trapesium = ½ (height(1) + height(2)) x base
= ½ (4.5 x 106 + 6.0 x 106 )(1.0 x 10-2)
= 5.25 x 104 J
9.1 Work and Energy
It is very easy to say what energy can do, but
very difficult to define exactly what energy is.
The concept of WORK was
developed BY PHYSICISTS as a
The relation between work and
means of quantifying and
energy is summarised by one
measuring ENERGY.
simple but powerful statement:
WORK DONE = ENERGY TRANSFERRED
If work has been done on an object, the amount of
energy it has MUST have increased.
By how much ?
By exactly the amount of work done on the object.
If an object has done some work, the amount of
energy it has MUST have decreased.
By how much ?
By exactly the amount of work done by the object.
Questions
Work & Energy
42. How much energy is stored in the spring in question 41 when it has
been compressed by 2.0 cm
Work Done = ½ base x height
= ½ (2.0 x 10-2)(3.0 x 106) = 3.0 x 104 J
Since Work Done = Energy Transferred,
the energy stored in the spring = 3.0 x 104 J
9.2
Kinetic Energy
Kinetic Energy is the energy possessed by moving objects.
It is called the “Energy Of Motion”.
Kinetic Energy is a SCALER quantity.
Mathematically:
K.E. = ½mv2
Where:
K.E. = Kinetic Energy (Joule)
m = mass (kg)
v = speed (ms-1)
Arrow has K.E.
due to its motion
Horse has K.E.
due to its movement
Gears have K.E.
due to their rotation
9.3 Gravitational Potential
Energy
Gravitational Potential Energy, often just called Potential Energy, is the
energy possessed by an object due to its position.
It is called the “Energy of Position”
Where:
Potential Energy is a SCALAR quantity.
P.E. = Potential Energy (Joules)
m = mass (kg)
Mathematically:
g = Grav. Field Strength (Nkg-1)
P.E. = mgh
h = height (m)
Potential Energy needs a zero point for the measurement of the height, h.
The zero point is usually, but not always, the surface of the Earth.
The zero point needs to be known for the calculation to have meaning.
The man has P.E. due to his
height above the ground
Skier has P.E. while he is on the
slope + KE due to his speed
Questions
Kinetic & Potential Energy
43. A cyclist is riding her bike along a flat road. She and her bike have a mass of
105 kg. she is travelling at a constant speed of 15 ms-1.
(a) Calculate her Kinetic Energy
KE = ½ mv2 = ½ (105)(15)2 = 1.18 x 104 J
She accidentally rides over a 15 m cliff.
(b) What is her potential energy at the top of the cliff ? (take g = 10 ms-2)
PE = mgh = (105)(10)(15) = 1.58 x 104 J
(c) If all the PE she had at the top of the cliff is converted to KE at the bottom,
calculate her vertical speed just before she hits the ground.
PETOP = KEBOTTOM  1.58 x 104 = ½ (105) v2  v = 17.3 ms-1
9.4 Hooke’s Law
Developed by English scientist
Robert Hooke in 1676, the law
states that the Restoring Force
in an elastic material is directly
proportional to its extension.
The spring constant (k) is a
measure of the nature or
quality of the elastic material.
The higher its value the greater
is the restoring force for a
given extension.
Mathematically:
F = - kx
Where:
F = Restoring Force (N)
k = Spring Constant (Nm-1)
x = Extension (m)
The negative sign in the
equation indicates that
the restoring force and
the extension are in
opposite directions.
Questions
Hooke’s Law
44. A spring of length 100 cm and spring constant 2.5 x 102 Nm-1 hangs
vertically from a retort stand. A total mass of 15.6 kg is hung from the spring.
Calculate the extent of the spring’s extension under this load. (Take g = 10
Nkg-1)
F = -kx: Force produced by hanging mass = mg = (15.6) (10) = 156 N.
So Restoring Force = 156 N
x = -F/k = -156/(2.5 x 102) = -0.62 m (could do this calculation without
negative sign) by using applied force in Hooke’s Law equation)
45. Calculate the amount of elastic potential energy stored in the extended
spring in question 44.
EPE = ½ kx2 = ½ (2.5 x 102) (0.62)2 = 48.1 J
9.5 Energy Transfers
The Law of Conservation of Energy says:
ENERGY CANNOT BE CREATED OR DESTROYED BUT ONLY TRANSFERRED
FROM ONE FORM TO ANOTHER.
When doing problems concerning energy and energy transfers, it is assumed
that the transfers are 100% efficient, meaning no energy losses occur.
For instance, a roller coaster will have a large
Gravitational Potential Energy component at its
highest point most of which will have been converted
to Kinetic Energy at its lowest point and in
calculations a direct equality can be made between
P.E. at the top and K.E at the bottom
In real life, these types of conversion processes are
never 100% efficient.
Friction and the production of heat and sound mean
significant losses occur.
The efficiency of energy transfer processes can be calculated from:
% Eff = Energy Out x 100
Energy In
1
9.6 Elastic Potential Energy
Energy stored in springs is called elastic potential energy
•
•
Elastic materials store energy when
they are deformed and release that
energy when they return to their
original condition.
The amount of energy stored can be
found from the Elastic Potential
Energy Formula:
ES = ½kx2
where
ES = Elastic P. E. (J )
k = Spring Constant (Nm-1)
x = extension and or
compression (m)
“REGULAR” ELASTIC BEHAVIOUR
Slope = Spring
Force
constant (k)
F
Extension
x
Area = ½Fx = ½kx2
= Energy stored
up to extension x
The stored energy can be used to
increase the kinetic energy of the
“arrows” such as used in cross
bows
Questions
9.7 Energy Transfers & Heat
In our friction riddled world, no energy transfer process is 100% efficient.
The end result of the action of these frictional effects is
the production of what is called “low grade,
unrecoverable heat”
This means the heat energy cannot
be harnessed to do any useful
work, such as boil water or run a
pump.
Large sources of this kind of heat are the
exhausts from fossil fuelled transport such
as cars, trucks and trains and from the
electricity generation industry.
9.8
Power
Power is the time rate of doing work, and since
work and energy are equivalent is also the time
rate of energy transfer.
Mathematically:
P = W/t = E/t
Since
P = W/t and W = F.d, we can say
P = F.d/t but d/t = v so
P = F.v
Where:
P = Power (W, Watts)
W = Work (J)
E = Energy (J)
t = time (s)
So, the power for a body moving at
constant velocity can be found in a one
step calculation.
Questions
Power
46. A train of mass 1.5 x 104 kg is travelling at a constant speed of 70 kmh-1. If
the engine is providing a driving force of 3.0 x 107 N , at what power is the
engine operating ?
70 kmh-1 = 19.4 ms-1
Power = F.v = (3.0 x 107)(19.4) = 5.82 x 108 W
47. A roller coaster moves through its journey from A to F.
The coaster has no motor and is not powered after it leaves
point A. Its total mass is 650 kg. The heights above ground
of each portion of the track are given below.
Positio
n
A
E
C
B
D
F
Height above
Ground (m)
A
25.0
B
5.0
C
12.5
D
12.5
E
15.0
F
7.5
Energy Transformations
(a) Between which points is (i) the Kinetic Energy increasing, (ii) the Potential
Energy falling, (iii) the coaster accelerating
(i) AB, EF (ii) BC, DE (iii) All except CD
(b) At which point is the force exerted on the coaster by the track at its greatest ?
Calculate the Gravitational Potential Energy at Point A. (take g = 10 ms-2)
B (At this point the track must supply the reaction force to counteract the
weight of the coaster AND the centripetal force to make the coaster go round
the ‘corner”
(c) Calculate the Gravitational Potential Energy at Point A. (take g = 10 ms-2)
PE = mgh = (650)(10)(25) = 1.63 x 105 J
(d) Assuming no frictional losses, calculate the Kinetic Energy at point F
Loss in PE between A and F = gain in KE between A and F
PELOSS = mgh1 – mgh2 = (650)(10)(25 – 7.5) = 1.14 x 105 J
(e) Calculate the speed of the coaster at point F
At point F, KE = 1.14 x 105 = ½ mv2  v = 18.7 ms-1
Ollie Leitl 2008