MECHANICS
Motion in one dimension
Motion is a fundamental feature of the physical world we live in. Motion can generally be defined as the continuous change in position of an object. The study of motion is called kinematics. We will limit our study of kinematics to one-dimensional motion - that is, motion along a straight line (also called rectilinear motion).
This motion can be in both directions in the same straight line. Two dimensional motion is motion along two axes (eg horizontal and vertical). Three dimensional motion adds a third axis (eg: left and right; up and down; towards or away from the observer). Some examples of one-dimensional motions are:
 a car moving on a straight road
 a person walking down a road
 a sprinter running a 100 metre race
 dropping a pencil
 throwing a ball straight up
In order to fully describe the motion of objects we will need to understand the following terms: position, distance, displacement, speed, velocity and acceleration. For example, a modern formula one car is capable of reaching speeds in excess of 340 kilometres per hour. These cars can accelerate from 0 to 100 km.h
-1 in under 3 seconds. The average length or distance of a formula one circuit is approximately 3,5km.
Frames of Reference
When we describe the motion of something, we specify how it moves relative to something else. A frame of reference therefore describes a viewing point relative to the object which is being observed. An example of this would be a person in a car (the observer) standing still on the side of a road and seeing another car (the object under observation) moving past at 60 km.h
-1 . Relative to each other, the cars are moving at a speed of 60 km.h
-1 to each other. If, however, the other car were to be travelling at 40 km.h
-1 in the same direction as the other car, they would be travelling at 20 km.h
-1 relative to each other. Similarly if they were travelling in opposite directions, the speed would be 100 km.h
-1 to each other. In order to understand this better, it is a good idea to let one direction be positive and the other negative.
Distance and displacement
Distance
Distance is measured in metres, m. Everyday distance measuring instruments include metre sticks and tapes.
Displacement
To describe accurately how the position of an object has changed, we need to specify the distance it has moved and the direction in which it has moved. Together, these two pieces of information are called the displacement .
Displacement is therefore defined as the length and direction of the straight line joining the start and end point of an objects motion.
Vectors and scalars
In physics, quantities that have both a size and a direction (such as displacement) are called vectors . The size of a vector is also called its magnitude. Quantities that have magnitude but no direction, such as distance, time, or mass, are called scalars .
For motion in a straight line we could use + or – to indicate direction. Another method is to simply state the direction in words, e.g. Graham walks 70 m west would represent a displacement vector (both magnitude and direction).
Example
A person walks 2 m East, then 4 m West followed by 6 m East. Calculate the distance walked and the displacement of the walker.
Distance = 2 + 4 + 6 = 12 m
Displacement = 2 + (-4) + 6 = 4 m East (note that a positive answer would give the direction as East and a negative answer as West).
1

Speed
An object in motion will cover a certain distance in a given time. Speed is a measure of how fast something is moving. It is the rate at which distance is covered, and is always measured in terms of a unit of distance divided by a unit of time.
Average speed
The speed of a car is not necessarily constant during its trip – the speed usually varies somewhat. The average speed is defined as follows: total distance covered
Average speed = time taken
If distance is measured in metres (m) and time in seconds (s), speed is measured in metres per second (m/s or m.s
-1 ). For example, if we drive a distance of 80 m in 4 s, our average speed will be 20 m.s
-1 .
If we know average speed and time of travel, distance traveled can easily be calculated. A simple rearrangement of the definition above gives
Total distance covered = average speed X time
If your average speed is 80 kilometres per hour on a 4 hour trip, for example, you will cover a distance of 320 kilometres.
Example
A boy travels from home to school, a distance of 10 km in 1 hour.
(i) Calculate his average speed in m.s
-1 .
(ii) Convert this speed into km.h
-1 . total distance covered
(i) Average speed = time taken
1km = 1000m
=
10000 m
3600 s
= 2,8 m.s
-1
1h = 60 X 60 s = 3600s
(ii) 2,8 m.s
-1 =
2 , 8
1000 x 60 x 60

10,1 km.h
1
(note that rounding off gives the 10,1 and not 10)
Example
If a taxi travels at an average speed of 80 km.h
-1 , how far would it travel in 30 minutes. Give your answer in kilometers (km).
Average speed = total distance covered time taken
Total distance covered = average speed X time taken
= 80 km.h
= 40 km
-1 X 0,5 h
NB: To convert from m.s
-1 to km.h
-1 divide by 1000 and multiply by 3600.
To convert from km.h
-1 to m.s
-1 multiply by 1000 and divide by 3600.
30 minutes = 0,5 hours
Activity 1 : Average speed calculations
1.
2.
What is the average speed of a car that covers 200 m in 8 s?
If a car moves with an average speed of 70 km.h
-1 . What distance will it cover in 3 hours?
2

Instantaneous speed
The average speed of an object does not give any indication of the different speeds and variation that may take place during shorter time intervals. The speed at a specific moment during motion is called the instantaneous speed and is determined by calculating the average speed for a very short time interval. The speed registered on the speedometer of a car is its instantaneous speed. The instantaneous speed is the same as the instantaneous velocity.
Velocity
We often use the words speed and velocity interchangeably. In science, however, we draw a distinction between the two words. Speed is a scalar quantity and is described by magnitude only. Velocity is a vector quantity and is described by both magnitude (how much) and direction (which way). When we say that a car travels at a rate of 80 kilometres per hour, we are specifying the speed of the car. When we say that something travels at 80 kilometres per hour to the south, we are specifying its velocity.
A formula one driver would be concerned with his speed – how fast he is moving. An aircraft pilot would be concerned with her velocity – how fast and in what direction she is moving.
Average velocity is defined as follows: displaceme nt
Average velocity = time taken
This formula for average velocity can be written in symbol form:
x v

x is the symbol for displacement (measured in m) t is the symbol for time (measured in s) v is the symbol for average velocity (measured in m.s
-1 )
A bar above a symbol represents
“average” as in average velocity v
Velocities in the positive direction are positive. Velocities in the negative direction are negative.
The diagram above does not mean that velocities to the left are always negative. Whatever direction you choose to be positive in a situation, velocities that way are positive, velocities the other way are negative.
Average velocity and instantaneous velocity
We distinguish between average velocity and instantaneous velocity as we do for speed. The words speed and velocity alone can be assumed to mean instantaneous speed and instantaneous velocity, respectively.
Acceleration
We can change the velocity of an object by changing its speed, or its direction, or by changing both its speed and direction. Motion with changing velocity is called accelerated motion. Acceleration is simply tells us how quickly the velocity of an object changes. We define acceleration as the rate of change in velocity
Acceleration = change in velocity time taken
Acceleration is a vector quantity.
3

This formula for acceleration can be written in symbol form: a =
Δ
Δ v t
The symbol Δ (Greek letter delta) is often used in science to represent
“change in” a quantity.
If velocity is measured in metres per second (m.s
-1 ) and time is measured in seconds, then acceleration is measured in “meters per second squared” or “meters per second per second”; written as m/s 2 or m.s
-2 . Note that as acceleration is a vector quantity and that we are dealing with one dimensional motion, acceleration can be either negative or positive. If, for example, a vehicle is moving in the forward, positive, direction and the velocity is increasing, its acceleration is positive. If however, it is moving in the forward direction, but it is slowing down, then its acceleration is negative (this can also be called ‘deceleration’). It could also be speeding up in the negative direction and that would also give it a negative acceleration.
Another way of representing the equation for acceleration:
If we assign the symbol v i
for initial velocity and v f
for final velocity, then
Δv = v f
– v i
and the equation for acceleration becomes a
 v f
-
 t v i
Example
A cyclist changes velocity from 2 m.s
-1 to 10 m.s
-1 in 4 seconds. What is her acceleration? v i
= 2 m.s
-1 v f
= 10 m.s
∆t = 4 s
-1 a = ?
Solution: a
 v f
-
 t v i
∴ a

 a

10 2
2
4
m.s
-2
Example
The following table shows the change in speed of a car after each second:
Time (s)
0
Velocity (m.s
0
-1 )
1 4
2 8
3 12
4 16
The car’s velocity increases from zero to 16 m.s
-1 in 4 s. So:
The velocity of the car is increasing by
4 m.s
-1 every second. The car therefore has an acceleration of 4 m.s
-2
.
Average acceleration: a
 v f
 t v i a

16 0
4
= 4 m.s
-2
4

3.
2.
1.
4.
Measuring motion using a ticker-timer
The tickertimer is a simple “timing device” used in the laboratory to investigate motion. A length of paper tape is attached to a trolley; the tape is threaded through the ticker-timer which punches carbon dots on the tape at regular intervals. A typical timer punches 50 dots per second - we usually refer to the frequency (f) of the timer as being 50 dots per second or 50 hertz (Hz).
The dots on a certain section of the tape can be analysed and information of the trolleys motion can be calculated. For example, the further the dots are, the faster the trolley is moving.
Here are a few examples of ticker-tape for a trolley moving to the left.
Constant speed the distance between the dots remain the same start
Higher constant speed start
Acceleration start

Deceleration

 the distance between the dots increases







 the distance between the dots is greater than before



 

 the distance between the dots decreases


 start
The period of a ticker-timer
      
If we know the period of a ticker-timer, we can then calculate the time between two successive dots on a length of tape. This is usually referred to as the period of the timer (symbol T, measured in seconds).
1
So, the distance traveled by the trolley from one dot to another (one space) is equal to
50 frequency

1 period period

1 frequency
second (0,02s). f

1
T
T

1 f

1
50

0,02s
A given time interval is calculated by multiplying the number of spaces between the dots by T. The first dot is the starting point and is therefore not counted.
For example:
A ten-tick interval will therefore have ten spaces between eleven dots as indicated above. The time interval of ten spaces is: t = 10

0,02 = 0,2 s.
5

Activity 3 : Ticker-tape analysis
A trolley runs down a slope pulling a ticker-tape behind it through a ticker-timer. The frequency of the tickertimer is 50 Hz. The ticker-tape is shown below and is marked at intervals which are 10 periods apart .
Now use the information on the ticker-tape to
50 mm 40 mm 30 mm

E

D

C

B
20 mm

A
Direction of Motion
1.1 explain whether the ACCELERATION of the trolley is the same throughout the motion or not.
1.2 determine the average velocity of the trolley between points A and C.
1.3 determine the average velocity of the trolley between points C and E and hence
1.4 calculate the acceleration of the trolley.
Motion Graphs
Graphs can be useful in describing the motion of a body. A graph is used to illustrate the mathematical relationship between two quantities. In this particular section, we deal with three types of graphs:
 displacement versus time (or distance versus time)
 velocity versus time (or speed versus time)
 acceleration versus time
Consider the interpretation of distance versus time graphs. The gradient (or slope) of a distance-time graph tells us the speed.



 y
Gradient
 y x
 x

A steep gradient means a high speed.

A zero gradient (a horizontal part on the graph) means that the object is not moving.

If the line is straight (not necessarily horizontal), it means a constant speed.

A curved line means the object is accelerating or decelerating.
DECELERATING gradient decreasing
STOPPED zero gradient
FAST - steep gradient
SLOW gentle gradient
STEADY SPEED - gradient constant
(graph is straight)
Time
ACCELERATING - gradient increasing
Time
6

The following table gives a detailed summary of the mathematical interpretation of graphs of motion.
ACCELERATED MOTION
MATHEMATICAL
RELATIONSHIP
CONSTANT VELOCITY
MOTION
( a = 0 m.s
-2 )
POSITIVE
ACCELERATION
NEGATIVE
ACCELERATION
Distance/time s s s
 x
Speed/time
0 v v
 t
Gradient = speed
 x

 t t
 x
 t
0
Gradient of the tangent at a point = speed v

 x
 t t v
 x
 t
0 t
Gradient of the tangent at a point = speed v

 x
 t v
 v  v
Displacement/time
0 t
Grad ient = acce lerat ion a


 v t

0 m.s
2
Area under g raph = distance
A = l

b
 x
0
Gradient = velocity

 v


 x t
 t

 x t
 x
 t
0 t
Grad ient = acce lerat ion a


 v t
Area under g raph = distance
A = ½

b

h

 x
 t
0 t
Gradient of the tangent at a point = velocity

 v


 x t
 x
 t
0 t
Grad ient = acce lerat ion a


 v t
Area under g raph = distance
A = ½

b

h

 x
 t
0 t
Gradient of the tangent at a point = velocity

 v

 x
 t
7

MATHEMATICAL
RELATIONSHIP
Velocity/time
Acceleration time
CONSTANT VELOCITY
MOTION ( a = 0 m.s
-2 )
 v
0 t
 a
+
 a


 t v

0 m.s
2
Area under g raph = displacement
A = l

b
0
- t
Area under g raph =
 v
(change in velocity) = 0
A = l

b = 0
POSITIVE
ACCELERATION
 v
ACCELERATED MOTION
NEGATIVE
ACCELERATION
 v

 v
 t
0 t
 a


 t v
Area under g raph = displacement
A = ½

b

h
 a
+

 v
 t
0 t
 a


 t v
Area under g raph = displacement
A = ½

b

h
 a
+
0
0
- t
Area under g raph =  v
(change in velocity)
A = l

b (positive)
- t
Area under g raph =  v
( change in velocity)
A = l

b (negative)
8

Examples:
The following questions refer to the accompanying graph that represents the relationship between the velocity of an object moving in a straight line and time.
V
(m.s
-1 ) 4
1.
1.1
1.2
0
-4
1 2 3 4 5 6 t (s)
1.3
1.4
Calculate the: acceleration during the first two seconds. acceleration between 2 s and 4 s. acceleration during the last two seconds. total displacement of the object after six seconds.
2. Sketch the displacement-time graph for the first six seconds clearly indicating, using values, the points at which a change in motion occurs.
3. Sketch the acceleration-time graph for this motion.
Answer:
1.1
1.2
1.3 a

 v
 t

4
2

2 m.s
2
a = 0 m.s
-2 a

 v
 t

2
8

4 m.s
2
, forward
, reverse
1.4 Δx = ½bh + lb + ½bh - ½bh
= (½ 
2

4) + (4
 2) + (½ 
1

4) – (½ 
1

4)
= 4 + 8 + 2 – 2
= 12 m
2. X
(m)
14
12
10
8
6
4
2
0
0 1 2 3 4 5
3.
 a
(m.s
-2
)
2
6 t (s)
0
-4
1 2 3 4 5 6 t (s)
9

Activity 4 : Graphing skills
Part 1:
1. Plot a graph for each of the tables below.
2. On your graphs, mark the places where the object is moving slowly, where it's moving fast, and where it is stationary. If a line is curved, just write "accelerating" or "decelerating" at that part of the graph.
3. Work out the speeds for each part of your graphs, and write them on the graph.
4. Finally, work out the average speed for each of your graphs.
(remember that average speed = total distance

total time)
Time
(seconds)
0
Distance
(metres)
0
Time
(seconds)
0
Distance
(metres)
5
50
Part 2:
5
10
15
20
25
30
35
40
45
1
2
3
4
4
7
10
13
16
19
2
4
6
8
10
12
14
16
18
20
10
15
20
25
35
45
55
60
62
63
The table below records displacement and velocity data measured from a rocket as it was launched. time (s)
0.0
0.8
1.6
2.4
3.2
4.0
4.8 displacement (m)
0.0
2.8
8.3
15.2
23.9
35.0
48.3 velocity (m.s
0.0
5.2
7.8
9.8
12.4
15.2
18.4
-1 )
5.6 64.4 20.7
Plot graphs of the rocket’s displacement and velocity against time as it lifts off from the launch pad. Choose suitable scales for the axes on each graph so that each graph fills most of a sheet of graph paper.
Remember, you can plot points more accurately on a large graph than on a small one.
Draw the best straight line through the points on the velocity-time graph. Measure the gradient of this line to f ind the rocket’s acceleration.
10

Equations of motion
A number of simple problems of motion can be solved using four basic equations:
The four equations of motion where acceleration is constant are: v f
 v i
 a
 t
 x
 v i
 t

1
2 at 2
Δx  displacement in m v i
 initial velocity in m.s
-1 v f
 final velocity in m.s
-1 a  acceleration in m.s
-2
∆t  time in s v f
2  v i
2 
2 a
 x
 x

( v i
 v f
)
 t
2
The list below represents a summary of the physical quantities used in the equations of motion given on the information sheet.
SYMBOL PHYSICAL QUANTITY
UNIT OF
MEASUREMENT
UNITS SYMBOL
Δx
Displacement metres m v i
Initial Velocity metres per second m.s
-1
V f
Final Velocity metres per second m.s
-1
∆t Time seconds s
3.
4. a Acceleration metres per second squared m.s
-2
Note the following points when using equations of motion:
1.
2.
Equations of motion apply to rectilinear motion, i.e. motion in a straight line.
The positive direction should always be stated because all the quantities except time are vectors.
The direction in which the object is initially moving, or about to move, is usually taken as being positive.
If an object starts from rest, v
If the velocity of an object is increasing, a is positive, and if the velocity of an object is decreasing, a is negative. i
= 0 m.s
-1 , or if an object is brought to rest v f
= 0 m.s
-1 .
5. Most problems dealing with equations of motion will expect you to identify which variables are known and then substitute them into one of the four equations to calculate the unknown value. A typical
 method to follow could be:

Start by making a list of the known variables.
Then decide which equation you should use to calculate the unknown variable from the known variables.

Write down the equation, substitute the values from the problem into it.

Re-arrange to make the unknown variable the subject of the formula and use your calculator to work out the result.
11

Example:
A body travelling at 30 m.s
-1 is decelerated by 3 m.s
-2 . Calculate the: a. b. c.
Answer: distance travelled in 8 s. velocity after travelling 144 m. time taken for the body to stop.
Take the forward direction as positive. a. Δx = v i
∆t + ½at 2
= 30
 8 + ½ 
-3

8 2
= 240 – 96
= 144 m
Δx v
V i a f
∆t
?
30 m.s
-1
-3 m.s
-2
8 s b. v f
2 = v i
2 + 2a ∆x

v 2 = 30 2 + 2

-3

144
= 36

v = 6 m.s
-1
OR
v f
= v i
+ a ∆t
= 30 + (-3)
= 30 - 24
= 6 m.s
-1
X 8
Δx v v i f a
∆t
144 m
30 m.s
?
-3 m.s
8 s
-1
-2 c. v f
= v i
+ a ∆t

0 = 30 – 3∆t

t = 10 s
Δx v i
30 m.s
-1 v f a
∆t
0
-3 m.s
?
-2
Note: If in c. above, it took 1 s for the car to stop, its acceleration would be:
v f
= v i
+ a ∆t

0 = 30 + a x 1

a = - 30 m.s
-2 – in other words an acceleration with a high value in opposite direction to the original direction of travel. It is for this reason that air bags in cars are a good idea – they increase the amount of time it takes for a passenger to come to a stop, thereby reducing the negative effects of coming to a stop too quickly.
Activity 5: Equation of motion calculations
1. The brakes of an aeroplane give it a uniform acceleration of -5 m.s
-2 . Calculate the minimum length of runway needed to bring the plane to a stop if it touches down at 75 m.s
-1 .
2. A car travelling at 30 m.s
-1 is brought uniformly to rest in a distance of 50 m. Calculate the time taken to come to rest.
3. a) b)
A car accelerates in a straight line from a speed of 10 m.s
-1 to a speed of 30 m.s
-1 in 5,0 s.
What is the car’s average acceleration?
Assuming the car’s acceleration is uniform, calculate the distance it travels while its speed is changing.
4. In order to stop at a traffic light, a bus decelerates uniformly from a speed of 65 km.h
-1 (18 m.s
-1 ) to
36 km.h
-1 (10 m.s
-1 ) in a distance of 40 m. Determine the total distance travelled by the bus from when the driver first applied the brakes (at 18 m.s
-1 ) to when the bus comes to a stop.
12

Gravity and mechanical energy
Gravitational force
The Earth exerts a force of attraction on all objects. This force is called the force of gravity. In reality, any two objects (masses) will have a force of attraction between them which causes them to be pulled towards one another.
This force of gravity depends on the size of the masses and their distances apart. The force is large when the masses are big and close together.
Near the Earth’s surface, there is a force of 10 newtons (N) on each kilogram of mass. The Earth’s gravitational field strength is 10 newtons per kilogram
(N.kg
-1 ).
Acceleration due to gravity
A lead weight and a feather will both fall downwards because of gravity.
However, the feather is slowed down much more by the air. If we were to seal the two objects in a container and remove all the air, both objects would fall with the same downward acceleration of 9,8 m.s
-2 i.e. for each second that the objects fall, their respective speeds increase by 9,8 m.s
-1 . This is called the acceleration due to gravity or the acceleration of free fall and is also represented by the symbol g.
So g has two meanings:

Gravitational field strength 10 N.kg
-1

Acceleration due to gravity 10 m.s
-2 .
Weight
Weight is another name for the Earth’s gravitational force on an object. Since weight is a force, it is measured in newtons (N). Near the Earth’s surface, an object of mass 1 kg will have a weight of approximately 9,8 N. Greater masses have greater weights.
What is the difference between mass and weight?
Your mass is a measure of how many particles there are in your body. It does not matter where you are in the Universe, your mass does not change.
Your weight is a measure of the pull of gravity on your body. Your weight depends upon what planet you are standing on. You would weigh less on Mercury than on Earth because Mercury will exert a smaller
Mass = 15 kg gravitational force on your body than Earth. weight = mass X g
In symbols : w = mg
15kg
Weight = 150 N
13





Practical : Determination of “g” using the Pendulum and plumb-bob method:


Set up apparatus on a supporting board as shown.
Ensure that the metre-stick can swing freely with as little friction as possible.
Pull the metre-stick sideways.
Pass thread supporting plumb-bob through eyescrew.

Pull thread to raise plumb-bob so that its middle is
 near the top end of the metre-stick. Tie the thread to the eye-screw.
Make sure that the metre-rule and plumb-bob hang in the same plane and free of the supporting board.


Blacken the wire surrounding the plumb-bob using a candle.
Mark the initial position of the blackened wire on the board.




Burn the supporting thread.
A mark is left where the metre-stick and the plumbbob strike.
Measure the distance between the initial and final marks.
Repeat a few times to determine the average value for s.
Remove the thread from the metre-stick. Using a stopwatch, determine the time taken for 20 swings.
Calculate the time taken for a quarter swing.

Using the formula: g =
2 s calculate the value of g. t
2
Example:
In an experiment to determine the value of “g” by the pendulum and plumb-bob method, the following results were obtained: average distance through which plumb-bob fell = time taken for 36 swings of the pendulum
Calculate the va lue of “g”.
Time taken for ¼ swing: t

60
36

4

0,42 s
 x

1
2 gt
2 ∴ g

2
 x t
2

2

0 , 866
0 , 42
2
86,6 cm
= 60,0 s

9 , 82 m.s
2
EQUATIONS OF MOTION APPLIED TO FREE FALL
Since falling objects experience accelerated motion, the equations of motion can be applied: v = u +g ∆t v f
2 = v i
2 + 2g ∆x
∆x = u∆t + ½g∆t 2 .
When applying equations of motion, note the following: a. b. c. d.
The time taken to rise to the maximum height is equal to the time taken to fall.
At any particular height, measured from the point of projection, the upward velocity equals the downward velocity.
If the object falls from rest, u = 0 m.s
-1 .
For convenience, choose the direction of motion as the positive direction. For upward motion, the values of v i
, v f
, ∆x and ∆t are positive and g is negative. For downward motion, the values of v i
, v f
,
∆x, ∆t and g are positive.
14

Example
A stone is thrown vertically upwards from the ground with an initial velocity of 30 m.s
-1 .
How long does it take for the stone to reach its highest point? (a)
(b) What is the maximum height reached by the stone?
(c) After how many seconds will it strike the ground again.
Solution
(a) v f
= v i
+g ∆t
0 = 30 – 10 x∆ t
10 ∆t = 30
∆t = 3 s
(b) ∆x = v i
∆t + ½g∆t 2
= 30 x 3 + ½ x – 10 x 3 2
= 45 m
∆x v v g i f
∆t v v g i f
30 m.s
0
-1
-10 m.s
?
∆x
∆t
?
-2
30 m.s
0
3 s
-1
-10 m.s
-2
(c) ∆t = 2 x 3 = 6 s (The time taken to rise to the maximum height is equal to the time taken to fall).
Activity 6 : Free-fall calculations
1. a) b)
A stone is dropped from a window 10 m above the ground.
How long does it take the ball to reach the ground?
How fast is the ball travelling when it hits the ground?
(Assume that the acceleration due to gravity, g, = 10 m.s
-2 ).
2. A gymnast jumps vertically into the air with an initial speed of 4,0 m.s
-1 . a) b)
How high does the gymnast jump?
How long is the gymnast in the air?
(Assume that the acceleration due to gravity g = 10 m.s
-2 )
Gravitational potential energy
Gravitational potential energy is the energy stored in an object as the result of its vertical position i.e. height.
The energy is stored as the result of the gravitational attraction of the Earth on the object. The gravitational potential energy of an object is dependent on two variables - the mass of the object and the height to which it is raised. The more massive the object, the greater the gravitational potential energy. The higher that an object is elevated, the greater the gravitational potential energy. These relationships are expressed by the following equation:
Gravitational potential energy = m X g X h
E
P
= mgh
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the acceleration of gravity (approximately 10 m.s
-2 on Earth). Energy is measured in joules (J).
Energy is a scalar quantity : it has magnitude (size) but no direction. So you do not have to specify direction when doing calculations concerning energy.
Example:
Calculate the potential energy of a 2 kg stone raised to a height of 4 m above the ground.
Solution:
E
P
= mgh
= 2 x 10 x 4
= 80 J
15

Kinetic Energy
The kinetic energy of an object is the energy it possesses due to its motion. All moving objects have kinetic energy. For an object of mass m and velocity v:
Kinetic energy = ½
X m X v 2
E
K
= ½ m v 2
Example:
Calculate the kinetic energy of a 2 kg stone which has reached a speed of 3 m.s
-1.
Solution:
E
K
= ½ m v 2
= ½ x 2 x 3 2
= 9 J
Mechanical Energy
Mechanical energy is the energy which is possessed by an object due to its motion or its stored energy of position. Mechanical energy can be either kinetic energy (energy of motion) and / or potential energy (stored energy of position). Objects have mechanical energy if they are in motion and/or if they are at some position relative to a zero potential energy position (for example, a brick held at a vertical position above the ground or zero height position). A moving car possesses mechanical energy due to its motion (kinetic energy). A moving soccer ball possesses mechanical energy due to both its high speed (kinetic energy) and its vertical position above the ground (gravitational potential energy). A book at rest on the top of a shelf possesses mechanical energy due to its vertical position above the ground (gravitational potential energy).
An object which possesses mechanical energy is able to do work. In fact, mechanical energy is often defined as the ability to do work. Any object which possesses mechanical energy - whether it be in the form of potential energy or kinetic energy - is able to do work. That is, its mechanical energy enables that object to apply a force to another object in order to cause it to be moved.
Conservation of mechanical energy
When energy changes from one form to another, we say that energy is transformed. During the transformation of energy, the total amount of energy remains the same.
The law of conservation of energy states: Energy cannot be created or destroyed, it may be transformed from one form to another.
In the absence of friction, we can consider only potential and kinetic energies. The sum of the potential and kinetic energies of an object is always constant. For an object falling from rest this means that the total mechanical energy will never exceed the potential energy at its maximum height nor the kinetic energy at the point of impact.
Consider a stone of mass 1 kg being dropped from a height of 10 m. We could easily calculate both the gravitational potential energy and the kinetic energy after each metre fallen by the stone. To calculate the kinetic energy, we need to use the equation of motion v f
2 = v i
2 + 2g ∆x, where s will be given by the actual distance fallen.
For example, at the 9 m mark, the total distance fallen will be 1 m:
v f
2 = v i
2 + 2g ∆x
= 0 + 2 x 10 x 1
v f
2 = 20
E
K
= ½mv 2

E

E
K
= ½ x 1 x 20
K
= 10 J
16

6 m
5 m
4 m
3 m
2 m
10 m
9 m
8 m
7 m
E
P
= mgh
100 J
90 J
80 J
70 J
60 J
50 J
40 J
30 J
20 J
E
K
= ½mv 2
0 J
10 J
20 J
30 J
40 J
50 J
60 J
70 J
80 J
E
P
+ E
K
100 J
100 J
100 J
100 J
100 J
100 J
100 J
100 J
100 J
10 m
1 kg
1 m 10 J 90 J 100 J
0 m 0 J 100 J 100 J
This example clearly illustrates the principle of conservation of mechanical energy. We also note that the potential energy at the top equals the kinetic energy at the bottom.
Example
A 5kg stone slides down a smooth slope (frictionless). Calculate the speed of the stone when it reaches the bottom of the slope.
Solution
This problem can be solved using the conservation of mechanical energy.
4 m
5 kg
At the top of the slope the stone possesses gravitational potential energy. When it reaches the bottom, all this potential energy is converted to kinetic energy.
At the top: E
P
= mgh = 5 x 10 x 4 = 200 J
At the bottom: E
K
= 200 J
 ½mv 2
 ½ x 5 x v 2
= 200 J
= 200
 v = 8,94 m.s
-1
Activity 7 : Mechanical energy calculations
1 A skydiver of mass 80 kg falls freely for 100 m from a stationary helicopter.
1.1 Ignoring air friction, calculate the skydiver's loss of gravitational potential energy.
1.2 Without doing a further calculation, write down the skydiver's kinetic energy when he is 100 m below the helicopter.
1.3 NAME the principle or law which you used to answer 1.2
1.4 Calculate the velocity of the skydiver when he is 100 m below the helicopter (correct to 2 decimal places).
17

2 A 'bungee' jumper, mass 70 kg, stands on a platform 12 m above the ground as shown in sketch A.
He jumps and a 8 m long rope stabilises him as in sketch B.
Disregard all effects of elasticity in the rope and determine:
2.1 the potential energy of the jumper at a height of 12 m.
2.2 the kinetic energy of the jumper the moment he has fallen through
8 m, i.e. just before the rope starts to stabilise him.
REVISION QUESTIONS
1.
(a)
(b)
(c)
A taxi, starting from rest, has a constant acceleration of 3 m.s
-2 . Calculate: the distance travelled by the taxi in 4 s; the velocity of the taxi after 4 s; the distance travelled by the taxi by the time the taxi reaches a speed of 25 m.s
-1 .
2. In order to take off, an aircraft must move on the ground for 15 s at a uniform acceleration of 8 m.s
Calculate the minimum length of runway needed.
-2 .
3. A ship is slowed down uniformly from a speed of 15 m.s
Calculate the magnitude of the deceleration.
-1 to 5 m.s
-1 over a distance of 180 m.
4. The velocity-time graph of a moving object is given below
4.1
4.2
At which point/s is the velocity zero?
Between which points is the velocity uniform? v
(m.s
-1 )
5
B C
0
3 6
D
9
4.3
4.4
- 5
E
Use the graph to determine the total distance covered by the object.
What is the magnitude of the displacement of the object?
18

5. The following tapes were all made by a ticker timer operating at a constant frequency.
DIRECTION OF
MOTION
A    
B        
C       
D   
Which of the tapes represent
5.1
5.2 an increasing speed? a constant speed?
5.3
5.4
5.5 a decreasing speed? the lowest average speed? the highest average speed?
6. Ticker-timer tape is fixed to a piece of lead. The tape passes through a ticker timer. The lead is allowed to fall and the tape is marked with a dot every 1/50 second, as shown:
A
           
0 2 8 18 32 50 72 98 128 162 200 242
9.
8.
Distances of dots from A in mm.
6.1
6.2
Find the average speed of the lead over the first 1/10 second (in m.s
Find the average speed of the lead over the second 1/10 second.
-1 ).
Calculate the acceleration of the falling lead. 6.3
7. An object has a mass of 7 kg. What is its gravitational potential energy
(a)
(b)
3 m above the ground
5 m above the ground?
An object of mass 8 kg has a speed of 4 m.s
-1 .
(a)
(b)
What is its kinetic energy?
What is its kinetic energy if its speed is doubled?
A stone of mass 0,4 kg has 120 J of kinetic energy. What is the speed of the ball?
10. A stone of mass 0,6 kg is dropped from the top of a cliff and hits the ground below at a speed of 12 m.s
-1 .
(a)
(b)
(c)
Calculate is the kinetic energy of the stone just before it hits the ground?
What was the stone’s gravitational potential energy before it was dropped?
Calculate the height from which the stone was dropped.
19