Top Banner
Physics Module Form 4 Chapter 2 Forces & Motion GCKL 2011 2-1 LINEAR MOTIONs Physical Quantity Definition, Quantity, Symbol and unit Distance, s Distance is the …… Quantity: SI unit : .. Displacement, s (a) The distance in .. …. (b) the distance between …. ….direction. (c) The distance of its final ….. specified . Quantity: SI unit: Speed,v Speed is the Speed = Quantity: SI unit: Velocity, v Velocity is the Velocity = Direction of velocity is Quantity : SI unit: Average speed v = Example: A car moves at an average speed / velocity of 20 ms -1 On average, the car moves a distance/ displacement of Average velocity Displacement TotalTime v 2.1
41
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-1

LINEAR MOTIONs

Physical Quantity Definition, Quantity, Symbol and unit

Distance, s

Distance is the ……

Quantity: … SI unit : ..

Displacement, s

(a) The distance in .. ….

(b) the distance between ….

….direction.

(c) The distance of its final …..

specified ….

Quantity: SI unit:

Speed,v

Speed is the Speed =

Quantity: SI unit:

Velocity, v

Velocity is the

Velocity =

Direction of velocity is

Quantity : SI unit:

Average speed

v =

Example: A car moves at an average speed / velocity of 20 ms

-1

On average, the car moves a distance/

displacement of

Average velocity

Displacement

TotalTimev

2.1

Page 2: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-2

Uniform speed Speed that remains the same in

Uniform velocity Velocity that remains

An object has a non-

uniform velocity if

(a) The direction of motion changes or the motion is not linear.

(b) The magnitude of its velocity changes.

Acceleration, a

v ua

t

Unit: ms-2

Acceleration is positive

When the velocity of an object

Acceleration is defined as the

Change in velocityAcceleration=

Time taken

Final velocity,v - Initial velocity,u =

Time taken,t

The velocity of an object increases from an initial velocity, u, to a higher final

velocity, v

Deceleration

acceleration is negative.

The rate of decrease in speed in a specified direction.

Zero acceleration An object moving at a constants velocity, that is,

Constant acceleration Velocity increases at a uniform rate.

When a car moves at a constant or uniform acceleration of 5 ms -2

, its velocity

1. Constant =

2. increasing velocity =

3. decreasing velocity =

4. zero velocity =

5. negative velocity = object moves at opposite direction

6. zero acceleration =

7. negative acceleration = deceleration

Page 3: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-3

Speed Velocity

The rate of change of

distance

The rate of change of

displacement

Scalar quantity Vector quantity

Comparisons between distance and displacement Comparisons between speed and velocity

Fill in the blanks:

1. A steady speed of 10 ms -1

= A distance of __________________________________________

2. A steady velocity of -10 ms -1

= A displacement of _________________________________

3. A steady acceleration of 4 ms -2

= Speed _____________________________________________

4. A steady deceleration of 4 ms -2

= ____________________________________________________________

5. A steady velocity of 10 ms -1

= A displacement of 10 m is travelled every 1 second to the right.

Distance Displacement

Total path length

travelled from

one location to

another

The distance between

two locations

measured along the

shortest path

connecting them in

specific direction

Scalar quantity

It has magnitude but no

direction

SI unit SI unit :

Page 4: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-4

Example 1

Every day Rahim walks from his house to the junction

which is 1.5km from his house.

Then he turns back and stops at warung Pak Din which is

0.5 km from his house.

(a) What is Rahim’s displacement from his house,

• when he reaches the junction.

• when he is at warung Pak Din.

(b) After, Rahim walks back to his house. breakfast

When he reaches home,

(i) What is the total distance travelled by

Rahim?

(ii) What is Rahim’s total displacement from

his house?

Example 2

Every morning Amirul walks to Ahmad’s house

which is situated 80 m to the east of Amirul’s house.

They then walk towards their school which is 60 m

to the south of Ahmad’s house.

(a) What is the distance travelled by Amirul and his

displacement from his house?

(b) If the total time taken by Amirul to travel from

his house to Ahmad’s house and then to school

is 15 minutes, what is his speed and velocity?

Speed =

Velocity =

Example 3

Salim running in a race covers 60 m in 12 s.

(a) What is his speed in ms-1

(b) If he takes 40 s to complete the race, what is his

distance covered?

Example 4

An aeroplane flies towards the north with a

velocity 300 km hr -1

in one hour. Then, the plane

moves to the east with the velocity 400 km hr -1 in

one hour.

(a) What is the average speed of the plane?

(b) What is the average velocity of the plane?

Page 5: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-5

(c) What is the difference between average speed

and average velocity of the plane?

Example 5

The speedometer reading for a car travelling due north

shows 80 km hr -1

. Another car travelling at 80 km hr -1

towards south. Is the speed of both cars same? Is the

velocity of both cars same?

A ticker timer

Use:

1 tick = time interval

The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2

consecutive dots is

1 tick =

Page 6: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-6

Relating displacement, velocity, acceleration and time using ticker tape.

VELOCITY FORMULA

Time, t = 10 dicks x 0.02 s

= 0.2 s

displacement, s = x cm

velocity =

ACCELERATION

Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or

t = (50 – 10) ticks x 0.02 s = 0.8 s

Initial velocity, u =

final velocity, v =

acceleration, a =

TICKER TAPE AND CHARTS TYPE OF MOTION

Distance between the dots increases uniformly

Page 7: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-7

- Distance between the dots decrease uniformly

Example 6

The diagram above shows a ticker tape chart for a

moving trolley. The frequency of the ticker-timer

used is 50 Hz. Each section has 10 dots-spacing.

(a) What is the time between two dots?

(b) What is the time for one strips?

(c) What is the initial velocity?

(d) What is the final velocity?

(e) What is the time interval to change from initial

velocity to final velocity?

(f) What is the acceleration of the object?

a = t

uv 2

THE EQUATIONS OF MOTION

2

2 2

1

2

2

v u at

s ut at

v u as

u =

v =

t =

s =

a =

Page 8: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-8

MOTION GRAPHS

DISPLACEMENT – TIME GRAPH

Velocity is obtained from A – B : gradient of the graph is

B – C : gradient of the graph =

object is

C – D : gradient of the graph

The object

VELOCITY-TIME GRAPH

Area below graph

Positive gradient

Negative gradient

Zero gradient

GRAPH s versus t v versus t a versus t

Zero

velocity

Negative

constant

velocity

Positive

Constant

velocity

2.2

Page 9: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-9

GRAPH s versus t v versus t a versus t

Constant

acceleration

Constant

deceleration

Example 1: Example 2:

Based on the s-t graph above:

(a) Calculate the velocity at

(i) AB (ii) BC (iii) CD

(b) Describe the motion of the object at:

(i) AB (ii) BC (iii) CD

(c) Find

(i) total distance

(ii) total displacement

(d) Calculate

(i) The average speed

(ii) The average velocity of the

moving particle

10

20

0 10 20 30 40 time/

s

velocity/ m s-1

(a) Calculate the acceleration at:

(ii) JK (ii) KL (iii) LM

(b) Describe the motion of the object at:

(ii) JK (ii) KL (iii) LM

(c) Calculate

(iii) The total displacement

(iv) The average velocity

Page 10: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-10

INERTIA

Inertia The inertia of an object is the tendency of the object

Newton’s first law Every object

Relation between inertia

and mass

The larger the mass,

SITUATIONS INVOLVING INERTIA

SITUATION EXPLANATION

EEEEEEEEJNVJLKN

DNFLJKVNDFLKJNB

VJKL;DFN BLK;XC

NB[F

NDPnDSFJ[POJDE]O-

JBD]AOP[FKBOP[DF

LMB NOPGFMB

LKFGNKLB

FGNMNKL’ MCVL

BNM’CXLB

NFGNKEPLANATION

When the cardboard is pulled away quickly, the coin drops straight into

the glass.

Paste a picture

Chilli sauce in the bottle can be easily poured out if the bottle is moved

down fast with a sudden stop. The sauce inside the bottle moves

together with the bottle.

When the bottle stops suddenly,

Paste a picture

Body moves forward when the car stops suddenly The passengers were in a

state of motion when the car was moving.

When the car stopped suddenly,

Paste a picture

A boy runs away from a cow in a zig zag motion. The cow has a large inertia

2.3

Page 11: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-11

The head of hammer is secured tightly to its handle by

knocking one end of the handle, held vertically, on a hard

surface.

This causes the hammer head to continue on its

downward motion

when the handle has been stopped, so that the top

end of the handle is slotted deeper into the hammer

head.

• The drop of water on a wet umbrella will fall when the

boy rotates the umbrella.

• This is because the drop of water on the surface of the

umbrella moves simultaneously as the umbrella is rotated.

• When the umbrella stops rotating, the inertia of

the drop of water will continue to maintain its

motion.

Ways to reduce the negative

effects of inertia

1. Safety in a car:

(a)Safety belt secure the driver to their seats.

When the car stops suddenly, the seat belt provides

the external force that prevents the driver from

being thrown forward.

(b)Headrest to prevent injuries to the neck during rear-

end collisions. The inertia of the head tends to

keep in its state of rest when

the body is moved suddenly.

(c)An air bag is fitted inside the steering wheel.

It provides a cushion to prevent the driver from

hitting the steering wheel or dashboard during a

collision.

2. Furniture carried by a lorry normally are tied up together by

string.

When the lorry starts to move suddenly, the furniture are

more difficult to fall off due to their inertia because

their combined mass has increased.

Page 12: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-12

Relationship between mass

and inertia

• Two empty buckets which are hung with rope from the

ceiling.

• One bucket is filled with sand while the other bucket is

empty.

• Then, both pails are pushed.

• It is found that

Push and compared to the bucket with sand.

• The bucket filled with sand offers more resistance to

movement.

• When both buckets are oscillating and an attempt is made

to stop them, the bucket filled with sand offers

Page 13: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-13

MOMENTUM

Definition Momentum =

SI unit:

Principle of

Conservation of

Momentum

In the absence of an external force,

Elastic Collision Inelastic collision

ƒ Both objects move

ƒ Momentum

ƒ Kinetic energy

Total energy

ƒ The two objects

ƒ Momentum

ƒ Kinetic energy.

ƒ Total energy

Total Momentum Before =

m1u

1 + m2u

2 = m1 v

1 + m2 v

2

Total Momentum Before =

m1 u

1 + m

2 u

2 = ( m1 + m

2 ) v

Explosion

Paste a picture

Before explosion both object

Total Momentum

before collision is

zero

Total Momentum after

collision :

m1v

1 + m2v

2

2.4

Page 14: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-14

From the law of conservation of momentum:

Total Momentum = Total Momentum

Before collision after collision

0 = m1v

1 + m2v

2

m1v

1 = - m

2v

2

Negative sign means

EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum)

Paste a picture

When a rifle is fired, the bullet of mass m,

moves with a high velocity, v. This creates a

momentum in the forward direction.

From the principle of conservation of

momentum,

Paste a picture

Application in the jet engine:

The launching of rocket

Mixture of hydrogen and oxygen fuels

These high speed hot gases produce

By conservation of momentum,

Page 15: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-15

Paste a picture

In a swamp area, a fan boat is used.

The fan produces a high speed movement of air

backward. This produces a large momentum

backward.

By conservation of momentum, an equal but opposite

momentum is produced and acted on the boat. So the

boat will move forward.

Paste a picture

A squid propels by expelling water at high velocity.

Water enters through a large opening and exits

through a small tube. The water is forced out at a

high speed backward.

Total Mom. before= Total Mom. after

0 =Mom water + Mom squid

0 = mwv

w + msvs

-mwv

w = msvs

The magnitude of the momentum of water and

squid are equal but opposite direction.

This causes the quid to jet forward.

Example

Car A of mass 1000 kg moving at 20 ms -1

collides with a car B of mass 1200 kg moving at

10 m s -1

in same direction. If the car B is

shunted forwards at 15 m s -1

by the impact,

what is the velocity, v, of the car A immediately

after the crash?

Example Before collision After collision

MA = 4 kg

MB

= 2 kg

UA = 10 ms

-1 r i g h t

UB = 8 ms

-1 l e f t V

B 4 ms-1

right

Calculate the value of VA .

Page 16: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-16

Example

A truck of mass 1200 kg moving at

30 ms-1

collides with a car of mass

1000 kg which is travelling in the opposite

direction at 20 ms-1

. After the collision, the two

vehicles move together. What is the velocity of

both vehicles immediately after collision?

Example A man fires a pistol which has a mass of 1.5 kg.

If the mass of the bullet is 10 g and it reaches a

velocity of 300 ms -1

after shooting, what is the

recoil velocity of the pistol?

Page 17: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-17

FORCE

Balanced Force

When the forces acting on an object are

balanced,

Effect : the object

[velocity ]

or

moves

[ a = ]

Example:

Weight, W = Lift, U Thrust, F = drag, G

Unbalanced Force/ Resultant Force

When the forces acting on an object are not balanced,

there must be

The net force is known as

Effect : Can cause a body to

-

Newton’s Second Law of Motion

The acceleration produced by a force on an object is

Force = Mass x Acceleration

F = ma

2.5

Page 18: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-18

Experiment to Find The Relationship between Force, Mass & Acceleration

Relationship

between

a & F a &

m

Situation

Both men are pushing the same mass

but man A puts greater effort. So he

moves faster.

Both men exerted the same strength.

But man B moves faster than man A.

Inference The acceleration produced by an

object depends on the net force

applied to it.

Hypothesis The acceleration of the object

increases when the force applied

increases

Variables:

Manipulated :

Responding :

Constant :

Force

Acceleration

Mass

Apparatus and

Material

Ticker tape and elastic cords, ticker timer, trolleys, power supply and friction

compensated runway and meter ruler.

Page 19: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-19

Procedure :

- Controlling

manipulated

variables. -Controlling

responding

variables. -

Repeating

experiment.

An elastic cord is hooked over the

trolley. The elastic cord is stretched

until the end of the trolley. The

trolley is pulled down the runway

with the elastic cord being kept

stretched by the same amount of

force

An elastic cord is hooked over a

trolley.

Determine the acceleration by

analyzing the ticker tape.

Acceleration

Acceleration v u

at

Determine the acceleration by analyzing

the ticker tape.

Acceleration v u

at

Repeat the experiment by using two

, three, four and five elastic cords

Repeat the experiment by

Tabulation of

data

Force, F/No of

elastic cord

Acceleration, a/ ms-2

1

2

3

4

5

Mass, m/

no of

trolleys

Mass,

m/g

1/m,

g-1

Acceleration/

ms-2

1

2

3

4

5

Analysing

Result

Page 20: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-20

1. What force is required to move a 2 kg object

with an acceleration of 3 m s-

2

, if

(a) the object is on a smooth surface?

(b) The object is on a surface where the

average force of friction acting on the

object is 2 N?

2. Ali applies a force of 50 N to move a 10 kg

table at a constant velocity. What is the

frictional force acting on the table?

3. A car of mass 1200 kg travelling at 20 ms -1

is brought to rest over a distance of 30 m.

Find

(a) the average deceleration,

(b) the average braking force.

4. Which of the following systems will

produce maximum acceleration?

Page 21: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-21

IMPULSE AND IMPULSIVE FORCE

Impulse The change of

Unit :

m =

u =

v =

t = Impulsive Force The rate of change

change of momentum

time

mv mu

t

Unit =

Effect of time Impulsive force

is

Longer period of time →Impulsive force

Shorter period of time →

Situations for Reducing Impulsive Force in Sports

Situations Explanation

Thick mattress with soft surfaces are used in events such as high jump

so that

Goal keepers will wear gloves to

A high jumper will bend his legs upon landing. This is to

so as to

A baseball player must catch the ball in the direction of the motion of

the ball. Moving his hand backwards when catching the ball prolongs

the time for the momentum to change so as to reduce the impulsive

force.

2.6

Page 22: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-22

Situation of Increasing Impulsive Force

Situations Explanation

A karate expert can break a thick wooden slab with his bare hand

that moves at a very fast speed. The short impact time results in

A massive hammer head moving at a fast speed is brought to rest

upon hitting the nail within a short time interval.

A football must have enough air pressure in it so

Pestle and mortar are made of stone. When a pestle is used to pound

chillies the hard surfaces of both the pestle and mortar cause the pestle

to be stopped in a very short time. A large impulsive force is resulted

and thus causes these spices to be crushed easily.

Example 1

A 60 kg resident jumps from the first floor of a burning house.

His velocity just before landing on the ground is 6 ms-1.

(a) Calculate the impulse when his legs hit the ground.

(b) What is the impulsive force on the resident’s legs if he

bends upon landing and takes 0.5s to stop?

(c) What is the impulsive force on the resident’s legs if

he does not bend and stops in 0.05 s?

(d) What is the advantage of bending his legs upon landing?

Example 2

Rooney kicks a ball with a force of 1500 N. The time of

contact of his boot with the ball is 0.01 s. What is the impulse

delivered to the ball? If the mass of the ball is 0.5 kg, what is

the velocity of the ball?

Page 23: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-23

SAFETY VEHICLE

Component Function

Headrest

Air bag

Windscreen

Crumple zone

Front

bumper

Absorb the shock from the accident. Made from steel, aluminium, plastic or

rubber.

ABS Enables drivers to quickly stop the car without causing the brakes to lock.

Side impact bar

Seat belt

Safety features in vehicles

2.7

Page 24: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-24

GRAVITY

Gravitational

Force

Objects fall because they are

This force is known as the

The earth’s gravitational force

Free fall An object is falling freely when it is falling under the force of gravity

only.

An object falls freely only

In vacuum,

They fall with

Acceleration due to

gravity, g Objects dropped

Gravitational field The gravitational field is the region around the earth in which an object

experiences a force towards the centre of the earth. This force is the

gravitational attraction between the object and the earth.

The gravitational field strength is defined as the gravitational force which acts

on a mass of 1 kilogram.

g = m

F Its unit is N kg

-1.

2.8

Page 25: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-25

Gravitational field strength, g = 10 N kg-1

Acceleration due to gravity, g = 10 m s-2

The approximate value of g can therefore be written either as 10 m s-2

or as 10 N kg-1

.

Weight The gravitational force acting on the object.

Weight = mass x gravitational acceleration

W = mg SI unit : Newton, N and it is a vector quantity

Comparison

between weight

&

mass

Mass Weight

The mass of an object is the

amount of matter in the object

The weight of an object is the force of

gravity acting on the object.

Constant everywhere Varies with the magnitude of gravitational

field strength, g of the location

A scalar quantity A vector quantity

A base quantity A derived quantity

SI unit: kg SI unit : Newton, N

The difference

between a

fall in air and

a free fall in a vacuum

of a coin and a

feather. Both the coin and the

feather are released

simultaneously from

the same height.

At vacuum state: There is no air

resistance.

The coin and the feather will fall

freely.

Only gravitational force

acted on the objects. Both will fall

at the same time.

At normal state: Both coin and feather

will fall because of gravitational force.

Air resistance effected by the surface area of

a fallen object.

The feather that has large area will have

more air resistance.

The coin will fall at first.

Page 26: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-26

(a) The two spheres are falling

with an acceleration.

The distance between two

successive images of the sphere

increases showing that the two

spheres are falling with increasing

velocity; falling with an

acceleration.

The two spheres are falling down with

the same acceleration

The two spheres are at the same level

at all times. Thus, a heavy object and

a light object fall with the same

gravitational acceleration

Gravitational acceleration is

independent of mass

Two steel spheres

are falling under

gravity. The two

spheres are dropped

at the same time

from the same

height.

Motion graph for free fall object

Free fall object Object thrown upward Object thrown upward and fall

Example 1

A coconut takes 2.0 s to fall to the ground. What

is

(a) its speed when it strikes the ground

(b) ) the height of the coconut tree

Page 27: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-27

FORCES IN EQUILIBRIUM

Forces in

Equilibrium

When an object is in equilibrium,

Newton’s 3rd

Law

Examples( Label the forces acted on the objects)

Paste more picture

Paste more picture

Resultant

Force

A single force that

Addition of Forces

Resultant force, F = +

Resultant force, F = +

2.9

Page 28: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-28

Two forces acting at a point at an angle [Parallelogram method]

STEP 1 : Using ruler and protractor, draw

the two forces F1 and F2 from a point.

STEP 3

Draw the diagonal of the parallelogram. The

diagonal represent the resultant force, F in

magnitude and direction.

scale: 1 cm = ……

STEP 2

Complete the parallelogram

Resolution of Forces A force F can be resolved into components

which are perpendicular to each other:

(a) horizontal component , FX

(b) vertical component, FY

Fx = F cos θ

Fy = F sin θ

Inclined Plane

Component of weight parallel to the plane = mg sin θ

Component of weight normal to the plane = mg cos θ

Page 29: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-29

Find the resultant force

(d) (e)

Page 30: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-30

Lift

Stationary Lift

Lift accelerate upward

Lift accelerate downward

Resultant Force = Resultant Force = Resultant Force =

The reading of weighing

scale =

The reading of weighing

scale =

The reading of weighing

scale =

Pulley

1. Find the resultant force, F

2. Find the moving mass, m

3. Find the acceleration, a

4. Find string tension, T

Page 31: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-31

WORK, ENERGY, POWER & EFFICIENCY

Work

Work done is

W = Fs W = , F = s =

The SI unit of work is the joule, J

1 joule of work is done when

The displacement, s of the object is in the direction of the force, F

The displacement , s of the

object is not in the direction of

the force, F W = Fs

s F

W = F s

W = (F cos θ) s

Example 1

A boy pushing his bicycle with a

force of 25 N through a distance

of 3 m.

Calculate the work done by the

boy.

Example 2

A girl is lifting up a 3 kg

flower pot steadily to a height

of 0.4 m. What is the work done by the

girl?

Example 3

A man is pulling a crate of fish

along the floor with a force of

40 N through a distance of 6 m.

What is the work done

in pulling the crate?

2.10

Page 32: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-32

Concept D

ef

in

iti

on

Formula & Unit

Power The rate at which work is

done,or P =

W

t

p = power, W = work / energy

t = time

Energy Energy is the capacity to do work.

Potential Energy Gravitational potential energy is

the energy of an object due to

its higher position in the

gravitational field.

m =

h =

g = E =

Kinetic Energy

Kinetic energy is the energy of an

object due to its motion.

m =

v =

E =

No work is done when:

A waiter is carrying a tray of

food. The direction of motion of

the object is perpendicular to

that of the applied force.

A waiter is carrying a tray of

food and walking

No force is applied on the object

in the direction of displacement

(the object moves because of its

own inertia)

A satellite orbiting in space.

There is no friction in space. No

force is acting in the direction of

movement of the satellite.

Page 33: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-33

Principle of Conservation of

Energy

Energy can be changed from one form to another, but it cannot

be created or destroyed.

The energy can be transformed from

Example 4

A worker is pulling a wooden block of weight, W, with a force

of P along a frictionless plank at height of h. The distance

travelled by the block is x. Calculate the work done by the

worker to pull the block.

Example 5

A student of mass m is climbing

up a flight of stairs which has

the height of h. He takes t

seconds..

What is the power of the student?

Example 6

A stone is thrown upward with initial velocity of

20 ms-1

. What is the maximum height which can be reached by the stone?

Example 7

A ball is released from point A of height

0.8 m so that it can roll along a curve frictionless track. What is the

velocity of the ball when it reaches point B?

Page 34: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-34

Example 8

A trolley is released from rest at

point X along a frictionless track.

What is the velocity of the trolley at

point Y?

Example 9

A ball moves upwards along a

frictionless track of height 1.5 m

with a velocity of 6 ms-1

. What is

its velocity at point B?

Example 10

A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the

slope, he does work to overcome friction of 140 J. What is his velocity at the end of the slope?

Page 35: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-35

ELASTICITY

Elasticity

A property of matter that enables an object

No external force is applied.

Molecules are at their equilibrium separation.

Intermolecular force is equal zero.

Compressing a solid causes its molecules

Repulsive intermolecular force

Stretching a solid

Stretching a wire by an external

force:

Its molecules are

When the external force is removed:

The attractive intermolecular forces

2.11

Page 36: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-36

Hooke’s Law

The extension of a spring

F = k x where

F=

x =

k =

Force extension graph

Based on the graph:

Relationship between F & x :

The gradient of the graph represent =

Area under the graph

= elastic potential energy = ½ F x = ½ k x2

The elastic limit of a spring

The maximum force that

If a force stretches a spring beyond its elastic limit, the spring

Force constant of the spring, k

The force required to produce one unit of extension of the

spring.

k is a measurement of the stiffness of the spring

The spring with a larger force constant is

A spring with a smaller force constant is

Page 37: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-57

Factors that effect elasticity

Factor Change in factor How does it affects the elasticity

Length Shorter spring

Longer spring

Diameter of spring wire Smaller diameter

Larger diameter

Diameter spring Smaller diameter

Larger diameter

Type of material Springs made of different materials

Elasticity changes according to the type of material

Arrangement of the spring

In series

The same load is applied to each spring.

Tension in each spring = W Extension of

each spring = x

Total extension = 2x

If n springs are used: The total

extension = n x

In parallel

The load is shared equally among the springs.

Tension in each spring = W

2

Extension of each spring = x

2

If n springs are used:

The total extension = n

x

Example 1

The original length of each

spring is 10 cm.

With a load of 10 g, the extension

of each spring is 2 cm.

What is the length of the spring

system for (a),

(b) and (c)?

Page 38: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-58

Example 2 Diagram below represent a 50 cent coin and a leaf falling in a vacuum container. The coin is heavier than the leaf.

Using the diagram shown and the information given about the weight of the two objects, compare the mass of the coin and the leaf, the time taken to fall, the position of the coin and the leaf and finally deduce the physical quantity which causes the objects to fall.

coin

leaf

Mass of the coin

Time taken to fall in a vacuum

Position of the coin and the leaf

Coin and leaf of different mass reach the bottom of the container at the same time. Coin and leaf fall down due to gravitational force. The magnitude of gravitational pull is constant. It does not depend on the mass Example 3

Diagram 10 shows a student trying to launch a water rocket.

You are required to give suggestion on how to design a water rocket for National

Competition. Based on your knowledge on forces, motion and properties of materials,

explain your suggestion based on the following aspect:

(i) material used

(ii) shape of the rocket

(iii) suitable angle to launch the rocket

(iv) volume of water in the rocket

(v) added structure for the motion of the rocket

Page 39: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-59

Water Rocket

Aspect Structure Explanation

Acceration of the Rocket Make from light material

Structure of the Rocket Aerodynamic

Upthrust force Fill with erated water or gasy

drink with water

Stability of the rocket during flight Use plasticine to make the head

of the rocket

Add fins at the back portion of

the rocket

Example 4

Diagram 4.1 shows a cradle with a baby in it is oscillating vertically. Diagram 4.2

shows another identical cradle with a heavier baby in it is oscillating vertically. It

is observed that the cradle with a heavier mass baby oscillates at a higher

frequency.

Design an experiment to test the hypothesis using spring, slotted weight and other

suitable apparatus.

4 (a) Inference : The extension of the spring depends

1

(b) Hypothesis : As the 1

c (i) Aim : To investigate the relationship between 1

Variables :

Page 40: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-60

c(ii) Manipulated variable :

Responding variable :

1

Constant variable : 1

c(iii) Apparatus : Metre rule, retort stand with clamp ,steel spring, slotted weight and

pin.

1

c(iv)

Set-up the apparatus

1

c (v)

Method of controlling the manipulated variables :

1. Arrangement the apparatus as shown in the diagram.

2.Mark the initial

1

Method of measuring the responding variables :

Record

Extension of spring :

Measure the

1

Repeat the 1

Page 41: 2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2-61

c ( vi) Tabulate Results

Initial length , l0 = cm

Mass of the Slotted weight , m / g 40 80 120 160 200

Weight of slotted mass / N 0.4 0.8 1.2 1.6 2.0

Length of the spring ,l / cm

Extension of the spring x = l- l0

1

(vii)

1

Total 12