2 0 Forces and Motion1

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Physics Module Form 4 Chapter 2 Forces & Motion GCKL 2011

2-1

L I N E A R M O T I O N

Physical Quantity Definition, Quantity, Symbol and unit

Distance, sDistance is the total path length travelled from one location to another.Quantity: scalar SI unit: meter (m)

Displacement, s

(a) The distance in a specified direction.

(b) the distance between two locations measured along the shortest path

connecting them in a specific direction.

(c) The distance of its final position from its initial position in a

specified direction.

Quantity: vector SI unit: meter (m)

Speed,vSpeed is the rate of change of distance

Speed =time

ceDis tan

Quantity: scalar SI unit: m s-1

Velocity, v

Velocity is the rate of change of displacement.

Velocity = time

ntDisplaceme

Direction of velocity is the direction of displacement

Quantity : Vector SI unit: m s -1

Average speedv =

TotalTime

tTotalDis tan

Example: A car moves at an averagespeed / velocity of 20 ms -1

On average, the car moves a distance/

displacement of 20 m in 1 second for the

whole journey.

Average velocity Displacement

TotalTimev

Uniform speed Speed that remains the same in magnitude without considering its direction

Uniform velocity Velocity that remains the same in magnitude and direction

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 When the velocity of an object increases, the object is said to be accelerating.

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v ua

t

Unit: ms

-2

Acceleration is positive

Acceleration is defined as the rate of change of velocity

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.

The velocity of an object decreases from an initial velocity, u, to a lower final

velocity, v.

Zero acceleration An object moving at a constants velocity, that is, the magnitude and direction of

its velocity remain unchangedis not accelerating

Constant acceleration Velocity increases at a uniform rate.

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

, its velocity

increases by 5 ms-1

for every second that the car is in motion.

1. Constant = uniform

2. increasing velocity = acceleration

3. decreasing velocity = deceleration

4. zero velocity = object at stationary / at rest

5. negative velocity = object moves in opposite direction6. zero acceleration = constant velocity

7. negative acceleration = deceleration


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Speed Velocity

The rate of change of

distance

The rate of change of

displacement

Scalar quantity Vector quantity

It has magnitude but

no direction

It has both magnitude

and direction

SI unit : m s-

SI unit : m s-

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 10 m is travelled every second.

2. A steady velocity of -10 ms-1

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

3. A steady acceleration of 4 ms-2

= Speed goes up by 4 ms-1

every 1 second.

4. A steady deceleration of 4 ms -2 = speed goes down by 4 ms-1 every 1 second

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 Vector quantity

It has magnitude but no

direction

It has both magnitude

and direction

SI unit meter SI unit : meter


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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 Rahims displacement from his house

when he reaches the junction. 1.5 km to the

right

When he is at warung PakDin. 0.5 km to the

left.

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

w hen he reaches home,

(i) what is the total distance travelled by

Rahim?

(1.5 + 1.5 + 0.5+0.5 ) km = 4.0 km(ii) what is Rahims total displacement from

his house?

1.5 +( -1.5) +(- 0.5 )+0.5 km = 0 km

Example 2

Every morning Amirul walks to Ahmadshouse

which is situated 80 m to the east of Amiruls house.

They then walk towards their school which is 60 m

to the south ofAhmads house.

(a)What is the distance travelled by Amiruland his displacement from his house?

Distance = (80 +60 ) m = 140 m

Displacement = 100 m

tan =60

80=1.333 = 53.1

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

from his house to Ahmads house and then

to school is 15 minutes, what is his speed

and velocity?

Speed =140

15 60

m

s=0.156 in ms

-1

Velocity =100

15 60

m

s= 0.111 ms

-1

Example 3

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

(a) What is his speed in ms-1

Speed =

s

m

12

60= 5 ms-1

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

distance covered?

distance covered = 40 s 5 ms-1

= 200 m

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?

Average speed = (300 km hr-1

+

4 00 km hr-1

) / 2 = 350 km hr-1

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

Average velocity = 250 km hr-1


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Tan =300

400 = 1.333 =

(c)What is the difference between average speed and

average velocity of the plane?

Average speed is a scalar quantity.

Average velocity is a vector quantity

Example 5

The speedometer reading for a car travelling due north

shows 80 km hr-1

. Anothercar travelling at 80 km hr-1

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

velocity of both carssame?

The speed of both cars are the same but the velocity ofboth cars are different with opposite direction

A ticker timer

Use: 12 V a.c. power supply 1 tick = time interval between two dots. The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2

consecutive dots is 1/50 = 0.02 s.

1 tick = 0.02 s


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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 = (51) x 0.2 s = 0.8 s ort = (5010) ticks x 0.02 s = 0.8 s

Initial velocity, u =

final velocity, v =

acceleration, a =

TICKER TAPE AND CHARTS TYPE OF MOTION

Constant velocity

slow moving

Constant velocity

fast moving

Distance between the dots increases uniformly the velocity is of the object is increasing uniformly The object is moving at a uniform / constant

acceleration.


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- Distance between the dots decrease uniformly

- The velocity of the object is decreasing Uniformly

- The object is experiencing uniform / constant

decceleration

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.

Time = 1/50 s = 0.02 s

(b) What is the time for one strips.

0.02 s 10 = 0.2 s

(c) What is the initial velocity

2 cm / 0.2 s = 10 ms-1

(d) What is the final velocity.

12 cm / 0.2 s = 60 ms-1

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

velocity to final velocity?

( 11 - 1) 0.2 s = 2 s

(f) What is the acceleration of the object.

a =t

uv =

2

1060 ms-2 = 25 ms-2

THE EQUATIONS OF MOTION

2

2 2

1

2

2

v u at

s ut at

v u as

u = initial velocity

v = final velocity

t = time taken

s = displacement

a = constant acceleration


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M O T I O N G R A P H S

DISPLACEMENT TIME GRAPH Velocity is obtained from the gradient of the graph.

A

B : gradient of the graph is positive and constantvelocity is constant.

BC : gradient of the graph = 0

the velocity = 0, object is at rest.

CD : gradient of the graph negative and constant.

The velocity is negative and object moves

in the opposite direction.

VELOCITY-TIME GRAPH Area below graph Distance / displacement

Positive gradient Constant Acceleration

(AB)

Negative gradient Constant Deceleration

(CD)

Zero gradient Constant velocity /

zero acceleration

(BC)

GRAPH s versus t v versus t a versus t

Zero

velocity

Negative

constant

velocity

Positive

Constant

velocity

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GRAPH s versus t v versus t a versus t

Constant

acceleration

Constant

deceleration


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Example 1

Contoh 11

Based on the s

t graph above:(a) Calculate the velocity at

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

(i) 5 ms-1

(ii) 0 ms-1

(iii) - 10 ms-1

(b) Describe the motion of the object at:

(i) AB (ii) BC (iii) CD(i) constant velocity 5 ms

-1

(ii) at rest / 0 ms-1

(iii) constant velocity of 10 ms

-1in opposite

direction

(c)Find:

(i) total distance 50 m + 50 m = 100 m

(ii) total displacement 50 m + (- 50 m) = 0

(d) Calculate

(i) the average speeds

m

35

100= 2.86ms

-1

(ii) the average velocity of the movingparticle.

0

Example 2

(a) Calculate the acceleration at:

(i) JK (ii) KL (iii) LM(i) 2 ms

-2(ii) -1 ms

-2(iii) 0 ms

-1

(b) Describe the motion of the object at:

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

(i)constant acceleration of 2 ms-2(ii) constant deceleration of 1ms

-2

(iii) zero acceleration or constant velocity

Calculate the total displacement.

Displacement = area under the graph

= 100 m + 150 m + 100 m + 25 m= 375 m

(c) Calculate the average velocity.

Average velocity = 375 m / 40 s

= 9.375 ms-1


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I N E R T I A

Inertia The inertia of an object is the tendency of the object to remain at rest or, if

moving, to continue its motion.

Newtons first law Every object continues in its state of rest or of uniform motion unless

it is acted upon by an external force.

Relation between inertia

and mass

The larger the mass, the larger the inertia

SITUATIONS INVOLVING INERTIA

SITUATION EXPLANATION

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

the glass.

The inertia of the coin maintains its state at rest.

The coin falls into the glass due to gravity.

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, the sauce continues in its state of

motion due to the effect of its inertia.

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, the inertia in the passengers made them

maintain their state of motion. Thus when the car stop, the passengersmoved forward.

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

making it difficult to change direction.

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The head of hammer is secured tightly to its handlebyknocking one end of the handle, held vertically, on a hard

surface.

This causes the hammer head to continue on itsdownwardmotion.

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 isrotated.

When the umbrella stops rotating, the inertia of

the drop of water will continue to maintain its

motion.

Ways to reduce the negative

effectsof 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 tokeep 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 theircombinedmass has increased.


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M O M E N T U M

Definition Momentum = Mass x velocity = mv

SI unit: kg ms-1

Principle of Conservation of Momentum In the absence of an external force, the total

momentum of a system remains unchanged.

ElasticCollision Inelastic collision

Both objects move independently at their

respective velocities after the collision.

Momentum is conserved.

Kinetic energy is conserved.

Total energy is conserved.

The two objects combine and move together

with a common velocity after the collision.

Momentum is conserved.

Kinetic energy is not conserved.

Total energy is conserved.

Total Momentum Before = total momentum after

m1u

1+ m

2u

2= m

1v

1 + m2 v2

Total Momentum Before = Total MomentumAfter

m1u

1+ m

2u

2 = ( m1 + m2 ) v

Explosion

Before explosion both object stick together and at

rest. After collision, both object move at opposite

direction.

Total Momentum

before collision is

zero

Total Momentum after

collision :

m1v

1 + m2v2

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From the law of conservation of momentum:

Total Momentum = Total Momentum

Before collision after collision

0 = m1v

1 +m2v2

m1v

1= - m

2v

2

Negative sign means opposite direction

EXAMPLESOF EXPLOSION (Principle Of Conservation Of Momentum)

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, an equal but opposite

momentum is produced to recoil the riffle

backward.

Application in the jet engine:

A high-speed hot gases are ejected from the back

with high momentum.

This produces an equal and opposite

momentum to propel the jet plane forward.

The launching of rocket

Mixture of hydrogen and oxygen fuels burn

explosively in the combustion chamber.

Jets of hot gases are expelled at very high

speed through the exhaust.

These high speed hot gases produce a large

amount of momentum downward.

By conservation of momentum, an equal but

opposite momentum is produced and acted on

the rocket, propelling the rocket upwards.


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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.

A squid propels by expelling water at high velocity.

Water enters through a large opening and exits

through a small tube. Thewater 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

collideswith a car B of mass 1200 kg moving at

10 m s-1

in same direction. Ifthe 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?

1000 kg x 20 ms-1

+ 1200 kg x 10 ms-1

=

1000 kg x v+ 1200 kg x 15 ms

-1

v= 14 ms-1

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 VB 4 ms-1 right

Calculate the value of VA .

[4 x 10 + 2 x (-8)]kgms-1

=[ 4 x v+ 2 x 4] kgms

-1

VA = 4 ms

-1right


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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?

1200 kg x 30 ms-1

+ 1000 kg x (-20 ms-1

)

= ( 1200 kg + 1000kg) v

v = 7.27 ms-1

to the right

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?

0 = 1.5 kg x v

+ 0.01 kg x 300ms-1

v = -2 ms-1

Or

it recoiled with 2 ms-1

to the left


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F O R C E

Balanced Force

When the forces acting on an object are

balanced, they cancel each other out. The

net force is zero.

Effect : the object is at rest

[velocity = 0]

or

moves at constant velocity

[ a = 0]

Example:

Unbalanced Force/ Resultant Force When the forces acting on an object are not balanced,

there must be a net force acting on it.

The net force is known as the unbalanced force or

the resultant force.

Effect : Can cause a body to

- change it state at rest (an object will

accelerate

- change it state of motion (a moving object

will decelerate or change its direction)

Newtons Second Law of Motion The acceleration produced by a force on an object isdirectly proportional to the magnitude of the net force

applied and is inversely proportional to the mass of the

object. The direction of the acceleration is the same as

that of the net force.

Force = Mass x Acceleration

F = ma

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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 forceapplied to it.

The acceleration produced by an

object depends on the mass

Hypothesis The acceleration of the object

increases when the force applied

increases

The acceleration of the object

decreases when the mass of the

object increases

Variables:

Manipulated :

Responding :

Constant :

Force

Acceleration

Mass

Mass

Acceleration

Force

Apparatus and

Material

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

compensated runway and meter ruler.


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Procedure :

- Controlling

manipulated

variables.

-Controlling

responding

variables.

-Repeatingexperiment.

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. 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

Determine the acceleration by

analyzing the ticker tape.

Acceleration

Accelerationv u

a

t

Determine the acceleration by analyzing

the ticker tape.

Accelerationv u

a

t

Repeat the experiment by using two

, three, four and five elastic cords

Repeat the experiment by using two,

three, four and five trolleys.

Tabulation of

data

Force, F/No ofelastic cord

Acceleration, a/ ms-2

1

2

3

45

Mass, m/no of

trolleys

Mass,m/g

1/m,g-1

Acceleration/ms-2

1

2

34

5

Analysing

Result


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1. What force is required to move a 2 kgobject 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 theaverage force of friction acting on the

object is 2 N?

(a) force = 6 N

(b) net force = (62) N

= 4 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?

Answer: 50 N

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.

(a) u = 20 ms-1

v = 0 s = 30 m a = ?

a = - 6.67 ms-2

(b) force = 1200 x 6.67 N

= 8000 N

4.

Which of the following systems willproduce maximum acceleration?D


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I M P U L S E A N D I M P U L S I V E F O R C E

Impulse The change of momentum mv - mu

Unit : kgms-1

or Ns

m = mass

u = initial velocity

v = final velocity

t = timeImpulsive Force The rate of change of momentum in a

collision or explosion

Impulsive force =

change of momentum

time

mv mu

t

Effect of time Impulsive force

is inversely

proportional totime of contact

Longer period of time Impulsive force decrease

Shorter period of time Impulsive force increase

Situations for Reducing Impulsive Force in Sports

Situations Explanation

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

so that the time interval of impact on landing is extended, thus

reducing the impulsive force. This can prevent injuries to the

participants.

Goal keepers will wear gloves to increase the collision time. This

will reduce the impulsive force.

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

time of impact in order to reduce the impulsive force acting on his legs.This will reduce the chance of getting serious injury.

A baseball player must catch the ball in the direction ofthemotion 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.

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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 alarge impulsive force on the wooden slab.

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

upon hitting the nail within a short time interval.

The large change in momentum within a short time interval

produces a large impulsive force which drives the nail into the

wood.

A football must have enough air pressure in it so the contact time is

short. The impulsive force acted on the ball will be bigger and the

ball will move faster and further.

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 residents legs if he

bends upon landing and takes 0.5 s to stop?

(c) What is the impulsive force on the residents legs if

he does not bend and stops in 0.05 s?

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

Answer:

(a) Impulse = 60 kg x ( 6 ms-1

- 0 )

= 360 Ns

(b)Impulsive force = sNs

5.0

360

=7200 N

(c)He experienced a greaterImpulsive force of 7200 N and he

might injured his legs(d)Increase the reaction time so as to

reduce impulsive force

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?

(a) Impulse = 1500Nx 0.01 s

= 15 Ns

(b) velocity =kg

Ns

5.0

15= 30 ms

-1


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S A F E T Y V E H I C L E

Component Function

Headrest To reduce the inertia effect of the drivers head.

Air bag Absorbing impact by increasing the amount of time the drivers head to come to the

steering. So that the impulsive force can be reduce

Windscreen To protect the driver (shattered proof)

Crumple zone Can be compressed during accident. So it can increase the amount of time the car

takes to come to a complete stop. So it can reduce the impulsive force.

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 Prevents the collapse of the front and back of the car into the passenger

compartment. Also gives good protection from a side impact

Seat belt To reduce the effect of inertia by avoiding the driver from thrown forward.

Crash resistant door

pillars

Anti-lock brake system

(ABS)

Traction control Front bumper

Windscreen

Air bags

Head rest

Crumple zones

2.7

Safety features in vehicles


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G R A V I T Y

Gravitational

Force

Objects fall because they are pulled towards the Earth by the force of gravity.

This force is known as the pull of gravity or the earths gravitational force.

Theearths gravitational force tends to pull everything towards its centre.

Free fall An object is falling freely when it is falling under the force ofgravityonly.

A piece of paper does not fall freely because its fall is affected by airresistance.

An object falls freely only in vacuum. The absence of air meansthere is no air resistance to oppose the motion of the object.

In vacuum, both light and heavy objects fall freely. They fall with the same acceleration i.e. The acceleration

due to gravity, g.

Acceleration due to

gravity, g Objects dropped under the influence of the pull of gravity with

constant acceleration.

This acceleration is known as the gravitational acceleration, g. The standard value of the gravitational acceleration, g is 9.81 m s-2.

The value of g is often taken to be 10 m s-2

for simplicity.

The magnitude of the acceleration due to gravity depends on thestrength of the gravitational field.

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

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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 = massx 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.


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2 - 26

(a) The two spheres are falling

with an acceleration.

The distance between two

successive images of the sphereincreases 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 anda 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 sameheight.

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. Whatis

(a) its speed when it strikes the ground(b) ) the height of the coconut tree

(a) t = 2 s u = 0 g = 10 v = ?

v = u + gt = 0 + 10 x 2 = 20 ms-1

(b) s = ut + at2

= 0 + (10) 22

= 20 m


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F O R C E S I N E Q U I L I B R I U M

Forces in Equilibrium When an object is in equilibrium, the resultant force acting on it is zero.

The object will either be

1. at rest2. move with constant velocity.

Newtons 3rd

Law Action is equal to reaction

Examples(Label the forces acted on the objects)

Paste more picture

Paste more picture

Resultant Force A single force that represents the combined effect of two of more forces

in magnitude and direction.

Addition of Forces

Resultantforce,F = F1 + F2

Resultantforce,F = F1 + - F2

2.9


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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 ofForces

A force F can be resolved into components

which areperpendicularto each other:

(a) horizontal component , FX

(b) vertical component, FY

Fx = F cos

Fy = F sin

InclinedPlane

Component of weight parallel to the plane = mg sin

Component of weight normal to the plane = mg cos


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2 - 29

Find the resultant force

(d) (e)

17 N

5 N

FR

7N


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2 - 30

Lift

StationaryLift Lift accelerate upward Lift accelerate downward

ResultantForce = ResultantForce = ResultantForce =

The reading of weighing

scale =


scale =


scale =

Pulley

1. Find the resultant force, F 40 -30 = 10 N 30-2 = 28 N

2. Find the moving mass, m 4 + 3 = 7 kg 3+ 4 = 4 kg

3. Find the acceleration, a 40 -30 = (3+4)a

10 = 7 a

a =10/ 7 ms-2

30 -2 = (4+3 )a

28 = 7a

a = 4 ms-2

4. Find string tension, T T- 3 (10) = 3 a

T = 30 + 3 (10/7)

=240 /7 N

30T = 3 (a)

T =30- 12

= 18 N


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WORK, ENERGY, POWER & EFFICIENCY

Work Work done is the product of an applied force and the

displacement of an object in the direction of the applied

force

W = Fs W = work, F = force s = displacement

The SI unit of work is the joule, J

1 joule of work is done when a force of 1 N moves an

object 1 m in the direction of the force

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

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. 75 Nm

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? 12 Nm

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?

40 N cos 50x 6 Nm

2.10


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Concept D

Formula & Unit

Power The rate at which work is done,

or the amount of work done per

second.

P =W

tp = power, W = work / energy

t = time

Energy Energy is the capacity to do work. An object that can do work has energy Work is done because a force is applied and the objects move.

This is accompanied by the transfer of energy from one object

to another object.

Therefore, when work is done, energy is transferred from oneobject to another.

The work done is equal to the amount of energytransferred.

Potential Energy Gravitational potential energy is

the energy of an object due to

its higher position in the

gravitational field.

m = mass

h = height

g = gravitational acceleration

E = mgh

Kinetic Energy Kinetic energy is the energy of an

object due to its motion.

m = mass

v = velocity

E = mv2

No work is done when:

The object is stationary.

A student carrying his bag while

waiting at the bus stop

The direction of motion of the

object is perpendicular to that of

the applied force.

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.


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Principle of Conservation of

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

be created or destroyed.

The energy can be transformed from one form to another, total

energy in a system is constant.

Total energy before = total energy after

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.

[Px = Wh]

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?

[t

mgh

Example 6

A stone is thrown upward with

initial velocity of 20 ms-1

.

What is the maximum heightwhich can be reached by thestone?

[ 10m ]

Example 7

A ball is released from point A of height 0.8 m so that it can rollalong a curve frictionless track. What is the velocity of the ball

when it reaches point B?

[4 ms-1

]


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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?

[ v2

= 30( ms-1

)2]

[v = 5.48 ms

-1]

Example 9

A ball moves upwards along a

frictionless track of height 1.5 m

with a velocity of6 ms-1

. What is

its velocity at point B?

[v2

= 30( ms-1

)2

v = 5.48 ms-1]

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?

[6 ms-1

]

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