chapter 2 force and motion(teacher)

7/27/2019 Chapter 2 force and motion(Teacher)

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NAME : ENCIK MOHAMED TARMIZI ABD KADIR

FORM 4 : 4 CERDIK, 4 CERDAS, 4 PINTAR


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LEARNING OUTCOMES :

By the end of this lesson, you should be able to:

2.1 Linear Motion* Define distance and displacement

* Define speed and velocity* Define acceleration and deceleration

* Calculate speed and velocity

* Calculate acceleration and deceleration

* Solve problems on linear motion with uniform acceleration

2.2 Motion Graphs* Plot and interpret displacement-time and velocity-time graphs

* Deduce from shape of displacement-time graph* Determine distance, displacement and velocity from a displacement-time graph

* Deduce from the shape of a velocity-time graph* Determine distance, displacement, velocity and acceleration from v-tgrph

* Solve problems on linear motion with uniform acceleration

2.3 Inertia * Explain what inertia is

* Relate mass to inertia

* Give examples of situations involving inertia* Suggest ways to reduce the negative effects of inertia

2.4 Momentum and Conservation of Momentum

* Define the momentum of an object

* Define momentum as the product of mass and velocity, p = mv

* State the principle of conservation of momentum* Describe applications of conservation of momentum

* Solve problems involving momentum

2.5 Effect of Forces* Describe the effects of balanced forces acting on an object

* Describe the effects of unbalanced forces acting on an object

* Determine the relationship between forces, mass and acceleration, F = ma* Solve problem using F = ma

2.6 Impulse and Impulsive Force* Explain what an impulsive force is

* Give examples of situations involving impulsive force

* Define impulse as a change of momentum, Ft = mv - mu

* Define impulsive force as the rate of change of momentum in a collision

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* Explain the effect of increasing / decreasing time of impact on the impulsive force

* Describe situations where an impulsive force needs to be reduced

* Describe situations where an impulsive force is beneficial* Solve problems involving impulsive forces

2.7 Safety Features in the Design of Vehicles* Describe the importance of safety features in vehicles

2.8 Gravity * Explain acceleration due to gravity

* State what a gravitational field is

* Define gravitational field strength

* Determine the value of acceleration due to gravity* Define weight as the product of mass and acceleration due to gravity, W = mg

* Solve problems involving acceleration due to gravity

2.9 Analyzing a scientific investigations* Describe situations where forces are in equilibrium* State what a resultant force is* Add two forces to determine the resultant force

* Resolve a force into two effective component forces

* Solve problems involving forces in equilibrium

2.10 Work, Energy, Power and Efficiency * Define work as the product of applied force and displacement of an object, W= Fs

* State that when work is done, energy is transferred from one object to another* Define kinetic energy and state that Ek = mv

2

* Define gravitational potential energy and state that Ep= mgh

* State the principle of conservation of energy * Define power

* Explaine what efficiency of a device is

* Solve problems involving work, energy, power and efficiency

2.11 The Importance of Maximising the Efficiency of Devices* Recognise the importance of maximising efficiency of devices in conserving

resources

2.12 Analyzing a scientific investigations* Define elasticity* Define Hookes Law

* Define elastic potential energy and state that Ep= kx2

* Describe applications of elasticity* Solve problems involving elasticity

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2.1 : Linear Motion : Motion along a straight line.

Examples:

a car moving on a straight road a person walking down a hallway

a sprinter running on a straight race course

dropping a pencil

throwing a ball straight up

a glider moving on an air track

and many others...

1.Distance : The total path length travelled from one location to another.It is a scalar quantity. The SI unit is metre (m)

2. Displacement : The distance between two locations measured along the shortest pathconnecting them, in a specified direction.

It is a vector quantity The SI unit is metre (m)

Example 1:

If Mawi takes 10s to finish each of following paths AB, find the total distance anddisplacement travelled .

8m

6m1m

2m figure 2.1

A B

(i) Total distance travelled : 2m +1m + 6m + 8m + 8m = 25m

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(ii) Displacement : A B

= 1m + 8m = 9m

Speed and Velocity

Speed Velocity

Definition The distance travelled per unit time or

the rate of change of distance

The speed in a given direction. Or

The rate of change of displacement.

Quantity Scalar Vector

Symbol V v

SI Unit ms-1 ms-1

Formula Speed = Change of distance

Time taken

Average speed = total distance travelled

Total time taken

Velocity = Change of displacement

Time taken

Average velocity = total displacementTime taken

Examples:

1.A car travels between 2 towns 60 miles apart in 2 hours. What is its average speed?

Answer:

average speed = distance/time Therefore, the average speed of the car is 60 miles/2 hours= 30 miles/hour.

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Acceleration : is define as the rate of change of velocity.It is a vector quantity.The symbol : a

SI Unit : meters per second or ms-2

Acceleration = Change of velocity

Time taken

= Final velocity (v) Initial velocity (u)

Time taken (t)

a = v - u

t

Example :

A car's velocity changes from 10 m/s to 2 m/s in 4 seconds. What is its acceleration?

Solution:

The car's change in velocity = ending velocity - starting velocity = 2 m/s - 10 m/s = -8m/s. Its acceleration = its change in velocity divided by the time taken = (-8 m/s)/(4 s) =

-2 m/s/s.

The acceleration is negative if the velocity decreases with time.

Decleration : when the objects experiences negative acceleration

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There are 3 things to remember:

The acceleration of an object is its change in velocity divided by the time.

If an object is moving in the positive direction, its velocity is positive. If it is

moving in the negative direction, its velocity is negative. Change in velocity is the ending velocity minus the starting velocity.

Problem solving

A car is moving in a staright road at a velocity of 10 ms-1.It then speed up uniformly to a

velocity of 30 ms-1 in 4s. immediately after this ,the brake is applied and the car slowsdown uniformly and stops in the next 5s. Calculate the acceleration of the car in the first

4s and the last 5s.

Solution

In the first 4s, u = 10ms-1, v = 30ms-1 , t=4s

Acceleration, a = v-ut

= 30 - 104

= 5.0 ms-2

In the last 5s, u = 30 ms-1, v = 0ms-1, t =5s

Acceleration , a = v-u

t

= 0 - 305

= -6ms-2

The negative sign means that negative acceleration or deceleration.

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Ticker timer

Figure 2.2

1. The ticker timer as shown in figure is used in the laboratory to study motion.2. It can print dots on a tape at a steady rate.

3. It si connected to an a.c . power supply of 50Hz.

4. The time taken to make 50 dots on the ticker tape is 1s. Hence , the time intervalbetween 2 consecutive dots (tick) is 1/50 = 0.02s.

Displacement, s

. . . . . . . .

1 tick = 0.02s. figure 2.3

Velocity,v = Displacement = s

time t

Accleration a, = v-u

t

** Hands on Activity 2.2 &2.3 : Practical book

8

Ticker tape a.u 12 v


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Ticker tapes for objects moving with a constant velocity, an accelerated motion are

shown in the figurebelow.

(a) Constant velocity

(b) Constant acceleration

(c) Constant deceleration

To determine acceleration from the ticker tape.

1.

. U . . . . . . v

Figure 2.4

0.4cm 2.4 cm

Initial velocity , u = s/t = 0.4/0.02 = 20 cms-1

Final velocity, v = s/t = 2.4/0.02 = 120 cms-1

Time taken = (7-1)(0.02s) = 0.12 s

Therefore ,the accleration of the motion, a = v-u = 120 - 20 = 833.33ms-2

t 0.12

Length/cm

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10

8

6

4

2

0.2 0.4 0.6 0.8 1.0 time/s

Figure 2.6

Solution:

Activity 2 : Draw a tape chart following each situation

10

Figure 2.5 shows a tape chart consisting of 5

tick strips. Each strips containing 10 ticks).

Determine the accleration of this motion.(dots formed by a 50 Hz ticker timer)

Initial velocity ,u = s/t = 2 / (10 x 0.02s) = 2 / 0.2s = 10 cms-1

Final velocity , v = s/t = 10 / (10 x 0.02) = 10 / 0.2 = 50 cms-1

Time taken, t = (5 1 ) x 0.2s = 0.8 s

Acceleration, a = v-u = 50 - 10 = 40 / 0.8 = 50 cms-2

T 0.8

u

v


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1. Constant velocity

Distance/ cm

Time /s

2. Constant accelaration

Distance/ cm

Time /s

3. Constant deceleration

Distance/ cm

Time /s

Equations of Motion

1.The equations used in linear motion involve two kinds of motion ;

a) Motions with constant velocity (zero accelerations).

b) Constant change in velocity (constant accleration).

Quantity Symbol Unit

Displacement s Metre (m)

Initial velocity u ms-1

Final velocity v ms-1

Accleration a ms-2

Time t Second (s)

Motion with constantvelocity

Motion with constantacceleration

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Vav = s

t

Vav = u + v

2

S = vav x t S = (u + v) x t

2

v = u + at

s = ut + at2

v2 = u2 + 2 as

Work example:

1.A sprint cyclist starts from rest and accelerates at 1 ms-2 for 20 seconds. He then travelsat a constant speed for 1 minute and finally decelerates at 2 ms-2 until he stops. Find hismaximum speed and the total distance covered in metres.

Solutions:

First stage: u = 0 a = 1ms-1 t = 20s

v= ?v = u + at

= 0 + 1(20)

= 20 ms-1

Distance moved s= ut + at2

= 0(20) + (1)(20)2

= 200m

Third stage

u= 20 ms-1 , v = 0, a = -2ms-1

v2 = u2 + 2 as

Distance moved. s = v2 - u2

2a

= 0 202

2(-2)

= 100 m

2. A car start from rest and accelerates uniformly along road for 10s and reaches a speedof 40 ms-1.Then, it moves at this speed for a while before arriving at a hill slope. The

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Second stage

u= 20 ms-1 (Constant), t = 60s

Distance moved, s = ut

= 20 x 60= 1200 m

Answer:

Maximum speed = 20ms-1

Total distance covered = 200m +1200m +100m = 1500m


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driver stops the engine and let the car move up the hill slope freely until it stops. The

distance traveled along the hill slope is 0.5 km.Calculate

(a) the acceleration of the car in the first 10s.

(b) the distance traveled along the straight road

(c) the deceleration of the car along the hill slope(d) the time taken for the car to stop while moving the hill slope.

(a) Acceleration a = v u = 40 0 = 4ms-2

t 10

(b) The distance traveled along straight road = v

2

= u

2

+ 2as(40)2 = 0 + 2 (4)s

1600 = 8s200m = s

(c ) The deceleration = v2 = u2 + 2as

0 = 402 + 2a( 0.5 x 100)

0 = 1600 + 100a

a = -1600 = 1.6 ms-1

100

(d) Time taken = V = u + at0 = 40 + -1.6 t

t = 40 = 25s

1.6

2.2 Analysing motion Graph

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Check your answer:

a) = 4ms-2, b) = 200m, c) = 1.6ms-2, d) t =25s


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1.There are three types of motion graphs:

(a) Displacement time graph(b) Velocity time graph

(c) Acceleration time graph

The Displacement time Graph

Displacement / m

100 A B

80

Figure 2.760

40

20 CO

10 20 30 40 60 80 Time/s

1. We can determine distance, displacement and velocity from a

displacement time graph.2. The gradient of straight line in the displacement-time graph is the average

velocity ,v of the motion.

3. Referring to figure 2.7 :The gradient of OA = 100 / 20 v = 5 ms-1

The gradient of AB = 0 v = 0 ms-1

The gradient of BC = - 100 / 30 v = - 3.33ms-1

*If the gradient is constant it indicates that the velocity is zero.

4. Negative gradient shows that the velocity is negative, which means that

the direction of motion has reversed.

**Hands-on activity 2.4: Practical book.Activity :

Fill in the blanks:

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Graph of s aginst t Explanation

S/m

t/s

The displacement of the object from a fixed point isconstant .There fore , the velocity of the object is

zero

S/m

t/s

The gradient of the graph is the rate of change ofdisplacement which is the velocity of the

object.The gradient of the the graph is constant.

Therefore, the velocity of the object is constant.

S/m

t/s

The gradient of the graph increases with time.Therefore, the velocity of the object increases with

time.

Examples:

The Velocity time graph

Velocity/ms-1

20 A B

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a car moving with a changing, rightward (+)velocity that is, a car that is moving

rightward and speeding up or accelerating


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16

12

8

4 figure 2.8C

O

0 5 10 15 20 25 30 time/s

1.The graph shows the car moving from rest and increase its velocity to amaximum of 20 ms-1 in 5s. The car maintains that velocity for 10s before it slow

down to a stop.

2.We can determine distance, displacement, velocity and acceleration from a

velocity-time graph.

3.The gradient of the straight line in the velocity-time graph represents the

acceleration .

4. If the gradient is constant, indicating that the acceleration is uniform.

5. If the zero gradient , the acceleration is zero .

6.If the gradient is negative, it means that the acceleration is also negative or

the car is decelerating .

7 .The area under the velocity-time graph represents the displacement. .

Activitiy:From the graph in figure 2.8 ,

Determine

16

A graph of velocity

against time for a car

moving a straight line.


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i) acceleration from t=0s to t=5s. ii)acceleration from t= 5s to t=15s.

= 20 / 5 = 4 ms-2 = 0 ms-2

iii)acceleration from t=15s to t=30s

= - 20 / 15 = - 1.33 ms-2

iv) Displacement from O to A = area under the graph

= x 5 x 20 = 50 m

v) Displacement from A to B = 10 x 20 = 200m

vi) Displacement from B to C = x 20 x 15 = 150 m

vii) The total displacement = 50 +200 + 150 = 400m

viii) Sketch the acceleration time graph

a /ms-2

4 t/s

05 15 30

-1.33

Problem1 - velocity versus time graphGiven is graph of motion of an object moving along a straight line. Determine from thisgraph the:

a) total distance traveled,

b) average velocity for this motion,

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c) maximum and minimum accelerations.

Solution:

a) Total distance traveled: = total area under the graph

= x ( 2 + 1)1 + x (1 + 3)2 + (3 x 3) + ( x 2 x 3)

= 1.5 + 4 + 9 +3 = 17.5 m

b) Average velocity = total distance

total time taken

= 17.5 m = 1.94ms-19 s

c) maximum accelerations = maximum gradient = 1/1 = 1ms-2 & 3-1 = 1 ms-2

2

Minimum accelerations = minimum gradient

= gradient between t =7s 9s

= - 3 = - 1.5 ms-1

2

Veocity time graph Acceleration time graph

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Table shows comparison between velocity-times graph and acceleration time

graph.


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The object is not moving.The gradient is

equal to zero. The acceleration is zero.

Since the object is not moving ,the

acceleration is zero.

Th

e object is moving with constant velocity.

Since the gradient is equal to zeroThe acceleration is zero

Constant velocity gives zero acceleration.

The object is moving with constant

velocity The acceleration is positive.

The velocity is decreasing uniformly with

time. The gradient of graph is negative.The acceleration is negative

a/ms-2

t/s

v/ms-1

t/s

a/ms-2

t/s

v/ms-1

t/s

a/ms-2

t/s

v/ms-1

t/s

a/ms-2

t/s

19

t/s

v/ms-1


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Problem 2:

A car which is moving with constant velocity of 30ms-1 passes a point O.The driverapplies brake and the car decelerates uniformly to stop in 5s. Then, the driver reverses his

car back to point O in uniform acceleration in 5s.Using graphical method, calculate

(a) the displacement and average velocity of the car

(b) the distance traveled and the average speed of the car.

Solution:

(a) total displacement = 0 m average velocity = 0 ms-1

(b) total distance traveled

= x 5 x30 + x 5 x30

=150 m

Average speed = total distance

Time taken

= 150

10s

= 15ms-1

5 10 t/s

20

30

v/ms-1

-30


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2.3 INERTIA

1. Inertia is the reluctant of a body to change its state of rest or motion.2. The characteristics of inertia can be described by Newton first law of motion.

3. Newtons First Law: States that every object will continue in its state of uniform

velocity or rest unless it is acted upon by an external force** Hands on Activity 2.5

2.3.1. Inertia and Mass

1. Mass is a measurement for the amount of inertia.

2. Relation between mass and inertia : Hands on activity 2.6 Practical book.3. A body of greater mass will have a greater inertia.

empty sand

The positive effect of inertia

1.

2. Blood rushes from your head to your feet while quickly stopping when riding on adescending elevator.

3.To dislodge ketchup from the bottom of a ketchup bottle, it is often turned upside down

and, thrusted downward at high speeds and then abruptly halted

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It is more difficult to push a bucket filled with sandthan an empty bucket because the sand filled with

sand has a greater inertia.

The head of a hammer can be tightened onto the wooden

handle by banging the bottom of the handle against a hardsurface.


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Ways to reduce the negative effects of Inertia.

Problem 1

Does a 2 kg rock have the inertia of a 1 kg rock?

Answer: The larger the mass , the larger its inertia.Therefore a 2kg rock has twice the

inertia of a 1 kg rock.Problem 2.

P

Heavy mass

Q

Force on lower string

Amassive block is suspended on a string P and slowly pulled by another string Q

attached to it from below ,as shown in figure.

i) which string has greater tension string P or Q? Pii) which spring is more likely to break? Piii) Which property is important here mass or weight? Weightiv) If the spring is snapped downward immedistely ,which string is more likely to

break? Qv) Is mass or weight important this time? mass

2.4 MOMENTUM

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Safety belt

headrest

Air bag

To prevent the driver from hitting the

steering wheel or dashboard during a

collision.


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1.The momentum of an object is defined as the product of its mass and velocity.

2. It is a vector quantity. Its direction is the same as the velocity and its unit is kgms-1 orNs.

Hands-on activity 2.7.

Example :

1. Calculate the momentum of the ball below:

(a) (b)

v = 20 ms-1 v = 60 ms-1

m = 1500 kg m = 3000 kg

RAJAH 1 p = m x v p = m x v

= 1500 x 20 = 3000 x ( -60 )

= 30000 kgms-1 = -180000 kg ms-1

2.

CONSERVATION OFMOMENTUM

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Momentum = mass x velocity

p = mv

Figure shows a 20g tennis ball hits thewall. Calculate the momentum of the

tennis ball before and after hits the wall.

Before:

P = mv

= 0.02 kg x 10 ms-1

= 0.2 kgms-1 @ Ns

After:

P = mv

= 0.02 kg (-5ms-1)= - 0.1kgms-1 @ Ns


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1.The principle of conservation of momentum : In the absence of an external force, thetotal momentum of a system remains unchanged.

2.This means the total momentum before a collision is equal to the total momentum after

a collision.

Types of Collision

(a) Completly inelastic collision

u1 u2 v

before ` after

(i) The colliding objects stick together after collision.

(ii) Total momentum and energy are conserved

(iii) Kinetic energy is not conserved.

(b) Elastic collision u1 u2 v1 v2

BEFORE ` AFTER

(i) The colliding objects spread away after collision

(ii) Total momentum and energy are conserved

(iii) Kinetic energy is conserved.

(c) Explosion

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m1

m2

m2

m1

m1 u1 + m2 u2 = ( m1 + m2 ) v

m1

m2

m2

m1

m1 u1 + m2 u2 = m1 v1 + m2 v2


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u= 0 ms-1 v1 v2

m1 m2

BEFORE ` AFTER

(i) The colliding objects spread away after explosion.

(ii) Total momentum and energy are conserved.

Worked example

1. A 15-kg medicine ball is thrown at a velocity of 20 m/s to a 60-kg person who

is at rest on ice. The person catches the ball and subsequently slides with the ball

across the ice. Determine the velocity of the person and the ball after the collision.

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m2

m1

0 = m1 v1 + m2 v20 = m1(- v1 ) + m2 v2

m1 v1 = + m2 v2


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2. A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car.

The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming thatmomentum is conserved during the collision, determine the velocity of the truck

after the collision.

(Discussion):Application of Conservation of Momentum

1. Launching of rockets.

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Momentum before = momentum after(inelastic)

P1 + P2 = P1 + P2

m1u1 + m2 u2 = (m1 + m2 )V

15(20) + 60 (0) = (15 + 60) V

300 = V75

4.0 ms-1 = V

Momentum before = momentum after (elastic)

P1 + P2 = P1 + P2

m1u1 + m2 u2 = m1v1 + m2v2

3000(10) + 1000(0) = 3000 v1 + 1000 (15)

30000 + 0 = 3000v1 + 15000

30000 15000 = v1

3000

V1 = 5 ms-1


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2. Jet engine

.

2.5 THE EFFECTS OF A FORCE

FORCE ( F )

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- is an action ofpushingorpullingan object.

- It can cause :

(i) a stationary object to move.

(ii) a moving object to accelerate or to slow down.

(iii) a moving object to change its direction of motion.

(iv) an object to change in size and shape.

- It is a vectorquantity.

- Its SI unit isNewton (N) orkgms-2

BALANCED FORCESWhen two or more externalforces acting on an object which produce no net

force. The object remain at a rest or moves with constant velocity.

Examples :

Remain at a rest Moves with constantvelocity

Notes:

28

Weight

Normal ReactionFriction

Thrust

BALANCED FORCES

a = 0 ms-2

Object at rest

v = 0 ms-1

Object in motion

v 0 ms-1

Stay at rest Stay in motion


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UNBALANCED FORCES

When two or more external forces acting on an object which are

Not balancedtherefore one netforce is produced.

This force known as the unbalancedforce or the resultantforce.

Also known as NewtonsSecondLaw of Motion.

Examples :

Footballer kicks a stationary ball, The engine of moving car is shut

the unbalanced force causes the ball down, the net force ( friction force )

to accelerate . causes the car to stop

Notes:

29

Hard Surface

UNBALANCED FORCES

a 0 ms-2


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The Relationship Between Force, Mass and Acceleration( Newtons Second Law of motion )

Experiment 2.2 and 2.3 : Practical Book ( pages : 29 / 31 )

The acceleration of an object is directly proportional to the magnitude of the

applied force, and is inversely propotional to the mass of the object.

The direction of the acceleration is the same. as the net force.

a F .. .. (1)

a 1/m (2)

a F

m

F m a

F = k ma (3)

( kis a constant )

By definition, a force of 1N acting on a mass of 1 kg, will produce anacceleration of 1 ms-2 . By substituting the values in equation (3), we have

1 N = k ( 1 kg ) ( 1 ms-2 )

30

a F a 1 m


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k = 1, therefore

Example :

5 N

A wooden block of mass 2 kg being pulled along a surface with a force of 5 N at a

constant velocity.

(a) What is the net force acting on the wooden block ?

0 N

(b) What is the magnitude of the frictional force acting on the wooden block ?

5 N ( to the left / opposite direction )

(c) The experiment is repeated with a force of 10 N. Calculate the acceleration of the

wooden block.F = ma

(10 5) = 2a5 = 2aa = 2.5 ms-2

(d) What is your assumption to answer the question in (c ) ?

The frictictional force is still the same

2.6 Impulse and Impulsive Force

1. Impulsive Force is the force acting on objects in a short periodof time

31

F = ma

Wooden blockm = 2 kg


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during collisions orexplosions.

Examples :

Hittinga nail using hammer A collision of cars Hittinga ball

F = mv mu

t

2. The SI units forImpulsive Force isNewton (N) orkgms-2

3. Impulse is the change ofmomentum

Ft = mv mu

4. The SI units for Impulse is Ns orkgms-1

Examples :

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u= 5 ms-1

1.

A ball of mass 0.4 kg moving along a smooth surface at a constant velocity 5 ms 1 as

shown in the figure above. It hits a wall elastically and moves back at the same speed.

If the time of impact of the ball with the wall is 0.1 s , calcullate:

(i) The impulsive forces acting on the ball

F = mv mut

= 0.4 ( -5 5 )0.1

= -40.0 N

(ii) The change of momentum of the ball

Ft = mv mu= 0.4 ( -5-5 )= -4.0 Ns

2. In one of the activities at the National Service camp, a participant of 55 kg is to climb

up 2.5 m wall from ground level and jump down. Based on the jump of the participant,calculate :

(i) The speed of the participant just before his legs touch the ground

Potential Energy = Kinetic Energy ( Tell the student )mgh = mv2

v = 2gh= 2 x 10 x 2.5= 7.07 ms-1

(ii) The change in momentum of the participant at this moment

Ft = mv mu= 55 ( 7.07-0 )= 388.85 kgms-1

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(iii) The impulse of the force on the participant if the time of action between

his legs and the ground is 0.03 s.

F = mv mu

t= 388.850.03

= 12961.67 N

(iv) The participant bends his leg when he reaches the ground and stops in

0.08s. Determine the impulse of the force acting on the participant.

F = 388.850.08

= 4860.63 N

(v) Why do the participant bend his leg on landing on the ground?

To increase the time of collision between the legs and the ground , sothat the force of impulse is reduced. Therefore, he can avoid a seriousinjury.

2.7 SAFETY FEATURES IN VEHICLES

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There are several safety features installed to a vehicle in order to reduce seriousinjury to the driver and passengers.

1.

Bumpers are installed at the back of vehicles to

lengthen the collision time and to reduce the

impulsive force.

2. Seat belts are fitted in the vehicles to lengthen the

collision time and to prevent driver and passengers from

hitting againstthe interior of the car during an accident.

3. An air bag is installed inside the steering wheel of a

vehicle to avoidthe driver and passengers from

direct impact with the steering wheel or windscreen.

The collision time will be lengthen and the

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Bumper


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impulsive force can be reduce and prevent

a serious injury.

2.8 GRAVITY

1.All objects are pulled towards the Earth by theforce of gravity.

2. This force is known as the Earths Gravitational Force

F = mg

where m the mass of the object

g the gravitational field strength

3. When an object falls without encounting any resistance from a height

towards the earth with an acceleration due to gravity, g the object is said to be

free falling.

Example:

Two objects of different mass are falling down in the air as shown in figure below. Based

on the diagram below, we can conclude that :

(i) The acceleration of the object is at the same rate due to the gravitational

force. We can call the situation as afree fallsituation.

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(ii) Object which has less mass, falls down slowerdue to the effect of

more air resistance acting on it.

(iii) If the experiment is carried out in vacuum, the objects will falling down at the

same rate .

Weight

1. The weight, W of an object is equal to the gravitational force acting on theobject.

W = mgwhere m the mass of the object

g the gravitational field strength

2. Weight is therefore aforce and the SI units for W isNewton (N) orkgms-2

3. W is a vector quantity.

Definitions :

The Gravitational Field of the Earth isthe region

around the Earth which an object is force

towards the centre of Earth

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The Gravitational Acceleration isthe acceleration of an

object object due to thepullof the gravitonal force.

The Gravitational Field Strength isthe gravitational

force acting on a mass ofan objectplaced at that point.

It can be calculated by using

gravitonal force. The SI units for gisNkg-1

2.9 FORCES IN EQUILIBRIUM

If the forces acting on any objects is in equilibrium, the objects remain at a rest

or moving at a constant velocity. This is because the forces are balanced

and resultant force iszero

Example :

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g = Wm


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Based on the Figure above, the aeroplane wil move at a constant velocity if ,

Lift Force Gravity Force

Thrust Force Drag Force

Examples:

1.

T1 Hint: Use the bold lines to form a triangle

300

T2

In the figure above, an object with a mass of 50 kg is balanced by two ropes with

a tension forces of T1 and T2 respectively.

a) What is the meaning of equilibrium of forces ?

The resultant force is equal to zero.

b) Draw a suitably vector diagram to show that the weight of object, W and the

tension forces T1 and T2 exist in equilibrium.

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50 kg

=

=


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T1

W (500 N)

300

T2

c) Determine the values of T1 and T2 .

T1 = 500 = 1000 Nsin 300

T2 = 500 = 866 N ( Also can use cos 300)tan 300

d) Among the ropes, T1 and T2 which is more likely to break ? Explain youranswer.

T1 because it has a higher tension.

RESULTANT FORCE

A Resultant Force is a single force that represents the combined effect of

two ormore forces with magnitude and direction.

Examples :

Calculate the Resultant Forces, FR for each diagrams.

1.

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F1 = 2N F2 = 5N FR = F1 + F2

= 7 N

2. F1 = 2N F2 = 5N FR = F2 - F1

= 3 N

TRIANGLE METHOD

1. The triangle method is recommended if the two forces areperpendicular

to each other.

2. Phythagoras theorem and trigonometry can be used to solve the problems.

Example:

1. The figure shows two forces F1 and F2 are acting perpendicular to each otherat one point. Calculate the value of resultant forces acting on it.

F1= 6N 6 N

F2= 8N 8N

FR= 82 + 62

= 10 N

= sin-1 0.6 =

PARALLELOGRAM METHOD

1. The Parallelogram Method is used if the two forces are notperpendicular

to each other.

2. This method uses scaled drawing.

Example :

1. The Figure below show a boat being pulled by two forces with the magnitued of 6 N

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and 8 N respectively. The angle between the two forces is 600 . Calculate the resultant

forces by using the parallelogram of forces method.

F1 = 6N

600

F2 = 8N

Solution:

Scale : 1cm : 1 N

RESOLUTION OF A FORCE

1. A single force can be resolved into two component forces which are

perpendicular to each other.

2. This is known as the resolution of force.

3. A force can be resolved into the horizontalforce Fx and the

verticalforce Fy .

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Fy F

Fx

Fx = F cos

Fy = F sin

Examples :

1.

The Figure shows a tourist is pulling a bag with a force of 12.0 N at angle an

angle of 600 to the horizontal floor. Calculate :

(i) The horizontal component of force, Fx= 12.0 cos 600 = 6.0 N(ii) The vertical component of force, Fy = 12.0 sin 600 = 10.4 N

2.

30

Fx

300

Fy W = 500N

The Figure above shows a wooden block with a weight, W = 500N resting on aplatform at an angle of 300 to the horizontal floor. Find :

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(i) The horizontal component of force, Fx = 500 sin 300 = 250 N(ii) The vertical component of force, Fy = 500 cos 300 = 433.0 N

RESULTANT FORCES and READING of a WEIGHING MACHINE in a LIFTNote : The value of R is equals to the reading on the scale of the weighing machine.

I - Lift at rest or moves up / down at a constant velocityR

F = maR mg = 0

a = 0 ms-2 R = mg

mg

II Lift moves up at an acceleration of a ms-2

R

F = ma a R mg = ma

R = mg + ma

mg

III Lift moves down at an acceleration of a ms-2

R

F = mamg R = ma

a R = mg - ma

mg

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IV Lift moves down when the cable of the lift breaks or Free Fall Situation

R

F = mamg R = ma

a = g = 10 ms-2 R = mg - mgR = 0 N

mg

Examples :

1. A student of mass 50 kg is standing on the platform of a weighing machimeplaced on the floor of a lift. What is the scale reading of the machine if

(a) the lift is stationaryF = maR mg = 0R = mg

= 50 (10)= 500 N

(b) the lift moves upward at an acceleration of 2.0 ms-2F = maR mg = maR = mg + ma

= 50 (10) + 50 (2)= 600 N

(c) the lift moves downward at an acceleration of 3.5 ms-2F = mamg R = ma

R = mg - ma= 50 (10) 50 ( 3.5)= 325 N

(d) the lift is free fallingF = maR mg = maR = mg- mg

= 0 N

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2. In the figure given, aFN force pulls a mass of 3 kg byusing a smooth pulley. If the acceleration of the mass is

2.0 ms-2 , find what is the value force,F?

F = maF 30

= 3( 2.0 )F = 36.0 N

F 30 N F N

3. Two weights of mass 2 kg and 3 kg are joined by a length of rope

which passes over a smooth pulley as shown below. If the systemis released from rest, what is the tension, T of the rope ?

F = maT F = ma

T T 30 20 = 5a 30 T = 3(2)10 = 5a T = 24 N

a = 2.0 ms-2

2 kg 3 kg 30 N ( For T can use 2 kg or 3 kg )

20 N 30 N

4.

In the above figure, the trolley of mass 500 gis placed on the smooth surface of a table.

What is the acceleration of the trolley when

the 2 kg weight is released ?

F = ma2 kg 20 = (2 + 0.5 )a

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3 kg

500 g


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Fx

300

a = 8.0 ms-2

20 N

5. F= 50N A small box of mass 500 g is placed on asmooth inclined plane as shown in the

following figure. What is the acceleration of

the box down the surface of the inclined

plane ?[ Pushing force F = 50 N] F = ma

300 50 - 5 sin 300 = 0.5aa = 95.0 ms-2

5 N

2.10 WORK, ENERGY, POWER and EFFICIENCY

1. Work is defined as the product of an applied force and the

displacement of an object in the direction of the applied force.

2. Work = Force x Displacement

W = F x s

where W Work

F Force Object

s Displacement

3. The SI units for work isNm orJ.

4. Work is a scalarquantity.

5. In general , work is the transferof energy. In physics we say that work is done on

an object when you transfer energy to that object.

Examples : Calculate the work done for each situations given below.

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Directionof Force

Direction ofmovement

a


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(i) (ii) F = 10 N

F = 10N

s = 10 cm

W = F s = 10 x 0.1 = 1.0 J s = 10 cm

(iii) W = Fs = 10 x 0 = 0 N

600 600

s = 10 cm

W = Fs = ( F cos )( s )= 10 cos 600 x 0.1= 0.5 N

4. No work is done if

a. The object is stationary

b. No force applied on the object in the direction of displacement

c. The direction of motion of the object isperpendicular

to the applied force

Area Under a Force Distance Graph

Force / N

Distance / m

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Work = Area under the graph

W = F x s

Work Done against the Force of Gravity

1. An externalforce must be applied to lift an object of weight , mgto the

height ofh metre.

2. Work = Force x Displacement

W = F x s = mg x h

W = mgh

ENERGY

1. Energy is the capacity to do work.

2. The SI units isNm orJ

3. Energy is a scalarquantity.

4. When energy changes from one form to another, work is done.

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Potential Energy

1. The Potential Energy of an object is the energy stored in the object because of

itsposition orstate.

2. There are two kinds of potential energy :

(i) Gravitational Potential Energy

(ii) Elastic Potential Energy

Gravitational Potential Energy

1. The Gravitational Potential Energy of any object is the energy stored in the

object because of its height above the earths surface.

2. The gravitational potential energy is equel to the work done to raise an object to aparticular height.

W = Ep = mgh

Ep Gravitational Potential Energy

Elastic Potential Energy

1. The Elastic Potential Energy of any object is the energy stored in the

object as a result ofstretching orcompressing it.

rubber ban

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2. The elastic potential energy is equel to the work done to extend the elastic object.

W = Ep = F x

Ep Elastic Potential Energy

F Force to extend the elastic object.x Extension of the elastic object

Kinetic Energy

1. The Kinetic Energy is the energy possessed by a movingobject.

2. The kinetic energy,Ek of an object of mass, m kg travelling at a velocity ofv ms-1

is given as :

W = Ek = m v2

Ek kinetic energy

m mass of the objectv velocity of the object

PRINCIPLE OF CONSERVATION of ENERGY

1. The Principle of conservation of energy states that the total energy in a system is

constant.

2. Therefore, energy cannot be created ordestroyed.

3. Energy can be converted from one form to another.

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v ms-1

m k


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POWER

1. Power is defined as the rate at which work is done or the amount of

work done persecond.

Power = Work done

Time taken

P = Wt

2. The SI unit of paower is Watt (W) orJs-1

3. When work is done, the object gains energy, therefore

Power = Change in energyTime taken

P = Et

4. Power is also can be expressed in terms offorce and velocity.

P = W

t

= F x s

t

= F x s

t

P = F x v where v - velocity

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EFFICIENCY

1. The efficiency of a device is defined as the percentage of the energy

input that is transformed into useful energy.

Efficiency, = Useful energy output x 100%Energy input

2. The efficiency also can be calculated in terms of useful power output and

power input.

Efficiency, = Useful power output x 100%Power input

3. Energy Convertors change the energy from one form to another.

4. Input energy is the energy supplied to the energy convertor.

5. The output energy can be classified into two types :

(i) the useful energy

(ii) the unwanted energy.

6. The efficiency of energy convertors is always less than 100 % because some

of the energy lost to the environment

7. When a device is operating at the maximum passible efficiency :

(i) The cost of operating the device is reduced.

(ii) The unwanted output energy is reduced.

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Convertors

Energy InputEinput

Useful EnergyEoutput

Unwanted Energy


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(iii) Less input energy is required to produce the same useful

ouput energy.

2.11 The Importance of Maximising the Efficiency of Devices1. The energy crisis can be lightened if efficiency use of devices is increased

2. Following are the examples of how to maximize the efficiency.

(a) When having a shower during hit weather, a lowerwater

temprature is needed . Hence less electrical energy is used.

(b) Thermostat should be used in air-conditioner to controlthe

temperature of a room. When the temperature is cool enough, electrical

energy supplied is automatically switched off.

(c) The fluorescent lamp is more efficient than the filament lamp,

because its lamp-light is brighter and use less electricity.

2.12 ELASTICITY

1. Elasticity is the ability of object to return to its original shape and

size after an applied external force is removed .

HOOKES LAW

1. Hookes Law state that the extension of a spring is directly proportional to the

applied force if the elastic limit is not exceeded.

Force, F (N)

Spring not obeying Hookes Law

Elastic limit

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Spring obeying Hookes Law

0 Extension, x (cm)

2. The elastic limit of a spring is defined as the maximumforce that can be

applied to a spring whereby the spring will be able to return to its original length

once the applied force is removed.

3. If a force stretches a spring beyond its elastic limit, the spring will

not return to its original length even though the force is removed.

We can say that the spring have apermanent extension.

4. For a spring that obeys Hookes Law,

F x

F = k x

whereF Force on the spring

x Extension ( x = Extended length Original length )

k Spring constant

Force, F (N)

b

0 Extension, x (cm)

a

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k= b = Gradient ofF-x graph.a

5. The SI units forkisNm-1 orkgs-2

6. The spring constant shows the stiffness of the spring.

7. The spring which has a larger value ofk, is harder to be extended.

Hence, it is said to be more stiff or also known as a stiff spring.

8. The spring which has a small value ofk, is easierto be extended.

Hence, it is said to be less stiff or also known as a soft spring.

Factors Affecting the Stiffness of a Spring

1. Physical Condition

(i) Length of spring

The longer the spring, the more elastic will the spring be.

Therefore, the stiffness of the spring is decreased.

(ii) Diameter of spring wire

The greater the diameter of the spring wire, the less elastic will

the spring be. Therefore, the stiffness of the spring is increased.

(iii) Diameter of spring coil

The larger the diameter of the spring coil, the more elastic will

the spring be. Therefore, the stiffness of the spring is decreased.

(iv) Material of spring

Different materials of springswill give a different elasticity. A steel

spring is less elasticity than a copper spring. Therefore, the

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stiffness of the spring is increased.

2. Spring System

Two springs can be connected in series or in parallel . The series system is

more elastic than theparallelsystem. Therefore, the stiffness of the

series spring is decreased.

3. Value of Spring Constant, k

A spring which is less elastic , has a highervalue of spring constant, k.

Therefore, the stiffness of the spring is increased, as the spring constant is

increased.

F (N)

Spring A ( k isgreater)

Spring B ( k is smaller)

x (cm)

ELASTIC POTENTIAL ENERGY

1. Elastic Potential Energy is the energy stored in a spring when it is

extendedorcompressed.

2. The elastic potential energy is a result of the work done to extend or

compress the spring.

F (N)

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x (cm)

Elastic Potential Energy = Area under the graph ofF-x

Ep = Fx

Substituting F = kx ( from Hookes Law )

Ep = kx xx

Ep = kx2

Application of Elasticity

The elastic property of springs is commonly used in daily life, for examples :

(i) Elasticity of mattress, enables sleeping in comfort.

(ii) Spring are used in ammeter to return the pointer to the zero mark on

the scale after a measurement has been taken.

(iii) The elastic strings of tennis or a badminton racket enable them to

rebound the ball or shuttle.

chapter 2 force and motion(teacher)

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