Bab 2 _2.3_2.4 pelajar

2.3 MEMAHAMI INERSIAUNDERSTANDING INERTIA

Jelaskan apakah inersia/Explain what inertia is

Inersia suatu objek ialah kecenderungan objek itu kekal dalam keadaan rehat atau terus bergerak

dalam keadaan gerakannya

The inertia of an object is the tendency of the object to remain at rest or, if moving, to continue its motion

• Suatuobjekberadadalamkeadaanrehatakancenderungkekaldalamkeadaan rehat .

An object in a state of rest tends to remain at rest .

• Suatuobjekyangberadadalamkeadaanbergerakcenderunguntukkekaldalamkeadaan gerakan .

An object in a state of motion tends to stay in motion .

Hukum Newton pertama/Newton’s first law:Setiapobjekakanterusberadadalamkeadaanrehatataukeadaangerakannyadenganhalajuseragam

kecuali ia dikenakan daya luar.

Every object continues in its state of rest or of uniform motion unless it is acted upon by an external force.

Semakinbesarjisim,semakinbesar /The larger the mass, the larger the • Duabaldikosongdigantungdengantalidarisiling. Two empty buckets are hung with rope from the ceiling.• Sebuahbaldidiisidenganpasirmanakalabaldiyanglainadalahkosong. One bucket is filled with sand while the other bucket is empty.• Kemudian,kedua-duabaldiditolak. Then, both buckets are pushed.

• Didapatibaldikosong itu ditolak berbanding dengan baldi yang diisi dengan pasir.

It is found that the empty bucket is to push compared to the bucket with sand.

• Baldiyangdiisidenganpasiradalahlebih untuk bergerak.

The bucket filled with sand is more to move.• Apabila kedua-duabaldi diayundandiberhentikan, baldi yangdiisi denganpasir lebih susahuntuk

diberhentikan. When both buckets are oscillating and an attempt is made to stop them, it is more difficult to stop the bucket

filled with sand.• Inimenunjukkanbaldidenganjisimyanglebihbesarmenghasilkanrintanganyanglebihuntukberubah

dari keadaan rehat atau dari keadaan gerakan. This shows that the bucket with a bigger mass offers a greater resistance to change from its state of rest or

from its state of motion.

• Olehitu,suatuobjekdenganjisimyangbesarmempunyaiinersiayanglebih .

So, an object with a larger mass has a inertia.

Hubung kait inersia dengan jisim/Relate mass to inertia

TaliRopes

BaldiBuckets

PasirSand

03 FIZ Form 4 Ch2 B 2F.indd 40 9/10/13 4:30 PM

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Aktiviti yang melibatkan inersia/Activities involving inertia

Apabilasekepingduitsyiling20sendikuiskearahtimbunanduitsyiling20senpadapermukaanyanglicin,duitsyiling di bawah dihentam keluar tanpa menggerakkan duit

syiling yang lain. Ini menunjukkan bahawa inersia bagi timbunan duit syiling di atas bercenderung untuk kekal

dalam keadaan rehat dan menentang gerakan.When a 20 cent coin is flicked towards a stack of 20 cent coins on a

smooth surface, the bottom coin is knocked off without moving

the rest of coins. This shows that the inertia of the stack of coins

above tends to remain at rest and resists motion.

Apabila kadbod ditarik keluar dengan cepat, duit syiling itu

terus jatuh ke dalam gelas. duit syiling itu

mengekalkannya dalam keadaan walaupun kadbod itu ditarik keluar.When the cardboard is pulled away quickly, the coin drops straight into the glass.

The of the coin maintains it in its position even when the cardboard is withdrawn.

LetakkansegelasairdiatassekepingkertasA4.Dengancepattarik keluar kertas itu secara mendatar. Apakah yang akanberlaku kepada gelas air itu?Place a glass of water on a piece of A4 paper. Suddenly you pull the paper horizontally. What happens to the glass of water?

TroliTrolley

PenghalangObstracle

Blok kayuWooden block

Sebuahblokkayudiletakkandiatassebuahtroliyangbergerakmenuruni landasan. Apabila gerakan troli itu dihalang olehsuatupenghalang,blokkayuituakankekaldalamkeadaan gerakan dan ia menggelongsor ke hadapan. Inersia blok kayu itu berkecenderung untuk mengekalkan keadaan gerakannya.A wooden block is placed on top of a moving trolley down a runway. When the motion of the trolley is stopped by an obstacle, the wooden

block will continue its state of motion and slide forward.

The inertia of the wooden block tends to keep its state of motion.


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Contoh situasi yang melibatkan inersia/Examples of situations involving inertia

Basyangpegun/The bus is stationary

Basbergeraksecaratiba-tibakedepanThe bus moves forward suddenly

badan penumpang masih kekal dalam keadaan rehat.

Ini menyebabkan badannya terhumban ke .

When the bus moves forward suddenly from rest, the of the passenger's body tends to keep him at rest. This causes his body

to be thrown .

Bassedangbergerak/The bus is moving

Basberhentisecaratiba-tibaThe bus stops suddenly

Penumpang dalam bas yang bergerak terhumban ke hadapan apabila bas itu berhenti secara tiba-tiba. Mengapa?Passengers in a moving bus will be thrown forward when the bus comes to a halt suddenly. Why?

Penumpang berada dalam keadaan apabila bas itu sedangbergerak.Apabilabas ituberhenti secara tiba-tiba,

inersia badan penumpang cenderung untuk terus bergerak ke

. Ini menyebabkan badan penumpang terhumban ke

.

The passengers are in a state of when the bus is moving. When the bus stops suddenly, the inertia of the passenger tends to

continue in its motion. This causes his body to be thrown

.

Aktiviti yang melibatkan inersia/Activities involving inertia

Sebuah buku ditarik keluar dari kedudukan tengahnya. Buku

di atasnya akan . Inersia

cuba perubahannya dari keadaan rehat,iaitu,apabilabukuditarikkeluar,buku-bukudiatastidakakanbergerakbersama-sama.A book is pulled out from its central position. The books on top will

drop . Inertia tries to the change from rest, that is, when the book is pulled out, the books on top do not follow suit.


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Penumpang di dalam bas akan terhumban ke belakang apabila bas yang pegun memecut ke hadapan. Mengapa?Passengers in a bus will be thrown backwards when a stationary bus starts to accelerate. Why?Apabila bas itu bergerakke depan secara tiba-tiba dari rehat,


Gerakan ke bawah yang cepatFast downward motion

SosSauce

Soscilidalambotolbolehdituangkeluardengansenang jikabotoldigerakkan turundengancepatdanberhentisecaratiba-tiba.Jelaskan.Chili sauce in the bottle can be easily poured out if the bottle is moved down fast with a sudden stop. Explain.

• Sos dalambotol bersama-samadengan botol semasa pergerakan kebawah.

The sauce in the bottle with the bottle during the downward movement.

• Apabilabotolituberhentisecaratiba-tiba, sos menyebabkan ia terus bergerak ke bawah dan mengakibatkan sos dituang keluar dari botol itu.

When the bottle is stopped suddenly, the of the sauce causes it to continue in its downward movement and thus the sauce is poured out of the bottle.

Gerakan ke bawah yang cepatFast downward motion

Kepalatukuldicantumdenganketatkepadapemegangnyadengan penghujungpemegangnya,secaramenegak,diataspermukaanyangkeras.

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

• Inimenyebabkan kepala tukulmeneruskan gerakan apabila gerakan pemegangitudiberhentikan.Denganini,hujungataspemegangituakandimasukkanlebih dalam ke dalam kepala tukul.

This causes the hammer head to continue on its motion when the motion of the handle is stopped. So that the top end of the handle is slotted deeper into the hammer head.

Titisan air pada payung basah akan jatuh apabila budak itu memusingkan payung itu.The water droplets on a wet umbrella will fall when the girl rotates the umbrella.

• Iniadalahdisebabkantitisanairpadapermukaanpayungitu secara serentak apabila payung itu dipusingkan.

This is because the water droplets on the surface of the umbrella simultaneously as the umbrella is rotated.

• Apabilapayung ituberhentimemusing, titisan air akan terus mengekalkan pergerakannya.

When the umbrella stops rotating, the of the of water droplets will continue in its original motion.

Seorangbudakmelarikandiridarilembudalamgerakanzig-zag.Mengapa?A boy runs away from a cow in a zig-zag motion. Why?


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Cadangan untuk mengurangkan kesan negatif inersiaSuggestions to reduce the negative effects of inertia

1 Keselamatandalamkereta:Safety in a car:

(a) Talipinggangkeledarmengekalkanpemandupadatempatduduknya.Apabilakeretaberhentisecara

mendadak,talipinggangitumengelakkanpemandudaripadaterhumbanke .A safety belt secures a driver to his seat. When the car stops suddenly, the seat belt prevents the driver from

being thrown .

(b) Alaskepalamencegahkecederaanlehersemasaperlanggarandaribelakang.Inersiakepalacenderung

untuk mengekalkannya keadaan apabilabadandigerakkansecaratiba-tibakedepan.A headrest prevents injuries to the neck during rear-end collisions. The inertia of the head tends to keep it

in its state of when the body is moved forward suddenly.

(c) Begudaradipasangdidalamstereng.Iamembekalkankusyen

untuk mengelakkan pemandu daripada pada stereng atau papan pesawat kereta semasa perlanggaran.An air bag is fitted inside the steering wheel. It provides a cushion

to prevent the driver from the steering wheel or dashboard during a collision.

2 Perabot yang diangkat oleh lori biasanya perlu dengan tali kepadabahagian-bahagian lori yang tertentu supaya apabilaloribergerakatauberhentidengantiba-tiba,perabotitutidakakanjatuh atau tidak akan terhumban ke depan.

Furniture carried by a lorry normally is by ropes to certain fixed parts of the lorry so that when the lorry moves or stops suddenly, the furniture will not fall or will not be thrown forward.

3 Empattangkikecildimanajisimmuatandibahagiantaratangki-

tangki tersebut akan mempunyai yang lebih kecil.

Ini akan mengurangkan pada setiap tangki yang disebabkanolehinersiajikaloritangkiituberhentidengantiba-tiba.

Four small tanks with distributed mass will have smaller .

This will greatly reduce the inertial on each tank if the tanker stops suddenly.

TaliRope

Treler dengan 4 tangki kecilTrailer with 4 small tanks

Kepala loriTractor Lori tangki

Tanker


Sebuah kapal minyak yang besar mengambil masa yang lebih panjang untuk memecut kepada lajumaksimumnya dan ia mengambil beberapa kilometer untuk berhenti walaupun propelernya telah diterbalikkan. Mengapa?A massive oil tanker (a very big ship) takes a long time to accelerate to its full speed and a few kilometers to come to a stop even though the engine has reversed its propeller to slow it down. Why?


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TujuanAim

Untuk mengkaji hubungan antara jisim dan inersia (tempoh ayunan)

To study the relationship between mass and inertia (period of oscillation)

RadasApparatus

BilahHacksaw,pengapit-G,jamrandikdanplastisin.Hacksaw blade, G-clamp, stopwatch and plasticine.

Pemboleh ubahVariables

ProsedurProcedure

1 Letakkansejumlahplastisin(berbentuksfera)denganjisim30gpadahujungbilahHacksaw.Place a lump of plasticine (sphere-shaped) with a mass of 30 g at the free end of the Hacksaw blade.

2 Sesarkan sedikit bilah Hacksaw dan lepaskannya supaya ia berayun secaramengufuk.Displace the Hacksaw blade slightly and release it so that it oscillates horizontally.

3 Tentukandanrekodkanmasayangdiambiluntuk10ayunanlengkap,t saat.Determine and record the time taken for 10 complete oscillations, t seconds.

4 Hitungkantempohayunan,T = t10

saat.

Calculate period of oscillation, T = t10

seconds.

5 Ulangilangkah1–4eksperimendenganjisim40g,50g,60gdan70g.Repeat steps 1 – 4 of the experiment with mass of 40 g, 50 g, 60 g and 70 g.

6 Lakarkan graf tempoh ayunan melawan jisim.Plot the graph of period of oscillation against mass.

KeputusanResults

Jisim / gMass / g

Masa untuk 10 ayunan, t/sTime for 10 oscillation, t/s T = t

10 s

t1 t2 tmin

30

40

50

60

70

Pengapit-G/G-clamp

Bilah HacksawHacksaw blade

Plastisin/Plasticine

EksperimenExperiment Inersia dan Jisim / Inertia and Mass


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Pemboleh ubah dimanipulasi/Manipulated variable

Pemboleh ubah bergerak balas/Responding variable

Pemboleh ubah dimalarkan/Fixed variable:

AnalisisAnalysis

Lakarkan graf Tmelawanjisim,m.Plot the graph T against mass, m.

PerbincanganDiscussion

1 Nyatakan kuantiti yang digunakan untuk mewakili inersia dalam aktiviti ini.State the quantity used to represent inertia in this activity.

2 Apakahhubunganantaratempohayunansuatuobjekdenganinersianya?What is the relationship between the period of oscillation of an object and its inertia?

3 Daripadagraf,nyatakanhubunganantaraFrom the graph, state the relationship between

(a) tempoh ayunan dengan jisim objek.period of oscillation and mass of object.

(b) inersia suatu objek dan jisimnya.inertia of an object and its mass.


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Pemain IPlayer I

Pemain IIPlayer II

2.4 MENGANALISIS MOMENTUMANALYSING MOMENTUM

Definisi momentum sebagai hasil darab jisim dan halajuDefine momentum as the product of mass and velocity

Momentum = jisim × halaju Momentum = mass × velocity

UnitSI: kg m s-1 atau N s (Newton saat)

SI unit: kg m s-1 or N s (Newton second)

Momentum adalah suatu kuantiti vektor .Arahmomentummengikutarah halaju .

Momentum is a vector quantity. The direction of the momentum follows the direction of the velocity .

Dalampermainanbolasepak,seorangpemainberjisim70 kg bergerak dengan halaju 4 m s-1 dan seorang pemain yang lain yang berjisim 75 kg bergerakdengan 3 m s-1 menghala antara satu sama seperti yangditunjukkan.Hitungkanmomentumkedua-duapemainitumasing-masing.In a football game, a player of mass 70 kg is moving with velocity of 4 m s–1 and the other player of mass 75 kg ismoving with 3 m s-1 towards each other as shown. Calculate the momentum of the two players respectively.Penyelesaian/SolutionMomentum pemain I/Momentum player I = m1v1=(70kg)(4ms–1)=280kgms-1

Momentum pemain II/Momentum player II = m2v2=(75kg)(–3ms–1)=–225kgms-1

Nenek (m = 80 kg) menggelongsor sekeliling gelangganggelongsordenganhalaju6ms–1.Tiba-tibadiaberlanggardenganBobby(m=40kg)yangberadadalamkeadaanrehat.HitungkanmomentumnenekdanBobbymasing-masing.Granny (m = 80 kg) whizzes around the ring with a velocity of 6 m s–1. Suddenly she collides with Bobby (m = 40 kg) who is at rest. Calculate the momentum of granny and Bobby respectively.

Penyelesaian/SolutionMomentum nenek/granny = m1v1=(80kg)(6ms–1)=480kgms–1

MomentumBobby=m2v2=(40kg)×(0ms–1)=0kgms–1 (dalam keadaan rehat / at rest)

Contoh/Example 1

Contoh/Example 2

NenekGranny Bobby

Nyatakan prinsip keabadian momentum/State the principle of conservation of momentum

Tanpakehadirandayaluar,jumlahmomentumdalamsuatusistemkekaltidakberubah.In the absence of an external force, the total momentum of a system remains unchanged.Jumlah momentum sebelum perlanggaran/letupan = Jumlah momentum selepas perlanggaran/letupan Total momentum before collision/explosion = Total momentum after collision/explosion


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Aktiviti/Activity 1

Aktiviti/Activity 2 Aktiviti/Activity 3

Rajah di sebelah menunjukkan dua orang adik-beradik yang sedangmenggelongsor.Abangbergerakdanberlanggardenganadiknyayangberadadalamkeadaanrehat.Apakahgerakanmerekaselepasperlanggaran?The diagram on the right shows two brothers skating. The elder brother moves and collides with his younger brother who is at rest. What is their movement after the collision?

Selepasperlanggaran,/After collison,

Laju abang berkurang ./The speed of the elder brother decreases .

Laju adik bertambah ./The speed of the younger brother increases .

Momentum abang berkurang ./Momentum of the elder brother decreases .

Momentum adik bertambah ./Momentum of the younger brother increases .

Adakahjumlahmomentumsebelumperlanggaransamadenganjumlahmomentumselepasperlanggaran?Is the total momentum before collision equal to the total momentum after collision?

Ya/Yes.

Menjentiksekepingduitsyiling20sen,A secara terus kepadasekepingduitsyiling20sen,B yang lain.Flick a 20-cent coin, A, directly to another 20-cent coin, B.

A B

(a) Apakahyangberlakukepadagerakankedua-duaduit syiling selepas perlanggaran?

What happens to the motion of both coins after collision?

DuitsyilingAberhenti,duitsyilingB bergerak.

Coin A stops, coin B moves.

(b) Apakah yang berlaku kepada momentum duitsyiling A selepas perlanggaran?

What happens to the momentum of coin A after collision?

Momentum duit syiling A dipindahkan kepada

duit syiling B selepas perlanggaran.

Momentum of coin A is transferred to coin B after

collision.

Menjentiksekepingduitsyiling20sen,A secara terus kepadaduitsyiling20senB dan C.Flick a 20-cent coin A, directly to 20-cent coins B and C.

A B C (a) Gambarkan gerakan semua duit syiling selepas

perlanggaran. Describe the motion of all the coins after collision.

DuitsyilingA/Coin A:Berhenti/Stop

DuitsyilingB/Coin B:Rehat/At rest

DuitsyilingC/Coin C: Bergerakkekanan/Moves to the right

(b) Apakah yang berlaku kepada momentum duitsyiling A selepas perlanggaran?What happens to the momentum of coin A after collision?

Momentum duit syiling A dipindahkan ke duit

syiling B dan duit syiling C.

Momentum of coin A is transferred to coin B and to

coin C.


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Aktiviti/Activity 4

Aktiviti/Activity 5

Rajahdisebelahmenunjukkansebijibolakeluli,A ditarik dan dilepaskan. The diagram on the right shows a steel ball, A is pulled and released.(a) Bola itu akan berlanggar dengan empat biji bola yang lain. Ini akan

menyebabkanbolaterakhir,Ebergerakkeketinggianyang sama dengan ketinggian bola A.

The ball will collide with the other four balls. This will cause the last ball, E to

move to the same height as ball A.

Adakahmomentumdiabadikan?Is the momentum conserved?

Ya/Yes

(b) Apakahyangakanberlakujikakedua-duabolaAdanBditarikdankemudiandilepaskan? What will happen if two balls A and B are pulled and then released?

BolaD dan E akan bergerak ke ketinggian yang sama dengan bola A dan Bmasing-masing.BolaC akan

berada dalam keadaan rehat.

Balls D and E will rise to the same heights of balls A and B respectively. Ball C is at rest.

E D C B A

Seorangbudakperempuanberdiridalamkeadaanrehatdiataspapanluncur.Dia membalingkan bola ke hadapan. Bola itu bergerak ke kiri. Budakperempuan bergerak ke kanan.A girl is standing at rest on the skateboard. She throws the massive ball forward. The ball moves to the left. The girl moves to the right.

• Momentumbolasebelumbalingan=0

Momentum of the ball before the throw = 0

• Momentumbudakperempuansebelumbalingan/Momentum of the girl before the throw = 0

• Jumlahmomentumselepasbalingansamadenganjumlahmomentumsebelumbalingan = 0

Total momentum after the throw is equal to total momentum before the throw = 0

• Jumlahmomentumselepasbalingan=momentumbola+momentumbudakperempuan=0

Total momentum after the throw = momentum of the ball + momentum of the girl = 0

Jadi, selepasbalingan,magnitudmomentumbudakperempuanadalah sama dengan magnitud

momentum bola tetapi dalam arah bertentangan .

Therefore, after the throw, the magnitude of the momentum of the girl is equal to the magnitude of the momentum

of the ball but in the opposite direction.


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Perlanggaran kenyal/Elastic collision Perlanggaran tak kenyal/Inelastic collision

m1 m2 m1 m2

u1 u2

Sebelum perlanggaran Before collision

Selepas perlanggaranAfter collision

v1 v2

m1 m2 m1 m2

u1 u2 vm1 + m2



• Kedua-duaobjekbergeraksecara berasingan denganhalajumasing-masingselepasperlanggaran.

Both objects move separately at their respective velocities after the collision.

• Jumlah momentum diabadikan. Total momentum is conserved.

• Jumlah tenaga diabadikan. Total energy is conserved.

• Tenaga kinetik diabadikan. Kinetic energy is conserved.

• Kedua-duaobjekbergabungdanbergerakbersama

dengan satu halaju sepunya selepas perlanggaran.The two objects combine and move together with a

common velocity after the collision.

• Jumlah momentum diabadikan. Total momentum is conserved.

• Jumlah tenaga diabadikan. Total energy is conserved.

• Tenaga kinetik tidak diabadikan. Kinetic energy is not conserved.

m1 m2 m1 m2

u1 u2 v1 v2



Tuliskan persamaan yang menghubungkaitkan jumlah momentum sebelum perlanggaran dengan jumlah momentum selepas perlanggaran:Write equation which relates the total momentum before collision with the total momentum after collision:

m1u1 + m2u2 = m1v1 + m2v2

m1 m2 m1 m2

u1 u2 v



Tuliskan persamaan yang menghubungkaitkan jumlah momentum sebelum perlanggaran dengan jumlah momentum selepas perlanggaran:Write equation which relates the total momentum before collision with the total momentum after collision:

m1u1 + m2u2 = (m1 + m2)v

Letupan/Explosion

Troli pegunStationary trolleys

m2 m1

m2 m1

v2 v1

pin

Sebelum letupan/Before explosion

Selepas letupan/After explosion

Sebelum letupan, kedua-duaobjek bercantum bersama dan berada dalam keadaan rehat. Selepas letupan, kedua-dua objek

bergerak pada arah yang bertentangan .

Before explosion, both the objects stick together and are at rest. After

explosion, both objects move at opposite directions.

Jumlah momentum sebelum

letupan adalah sifar .The total momentum before

explosion is zero .

Jumlah momentum selepas

letupan = m1v1+m2v2

Total momentum after explosion

= m1v1 + m2v2


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(Catatan: v2 bernilai negatif)(Remarks: V2 has a negative value)

Apakah yang dimaksudkan dengannilai negatif bagi v2?Why does v2 have a negative value?

Arahbertentangan

Opposite direction

Daripadaprinsipkeabadianmomentum:From the principle of conservation of momentum:

Jumlah momentum sebelum perlanggaranTotal momentum before collision

= Jumlah momentum selepas perlanggaranTotal momentum after collision

Terbitkan persamaan untuk letupan:Derive an equation for explosion:0=m1v1+m2v2m1v1 = –m2v2

Huraikan aplikasi prinsip kebadiaan momentumDescribe applications of the principle of conservation of momentum

m1 + m2

(a) Sebelum letupan Before explosion

(b) Selepas letupan After explosion

Pegun/Stationaryu = 0

m1

PeluruBullet

m2

v2

v1

Catatan/Remarks: Jisim senapang/Mass of riffle = m1 Jisim peluru/Mass of bullet = m2

Selepasletupan/After explosion: v1 = Halaju senapang/Velocity of riffle v2 = Halaju peluru/Velocity of bullet

• Apabila sepucuk senapang ditembak, peluru yang berjisimm2 bergerak dengan halaju tinggi, v2. Ini menghasilkan suatu

momentum ke arah hadapan . When a rifle is fired, the bullet of mass m2 moves with a high velocity, v2.

This creates a momentum in the forward direction.• Daripada prinsip keabadian momentum, suatu momentum yang

sama tetapi bertentangan arah dihasilkan supaya senapang itu

tersentak ke belakang . From the principle of conservation of momentum, an equal but opposite

momentum is produced to recoil the rifle backward .

RoketRocket

Gas panasHot gas

Pelancaran roket/The launching of rocket• Campuranbahanapihidrogendanoksigenterbakardenganletupan

dalam kebuk pembakaran. Gas panas dalam jet itu dipancutkan

dengan kelajuan yangsangattinggimelaluiekzos. A 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.

• Kelajuan tinggi gas panas ini menghasilkan momentum yang

besar ke bawah .

This high-speed hot gas produces a large momentum .• Dengan prinsip keabadian momentum, suatu momentum yang sama tetapi bertentangan arah dihasilkan dan menggerakkan

roket itu ke atas . By the principle of conservation of momentum, an equal but opposite

momentum is produced and propels the rocket upwards .


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Huraikan aplikasi prinsip kebadiaan momentum linearDescribe applications of the principle of conservation of linear momentum

Gas panasHot gas

Jet

Aplikasienjinjet:Application in the jet engine:• Suatu gas panas yang berkelajuan tinggi dipancut keluar dari

belakang dengan momentum tinggi .

A high-speed hot gas is ejected from the back with high momentum .

• Inimenghasilkanmementumyang sama tetapi bertentangan

arah untuk menolak jet bergerak ke hadapan.

This produces an equal and opposite momentum to propel the jet plane forward.

Gerakan udara ke belakangMovement of air backwards

Bot berkipasFan boat

Dalamkawasanpaya,suatubotberkipasdigunakan.In a swamp area, a fan boat is used. • Kipasitumenghasilkangerakanudaraberkelajuantinggike

belakang . Ini menghasilkan suatu momentum yang besar ke belakang.

The fan produces a high speed movement of air backwards . This

produces a large momentum backwards.

• Dengankeabadianmomentum,suatumomentumyang sama tetapi bertentangan arah dihasilkan dan ditindakkan ke atas bot

itu.Jadi,botituakanbergerakke hadapan .

By conservation of momentum, an equal but opposite momentum

is produced and acts on the boat. So the boat will move forward .

SotongSquid

Seekorsotongbergerakdenganmengeluarkancecairpada halaju

yang tinggi.Airmasukmelalui pembukaanyangbesar dankeluar

melalui tiubyangkecil.Airdipaksakeluarpada kelajuan tinggi

ke belakang. Magnitud momentum air dan sotong adalah sama

tetapi pada arah yang bertentangan. Ini menyebabkan sotong itu

bergerak ke hadapan .

A squid propels by expelling a liquid 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. The magnitude of the momentum of

water and squid are equal but opposite in direction. This causes the

squid to jet forward .


13

Menyelesaikan masalah melibatkan momentum linearSolve problems involving linear momentum

KeretaAyangberjisim1000kgbergerakpada20ms–1 berlanggar dengan kereta Byangberjisim1200kgdanbergerakpada10ms–1dalamarahyangsama.Akibatnya,keretaB,bergerakkehadapanpada15ms–1.Berapakahhalaju,v, bagi kereta A sebaik sahaja selepas perlanggaran?Car A of mass 1 000 kg moving at 20 m s–1 collides with car B of mass 1 200 kg moving at 10 m s-1 in the same direction. If car B is shunted forwards at 15 m s–1 by the impact, what is the velocity, v, of car A immediately after the crash?

uA = 20 m s-1

m1 = 1 000 kg

A

uB = 10 m s-1

m2 = 1 200 kg

B

Penyelesaian/Solution Jumlah momentum sebelum perlanggaran = Jumlah momentum selepas perlanggaran Total momentum before collision = Total momentum after collision

(1000kg)(20ms–1)+(1200kg)(10ms–1) =(1000kg)v+(1200kg)(15ms–1) 20000kgms–1+12000kgms–1 =(1000kg)(v)+18000kgms–1

(1000kg)(v) =14000kgms–1

∴ v =14ms–1

Latihan/Exercises 1

Sebijibolayangberjisim5kgdibalingkanpadahalaju20kmj–1kepadaLilyyangberjisim60kgpadakeadaan rehat di atas ais. Lily menangkap bola itu dan kemudian menggelongsor dengan bola di atas ais. Tentukan halaju Lily dengan bola selepas perlanggaran. A 5 kg ball is thrown at a velocity of 20 km h–1 towards Lily whose mass is 60 kg at rest on ice. Lily catches the ball and subsequently slides with the ball across the ice. Determine the velocity of Lily and the ball together after the collision.

u1 = 20 km j-¹m1 = 5 kg

u2 = 0 km j-¹m2 = 60 kg

m1 = 5 kg

v = ?

m2 = 60 kg

Penyelesaian/Solution Jumlah momentum sebelum perlanggaran = Jumlah momentum selepas perlanggaran Total momentum before collision = Total momentum after collision

(5kg)(20kmj–1)+(60kg)(0kmj–1) =(5+60)kg× v (100+0)kgkmj–1 =(65kg)v ∴ v =1.54kmj–1

(v =1.54kmh–1)

Latihan/Exercises 2


14

Sebuahtrakyangberjisim1200kgbergerakpada30ms–1 berlanggar dengan sebuah kereta yang berjisim 1000kgyangbergerakdalamarahbertentanganpada20ms–1.Selepasperlanggaran,kedua-duakenderaanitubergerakbersama.Berapakahhalajukedua-duakenderaanitusebaiksahajaselepasperlanggaran?A truck of mass 1 200 kg moving at 30 m s–1 collides with a car of mass 1 000 kg which is traveling in the opposite direction at 20 m s–1. After the collision, the two vehicles move together. What is the velocity of both vehicles immediately after collision?

30 m s-¹ 20 m s-¹ v

(a) Sebelum perlanggaran Before collision

(b) Selepas perlanggaran After collision

Penyelesaian/SolutionJumlah momentum sebelum perlanggaran = Jumlah momentum selepas perlanggaran Total momentum before collision = Total momentum after collision

(1200kg)(30ms–1)+(1000kg)(–20ms–1)=(1200+1000)kg× v (36000–20000)kgms–1=(2200kg)v (2200kg)v=16000kgms–1

∴ v=7.27ms–1

Latihan/Exercises 3

Seorangmenembak sepucuk pistol yang berjisim 1.5 kg. Jika peluru itu berjisim 10 g danmempunyaihalaju300ms–1selepastembakan,berapakahhalajusentakanpistolitu?A man fires a pistol which has a mass of 1.5 kg. If the mass of the bullet is 10 g and it has a velocity of 300 m s–1 after shooting, what is the recoil velocity of the pistol?

1.5 kg 10 g

300 m s-¹v

Pegun/Stationary

(a) Sebelum tembakan Before shooting

(b) Selepas tembakan After shooting

Penyelesaian/Solution Jumlah momentum sebelum tembakan = Jumlah momentum selepas tembakan Total momentum before explosion = Total momentum after explosion

0kgms–1=(1.5kg)(v)+(0.010kg)(300ms–1) (1.5kg)(v)=–3.0kgms–1

∴ v=–2.0ms–1

Latihan/Exercises 4


15

Bayangkanandaberlegardi sebelahkapalangkasapada orbit bumi dan rakan anda yang sama jisim bergerak dengan 4 km j–1 (dengan merujuk kepada kapal angkasa) melanggar anda. Jika dia memeganganda,berapakahkelajuanandabergerak(dengan merujuk kepada kapal angkasa itu)?Imagine that you are hovering next to a space shuttle in earth orbit and your buddy of equal mass who is moving at 4 km/hr (with respect to the ship) bumps into you. If she holds onto you, how fast do you move (with respect to the ship)?

Sebelum perlanggaranBefore collison

m

Dalam gerakanIn motion

u1 = 4 km j–1

Dalam keadaanrehatAt rest

u2 = 0 km j–1

m

Dalam gerakan bersamapada laju yang sama

In motion togetherat the same speed

mm

Selepas perlanggaranAfter collison

Penyelesaian/Solution


=

Jumlah momentum selepas perlanggaranTotal momentum after collision

(mkg)(4kmj–1)+(mkg)(0kmj–1) = (m+m) kg × v(4m) kg km j–1+0=(2m) kg × v

∴ v = (4m) kg km j–1

(2m) kg =2kmj–1

Seekorikanyangbesaryangberjisim3mbergerakdengan2ms–1 bertemu seekor ikan kecil yang berjisim m dalam keadaan rehat. Ikan besar itu menelan ikan kecil dan meneruskan gerakan dengan kelajuan yang berkurang. Jika jisim ikan besar adalah tiga kali ganda jisim ikan kecil,berapakah halaju ikan besar selepas menelan ikan kecil itu?A large fish of mass 3 m is in motion at 2 m s–1 when it encounters a smaller fish of mass m which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish after swallowing the smaller fish?

3m m

Dalam gerakan/In motionu1 = 2 m s–1

RehatAt rest

Sebelum perlanggaranBefore collison

Selepas perlanggaranAfter collison



=

Jumlah momentum selepas perlanggaranTotal momentum after collision

[(3m) kg ×(2ms–1)]+0=(3m +m) kg × v (6m) kg m s–1 =(4m) kg × v ∴ 4v =6ms–1

v =1.5ms–1

Huraikan apa yang dilakukan oleh penjaga gol sebelum dia menendang bola itu.Describe what the goalkeeper does before kicking the ball.•Penjagagolituakanmengambilbeberapalangkahkebelakangdankemudian

berlari ke hadapan untuk menendang bola itu.

The goalkeeper takes a few steps backwards and then runs forward to kick the ball.

•Bolaituakanbergerak lebih jauh/lebih cepat apabila ditendang semasa berlari berbanding dengan tendangan dari kedudukan pegun.

The ball goes further / faster when kicked while running compared to kicking from a standing position.

• Iniadalahdisebabkanseorangpemainbolasepakyangberlarimempunyaimomentumyang besar dan

momentumnya dipindahkan kepada bola.

This is because a running football player has a large momentum and his momentum is transferred to the ball.

Latihan/Exercises 5 Latihan/Exercises 6


16

TujuanAim

Untuk menunjukkan jumlah momentum bagi suatu sistem tertutup adalah malar dalam perlanggaran tak kenyal.To show that the total momentum of a closed system is constant in an inelastic collision.

RadasApparatus

Jangkamasadetik,pitadetik,plastisin,pita selofan, troli, landasan,bekalankuasaa.u.12V.Ticker timer, ticker tape, plasticine, cellophane tape, trolleys, runway, 12 V ac power supply.

ProsedurProcedure

1 Dirikansatulandasandenganmengubahsuaikecerunannyasupayalandasanterpampasgeseran di mana troli boleh bergerak turun landasan dengan halaju malar.Set up a runway and adjust the slope to compensate for friction where the trolley moves down the runway with constant velocity.

2 Letakkan plastisin pada troli P dan Q supaya mereka akan melekat antara satu sama lain semasa perlanggaran.Fix plasticine on trolleys, P and Q so that they can stick together upon collision.

Pita detikTicker tape

Jangka masa detikTicker timer Landasan

terpampasgeseranFriction-

compensatedrunway

Blok kayu/Wooden block

Bekalan kuasaPower supply

Troli PTrolley P Plastisin

Plasticine Troli QTrolley Q

3 Pita detik diletakkan melalui jangka masa detik dan dilekatkan pada troli P.A ticker tape is passed through the ticker timer and is attached to trolley P.

4 Mulakan jangka masa detik dan tolakkan troli P supaya ia bergerak menuruni landasan dan berlanggar dengan troli Q,yangberadadalamkeadaanrehat.Start the ticker timer and give trolley P a push so that it will move down the runway and collide with trolley Q, which is at rest.

5 Daripadapitadetikyangdiperoleh,tentudanukurkanhalajuberikut.From the ticker tape obtained, determine and measure the following velocities.

(a) Halaju troli Psebelumperlanggaran,uP / Velocity of trolley P before collision, uP

(b) Halaju troli Qsebelumperlanggaran,uQ / Velocity of trolley Q before collision, uQ

(c) Halaju troli (P+Q)selepasperlanggaran,vVelocity of trolley (P + Q) after collision, v

6 Langkah-langkah 2 – 5 diulangi dengan jisim P dan jisim Q yang berbeza sepertiditunjukkan dalam jadual di bawah.Steps 2.5 are repeated for different masses of P and Q as shown in the table below.

Sebelum perlanggaranBefore collision


Jisim troli PMass of trolley P

kg

Jisim troli QMass oftrolley Q

kg

Halaju PVelocity of P

m s–1

Jumlah momentum

Total momentum kg m s–1

Halaju sepunyaCommon velocity

m s–1

Jumlah momentum

Total momentum kg m s–1

1 1

2 1

1 2

2 2

3 2

EksperimenExperiment Keabadian Momentum / Conservation of Momentum


17

KeputusanResult

Contoh: Perlanggaran tak kenyal antara dua troliExample: Inelastic collision between two trolleys

18.4 cm 9.2 cm



Perlanggaran berlaku di siniCollision

occurs here

Arah gerakan/Direction of motion

Kuantiti fizikPhysical quantity



Panjang10detik10-tick length

18.4cm 9.2cm

Masadiambiluntuk10detikTime taken for 10 ticks

0.2s 0.2s

HalajuVelocity

18.4cm0.2s

=0.92ms–1 9.2cm0.2s

=0.46ms–1

Jisim troli (jisim 1 troli = 1 kg)Mass of trolley (mass of 1 trolley = 1 kg)

1 kg 2kg

MomentumMomentum

(1kg)(0.92ms–1)=0.92kgms–1

(2kg)(0.46ms–1)=0.92kgms–1


1 Bandingkanjumlahmomentumsebelumperlanggarandanselepasperlanggaran.Compare the total momentum before collision and after collision.

Jumlah momentum sebelum dan selepas perlanggaran adalah sama.

The total momentum before collision and after collision are equal.

2 Nyatakan satu kesimpulan.State a conclusion.

Tanpakehadirandayaluar,jumlahmomentumsebelumperlanggaranadalahsama

dengan jumlah momentum selepas perlanggaran.

In the absence of any external force, the total momentum before collision is equal to total

momentum after collision.

3 Apakahtujuanutamamengubahsuailandasansupayalandasanterpampasgeseran?What is the main purpose of adjusting the runway so that it is friction-compensated?

Troli bergerak dengan halaju malar.

The trolley moves with constant velocity.


18

MODUL • Fizik TINGKATAN 4

© Nilam Publication Sdn. Bhd. 58

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UNIT

Huraikan kesan daya tidak seimbang yang bertindak ke atas suatu objekDescribe the effects of unbalanced forces acting on an object

Apabiladayayangbertindakkeatasobjek tidak seimbang , terdapatdaya bersih yang bertindak ke atasnya.

When the forces acting on an object are not balanced , there must be a net force acting on it.

Dayabersihdikenalisebagaidaya paduan yang bertindak ke atasnya.

The net force is known as the resultant force acting on it.

Dayatidakseimbang=dayabersih=dayayangdikenakan–dayageseranThe unbalanced force = net force = force applied – frictional force

Kesan:Bolehmenyebabkanbadanseseorang/Effect: Can cause a body to• bertukarkeadaanrehatnya(objekituakanmemecut)

change its state at rest (an object will accelerate)• bertukarkeadaangerakannya(suatuobjekyangbergerakakanmemecut/nyahpecutataumenukararahnya)

change its state of motion (a moving object will accelerate/decelerate or change its direction)

2.5 MEMAHAMI KESAN DAYAUNDERSTANDING THE EFFECTS OF A FORCE

Huraikan kesan daya seimbang yang bertindak ke atas objekDescribe the effect of balanced forces acting on an object

Daya seimbang/Balanced force

Apabiladaya-dayayangbertindakkeatassuatuobjekdalamkeadaan seimbang ,iaakanmembatalkan

antarasatusamalain.Dayabersihadalah sifar .

When the forces acting on an object are balanced , they cancel each other out. The net force is zero .

Kesan/Effect:

Objekberadadalamkeadaan rehat [halaju = 0] atau bergerak pada halaju malar [pecutan = 0]

The object is at rest [velocity = 0] or moves at constant velocity [acceleration = 0 ]

BeratWeight

Daya dikenakan oleh meja ke atas cawanForce exerted by table on the cup

Cawan itu berada dalam keadaan rehat. Daya

bersih yang bertindak ke atasnya adalah sifar . The cup stays at rest. The net force acting on it is

zero . W = R di mana/where W:Berat/Weight R : Tindak balas normal/Normal reaction

Daya angkat, U/Lift, U

Berat, WWeight, W

Tujahan, FThrust, F

Seretan, GDrag, G

Kapal terbang bergerak dengan halaju malar. Daya

bersih yang bertindak ke atasnya adalah sifar .The plane moves with constant velocity. The net force acting

on it is zero .W = U di mana/where W :Berat/Weight U :Dayaangkat/Lift

F = G di mana/where F : Tujahan/Thrust G:Seratan/Drag


19


© Nilam Publication Sdn. Bhd.59

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UNIT

Menentukan hubungan antara daya, jisim dan pecutan (F = ma)Determine the relationship between force, mass and acceleration (F = ma)

Hukum Gerakan NewtonKeduaNewton’s Second Law of Motion

F = 5 N

m = 25 kg

Pecutan yang dihasilkan oleh daya ke atas suatu objek adalah berkadar langsung dengan magnitud daya bersih yang dikenakan dan berkadar songsang dengan jisim objek itu.

The acceleration produced by a force on an object is directly proportional to the

magnitude of the net force applied and is inversely proportional to the mass of the object.

Daya=Jisim×Pecutan F = ma

Force = Mass × Acceleration F = ma

Hubungan antara a dan FRelationship between

a and F

a α FPecutan,a,berkadarlangsungdengandayayangdikenakan,FThe acceleration, a, is directly proportional to the applied force, F

a

0 F

Hubungan antara a dan mRelationship between

a and m

a α 1m

Pecutan, a, bagi suatu objek berkadar songsang denganjisimnya,mThe acceleration, a, of an object is inversely proportional to its mass, m

1m

a

0

Hubungan antaraRelationship between a α F a α m

SituasiSituation

A

B

Dua orang pemuda menolak jisim yangsama tetapi pemuda A menolak dengan daya yang lebih besar. Jadi dia bergerak dengan lebih cepat.Both men are pushing the same mass but man A pushes with a greater force. So he moves faster.

A

B

Dua orang pemuda mengeluarkan dayayang sama. Tetapi pemuda B bergerak dengan lebih cepat daripada pemuda A.Both men exerted the same force. But man B moves faster than man A.

HipotesisHypothesis

Semakinbesardaya,semakinbesarpecutan.

The larger the force, the greater the

acceleration.

Semakinbesarjisim,semakinkecilpecutan.

The greater the mass, the smaller the

acceleration.

EksperimenExperiment

Mencari hubungan antara daya, jisim dan pecutanFind the relationship between force, mass and acceleration


20



2

UNIT

Pemboleh ubah dimanipulasiManipulated variable

Daya/Force Jisim/Mass

Pemboleh ubah bergerak balasResponding variable

Pecutan/Acceleration Pecutan/Acceleration

Pemboleh ubah dimalarkanConstant variable

Jisim/Mass Daya/Force

Bahan dan radasMaterials and apparatus

Jangkamasadetikdanpitadetik,bekalankuasa,landasanterpampasgeseran,

pembaris,troli,takallicin(denganpengapit),talitakkenyal,pemberatberslot

Ticker timer and ticker tape, power supply, friction-compensated runaway, ruler, trolley,

smooth pulley (with clamp), inelastic string, slotted weights

RajahDiagram


Jangka masa detikTicker timer

Landasan terpampasgeseran

Friction-compensatedrunway

Tali tak kenyalInelastic string

Blok kayuWooden block

Bekalan kuasa a.u.a.c. power supply

Troli PTrolley P

Takal licinSmooth pulley

PemberatberslotSlottedweight

Susunanradasuntukmengkajihubunganantara(1) daya dan pecutan(2)jisimdanpecutan

Arrangement of apparatus to investigate the relationship between(1) Force and acceleration(2) mass and acceleration

ProsedurProcedure

1 Radas disusun seperti ditunjukkan dalam rajah di atas.The apparatus is set up as shown in the diagram above.

2 Sebuah troli berjisim 1.0 kg (jisimmalar) diletakkan di atas landasan. Pita detik dilekat pada troli itu.A trolley of mass 1.0 kg (constant mass) is placed on the runway. A length of ticker tape is attached to the trolley.

3 Jangka masa detik dihidupkan dan troli itu ditarik oleh pemberat yang mempunyai daya,F=10.0N. The ticker timer is switched on and the

trolley is pulled by a weight of force, F = 10.0 N.

4 Daripitadetikyangdiperoleh,pecutantroli dihitung dengan menggunakan

formula,a = (v – u)t

From the ticker tape obtained, the acceleration of the trolley is calculated by using the formula, a = (v – u)

t.

1 Radas disusun seperti ditunjukkan dalam rajah di atas.The apparatus is set up as shown in the diagram above.

2 Sebuahloridenganjisim,m=1.0kgdiletakkan di atas landasan. Pita detik dilekat pada troli itu.A trolley of mass, m = 1.0 kg is placed on the runway. A length of ticker-tape is attached to the trolley.

3 Jangka masa detik dihidupkan dan troli itu ditarik oleh pemberat (daya malar pemberatiniialah10N)The ticker timer is switched on and the trolley is pulled by a weight of constant force, 10 N

4 Daripitadetikyangdiperoleh,pecutantroli dihitung dengan menggunakan

formula,a = (v – u)t

.From the ticker tape obtained, the acceleration of the trolley is calculated by

using the formula, a = (v – u)t

.


21



2

UNIT

5 Langkah-langkah2–4diulangidenganmenambahkan pemberat berslot supaya F=15.0N,20.0N,25.0Ndan30.0N.Steps 2 – 4 are repeated by adding slotted weights to pull the trolley so that F = 15.0 N, 20.0 N, 25.0 N and 30.0 N.

5 Langkah-langkah2–4diulangidenganmelekat pemberat berslot pada troli supaya jisim troli,m = 1.5 kg, 2.0 kg,2.5kgdan3.0kg.Steps 2 – 4 are repeated by taping up slotted weights to the trolley to give m = 1.5 kg,2.0 kg, 2.5 kg and 3.0 kg.

Merekodkan dataRecording data

Daya, F/NForce, F/N

Pecutan, a/cm s–2

Acceleration, a/cm s-2

10.0

15.0

20.0

25.0

30.0

Jisim, m/kgMass, m/kg

Pecutan, a/m s–2

Acceleration, a/cm s–2

1.0

1.5

2.5

2.5

3.0

Menganalisis dataAnalysing data

Pecutan, a/cm s–2


0 Daya, F/NForce, F/N

Pecutan, a/cm s–2


0 Jisim, m/kgMass, m/kg

Menyelesaikan masalah menggunakan F = maSolve problems using F = ma

1 Hitungkan pecutan bagi blok di bawah:/Calculate the acceleration of the block:

(a) m = 2 kg

F = 8.0 N (c) m = 10 kg

F = 18 NF = 2 N

a = Fm = 8.0N

2kg a = (18–2)N

10kg = 16N

10kg

=4m s-2 =1.6ms-2

(b) m = 8 kg

F = 14 NF = 6 N (d) m = 12 kg

F = 10 NF = 5 NR = 5 N

a = (14+6)N

8kg =

20N8kg

a = (10–5–5)N12kg

=0

=2.5ms-2 =0ms-2


22



2

UNIT

Menyelesaikan masalah menggunakan F = maSolve problems using F = ma

2 Seorang lelaki menolak troli yang berisi kotak (jumlah jisim 5 kg) di ataspermukaanyanglicin.Jikadiamenggunakandaya30Nuntukmenolaktroliitu,apakahmagnituddanarahpecutantroliitu?A man pushes a trolley with a box (total mass 5 kg) on a smooth surface. If he uses a force of 30 N to move the trolley, what is the magnitude and direction of the acceleration of the trolley?

Penyelesaian/Solution:F = ma

a = 30N5kg

=6ms-2 ke kanan / to the right.

3 Sebuahobjekyangberjisim2kgditarikdiatastanahdengandaya5Ndanhalajumalar.An object of mass 2 kg is pulled on the floor by a force of 5 N and has a constant velocity.

(a) Berapakahdayageseranantaraobjekdantanah?What is the frictional force between the object and the floor?

(b) Hitungkanpecutanobjekitujikaobjekituditarikdengandaya17N.Calculate the acceleration of the object if the object is pulled by a 17 N force.

Penyelesaian/Solution:(a) R ialah daya geseran/R is the frictional force F1 – R = ma ∴ R = F1 – ma Olehkeranahalajumalar/Because the velocity is constant, a=0 ∴ R = F1–0 = F1

=5N(b) F2 – R = ma 17N–5N=(2kg) (a)

a = 12N2kg

=6ms–2

4 Sebuahbasberjisim2000kgbergerakdenganhalajuseragam40ms-1sejauh2500msebelumberehat.Hitungkan A bus of mass 2 000 kg travels at a uniform velocity 40 m s-1 for a distance of 2 500 m before it comes to rest. Calculate

(a) purata nyahpecutan bas itu./the average deceleration of the bus. (b) purata daya yang dikenakan oleh brek itu untuk membolehkan bas itu berhenti bergerak.

the average force applied by the brakes to bring the bus to a standstill.

Penyelesaian/Solution:(a) v2 = u2+2as (b) F = ma 0=(40ms–1)2+2a(2500m) =(2000kg)(–0.32ms–2) ∴5000a=–1600ms–2 =–640N a=–0.32ms-2 (Negatif bermaksud daya untuk menentang

gerakan/Negative means force to resist the motion)

F = 30 N

5 kg

F1 = 5 NR


23



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UNIT

2.6 MENGANALISIS IMPULS DAN DAYA IMPULSANALYSING IMPULSE AND IMPULSIVE FORCE

Menerangkan daya impuls/Explain what an impulsive force is

Daya yang besar yang bertindak dalam tempohmasa yang singkat semasa perlanggaran atau letupan

dikenali sebagai daya impuls .

A large force that acts over a short period of time during a collision or explosion is known as an impulsive force .

Daripadahubunganantaradaya,jisimdanpecutan:From the relationship between force, mass and acceleration:

F = ma = m ( v – ut

)

F = mv – mut

= perubahan momentum

t Unit = N = kg m s-2

m = jisim/mass u = halaju awal/initial velocity t = masa/time v = halaju akhir/final velocity

Dayaimpulsialahkadar perubahan momentum dalam perlanggaran atau letupan.

An impulsive force is the rate of change of momentum in a collision or explosion.

Mendefinisikan impuls/Define impulse

Impuls didefinisikan sebagai perubahan momentum /Impulse is defined as the change of momentum .

atau (momentum akhir – momentum awal) atau (mv – mu)or (final momentum – initial momentum) or (mv – mu)

Unit: kg m s-1 atau/or N s

DaripadaF = mv – mut

, Ft = mv – mu = perubahan momentum = impuls

From F = mv – mut

, Ft = mv – mu = change of momentum = impulse

Impuls ialah hasil darab antara daya dan masa .

The product of the force and the time is called the impulse.

Kesan peningkatan dan pengurangan masa perlanggaranThe effects of increasing and decreasing the time of collision

Daripadaformula,F = Perubahan momentumMasa

/From the formula, F = Change of momentumTime

.

Dayaimpulsberkadar songsang dengan masa sentuhan atau tindakan atau perlanggaran.

Impulsive force is inversely proportional to the time of contact or impact or collision.

Tempoh masa yang panjang/Longer period of time Dayaimpuls kecil / Impulsive force is small

Tempoh masa yang pendek/Shorter period of time Dayaimpuls besar / Impulsive force is large


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Terangkan situasi di mana daya impuls perlu dikurangkan dan kaedah untuk mengurangkannya

Explain situations where an impulsive force needs to be reduced and suggest ways to reduce it

SituasiSituations

PenjelasanExplanation

Peserta lompat tinggi mendarat di atas tilam yang tebal. Mengapa?A high jumper lands on the thick mattress. Why?

Masa tindakan yang panjang, maka daya impuls menjadi kecil.

The time of impact is longer, so the impulsive force is smaller.

Penjaga gol memakai sarung tangan untuk menangkap bola. Mengapa?Goal keepers wear gloves to catch a ball. Why?

Masa tindakan yang panjang, maka daya impuls menjadi kecil.


Ahli lompat jauh akan membengkokkan kaki semasa mendarat. Mengapa?A long jumper will bend his legs upon landing. Why?

Masa tindakan adalah lebih panjang, maka daya impuls menjadi kecil.

Ini mengurangkan kecederaan./The time of impact is longer, so the impulsive

force is smaller. This reduces injury.

Seorang pemain besbol mesti menangkap bola mengikut arah pergerakan bola. Mengapa?A baseball player must catch the ball in the direction of the motion of the ball. Why?

Masa tindakan lebih panjang, maka daya impuls menjadi lebih kecil.


Bahan yang mudah pecah seperti telur, kaca dan perkakasan elektrik mestilah dibungkus dalam bahan yang lembut dan boleh dimampatkan. Mengapa?Items that are fragile such as eggs, glass and electrical appliances must be packed in materials that are soft and compressible. Why?

Bahan yang lembut dan mudah dimampatkan menghasilkan masa perlanggaran yang panjang. Ia

menyerap hentakan. Jadi ia mengurangkan daya impuls.

Soft and compressible material provides longer time of impact. It absorbs the shock. So it can reduce the

impulsive force.

04 FIZ Form 4 Ch2 C 2F.indd 64 9/10/13 4:28 PM

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Terangkan situasi di mana daya impuls mendatangkan faedahExplain situations where an impulsive force is beneficial

SituasiSituations

PenjelasanExplanation

Peserta karate yang mahir boleh memecahkan kayu yang tebal dengan menggunakan sisi tangan yang bergerak dengan kelajuan yang sangat tinggi.A karate expert can break a thick wooden slab with his bare hand that moves at a very fast speed.

Tangan tersebut digerakkan pada halaju yang tinggi. Masa tindakan

adalah singkat. Menghasilkan daya impuls yang besar.

The hand is moved at high velocity. Time of impact is shorter. Produce large

impulsive force.

Kepala penukul yang besar bergerak pada kelajuan yang tinggi untuk memukul paku.A massive hammer head moving at a fast speed is brought to rest upon hitting the nail.

Masa tindakan adalah pendek apabila kepala penukul memukul pada

halaju yang tinggi. Ia akan menghasilkan daya impuls yang besar.

Time of impact is shorter when the hammer head is hit at high velocity. It

produces large impulsive force.

Daya impulsImpulsive force

Sebiji bola sepak mestilah mempunyai tekanan udara yang cukup tinggi.A football must have an air pressure that is high enough.

Sebiji bola sepak yang mempunyai tekanan udara yang cukup tinggi

akan mempunyai masa tindakan yang pendek. Maka daya impuls adalah

besar. Bola akan bergerak lebih jauh.

A football which has a high-enough air pressure will have a short time of

impact. So, the impulsive force is large. The ball will move further.

AntanPestle

Daya impulsImpulsive force

Lesung/Mortar

Antan dan lesung diperbuat daripada batu.A pestle and mortar are made of stone.

Masa tindakan adalah kecil. Maka daya impuls adalah besar.

Time of impact is shorter. So the impulsive force is large.


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Latihan/Exercises

1 Rosli yang berjisim 60 kg melompat dari tingkat pertama sebuah rumah yang terbakar. Halajunya sejurus sebelum mendarat ialah 6 m s-1.Rosli, with a mass of 60 kg, jumps from the first floor of a burning house. His velocity just before landing on the ground is 6 m s-1.

(a) Kirakan impuls apabila kakinya mencecah tanah.Calculate the impulse when his legs hit the ground.

(b) Berapakah daya impuls yang bertindak ke atas kaki Rosli jika dia membengkokkan kaki ketika mendarat dan mengambil masa 0.5 s untuk berhenti?What is the impulsive force on Rosli’s legs if he bends upon landing and takes 0.5 s to stop?

(c) Berapakah daya impuls yang bertindak ke atas kaki Rosli jika dia tidak membengkokkan kaki dan berhenti dalam 0.05 s?What is the impulsive force on Rosli’s legs if he does not bend and stops in 0.05 s?

(d) Apakah kebaikan membengkokkan kaki semasa mendarat?What is the advantage of bending his legs upon landing?


2 Rooney menyepak bola dengan kekuatan daya 1 500 N. Masa tindakan di antara kasut dan bola ialah 0.01 s. Berapakah impuls yang dikenakan kepada bola itu? Jika jisim bola itu ialah 0.5 kg, berapakah halaju bola tersebut?Rooney kicks a ball with a force of 1 500 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?

Penyelesaian/SolutionImpuls/Impulse = Ft = 1 500 N × 0.01 s = 15.0 N s = 15.0 kg m s-1

mv – mu = 15 kg m s-1 u = 0 (0.5 kg)v = 15 kg m s-1

v = 30 m s-1

3 Dalam satu perlawanan tenis, pemain memukul bola yang mempunyai jisim 0.2 kg yang menuju ke arahnya dengan halaju 20 m s-1. Bola itu memantul dengan halaju 40 m s-1. Masa yang diambil semasa perlanggaran antara bola dan raket tenis ialah 0.01 s.In a tennis match, a player hits an on-coming ball with mass of 0.2 kg and velocity of 20 m s-1. The ball rebounds with a velocity of 40 m s-1. The time taken in the collision between the ball and the tennis racket is 0.01 s.

u = –20 m s-1

v = 40 m s-1

t = 0.01 s0.2 kg

(a) Berapakah impuls yang dialami oleh bola itu?What is the impulse experienced by the ball?

(b) Berapakah daya impuls yang dikenakan ke atas bola tenis?What is the impulsive force exerted on the tennis ball?

Penyelesaian/Solution(a) Impuls

Impulse = m (v – u)

= (0.2 kg)[(40 – (–20)] m s–1

= 12.0 kg m s-1

= 12.0 N s

(b) F = 12 N s0.01 s

= 1 200 N

(a) mv – mu = m (v – u) = (60 kg) (0 – 6) m s–1 = –360 kg m s–1

Impuls adalah –360 kg m s–1 kerana momentumnya dikurangkan sehingga sifar. The impulse is –360 kg m s–1 because its

momentum is reduced to zero.

(b) F = mv – mut

= –360 kg m s–1

0.5 s = –720 N

(c) F = mv – mut

= –360 kg m s–1

0.05 s = –7 200 N(b) dan (c) : Tanda negatif bagi daya bermakna daya ini telah menyebabkan kehilangan momentum(b) and (c) : The negative sign to the force means that the force has caused a loss of momentum(d) Dengan membengkokkan kaki semasa

mendarat, dia akan meningkatkan masa tindakan dan mengurangkan daya impuls. Jadi ia dapat mengurangkan kecederaan.

By bending his legs upon landing, he will increase the time of impact and reduce the impulsive force. So it will minimise the injuries.


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2.7 KESELAMATAN KENDERAANVEHICLE SAFETY

Komponen/Component Fungsi/Function

Pelapik kepalaHeadrest Untuk mengurangkan kesan inersia terhadap kepala pemandu. Mengurangkan

kecederaan leher apabila kereta dilanggar daripada belakang.

To reduce the inertia effect on the driver’s head. Reduce neck injury when the car is

hit from behind.

Beg udaraAir bag Menyerap hentakan dengan menambahkan masa perlanggaran apabila kepala

pemandu terhentak ke stereng. Oleh itu daya impuls dikurangkan.

Absorbing impact by increasing the collision time when the driver’s head is thrown

towards the steering. So the impulsive force is reduced.

Cermin hadapan kereta/Cermin keselamatanWindscreen/safety glass

Kaca tahan pecah yang tidak akan mudah pecah kepada serpihan yang kecil dengan

mudah semasa perlanggaran. Mengurangkan kecederaan disebabkan oleh serpihan

kaca yang berselerak.

Shatterproof glass that will not break into small pieces easily during collision. This will

reduce injuries caused by scattered glass.

Pelapik kepalaHeadrest

Zon kemek belakangRear crumple zone

Cermin hadapan keretaWindscreen

Beg udaraAir bag

Zon remuk depanFront crumple zone

Bumper depanFront bumper Tali pinggang keselamatan

Safety beltBar hentaman sisiSide impact bar

Sistem brek anti kunciAnti-lock braking system (ABS)

SteringSteering


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Komponen/Component Fungsi/Function

Zone remuk (depan dan belakang)Crumple zone (front and rear)

Boleh dimampatkan ketika kemalangan. Jadi ia akan meningkatkan masa yang

diperlukan kereta untuk berhenti sepenuhnya. Maka ia akan kaca yang

mengurangkan daya impuls.

Can be compressed during an accident. So it can increase the time taken by the car to

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

Bumper depanFront bumper Menyerap hentakan akibat daripada kemalangan. Diperbuat daripada keluli,

aluminium, plastik, getah dan fiber komposit.

Absorb the shock from the accident. Made from steel, aluminium, plastic, rubber and

composite fibres.

Sistem brek anti kunciAnti-lock braking system (ABS)

Membolehkan pemandu memberhentikan kereta dengan segera tanpa

menyebabkan roda terkunci apabila brek ditekan secara tiba-tiba. Mengelakkan

kereta daripada menggelongsor.

Enables drivers to quickly stop the car without the wheels locking when the brake is

applied suddenly. Prevents the car from skidding.

Bar hentaman sisiSide impact bar Boleh dimampatkan ketika kemalangan. Jadi ia akan meningkatkan masa yang

diperlukan kereta untuk berhenti sepenuhnya. Maka ia akan mengurangkan

daya impuls.

Can be compressed during accident. So it can increase the time the car takes to come

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

Tali pinggang keselamatanSafety belt

Untuk mengurangkan kesan inersia dengan mengelakkan pemandu daripada

tercampak ke hadapan.

To reduce the inertia effect by preventing the driver from being thrown forward.

Papan pesawatDashboard Semasa perlanggaran, papan pesawat meroboh. Ini akan menyerap kesan

hentaman dengan meningkatkan masa perlanggaran antara kepala pemandu

dan stereng. Jadi ia mengurangkan daya impuls.

During collision, the dashboard collapses. This will absorb the impact by increasing the

time of collision between the driver's head and the steering. This will reduce the

impulsive force.


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2.8 MEMAHAMI GRAVITIUNDERSTANDING GRAVITY

Kekuatan medan graviti/Gravitational field strength

• Objekjatuhbebaskeranaditarikkearahpusatbumiolehdayatarikan graviti .

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

• Kekuatanmedangraviti= daya tarikan gravitijisim

. / Gravitational field strength = gravitational force

mass.

• DipermukaanBumi, At the surface of the Earth, Kekuatan medan graviti = 10 N kg–1 gravitational field strength = 10 N kg–1

= 10 m s–2 = 10 m s–2

• SetiapkilogramjisimpadapermukaanBumimengalamidayagravitisebanyak10Nyangbertindakkeatasnya. Earth kilogram of mass at the Earth's surface has a gravitational force of 10 N acting on it.

Aktiviti 1: Pecutan disebabkan gravitiActivity 1: Acceleration due gravity

Rajah di sebelah menunjukkan gambarfoto stroboskop bagi bola yang jatuh bebas dan graf halaju lawan masa bagi gerakannya. The diagram on the right shows a stroboscopic photograph of a free falling ball and its velocity-time graph.(a) Perhatikan gambarfoto dan terangkan halaju bola. Observe the photograph and describe the velocity of the ball.

Halaju bola itu meningkat dengan seragam.

The velocity of the ball increases uniformly.

(b) Apakah yang boleh anda simpulkan daripada graf kecerunan v – t? What can we deduce from the gradient of the v – t graph?

Kecerunan ialah pecutan bola itu.

The gradient is the acceleration of the ball.

(c) Terangkan gerakan bola tersebut. Describe the motion of the ball.

Bola tersebut bergerak dengan pecutan seragam.

The ball moves with constant acceleration.

Terangkan pecutan yang disebabkan oleh graviti, g/Explain acceleration due to gravity, g

Pecutan disebabkan oleh graviti, g, ialah pecutan bagi objek yang disebabkan oleh daya tarikan graviti.

Acceleration due to gravity, g, is the acceleration of an object due to the pull of the gravitational force .

Nilai piawai bagi pecutan graviti, g, ialah 9.81 m s-2. Nilai g yang sering digunakan ialah 10 m s-2.

Magnitud bagi pecutan yang disebabkan oleh graviti bergantung pada kekuatan medan graviti .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 the gravitational field strength .

V

t0


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Apakah jatuh bebas?/What is free fall?

Objek dikatakan 'jatuh bebas' apabila ia jatuh di bawah daya tarikan graviti sahaja.

An object is falling freely when it is falling under the gravitational force only.

Sehelai kertas tidak jatuh bebas kerana kejatuhannya dipengaruhi oleh rintangan udara .

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

Objek hanya jatuh bebas di dalam vakum . Ketiadaan udara bermaksud tiada rintangan udara yang

menentang pergerakan objek.

An object falls freely only in vacuum . The absence of air means there is no air resistance to resist the motion of the object.

Di dalam vakum, kedua-dua objek yang ringan dan berat jatuh bebas. Ia jatuh dengan pecutan graviti iaitu pecutan disebabkan oleh graviti, g.

In vacuum, both light and heavy objects fall freely. They fall with the gravitational acceleration, that is the acceleration due to gravity, g.

Aktiviti 2: Pecutan disebabkan graviti/Activity 2: Acceleration due gravity

Pegang dua biji batu yang berbeza saiz pada ketinggian yang sama, kemudian kedua-dua batu itu dijatuhkan serentak daripada ketinggian yang sama. Hold two stones of different sizes at the same height and then drop both stones simultaneously from the same height.(a) Huraikan bagaimana halaju berubah. Describe how the velocity changes.

Halaju meningkat dengan seragam.

The velocity increases uniformly.

(b) Bandingkan masa yang diambil untuk batu mencecah lantai. Compare the time taken for the stones to reach the floor.

Sama/same

(c) Adakah pecutan batu dipengaruhi oleh jisimnya? Is the acceleration of each stone influenced by its mass?

Jisim tidak mempengaruhi pecutan.

Mass does not affect the acceleration.


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Aktiviti 3: Yang mana satukah mencecah tanah dahulu?Activity 3: Which one reaches the ground first?

Bola golf dan sehelai kertas dipegang pada ketinggian yang sama dan dijatuhkan serentak.Hold a golf ball and a piece of paper at the same height and drop them simultaneously.(a) Objek yang manakah mencecah tanah dahulu? Which object reaches the floor first?

Bola golf.

The golf ball.

(b) Terangkan mengapa./Explain why.

Kertas mempunyai luas pemukaan yang besar.

Jadi lebih banyak rintangan udara yang

bertindak ke atasnya.

The paper has large surface area. As such, the air

resistance acting on it is big.

Ulangi dengan bola golf dan sehelai kertas yang direnyukkan.Repeat with a golf ball and a piece of paper which is crumpled.(a) Objek yang manakah mencecah tanah dahulu? Which object reaches the floor first?

Kedua-duanya mencecah tanah pada masa

yang sama.

Both reach the floor at the same time.

(b) Terangkan mengapa./Explain why.

Kedua-dua objek mempunyai saiz dan luas

pemukaan yang sama. Jisim tidak memberi

kesan kepada pecutan graviti.

Both objects have same size and surface area.

Mass does not affect gravitational acceleration.

Bola golfGolf ball

KertasPaper

Bola golfGolf ball

Kertas yang direnyukkanPaper which is crumpled

Aktiviti 4: Perbezaan antara jatuh bebas di atmosfera (udara) dan jatuh bebas di dalam vakum bagi duit syiling dan bulu pelepah.

Activity 4: The difference between free fall in atmosphere and free fall in a vacuum of a coin and a feather.

Duit syiling dan bulu pelepah dilepaskan daripada ketinggian yang sama serentak di dalam makmal.A coin and a feather are released from the same height simultaneously in the laboratory.

Pemerhatian/Observation

Duit syiling jatuh lebih cepat daripada bulu pelepah.

The coin falls faster than the feather.

Duit syiling dan bulu pelepah yang sama diletakkan di dalam satu tiub vakum dan kemudian dijatuhkan serentak pada ketinggian yang sama.The same coin and feather are put into a vacuum tube and then dropped simultaneously from the same height.

Pemerhatian/Observation

Kedua-dua objek mencecah ke bawah silinder

pada masa yang sama .Both objects reach the bottom of the cylinder at the

same time.


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g sebagai kekuatan medan graviti. g sebagai pecutan yang disebabkan gravitig as gravitational field strength. g as acceleration due to gravity

1 Sebuah objek yang jatuh bebas berdekatan dengan permukaan bumi akan memecut pada 10 m s-2.

An object falling freely near the earth's surface will accelerate at 10 m s-2.

2 Setiap kilogram jisim yang berdekatan dengan permukaan bumi mempunyai daya graviti 10 N yang bertindak ke atasnya.Each kilogram of mass near the earth’s surface has a gravitational force of 10 N acting on it.

Kekuatan medan graviti , g = 10 N kg-1/ Gravitational field strength, g = 10 N kg-1

Pecutan disebabkan graviti/Ac celeration due to gravity, g = 10 m s-2

Nilai g boleh ditulis sebagai 10 m s-2 atau 10 N kg-1. The approximate value of g can be written either as 10 m s-2 or as 10 N kg-1.

Penjelasan/Explanation

Rintangan udara yang besar bertindak ke atas

bulu pelepah kerana ia mempunyai luas permukaan yang besar.

A bigger air resistance acts the feather because

it has a large surface area .

Daya graviti pada duit syiling mampu untuk

mengatasi rintangan udara lebih baik daripada bulu pelepah.

The gravitational force on the coin is able to

overcome air resistance better than the feather.

Penjelasan/Explanation

Di dalam keadaan vakum, tiada rintangan udara.Hanya terdapat satu daya yang bertindak ke atas

objek iaitu daya graviti .

In vacuum, there is no air resistance . The only

force acting on both objects is the force of gravity .

Kedua-dua objek jatuh bebas dengan pecutan yang disebabkan graviti walaupun berbeza dari

segi jisim dan bentuk .

Both objects free fall with acceleration due to

gravity despite the differences in their mass and

shapes .

Aktiviti 5: Pecutan yang disebabkan gravitiActivity 5: Acceleration due to gravity

Tujuan/Aim

Menentukan pecutan disebabkan graviti/To determine the acceleration due to gravity.

Radas/Apparatus

Jangka masa detik, bekalan kuasa 12 V, bangku, pengapit-G, pemberat, pita detikTicker timer, 12 V ac power supply, stool, G-clamp, slotted weight, ticker tape.

Prosedur/Procedure

1 Potong sekeping pita detik lebih kurang 2.5 m panjang dan lalukan melalui jangka masa detik yang diapit kepada kerusi oleh pengapit-G.Cut a piece of ticker tape about 2.5 m long and pass through the ticker timer which is clamped to a stool using G-clamp.


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2 Sambungkan satu hujung pita pada pemberat 100 g.Attach one end of the tape to the 100 g slotted weight.

3 Hidupkan jangka masa detik dan pemberat dilepaskan supaya ia jatuh bebas.Switch on the ticker timer and release the slotted weight so that it falls freely.

4 Kaji pita itu untuk menentukan nilai bagi pecutan disebabkan oleh graviti, g.Analyse the tape to determine the value of the acceleration due to gravity, g.

Perbincangan/Discussion

1 Mengapakah sukar untuk menentukan pergerakan objek yang jatuh dengan hanya memerhatikannya jatuh?Why is it difficult to describe the motion of a falling object by just observing it fall?

Objek bergerak sangat laju.

The object moves very fast.

2 Apakah jenis pergerakan objek jika ia jatuh di bawah tarikan graviti?What is the type of motion of an object falling under the pull of gravity?

Pecutan seragam./Constant acceleration.

3 Mengapakah pergerakan pemberat boleh diandaikan sebagai jatuh bebas?Why is it that the motion of the slotted weight can be assumed to be a free fall?

Ringtangan udara yang kecil boleh diabaikan.

The small air resistance is negligible.

4 Apakah langkah yang akan anda ambil untuk mengurangkan geseran antara pita dan jangka masa detik?What steps did you take to minimise the friction between the ticker tape and the ticker timer?

Pegang pita detik dalam keadaan menegak dan lepaskannya. Pastikan ia jatuh mealui jangka masa

detik dengan lancar.

Hold the ticker tape vertically when releasing it. Make sure it slips through the ticker timer smoothly.

5 Terangkan mengapa perlu menjatuhkan pemberat daripada kedudukan yang tinggi.Explain the need for the slotted weight to be dropped from a high position.

Pengiraan akan menjadi lebih tepat kerana ralat eksperimen dikurangkan.

The calculation will be more accurate because experimental errors are reduced.

6 Tunjukkan bagaimana anda mengira g daripada pita.Show how you would calculate g from the tape.

u = s1

t1 v = s2

t2a = v – u

t

Pengapit-GG-clamp

Jangka masadetikTicker timer


Bekalankuasa, 12 VA.C. Powersupply, 12 V

BangkuStool

PemberatWeight

Kepingan polisterenaPolystyrene sheet


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7 Bandingkan nilai pecutan disebabkan oleh graviti daripada aktiviti ini dengan nilai yang sebenar. Berikan alasan yang munasabah bagi perbezaan di antara dua nilai tersebut.Compare the value of the acceleration due to gravity from this activity with the actual value. Give possible reasons for any difference in these two values.

Nilai daripada eksperimen adalah lebih rendah berbanding dengan nilai sebenar. Sebabnya ialah

rintangan akibat jangka masa detik. / The value from the experiment is lower than the actual value.

The reason is the resistance caused by the ticker timer.

8 Bandingkan nilai bagi g daripada eksperimen jika anda mengulangi eksperimen dengan menggunakan pemberat 200 g dan 300 g.Compare the values of g from the experiment if you repeat the experiment using 200 g and 300 g weights.

Keputusan sepatutnya sama.

The result should be the same.

9 Apakah yang boleh anda simpulkan tentang hubungan antara g dan jisim bagi objek yang jatuh?What can you conclude about the relationship between g and the mass of the falling object?

Jisim tidak mempengaruhi pecutan graviti, g.

Mass does not affect the gravitational acceleration, g.

Definisi beratDefinition of weight

Berat, W, bagi sesuatu objek ialah daya graviti yang dikenakan ke atasnya.The weight, W, of an object is the gravitational force acting on it.

Berat ialah daya dan diukur dalam unit Newton, N . Berat ialah kuantiti vektor .

Weight is a force and is measured in Newton, N . Weight is a vector quantity.

Sebiji batu yang berjisim m, dilepaskan jatuh bebas pada pecutan graviti, g.A stone of mass, m, is released and free falls with a gravitational acceleration of g.

Daya yang bertindak ke atas batu hanyalah berat, W, di mana ianya menuju ke arah bawah.The only force acting on the stone is its weight, W, which is downward.

Hukum gerakan Newton kedua:Newton’s second law of motion:F = ma di mana/where F = W , a = g

Oleh itu/Therefore:

W = mg

W = berat/weightm = jisim/massg = pecutan graviti/acceleration due to gravity unit bagi g ialah = m s-2

the unit of g is = m s–2

Berat/Weight = m × 10 N

Bumi/Earth

Jisim/Mass = m kg

Pecutan = gAcceleration = g


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Berat berubah, jisim tetapWeight changes but mass is fixed

Di Bulan, berat kita lebih ringan daripada di Bumi, ini kerana medan graviti di Bulan adalah lebih kecil .

On the Moon, our weight is less than that on Earth because

the Moon’s gravitational field is smaller than that of the Earth.Malahan di Bumi, berat kita berbeza sedikit dari suatu tempat ke tempat yang lain, kerana kekuatan medan graviti yang berbeza. Semakin jauh dari bumi, berat kita

semakin berkurang .Even on Earth, our weight can vary slightly from place to place, because the Earth’s gravitational field strength varies.

Moving away from the Earth, our weight decreases .Jika kita boleh pergi lebih jauh ke dalam ruang angkasa lepas dan bebas daripada sebarang tarikan graviti, berat

kita akan menjadi sifar .If we could go deep into space, and be free from any

gravitational pull, our weight would be zero .Sama ada di atas Bumi, Bulan atau di angkasa, jisim kita tetap tidak akan berubah.Whether on the Earth, Moon or deep in space, our mass does not change.

Nilai bagi pecutan graviti, gB, di bulan ialah 16

daripada nilai gE di bumi.The value of the gravitational acceleration, gB, on

the Moon is 16

the value of gE on the Earth.

Di angkasa lepas

Deep in space

Di permukaan

BulanOn Moon’s

surface

Di permukaan

BumiOn Earth’s

surface

JisimMass

100 kg 100 kg 100 kg

BeratWeight

0 N 16

(1 000) N 1 000 N

Perbezaan antara berat dan jisimDifference between weight and mass

Berat/Weight Jisim/Mass

Definisi Definition

Daya graviti yang bertindak ke atas objek.

The force of gravity acting on the object.

Jumlah jirim di dalam objek.

The amount of matter in the object.

Perubahan / Tiada perubahanChange / unchanged

Berat sesuatu objek berubah dengan kekuatan medan graviti pada sesuatu tempat.

The weight of an object changes with the gravitational field strength at the location.

Jisim sesuatu objek tidak berubah walau di mana-mana.

The mass of an object is unchanged anywhere.

Kuantiti asas atau kuantiti terbitanBase quantity or derived quantity

Kuantiti terbitan

A derived quantity

Kuantiti asas

A base quantity

Kuantiti vektor atau kuantiti skalarVector of scalar quantity

Kuantiti vektor

A vector quantity

Kuantiti skalar

A scalar quantity

Unit SI/SI unit Newton, N/Newton, N Kilogram, kg/Kilogram, kg


36



2

UNIT

Untuk objek yang jatuh dengan pecutan, g, berikut adalah persamaan-persamaan yang berkaitan:For an object falling with acceleration g, the following equations apply: 1 v = u + at di mana/where s = sesaran/displacement 2 s = ut + ½ at2 u = halaju awal/initial velocity 3 s = ½ (u + v)t v = halaju akhir/final velocity 4 v2 = u2 + 2as t = masa/time

a = g, pecutan graviti/acceleration due to gravity,

Nota/Notes: 1 Apabila sebuah objek jatuh bebas: a = g = 10 m s-2 (pecutan)

When an object fall freely: a = g = 10 m s-2 (acceleration) 2 Apabila sebuah objek dilambung ke atas: a = –g = –10 m s-2. (nyahpecutan)

When an object is thrown upwards: a = –g = –10 m s-2. (deceleration) 3 Pada kedudukan yang tertinggi, v = 0.

At the highest point, v = 0. 4 Jatuh ke bawah, v adalah positif./Downward direction, v is positive. 5 Arah ke atas, v adalah negatif./Upward direction, v is negative.

Contoh/Example

1 Sebiji batu jatuh daripada ketinggian 45 m.A rock falls from a height of 45 m.

(a) Berapa lamakah masa yang diambil oleh batu itu untuk mencecah ke tanah?How long does it take to reach the ground?

(b) Berapakah halaju batu itu semasa ia menghentam lantai?What is its velocity as it hits the ground?

Penyelesaian/Solution(a) u = 0, s = 45 m, g = 10 m s–2, t = ? s = ut + ½ gt2

45 m = 0 + ½ (10 m s–2)(t2) t2 = 9 s2

t = 3 s(b) v = u + gt = 0 + (10 m s–2)(3 s) = 30 m s–1

2 Sebiji bola dilambung ke atas daripada tanah dengan halaju 30 m s–1. Selepas beberapa lamakah bola itu akan menyentuh tanah semula?A ball is thrown upwards from the ground with a velocity of 30 m s–1. After how many seconds will it strike the ground again?

g = –10 m s–2

v = 0

u = 30 m s–1

Penyelesaian/SolutionUntuk gerakan ke atas,/For the upward motion, u = 30 m s–1, v = 0, g = –10 m s–2, v = u + gt ∴0 = 30 m s–1 + (–10 m s–2)(t) 10t = 30 s t = 3 s (gerakan ke atas/upward motion)Maka, masa untuk gerakan ke bawah juga mengambil 3 s. Oleh itu, ia mengambil masa 6 saat.The time taken for the downward motion is also 3 s.So it takes a total of 6 seconds.

Andaikan g = 10 m s-2 dan tiada rintangan udara. / Assume g = 10 m s-2 and there is no air resistance.

PecutanAcceleration

gTinggi/Height

hMasaTime

t

Objek dilambung ke atasObject thrown upwards

titik tertinggi, v = 0highest point, v = 0


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2

UNIT

Latihan/Exercises

Andaikan nilai g = 10 m s–2./Assume the value of g = 10 m s–2.

1 Amir menjatuhkan batu ke dalam perigi. Jika jarak antara bahagian atas perigi dan permukaan air ialah 20 m,Amir releases a stone into a well. If the distance between the top of the well and the water surface is 20 m,

(a) berapakah masa yang diambil oleh batu itu untuk sampai ke permukaan air?what is the time required for the stone to reach the surface of the water?

(b) berapakah halaju batu itu apabila ia terkena permukaan air?what is the velocity of the stone when it strikes the surface of the water?

Penyelesaian/Solution(a) u = 0, s = 20 m , g = 10 m s–2 , t = ? s = ut + ½ gt2 20 m = 0 + ½ (10 m s–2)(t2) t2 = 4 s2 t = 2 s(b) v2 = u2 + 2gs v2 = 0 + 2(10 m s–2)(20 m) ∴v = 20 m s–1

2 Suatu objek yang berjisim 5 kg dilepaskan dari sebuah bangunan setinggi 500 m. BerapakahAn object of mass 5 kg is released from a tall building of height 500 m. What is the

(a) berat objek itu?/weight of the object? (b) kekuatan medan graviti?/gravitational field strength? (c) masa yang diambil untuk sampai ke tanah?/time taken to reach the ground?

Penyelesaian/Solution(a) W = 5 kg × 10 m s–2 (c) u = 0, s = 500 m, g = 10 m s–2

= 50 N s = ut + ½ gt2

(b) g = 10 N kg–1 atau/or 10 m s–2 500 m = 0 + ½(10 m s–2)(t2) t2 = 100 s2 t = 10 s

2.9 MENGANALISIS KESEIMBANGAN DAYAANALYSING FORCES IN EQUILIBRIUM

Keseimbangan daya/Forces in equilibrium

1 Apabila suatu daya dikenakan terhadap objek dan ia mengekalkan keadaan pegun atau

bergerak dengan halaju seragam, maka objek itu dikatakan berada di dalam keadaan keseimbangan.

When forces act upon an object and it remains stationary or moves at a constant velocity , the object is said to be in a state of equilibrium.

2 Apabila keadaan keseimbangan berlaku, daya paduan yang bertindak ke atas objek itu

adalah sifar iaitu tiada daya bersih bertindak ke atasnya.

When equilibrium is reached, the resultant force acting on the object is zero , i.e. there is no net force acting upon it.

Menerangkan situasi di mana daya berada dalam keseimbanganDescribe situations where forces are in equilibrium

Perhatian/Note:Untuk (b), mengapa v2 = u2 + 2gs digunakan dan bukan v = u + gt?For (b), why v2 = u2 + 2gs is used and not v = u + gt?

Jawapan/Answer:Semua nilai yang diperlukan diberi dalam soalan.All the values required are given in the question.


38



2

UNIT

Menyatakan maksuddaya paduan

State what a resultant force is

Daya paduan: daya tunggal yang menunjukkan kesan daripada gabungan

dua atau lebih daya dalam magnitud dan arah

Resultant force: a single force that represents the combined effect of two or

more forces in magnitude and direction

Daya paduan ialah hasil tambah vektor bagi dua atau lebih daya yang bertindak ke atas objek. Dalam kes-kes berikut, jika F ialah daya paduan, maka,The resultant force is the vector sum of two or more forces which act on the object. In the cases below, if F is the resultant force, hence,

1F

2F

1F2F

Daya paduan/Resultant force = F = F1 + F2 Daya paduan/Resultant force = F = F1 – F2

Latihan/Exercises

1 Hitungkan daya paduan. Ke arah manakah objek itu bergerak?Calculate the resultant force. Which direction does the object move?

5 N12 N

Daya paduan, F/Resultant force, F = 5 N + 12 N = 17 N Arah ke kanan/To the right.

2 Hitungkan daya paduan. Ke arah manakah objek itu bergerak?Calculate the resultant force. Which direction does the object move?

5 N12 N

Daya paduan, F/Resultant force, F = 12 N – 5 N = 7 N Arah ke kanan/To the right.

3 Seekor kuda menarik kereta dengan daya 500 N. Seorang petani membantu kuda itu menolak

kereta itu dengan daya 200 N. Berapakah daya paduan?A horse pulls a cart with a force of 500 N. A farmer helps the horse by pushing the cart with a force of 200 N. What is the resultant force?Daya paduan, F/Resultant force, F = 500 N + 200 N = 700 N ke kanan/to the right

4 Seekor kuda menarik kereta dengan daya 500 N. Seorang petani menarik kereta itu pada arah

bertentangan dengan daya 200 N. Berapakah daya paduan?A horse pulls a cart with 500 N force. A farmer pulls the cart with a force of 200 N but in opposite direction. What is the resultant force?Daya paduan, F/Resultant force, F= 500 N – 200 N = 300 N ke kanan/to the right

200 N500 N

500 N200 N


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2

UNIT

5 Rajah 5.1 menunjukkan sebiji ladung berjisim 0.3 kg digantung dari siling.Diagram 5.1 shows a pendulum bob of mass 0.3 kg hung from the ceiling.Benang itu ditarik secara mengufuk dengan daya, F, supaya sudut antara benang dengan garis mencancang adalah 40°seperti ditunjukkan dalam Rajah 5.2.The thread is then pulled horizontally by a force, F, so that the thread makes an angle of 40° with the vertical line as shown in Diagram 5.2.

(a) Dalam ruang di bawah, lukis sebuah segi tiga keseimbangan daya bagi T, F dan berat ladung itu.In the space below, draw a triangle of forces in equilibrium for T, F and the weight of the bob.

T

F

BeratWeight3.0 N

40°

Perhatian/Note:Arah bagi tiga daya itu adalah berkitar.The directions of the three forces are cyclic.

(b) Hitung Daya, FCalculate the force, FDari segi tiga di atas,/From the triangle above,

F3.0 N

= tan 40°

F = 3.0 N tan 40° = 2.52 N

(c) Hitung tegangan, T, dalam benang ituCalculate the tension, T, in the string.Dari segi tiga di atas,/From the triangle above,

3.0 NT

= kos 40°

T = 3.0 Nkos 40°

= 3.92 N

Rajah 5.1/Diagram 5.1

SilingCeiling Benang

Thread

Ladung Pendulum Bob

Perhatian/Note: Soalan 5 di atas boleh dijawab dengan kaedah lukisan berskala. Question 5 above can be answered by scale-drawing.

40°

m = 0.3 kg

BeratWeight

T

F

Rajah 5.2/Diagram 5.2


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2

UNIT

Dua daya yang betindak pada sudut tertentu antara satu sama lainTwo forces acting at an angle to each other

Daya paduan bagi dua daya yang bertindak ke atas dua objek pada dua arah berbeza, boleh ditentukan

dengan menggunakan kaedah segi tiga daya atau kaedah segi empat selari daya.

The resultant force of two forces, which act on an object in two different directions, can be determined by a triangle

of forces or a parallelogram of forces.

Dua daya yang bertindak pada satu titik melalui satu sudut [kaedah segi empat selari]Two forces acting at a point at an angle [Parallelogram method]

Skala/Scale: 1 cm = N

LANGKAH 1/ STEP 1: Dengan menggunakan pembaris dan protaktor, lukis dua daya F1 dan F2 bermula dari satu titik, X.Using ruler and protractor, draw the two forces F1 and F2 from a point, X.

60 °C

F1

X

F2

LANGKAH 2/ STEP 2: Lengkapkan rajah segi empat selari.Complete the parallelogram.

F1

X

F2

LANGKAH 3/ STEP 3: Lukis pepenjuru (dari X) bagi segi empat selari bagi menunjukkan magnitud dan arah bagi daya paduan, F. Draw the diagonal (from X) of the parallelogram. The diagonal representsthe resultant force, F in magnitude and direction.

Daya paduan

Resultant force

F1

X

F2

F1

F2

60 °C

Contoh/Example

Rajah di bawah menunjukkan dua daya yang bertindak ke atas objek P.The diagram below shows two forces acting on object P.

5 N

12 N60°

F1

F2P

Tentukan daya paduan yang terhasil.Determine the resultant force produced.


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2

UNIT

Jawapan/Answer:Kaedah I/Method IKaedah segi empat selari dayaParrallelogram of forces method 1 Tentukan skala. Dengan menggunakan pembaris dan protaktor, lukis dua daya F1 dan F2 bermula dari

titik O.Set a scale. Using a ruler and protactor, draw the two forces, F1 and F2 from a point O.

2 Lengkapkan rajah segi empat selari./Complete the parallelogram.

3 Lukis pepenjuru bagi segi empat selari. Pepenjuru menunjukkan magnitud dan arah daya paduan, F.

Draw the diagonal of the parallelogram. The diagonal represents the resultant force, F in magnitude and

direction .

5 cm = 5 N

12 cm = 12 N

60°

O

Kaedah II/Method IIKaedah segi tiga dayaTriangle of forces method 1 Tentukan skala. Dengan menggunakan pembaris dan protaktor, lukis daya yang pertama, F1 dari titik O.

Set a scale. Using a ruler and protractor, draw the first force, F1, from a point O.

2 Lukis daya yang kedua F2 dari hujung atas F1.Draw the second force, F2 from the head of F1.

3 Lengkapkan segi tiga daya dengan melukis garisan dari pangkal F1 ke hujung F2.Complete a triangle of forces by drawing a line from the tail of F1 to the head of F2.

4 Pepenjuru menunjukkan magnitud dan arah daya paduan, F.The diagonal represents the resultant force, F, in magnitude and direction.

F2 = 12 N

F1 = 5 N

F = Daya paduan Resultant force

60°

O


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2

UNIT

1 Dengan menggunakan skala dan kaedah yang sesuai, tentukan daya paduan. (Perhatian: magnitud dan arah diperlukan).By using suitable scale and method, determine the resultant force (Note: Magnitudde and direction are required).

(a) (b) (c)

5 N

5 N

8 N

8 N

6 N

10 N

600 1200

2 Lengkapkan rajah untuk menunjukkan daya paduan. Complete the diagram to show the resultant force.

2 N

5 N

120°

Jawapan/Answers:

1 Skala/Scale: 1 cm : 1 N

(a)

8 N

6 N

37°

Daya paduan = 10.0 N

Resultant force

Latihan/Exercises

10 N pada sudut 37° dengan daya 8 N10 N at angle of 37° with the 8 N-force


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2

UNIT

(b)

8 N

10 N

260600

Daya paduan = 15.6 N

Resultant force

(c)

5 N

5 N

600

600

Day

a pa

duan

= 5

.0 N

Resu

ltant

forc

e

2 Skala/Scale: 2 cm : 1 N

(a)

2 N

5 N

120°16°

Daya paduan/Resultant force = 6.2 N

15.6 N pada sudut 26° dengan daya 10 N15.6 N at an angle of 26° with the 10 N-force

5 N pada sudut 60° dengan daya 5 N5 N at an angle of 60° with 5 N-force

6.2 N pada sudut 16° dengan daya 5 N6.2 N at an angle of 16° with 5 N-force


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UNIT

Daya dileraikan kepada komponen berkesanResolve a force into the effective components

Leraian Daya/Resolution Of Forces:

Daya F boleh dileraikan kepada dua komponen yang berserenjang/bersudut tegak antara satu sama lain:

A force F can be resolved into two components which are perpendicular/at 90° to each other:

(a) Fx: komponen mengufuk / horizontal component,

(b) Fy: komponen menegak / vertical component,

sin θ = Fy

F kos θ =

Fx

FFy = F sin θ Fx = F kos θ

Fy

θ

θ

F

Fx

Latihan/Exercises

1 Dapatkan komponen mengufuk dan komponen menegak daya tersebut. (Perhatian: Pertimbangkan magnitud daya yang positif sahaja.)Find the horizontal component and the vertical component of the force. (Note: Consider only the positive magnitudes of the forces.)(a) (b)

(c) (d)

Fx

75 N = sin 70°

∴ Fx = 75 N sin 70° = 70.48 N

FY

75 N = kos 70°

∴ FY = 75 N kos 70° = 25.65 N

Fx

6 N = kos 60°

∴ Fx = 6 N kos 60° = 3.0 N

FY

6 N = sin 60°

∴ FY = 6 N sin 60° = 5.20 N

Fx

200 N = kos 50°

∴ Fx = 200 N kos 50° = 128.6 N

FY

200 N = sin 50°

∴ FY = 200 N sin 50° = 153.2 N

Fx

5 N = sin 40°

∴ Fx = 5 N sin 40° = 3.21 N

FY

5 N = kos 40°

∴ FY = 5 N kos 40° = 3.83 N

Fy

Fx

200 N

500

500

400400

Fx

Fy

F = 5 N

0

Fy

Fx

700700

75 N

600

600

Fy

Fx

F = 6 N

0


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2

UNIT

2 Rajah menunjukkan troli yang berjisim 2 kg di atas permukaan licin, ditarik oleh daya 6 N pada sudut 60° dengan ufuk.The diagram shows a trolley of mass 2 kg on a smooth surface being pulled by a force of 6 N at an angle of 60° with the horizontal.

(a) Berapakah komponen mengufuk daya itu?What is the horizontal component of the force?

(b) Berapakah pecutan troli itu?/What is the acceleration of the trolley?


(a) Fx

6 N = kos 60° (b) F = ma, a = 3 N

2 kg = 1.5 m s-2

∴ Fx = 6 N kos 60° = 3.0 N

3 Rajah menunjukkan sebuah kereta sedang ditunda. Kabel itu mempunyai daya F, 5 000 N. The diagram shows a car being towed. The cable has a force F of 5 000 N.

(a) Tunjukkan dan labelkan/Indicate and label: • dayaF/the force F • dayakomponenmengufuk, Fx /the horizontal component force Fx

• dayakomponenmenegak,Fy /the vertical component force Fy

(b) Cari/Find • dayamengufukkabelyangmenggerakkankeretakehadapan.

the horizontal force of the cable which moves the car forward. • dayamenegakkabel./the vertical force of the cable.

Penyelesaian/Solution(a)

Fy

Fx

60°

F = 5 000 N

(b) Fx

5 000 N = kos 60°

∴ Fx = 5 000 N kos 60° = 2 500 N

FY

5 000 N = sin 60°

∴ FY = 5 000 N sin 60° = 4 330 N

4 Seorang pelancong menarik begnya dengan daya 100 N pada sudut 55° dari garis mengufuk. A tourist pulls his bag with a force of 100 N at an angle of 55° with the horizontal.

(a) Tunjukkan dan labelkan/Indicate and label: • dayaF/the force F • dayakomponenmengufuk, Fx /the horizontal component force Fx

• dayakomponenmenegak,Fy /the vertical component force Fy

(b) Cari/Find • dayamengufukbegyangmenggerakkannyakehadapan.

the horizontal force of the cable which moves it forward. • dayamenegakbeg./the vertical force of the bag.

Penyelesaian/Solution(a)

Fy

Fx

55°

F = 100 N

(b) Fx

100 N = kos 55°

∴ F = 100 N kos 55° = 57.36 N

FY

100 N = sin 55°

∴ FY = 100 N sin 55° = 81.92 N

600

6 N

Fx

2 kg

Permukaan licin/Smooth surface

600


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2

UNIT

Masalah yang melibatkan daya paduan dan prinsip leraian dayaProblems involving resultant force and the principle of resolution of forcesA Lif/Lift

Seorang budak lelaki berada di dalam sebuah lif. Dia berdiri di atas mesin penimbang. Berat budak lelaki, W, bertindak arah ke bawah budak itu dan tindak balas normal, R, bertindak arah ke atas.A boy is inside a lift. He is standing on a weighing machine. The weight of the boy, W acts downward on the boy and a normal reaction, R, acts upwards.

Lif yang pegun (atau bergerak ke bawah atau ke atas dengan

halaju seragam)Stationary lift (or moves upwards

or downwards with uniform velocity)

Lif bergerak ke atas dengan pecutan a m s-2

The lift moves upwards with an acceleration of a m s-2

Lif bergerak ke bawah dengan pecutan a m s-2

The lift moves downwards with an acceleration of a m s-2

R

W = mg

Mesin penimbangWeighing machine

R

a

W = mg


R

a

W = mg


Daya paduan, F = 0Resultant force, F = 0

Daya paduan, F arah ke atasResultant force, F is upwards

Daya paduan, F arah ke bawahResultant force, F is downward

F = R – mg = 0R = mg

R > mgF = maF = R – mg = maR = mg + ma

mg > RF = maF = mg – R = maR = mg – ma

Bacaan pada mesin penimbang = berat budak lelaki tersebutThe reading on the weighing scale = the weight of the boy

Bacaan pada mesin penimbang lebih besar daripada berat budak ituThe reading on the weighing scale machine is larger than the weight of the boy

Bacaan mesin penimbang lebih kecil daripada berat budak ituThe reading on the weighing scale machine is smaller than the weight of the boy

Perhatian : Dalam setiap kes di atas, R memberi bacaan pada mesin penimbang.Note : In each of the cases above, R gives the reading on the weighing scale.


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UNIT

Seorang budak lelaki berjisim 50 kg berada di dalam sebuah lif. A boy of mass 50 kg is inside a lift.(a) Kirakan berat budak lelaki itu,/Calculate the weight of the boy,(b) Kirakan bacaan yang ditunjukkan oleh mesin penimbang jika lif itu: Calculate the reading on the weighing scale if the lift: (i) pegun is stationary (ii) memecut ke atas dengan pecutan 2 m s-2

accelerates upwards with an acceleration of 2 m s-2

(iii) memecut ke bawah dengan pecutan 2 m s-2

accelerates downwards with an acceleration of 2 m s-2

(iv) bergerak ke atas dengan halaju seragam 1.5 m s-1

moves upwards with constant velocity of 1.5 m s-1


LifLift m = 50 kg

Mesin penimbangWeighing scale

B Takal/Pulley

1

3 kg

4 kg

Takal licinSmooth pulley

Berat/Weight,40 N Berat/Weight,

30 N

TT

3 kg

4 kg

Daya geseran, 2 NFrictionalforce, 2 N Takal licin

Smooth pulley

T

T

Berat/Weight,30 N

Cari daya paduan, FFind the resultant force, F

F = 40 N – 30 N = 10 N F = 30 N – 2 N = 28 N

Cari jisim yang bergerak, mFind the moving mass, m

m = 4 kg + 3 kg = 7 kg m = 4 kg + 3 kg = 7 kg

Cari pecutan, aFind the acceleration, a,

F = ma

a = 10 N7 kg

= 1.43 m s-2

F = ma, ∴28 N = (7 kg)(a)

a = 28 N7 kg

= 4 m s-2

(a) Berat budak = W Mass of the boy = mg = 50 kg × 10 m s–2

= 500 N(b) (i) R = W = 500 N (ii) R – mg = ma R = 500 N + (50 kg)(2 m s–2) = 600 N

(iii) mg – R = ma R = 500 N – (50 kg)(2 m s–2) = 400 N (iv) R – mg = ma Tetapi a = 0 (kerana halaju malar) but a = 0 (because constant velocity) ∴ R = mg = 500 N


48



2

UNIT

Cari tegangan benang, TFind the string tension, T

Kaedah I/Method IPertimbangkan jisim 4 kg sahaja (gerak ke bawah)Consider only the 4 kg-mass (moving downwards)40 N – T = m1 a40 N – T = (4 kg)(1.43 m s–2) ∴ T = 34.28 N

Kaedah II/Method IIPertimbangkan jisim 3 kg sahaja (gerak ke atas)Consider only the 3 kg-mass (moving upwards) T – 30 N = m2 a T = 30 N + (3 kg)(1.43 m s–2) T = 34.29 N

Kaedah I/Method IPertimbangkan jisim 3 kg sahaja (yang bergerak ke bawah)Consider only the 3 kg-mass (moving downwards)30 N – T = ma30 N – T = (3 kg)(4 m s–2) ∴ T = 30 N – 12 N = 18 N

Kaedah II/Method IIPertimbangkan jisim 4 kg sahaja (yang bergerak ke kanan)Consider only the 4 kg-mass (moving to the right) T – 2 N = ma ∴ T = 2 N + (4 kg)(4 m s–2) = 2 N + 16 N = 18 N

2 Troli yang berjisim 2 kg disambungkan dengan tali kepada pemberat berjisim 3 kg. Tali diletakkan di atas takal licin. Pemberat kemudiannya dilepaskan.A 2 kg-trolley is connected by a rope to a 3 kg-load. The rope passes over a smooth trolley. The load is then released.

(a) (i) Jika permukaan meja adalah licin, berapakah pecutan troli?If the surface of the table is smooth, what is the acceleration of the trolley?

(ii) Berapakah tegangan dalam tali?/What is the tension in the rope? (b) (i) Jika daya geseran antara troli dan permukaan meja ialah 10 N, berapakah pecutan troli?

If the frictional force between the trolley and the surface of the table is 10 N, what is the acceleration of the trolley?

(ii) Berapakah tegangan tali sekarang?What is the tension in the rope now?


(a) (i) F = ma 30 N = [(2 + 3) kg] [a] a = 6 m s–2 (ii) dari gerakan troli from the motion of the trolley T = ma = (2 kg)(6 m s–2) = 12 N

atau dari gerakan jisim 3 kg or from the motion of the 3 kg-mass 30 N – T = ma 30 N – T = (3 kg)(6 m s–2) T = 30 N – 18 N = 12 N

(b) (i) 30 N – 10 N = (3 + 2) kg × a 20 N = 5 kg × a a = 4 m s-2 (ii) dari gerakan troli from the motion of the trolley T – 10 N = ma T – 10 N = (2 kg)(4 m s-2) T = 8 N + 10 N = 18 N

atau dari gerakan jisim 3 kg or from the motion of the 3 kg-mass 30 N – T = m1 a 30 N – T = (3 kg)(4 m s-2) T = 30 N – 12 N = 18 N

MejaTable

Takal licinSmooth trolley

3 kg

30 N

2 kgT

T


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2

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2.10 MEMAHAMI KERJA, TENAGA, KUASA DAN KECEKAPANUNDERSTANDING WORK, ENERGY, POWER AND EFFICIENCY

Kerja dilakukan apabila daya membuatkan suatu objek bergerak. Semakin besar daya dan jarak sesuatu objek itu digerakkan, maka semakin besar kerja dilakukan.

Work is done whenever a force makes something move. The greater the force and the greater the distance moved, the more work is done.

Definisi kerja/Definition of work

Kerja yang dilakukan ialah hasil darab daya yang dikenakan dengan sesaran objek pada arah daya

dikenakan.

Work done is the product of an applied force and the displacement of the object in the direction of the applied

force.

Kerja, W = Fs , di mana F = daya, s = sesaran

Work, W = Fs , where F = force, s = displacement

Unit SI bagi kerja ialah joule, J

The SI unit of work is the joule, J

Kerja 1 joule dilakukan apabila daya 1 N menggerakkan objek sejauh 1 m dalam arah daya dikenakan.1 joule of work is done when a force of 1 N moves an object 1 m in the direction of the force.

sF

C Satah condong/Inclined plane

Sebuah bongkah kayu berjisim m kg diletakkan di atas satah condong yang membuat sudut θ dengan ufuk. Blok kayu dikenakan dengan daya-daya seperti yang disenaraikan di bawah:A block of wood of m kg which is placed on an inclined plane makes an angle θ with the horizontal. The block of wood is acted upon by the forces as listed below: (a) Wx, komponen berat yang selari dengan satah condong. Wx, the weight component which is parallel to the inclined plane.(b) Wy, komponen berat yang berserenjang dengan satah condong. Wy, the weight component which is perpendicular to the inclined plane.(c) tindak balas normal, N. the normal reaction, N.Ungkapkan setiap komponen di (a), (b), (c) dalam sebutan m, g dan θ, di mana g ialah pecutan akibat graviti.Express each of the components in (a), (b), and (c) in terms of m, g and θ, where g is acceleration due to gravity.

Penyelesaian/Solution(a) Wx = mg sin θ(b) Wy = mg kos θ(c) N = Wy = mg kos θ

θ

Wx Wy

N

θ

mg

05 FIZ Form 4 Ch2 D 2F.indd 89 9/10/13 4:14 PM

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2

UNIT

Pengiraan kerjaCalculation of work

Sesaran, s, bagi objek dalam arah sama dengan daya, FThe displacement, s, of the object is in the direction of the force, F

Sesaran, s, bagi objek pada arah yang tidak sama dengan arah daya, FThe displacement, s, of the object is not in the direction of the force, F

F

s

W = Fs

Fs

W = Fs

F

F1

s

θ

F1

F = kos θ

F1 = F kos θ ∴ W = F1 × s = (F kos θ) s

Seorang budak menolak basikalnya dengan daya 25 N melalui jarak 3 m. Kira kerja yang dilakukan oleh budak itu.A boy is pushing his bicycle with a force of 25 N through a distance of 3 m. Calculate the work done by the boy.

Penyelesaian/SolutionW = Fs = 25 N × 3 m = 75 J(Perhatian/Note: 1 N m = 1 J)

Contoh/Example 1

Seorang budak perempuan mengangkat sebuah pasu berjisim 3 kg pada ketinggian 0.4 m. Berapakah kerja yang dilakukan oleh budak perempuan itu?A girl lifts up a 3 kg flower pot steadily to a height of 0.4 m. What is the work done by the girl?

Penyelesaian/SolutionW = 3 kg × 10 m s–2 × 0.4 m = 12 JAtau/OrW = 3 kg × 10 N kg–1 × 0.4 m = 12 J

Contoh/Example 2

25 N

30 N

Seorang pekebun menolak mesin rumput dengan daya 50 N pada sudut 60 °C dengan ufuk. Berapakah kerja yang dilakukan untuk menolak mesin rumput pada jarak 100 m?A gardener pushes a lawn mower with a force of 50 N at an angle of 60 °C from horizontal. What is the work done in pushing the lawn mower through a distance of 100 m?

Penyelesaian/SolutionF

50 N = kos 60°

F = 50 N kos 60° ∴ W = F × 100 m = (50 N kos 60°) × (100 m) = 2 500 N

60°

60°

50 N

F

Contoh/Example 3

600

50 N


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2

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Tiada kerja dilakukan apabila:/No work is done when:

Objek berada dalam keadaan pegun .The object is stationary .

Arah gerakan objek adalah berserenjang dengan daya yang dikenakan.

The direction of motion of the object is perpendicular to that of the applied force.

Seorang pelayan membawa dulang makanan dan berjalan.A waiter is carrying a tray of food and walking.

Latihan/Exercises

Sesaran/Displacement = 50 cm

F = 80 N 1 Berapalah kerja dilakukan oleh daya 80 N itu?

How much work is done by the 80 N force?


W = 80 N × 0.5 m = 40 J

2 Ali menolak sebiji batu dengan mengenakan daya 200 N. Berapakah kerja yang dilakukannya?Ali pushes a big rock by applying a force of 200 N. How much work has he done?


W = sifar (pegun)/zero (stationary)

Pegun/Stationary

F = 200 N

3 Seorang lelaki menarik satu beban dengan menggunakan satah condong. Tinggi landasan condong itu ialah 80 cm. Berapakah kerja yang dilakukan oleh lelaki itu untuk menarik beban itu?A man pulls up a load using an inclined plane. The height of the inclined plane is 80 cm. How much work is being done by the man to lift the load?


s80 cm

30°

80 cms

= sin 30°

∴s = 80 cmsin 30°

= 0.8 m

sin 30°

F = 320 N

30°80 cm

W = Fs

= 320 N × ( 0.8 m

sin 30° )

= 512.0 J


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2

UNIT

5 Satu daya 25 N digunakan untuk mengangkat sebuah beg. Encik Rahim berjalan sejauh 20 m dengan memegang beg itu. Berapakah kerja yang dilakukan oleh Encik Rahim terhadap beg itu? Terangkan jawapan anda.A force of 25 N is used to lift a bag. Encik Rahim walks a distance of20 m holding the bag. How much work is being done by Encik Rahim with respect to the bag? Explain your answer.


Kerja dilakukan adalah sifar kerana daya adalah berserenjang dengan sesaran.

Work done is zero because the force is perpendicular to the displacement.

F = 25 N

Sesaran/Displacement = 20 m

4 Berapakah kerja yang dilakukan oleh Raju untuk mengangkat beban melalui jarak 4 m?How much work is done by Raju to lift the load through the displacement of 4 m?


W = 150 N × 4 m = 600 J

SesaranDisplacement= 4 m

F = 150 NF

Menyatakan bahawa apabila kerja dilakukan, tenaga dipindahkan dari satu objek ke satu objek yang lainState that when work is done, energy is transferred from one object to another

• Tenaga boleh ditakrifkan sebagai kebolehan untuk melakukan kerja. Energy can be defined as the ability to do work.• Satuobjekyangbolehmelakukankerjadikatakanmempunyaitenaga./An object that can do work has energy.• Tenagabolehwujuddalampelbagaibentuk,misalnya/Energy exists in many forms, for example:

(a) Tenaga keupayaan graviti/Gravitational potential energy

(b) Tenaga kinetik/Kinetic energy

(c) Tenaga keupayaan kenyal/Elastic potential energy

(d) Tenaga keupayaan elektrik/Electric potential energy

(e) Tenaga bunyi/Sound energy

(f) Tenaga mekanik/Mechanical energy

(g) Tenaga nuklear/Nuclear energy

• Kerja dilakukan apabila daya dikenakan ke atas objek dan objek itu bergerak. Ini diikuti dengan pemindahan tenaga dari satu objek ke objek lain.Work is done when a force is applied on an object and the object moves. This is followed by the transference of energy from one object to another.

• Oleh itu, apabila kerja dilakukan, tenaga dipindahkan dari satu objek ke satu objek yang lain.

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


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2

UNIT

Kirakan tenaga keupayaan graviti bagi setiap keadaan berikut.Calculate the gravitational potential energy for each of the following.

BeratWeight

PendulumBob

KotakBox

m = 2 kgm = 0.3 kg

m = 0.3 kg

h = 0.8 mh = 20 cm

h = 1.2 m

Penyelesaian/SolutionE = mgh E = mgh E = mgh = (2 kg) (10 m s–2) (0.8 m) = (0.3 kg) (10 m s–2) (0.2 m) = 0.3 kg × 10 m s–2 × 1.2 m = 16 J = 0.6 J = 3.6 J

Contoh/Example

Definisi tenaga keupayaan graviti/Define gravitational potential energy

Seorang budak perempuan membuat kerja apabila dia memanjat tangga sebuah papan gelongsor. Dia mempunyai tenaga keupayaan gravity apabila dia berada pada kedudukan tertinggi papan gelongsor itu.A girl does work when she climbs up the stairs of a sliding board. She has gravitational potential energy when she is at the top of the sliding board.

Tenaga keupayaan gravity ialah tenaga yang tersimpan dalam objek

disebabkan ketinggian (kedudukan) nya daripada permukaan bumi.The gravitational potential energy is the energy stored in the object because

of its height (position) above the earth’s surface.

Tenaga keupayaan graviti adalah sama dengan kerja dilakukan untuk menaikkan satu objek kepada satu ketinggian tertentu. Daya diperlukan

untuk menaikkan objek adalah sama dengan berat objek, F = mg.

The gravitational potential energy is equal to the work done to raise an object to a particular height. The force required to raise the object is the same

as the weight of the object, F = mg.

Jika jarak menegak yang dilalui oleh objek ialah h,/If the distance moved by the object is h,

Kerja dilakukan, W = F × s = mg × h = mgh

BolaBall

KedudukanakhirFinalposition

KedudukanawalInitialposition

Jisim/Mass= m kg

h

Tenaga keupayaan graviti,Gravitational potential Energy, EP = W

Work done, W = F × s = mg × h = mgh

(a) (b) (c)


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UNIT

Nota/Notes: 1 Tenaga keupayaan graviti bagi objek

bergantung padaThe gravitational potential energy of an object depends on the

(a) jisim objek, m

mass of the object, m

(b) kekuatan medan graviti, g

gravitational field strength, g

(c) perubahan ketinggian, h

change in height, h

2 "Kehilangan" tenaga keupayaan graviti tidak bergantung pada kecerunan tetapi bergantung pada jarak tegak .

ia bergerak.The "loss" of gravitational potential energy does not depend on the gradient of the slope but depends on the

vertical distance traversed.

Tenaga keupayaan graviti 100 JGravitational potential energy 100 J

Kerja dilakukanWork done in this path= mgh= 100 J

Kerja dilakukansepanjang landasanWork done alongthis path= 100 J

h m

A A A

B B B

Definisi tenaga kinetikDefinition of kinetic energy

Seorang budak lelaki yang sedang mengayuh basikal mempunyai tenaga kinetik. Apabila dia mengayuh lebih pantas, maka dia mempunyai tenaga kinetik yang lebih besar. Pada keadaan pegun, dia tidak mempunyai tenaga kinetik.A boy riding a bicycle possesses kinetic energy. When he rides faster, he will have more kinetic energy. When he is stationary, he does not have any kinetic energy.

Tenaga kinetik ialah tenaga yang diperoleh sesuatu objek disebabkan oleh gerakannya.

Kinetic energy is the energy of an object due to its motion.

Tenaga kinetik = 12

mv2

di mana m = jisim v = halaju

Kinetic energy = 12

mv2

where m = mass v = velocity


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2

UNIT

Contoh/Example

1 Sebiji bola berjisim 0.5 kg bergerak dengan halaju 4 m s-1. Hitungkan kerja dilakukan.A ball of mass 0.5 kg moves with velocity of 4 m s-1. Calculate the work done.


Kerja = Tenaga kinetik

= 12

mv2

= 12

(0.5 kg) (4 m s–1)2

= 4.0 J

Work done = Kinetic energy

= 12

mv2

= 12

(0.5 kg) (4 m s–1)2

= 4.0 J

2 Sebuah kereta berjisim 950 kg bergerak dengan halaju malar 20 m s-1 selama satu minit. Hitungkan kerja yang dilakukan oleh kereta itu dalam tempoh yang tersebut.A car of mass 950 kg moves at a constant velocity of 20 m s-1 for one minute. Calculate the work done by the car during this period.

m = 950 kg

v = 20 m s–1

Penyelesaian/SolutionKerja dilakukan = Tenaga kinetikWork done = Kinetic energy

= 12

mv2

= 12

(950 kg) (20 m s–1)2

= 190 000 J

Menyatakan prinsip keabadian tenaga/State the principle of conservation of energy

Prinsip keabadian tenaga menyatakan bahawa tenaga boleh berubah dari satu bentuk ke satu

bentuk yang lain, tetapi tidak boleh dicipta atau dimusnahkan.

The principle of conservation of energy states that energy can be changed from one form to another

form, but it cannot be created or destroyed.

Jumlah tenaga di dalam sistem diabadikan /The total energy in a system is conserved .

Jumlah tenaga sebelum berubah kepada bentuk tenaga yang lain = Jumlah tenaga selepas berubah kepada bentuk tenaga yang lain.

Total energy before conversion to other forms of energy = Total energy after conversion to other forms of energy.


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Aktiviti: Keabadian Tenaga/Activity: Conservation of Energy

1 (a) Sebiji bola dipegang pada ketinggian tertentu di atas lantai. Hold a ball at a certain height above the floor.

Apakah jenis tenaga yang terdapat pada bola?What is the energy gained by the ball?

Tenaga keupayaan graviti/Gravitational potential energy

(b) Bola dilepaskan. Release the ball. (i) Apakah jenis tenaga yang dipunyai oleh bola sejurus sebelum ia

menghentam lantai?What is the energy gained by the ball just before it hits the floor?

Tenaga kinetik/Kinetic energy

(ii) Dari manakah tenaga pada bola itu berasal?Where does the energy of the ball originate?

Tenaga keupayaan graviti/Gravitational potential energy

(iii) Apakah hubungan antara tenaga di (a) dengan tenaga di (b)?What is the relationship between the energy in (a) and the energy in (b)?

Sama/Equal

2 (a) Troli ditolak ke arah dinding untuk memampatkan spring. Push a trolley against a wall to compress a spring.

Apakah tenaga yang tersimpan di dalam spring?What is the energy stored in the spring?

Tenaga keupayaan kenyal

Elastic potential energy

(b) Troli dilepaskan supaya ia bergerak menjauhi dinding. Release the trolley so that it moves away from the wall.

Apakah yang berlaku kepada tenaga yang tersimpan di dalam spring?What happens to the energy stored in the spring?

Tenaga keupayaan kenyal ditukarkan kepada tenaga kinetik.

The elastic potential energy is transferred to kinetic energy.

BolaBall

Lantai/Floor

KetinggianHeight


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2

UNIT

Contoh/Example

1 Sebiji kelapa berjisim 1.2 kg jatuh dari satu ketinggian. Abaikan rintangan udara.A coconut of mass 1.2 kg drops from a height. Ignore air resistance.

(a) (i) Tentukan jarak buah kelapa itu jatuh dalam masa 2.0 s.Determine the distance the coconut falls in 2.0 s.

(ii) Tentukan halajunya, selepas 2.0 s.Determine its velocity after 2.0 s.

(b) (i) Berapakah jumlah kehilangan tenaga keupayaan graviti selepas 2.0 s?What is the loss of its gravitational potential energy after 2.0 s?

(ii) Berapakah tenaga kinetiknya?/What is its kinetic energy then? (c) Apakah yang boleh dikatakan tentang kehilangan tenaga keupayaannya dan tenaga kinetik yang diperoleh?

What can be said about the loss of its gravitational potential energy and the kinetic energy gained?


(a) (i) s = ut + 12

at2

= 0 + 12

(10 m s–2)(2.0 s)2

= 20.0 m (ii) v = u + gt v = 0 + (10 m s–2)(2.0 s) = 20.0 m s–1

(b) (i) E = mgh = (1.2 kg) (10 m s–2) (20.0 m) = 240 J (ii) Tenaga kinetik/Kinetic energy = 240 J(c) Tenaga keupayaan graviti yang hilang telah ditukarkan kepada tenaga kinetik. Lost of gravitional potential energy has been changed to kinetic energy.

2 Sebiji durian jatuh dari ketinggian 15 m. Berapakah halaju buah durian itu sejurus sebelum ia menghentam tanah? (Andaikan bahawa g = 10 m s-2)A durian falls from a height of 15 m. What is the velocity of the durian just before it hits the ground? (Assume that g = 10 m s-2)

Penyelesaian/Solution mgh = ½ mv2

v2 = 2gh = 2 × 10 m s–2 × 15 m v = 17.32 m s–1

3 Sebiji bola dilepaskan pada titik A dari ketinggian 0.8 m dengan menggunakan landasan licin. Berapakah halaju bola itu pada titik B?A ball is released at point A from a height of 0.8 m using a smooth inclined plane. What is the velocity of the ball at point B?

Penyelesaian/SolutionJumlah tenaga di A = Jumlah tenaga di B Total energy at A = Total energy at B mgh = ½ mv2

v2 = 2gh = 2 × 10 m s–2 × 0.8 m = 16 m2 s–2

∴ v = 4 m s–1

Sebelum kelapa jatuh/Before the coconut falls,Tenaga keupayaan graviti = mghGravitational potential energy = mghTenaga kinetik/Kinetic energy = 0

h

0.8 m

B

A

(g = 10 m s–2)


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Menerangkan kecekapan peralatanExplain what efficiency of a device is

Kecekapan = Kuasa outputKuasa input

× 100%

Efficiency = Output power

Input power × 100%

1 Seorang murid yang berjisim 45 kg mengambil masa 6 s untuk menaiki tangga yang mempunyai 36 anak tangga. Jika tinggi setiap anak tangga ialah 12 cm, kirakanA student of mass 45 kg takes 6 s to climb a flight of stairs that has 36 steps. If each step is 12 cm high, calculate the

(a) kerja yang dilakukan oleh murid ituwork done by the student

(b) kuasa murid itupower of the student

Penyelesaian/Solution(a) W = mgh = (45 kg) (10 m s–2) × (36 anak tangga × 0.12 m setiap anak tangga) (45 kg) (10 m s–2) × (36 steps × 0.12 m each step) = 1 944 J

(b) P = 1 944 J6 s

= 324 W

2 Sebuah motor mengangkat pemberat yang berjisim 1.5 kg pada ketinggian 1.0 m dalam masa 4.0 s. Berapakah kuasa motor itu?A motor lifting a weight having a mass of 1.5 kg up to a height of 1.0 m in 4.0 s. What is the power of the motor?


P = mgh

t

= 1.5 kg × 10 m s–2 × 1 m 4.0 s

= 3.75 W

Contoh/Example

12 cm

Definisi kuasaDefine power

Kuasa ditakrifkan sebagai kadar kerja dilakukan atau kadar tenaga ditukarkan .

Power is defined as the rate of work done or the rate of energy transformed .

Kuasa, P = KerjaMasa =

TenagaMasa

Power, P = WorkTime

= EnergyTime

Unit SI bagi kuasa ialah watt, W atau J s–1 / S.I unit of power is watt, W or J s–1 .

Kuasa 1 W dihasilkan apabila 1 J kerja dilakukan dalam masa 1 saat.A power of 1 W is generated when 1 J of work is done in 1 second.

(g = 10 m s–2)


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2

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3 Sebuah enjin petrol mempunyai kuasa output 96 kJ per minit. Berapakah kuasa input jika kecekapan enjin itu ialah 20%? A petrol engine has an output power of 96 kJ per minute. What is the input power if the engine efficiency is 20%?


Kuasa output/Output power = 96 × 103 J 60 s

= 1 600 W

Kecekapan = Kuasa output Kuasa input

× 100%

∴ 20% = 1 600 W Kuasa input

× 100%

Kuasa input = 1 600 W × 100% 20%

= 8 000 W

Efficiency = Output power Input power

× 100%

∴ 20% = 1 600 W Input power

× 100%

Input power = 1 600 W × 100% 20%

= 8 000 W

Latihan/Exercises

1 Sebuah troli dilepaskan dari keadaan rehat pada titik X. Berapakah halaju troli di titik Y?A trolley is released from rest at point X. What is the velocity of the trolley at point Y?

Penyelesaian/SolutionDari X ke Y, jarak tegak = 1.5 mFrom X to Y, the vertical distance = 1.5 m ∴ h = 1.5 m

mgh = 12

mv2

v2 = 2gh = 2 × (10 m s–2)(1.5 m) ∴ v = 5.48 m s–1

2 Sebiji bola bergerak di sepanjang permukaan mengufuk yang licin dengan halaju 6 m s-1. Bola itu kemudiannya bergerak naik ke atas satu satah condong licin. Ketinggian satah condong itu ialah 1.5 m. Berapakah halaju bola itu di titik B?A ball is moving along a smooth horizontal surface at a velocity of 6 m s-1. The ball then moves up a smooth inclined plane. The height of the inclined plane is 1.5 m. What is its velocity at point B?


Jumlah tenaga di A = Jumlah tenaga di B Total energy at A = Total energy at B

12

mvA2 = mgh + 1

2 mB

2

12

(6 m s–1)2 = (10 m s–2)(1.5 m) + 12

vB2

12

vB2 = (18 – 15) m2 s–2

vB = 2.45 m s–1

2.5 m

2 kg

1.0 m

X

Y

Z

vA = 6 m s–1

vB = ?

1.5 m

A

B

(g = 10 m s–2)


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UNIT

3 Sebiji bola tenis dilontar ke atas dengan halaju awal 20 m s-1. Berapakah tinggi maksimum yang

boleh dicapai oleh bola tersebut?A tennis ball is thrown upwards with an initial velocity 20 m s-1. What is the maximum height that the ball can achieve?

Penyelesaian/Solution ½ mv2 = mgh, h = v2

2g

= (20 m s–1)2 2(10 m s–2)

= 20 m

4 Nyatakan bentuk tenaga di titik/State the form (s) of energy at point

(a) P = Tenaga keupayaan graviti

Gravitational potential energy

(b) Q = Tenaga kinetik/Kinetic energy

(c) R = Tenaga kinetik + Tenaga keupayaan graviti

Kinetic energy + Gravitational potential energy

(d) S = Tenaga keupayaan graviti

Gravitational potential energy

5 Seorang budak lelaki berjisim, m, sedang berlari menaiki sebuah tangga. Dia mengambil masa, t, untuk sampai ke puncak. Berapakah kuasa budak itu? Beri jawapan anda dalam sebutan m, g, Y dan t.A boy of mass, m. runs up the stairs. He takes time, t, to reach the top. What is the power of the boy? Give your answer in terms of m, g, Y and t.


Kuasa = mgY

t Power =

mgY

t

6 Sebuah kereta dengan kecekapan 25% menghasilkan 3 000 J tenaga mekanikal setiap saat. Berapakah kuasa output enjin itu?A car engine with an efficiency of 25% produces 3 000 J of mechanical energy per second. What is the output power of the engine?



× 100%

25% = Kuasa output

3 000 W × 100%

Kuasa output = 25%100%

× 3 000 W

= 750 W


Input power × 100%

25% = Output power

3 000 W × 100%

Output power = 25%

100% × 3 000 W

= 750 W

KetinggianmaksimumMaximum height

Ketinggian minimumMinimum height

Bandul ringkasSimple pendulum

P

QR

S

X Y


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2

UNIT

7 Sebuah kren mengangkat beban 500 kg ke ketinggian

12 m dalam masa 8 s. Kuasa input ialah 45 000 W, berapakah kecekapan motor yang digunakan oleh kren

itu?A crane lifts a load 500 kg to a height of 12 m in 8 s. The power input is 45 000 W, what is the efficiency of the motor used in the crane?



× 100%

= 500 kg × 10 m s–2 × 12 m

8 s 45 000 W

× 100%

= 7 500 W45 000 W × 100%

= 16.67%


Input power × 100%

8 Seorang budak lelaki berjisim 30 kg sedang duduk di atas puncak satu papan gelongsor condong pada ketinggian 2.5 m dari tanah. Apabila budak lelaki itu menggelongsor menuruni papan gelongsor, kerja yang dilakukan untuk mengatasi geseran ialah 510 J.A boy of mass 30 kg sitting on the top end of an inclined sliding board at a height of 2.5 m from the ground. When the boy slides down the inclined board, the work done to overcome friction is 510 J.

2.5 m

Berapakah halaju pelajar itu sejurus sebelum dia menyentuh tanah?What is the velocity of the student just before he touches the ground?


mgh = 12

mv2 + 510 J

(30 kg)(10 m s–2)(2.5 m) = 12

(30 kg)v2 + 510 J

∴ v = 4 m s–1

12 m


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2

UNIT

9 Abu bersama basikalnya menuruni satu cerun bukit yang mempunyai ketinggian 3 m pada halaju

awalnya 2 m s-1 tanpa mengayuh. Pada kaki bukit, halajunya ialah 6 m s-1. Abu and his bicycle go down the slope of a hill of 3 m high at an initial velocity of 2 m s-1, without pedaling. At the foot of the hill, the velocity is 6 m s-1.

u = 2 m s–1

3 m

v = 6 m s–1

Diberi bahawa jisim Abu dengan basikalnya ialah 75 kg, cariGiven that the mass of Abu with his bicycle is 75 kg, find

(a) tenaga kinetik awal dipunyai oleh Abu dengan basikal. the initial kinetic energy of Abu and his bicycle. (b) tenaga keupayaan graviti awal yang dipunyai oleh Abu dan basikalnya.

the initial gravitational potential energy of Abu and his bicycle. (c) kerja dilakukan menentang geseran sepanjang cerun.

the work done against friction along the slope.


(a) Tenaga kinetik/Kinetic energy

= 12

mv2

= 12

(75 kg)(2 m s–1)2

= 150 J

(b) mgh = (75 kg)(10 m s–2)(3 m) = 2 250 J(c) Tenaga di puncak bukit = Tenaga di kaki bukit Energy at the top of the hill = Energy at the foot of the hill

2 250 J + 150 J = 12

(75 kg)(6 m s–1)2 + geseran/friction

Geseran/Friction = 1 050 J


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2

UNIT

10 Rajah menunjukkan atlit lompat bergalah berjisim 60 kg melompat melepasi palang pada ketinggian

5.0 m. J, K, L, M, N, O, P dan Q menunjukkan beberapa peringkat lompatan yang dibuat oleh atlit. Pada titik N, ketinggian atlit dari paras palang ialah 0.2 mThe diagram shows a pole vault jumper of mass 60 kg jumping over the bar of height 5.0 m. J, K, L, M, N, O, P and Q show the different stages of the jump made by the athlete. At point N, the athlete clears the bar with 0.2 m to spare.

GalahPole

J K

L

M

NO

P

Q

Tilam getah yang tebalA thick rubber maltress

(a) Mengapakah atlit diperlukan untuk memecut kepada halaju tertentu pada peringkat J ke K sebelum dia mula melompat?Why is the athlete required to accelerate from J to K to a certain velocity before he begins to jump?

Untuk menambahkan tenaga kinetik. Apabila halaju bertambah, maka tenaga kinetik bertambah.

To increase the kinetic energy. Kinetic energy increases with velocity.

(b) Terangkan mengapa galah itu perlu dibengkokkan di L.Explain why the pole has to be bent at L.

Untuk mendapatkan tenaga keupayaan kenyal yang maksimum.

To get maximum elastic potential energy.

(c) Kirakan tenaga keupayaan graviti bagi atlit pada titik N.Calculate the gravitational potential energy of the athlete at point N.E = mgh = (60 kg) × (10 m s–2) × (5.2 m) = 3 120 J

(d) Berapakah pecutan menegak atlit di peringkat P?What is the vertical acceleration of the athlete at stage P?

10 m s-2

(e) Mengapa sebuah tilam getah yang tebal diletakkan di kawasan di mana atlit mendarat?Why is a thick rubber mattress placed in the area where the athlete lands?

Menambahkan masa hentaman untuk mengurangkan daya impuls.

Increase the time of collision to reduce impulsive force.


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2

UNIT

2.11 MEMAHAMI KEKENYALANUNDERSTANDING ELASTICITY

Aktiviti 1: Idea tentang kekenyalan/Activity 1: Idea of elasticity

Definisi kekenyalanDefine Elasticity

Sifat bahan yang membolehkan objek kembali ke panjang dan bentuk asal

apabila daya yang dikenakan ke atasnya dialihkan.

A property of matter that enables an object to return to its original length and shape

when the force that acts on it is removed.

Tiada daya luar dikenakan. Molekul berada pada kedudukan asalnya . Daya antara molekul adalah sifar.

No external force is applied. The molecules are at their original positions. Intermolecular force is equal to zero.

Memampatkan pepejal menyebabkan molekulnya bergerak lebih rapat antara

satu sama lain. Daya tolakan antara molekul bertindak untuk menolak molekul kembali kepada kedudukan asalnya.

Compressing a solid causes its molecules to be moved closer to each other.

Repulsive intermolecular forces act to push the molecules back to their original positions.

Meregangkan pepejal menyebabkan molekulnya bergerak menjauhi antara

satu sama lain. Daya tarikan antara molekul bertindak untuk menarik kembali molekul kepada kedudukan asalnya.

Stretching a solid causes its molecules to be moved further from each other.

Attractive intermolecular forces act to pull back the molecules to their original positions.

Meregang wayar dengan daya luarStretching a wire by an external force:

• Molekulnyaakan menjauhi antara satu sama lain.

Its molecules will move further away from one another.

• Dayatarikanyangkuatakanbertindakdiantaramolekuluntuk menentang regangan yang dikenakan.

Strong attractive forces act between the molecules to oppose the stretching.

Apabila daya luar dialih/When the external force is removed:• dayatarikanantaramolekulmembawamolekulkembalikekedudukan

asalnya .

the attractive intermolecular forces bring the molecules back to their original positions.

• Wayaritu kembali ke kedudukan asalnya.

The wire returns to its original position.


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2

UNIT

TujuanAim

Untuk menyiasat hubungan antara daya dan pemanjangan springTo investigate the relationship between a force and the extension of a spring

RadasApparatus

Spring keluli, lima pemberat berjisim 50 g dan pemegang, pembaris meter, kaki retortSteel spring, five 50 g slotted weights and holder, metre rule, retort stand

HipotesisHypothesis

Pemanjangan spring bertambah dengan penambahan daya

Extension of a spring increases as the force increases.

Pemboleh ubah dimanipulasiManipulated variable

Daya/berat/jisim

Force/weight/mass

Pemboleh ubah bergerak balasResponding variable

Pemanjangan spring

Extension of a spring

Pemboleh ubah dimalarkanFixed variable

Diameter spring

Diameter of the spring

ProsedurProcedure

1 Tandakan kedudukan awal pin pada pembaris meter apabila tiada pemberat dilekatkan kepada spring, l0.Mark the initial position of the pin on the metre rule when no weight is attached to the spring, l0.

2 Gantungkan pemberat berjisim 50 g di bahagian hujung spring dan bandingkan kedudukan baru pin sekarang dengan kedudukan asalnya.Attach a slotted weight of 50 g to the end of the spring and compare the new position of the pin with its initial position.

3 Ukur pemanjangan spring, x = l – l0./Measure the extension of the spring, x = l – l0. 4 Ulangi eksperimen dengan jisim 100 g, 150 g, 200 g dan 250 g.

Repeat the experiment with mass 100 g, 150 g, 200 g and 250 g.

Merekodkan dataRecording data

Jadualkan data bagi m, F, l dan x./Tabulate data for m, F, l and x.

l0 = cm

Jisim/kgMass/kg

Daya/Force, FF = mg / N l / cm

Pemanjangan/Extension, xx = l – l0 / cm

Plotkan graf daya, F melawan pemanjangan spring, x.Plot a graph of Force, F against extension of the spring, x.

PembarismeterMetre rule

PemberatWeight

Pin/Pin

Spring/Spring

Aktiviti 2: Hubungan antara daya dan pemanjangan springActivity 2: Relationship between force and extension of a spring


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2

UNIT

Menganalisis dataAnalysis data

Daya/Force, F/N

0 Pemanjangan/Extension, x/cm


1 Daripada graf, apakah hubungan antara daya dengan pemanjangan spring, x?From the graph, what is the relationship between a force and the extension of the spring, x?

Daya, F, berkadar langsung dengan pemanjangan spring, x

The force, F, is directly proportional to the extension of the spring, x

2 Kira kecerunan graf. Tunjukkan bagaimana anda mendapat kecerunan daripada graf. Calculate the gradient of the graph. Show how you get the gradient from the graph.

Kecerunan = Daya

Pemanjangan spring Gradient = Force

Extension of spring

3 Langkah berjaga-jaga perlu diambil supaya spring itu tidak diregangkan dengan berat yang berlebihan. Terangkan mengapa.Precaution should be taken so that the spring is not stretched by excessive weights. Explain this.

Jika spring diregangkan dengan berat yang berlebihan, ia mungkin tidak akan

kembali ke panjang asal kerana telah melebihi had kenyal.

If the spring is stretched by too large a weight, it might not return to its original length

due to its exceeding the elastic limit.

Nyatakan Hukum HookeState Hooke’s Law

Pemanjangan spring berkadar terus dengan daya yang dikenakan asalkan

had kenyalnya tidak dilebihi.

The extension of a spring is directly proportional to the applied force provided the

elastic limit is not exceeded.

F = kx di mana/where

F = daya ke atas spring / force on the spring

x = pemanjangan spring / extension of the spring

k = pemalar spring / spring constant

Graf daya-pemanjangan springForce-extension graph

Daya/Force, F/N

Pemanjangan/Extension,x/cm

0

Berdasarkan graf/Based on the graph:• F berkadar terus kepada x. / F is directly proportional to x.

• Kecerunangraf= pemalar spring bagi spring, k.

The gradient of the graph = spring constant of the spring, k.• Luasdibawahgrafadalahsamadengankerjayangdilakukanuntukmemanjangkan

spring = tenaga keupayaan elastik

Area under the graph is equal to the work done to extend the spring = elastic potential energy • Kerjayangdilakukan=tenagakeupayaankenyal Work done = elastic potiential energy ½ Fx = ½ kx2


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2

UNIT

Had kenyal bagi springThe elastic limit of a spring

F/N

0 x (cm)

Had kenyalElastic limit

Berat maksimum yang boleh dikenakan kepada spring supaya spring itu masih

boleh kembali ke panjang asalnya apabila daya dialihkan.

The maximum weight that can be applied to a spring such that the spring will be able

to return to its original length when the force is removed.

Jika daya regangan spring melebihi had kenyal, spring tidak kembali ke panjang asal walaupun tiada lagi daya dikenakan ke atasnya.

If a force stretches a spring over its elastic limit, the spring cannot return to its original length even though the force no longer acts on it.

Hukum Hooke tidak dipatuhi lagi./The Hooke’s Law is not obeyed anymore.

Pemalar daya spring, kForce constant of a spring, k

Spring kerasSti� spring

SpringlembutSoft spring

F (N)

x (m)

75

12.5

0

Pemalar daya spring, k, ialah daya yang diperlukan untuk menghasilkan satu unit pemanjangan spring.

The force constant, k, of a spring is the force required to produce one unit of extension of the spring.

k = Fx

unit: N m-1

k ialah pengukuran kekerasan spring./k is a measurement of the stiffness of the spring.

• Spring dengan pemalar daya,k, yang tinggi adalah lebih susah untuk dipanjangkan dan dikatakan lebih keras.

The spring with a high force constant, k, is harder to extend and is said to be more stiff.

• Spring dengan pemalar daya, k, yang kecil adalah lebih mudah untuk dipanjangkan dan dikatakan lebih lembut atau kurang keras.

A spring with a small force constant, k, is easier to extend and is said to be softer or less stiff.

Faktor-faktor yang mempengaruhi kekenyalan/Factors that affect elasticity

FaktorFactor

Perubahan faktor/Change in factorBagaimana ia mempengaruhi kekenyalanHow it affects the elasticity

PanjangLength

Spring lebih pendek/Shorter spring Kurang kenyal/Less elastic

Spring lebih panjang/Longer spring Lebih kenyal/More elastic

Diameter dawai spring

Diameter of spring wire

Dawai lebih tebal/Thicker wire Kurang kenyal/Less elastic

Dawai lebih nipis/Thinner wire Lebih kenyal/More elastic

Diameter springDiameter spring

Diameter lebih kecil/Smaller diameter Kurang kenyal/Less elastic

Diameter lebih besar/Larger diameter Lebih kenyal/More elastic

Jenis bahanType of material

Spring diperbuat daripada pelbagai bahan. Perubahan kekenyalan mengikut jenis bahan yang digunakan.Springs are made of different materials. Elasticity changes according to the type of material.


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2

UNIT

Panjang asal spring ialah 5 cm. Dengan beban berjisim 20 g, spring memanjang kepada 7 cm.The original length of a spring is 5 cm. With a load of mass 20 g, the length of the spring is extended to 7 cm.Tentukan/Determine(a) pemanjangan spring dengan beban 40 g. the extension of the spring with a load 40 g.(b) panjang spring dengan beban 60 g. the length of the spring with a load of 60 g.

Penyelesaian/Solution(a) 20 g → 7 cm – 5 cm = 2 cm 40 g → 4 cm

(b) 20 g menghasilkan pemanjangan 2 cm 20 g gives an extension of 2 cm ∴ 60 g → pemanjangan/extension 6 cm ∴ panjang spring dengan beban 60 g = 5 cm + 6 cm = 11 cm length of spring with 60-g load = 5 cm + 6 cm = 11 cm

Contoh/Example 1

Susunan spring yang serupaArrangement of identical springs

Secara bersiri/In series

Spring serupaIdentical springs

BebanLoad

Beban yang sama dikenakan kepada setiap springThe same load is applied to each spring.Tegangan dalam setiap spring = TTension in each spring = TPemanjangan spring = xExtension of each spring = xJumlah pemanjangan = 2xTotal extension = 2x

Jika bilangan spring digunakan = n,Jumlah pemanjangan = nxIf number of springs used = n, The total extension = nx

Secara selari/In parallel

Spring serupaIdentical springs

BebanLoad

Beban dikongsi bersama sesama springThe load is shared equally among the springs.

Tegangan dalam setiap spring = T2

Tension in each spring = T2

Pemanjangan setiap spring = x2

Extension of each spring = x2

Jika bilangan spring digunakan = n,If number of springs used = n,

Jumlah pemanjangan = xn

The total extension = xn


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2

UNIT

Spring A memanjang 2 cm apabila ia digantung dengan pemberat 10 g. Spring B memanjang 4 cm apabila ia digantung dengan pemberat 10 g. Cari jumlah regangan pada setiap sistem spring seperti ditunjukkan dalam rajah sebelah.Spring A extends by 2 cm when a 10 g weight is hung on it. Spring B extends by 4 cm when a 10 g weight is hung on it. Find the total extension in each of the spring systems shown in the diagrams on the right.

Penyelesaian/Solution (a) A: 10 g → 2 cm B: 10 g → 4 cm 20 g → 4 cm 20 g → 8 cm Jumlah pemanjangan/Total extension = 4 cm + 8 cm = 12 cm(b) 10 g → 1 cm 50 g → 5 cm Pemanjangan sistem/Extension in the system = 5 cm(c) Sistem B/System B : 10 g → 2 cm ∴ 40 g → 8 cm A : 10 g → 2 cm ∴ 40 g → 8 cm ∴ Pemanjangan sistem/Extension in the system = 8 cm + 8 cm = 16 cm

Contoh/Example 3

A

20 g

50 g

40 g

A A

A

B

B

B

(a) (b) (c)

(a) Panjang asal spring ialah 10.0 cm. Apabila ia diregangkan dengan daya 6 N, ia memanjang kepada 13.0 cm. Apakah pemalar spring?The original length of a spring is 10.0 cm. When it is stretched by a force of 6 N, it extends to 13.0 cm. What is the spring constant?x = 13 cm – 10 cm = 3 cm

k = Fx

= 6 N3 cm

= 2 N cm–1

(b) Dua spring yang serupa disambungkan secara bersiri seperti ditunjukkan dalam rajah (b).Two identical springs are connected in series as shown in diagram (b).

(i) Berapakah jumlah panjang spring jika diregangkan oleh daya 12.0 N?What is the total length of the spring if stretched by 12.0 N force?6 N → 3 cm Jumlah panjang/Total length12 N → 6 cm = 10 cm + 10 cm + 6 cm + 6 cm = 32 cm

(ii) Berapakah pemalar spring bagi sistem spring di (b)?/What is the spring constant of the spring system in (b)?

k = Fx

= 12 N12 cm

= 1 N cm–1

(c) 2 spring yang serupa disambungkan secara selari seperti ditunjukkan dalam rajah (c).2 identical springs are connected in parallel as shown in diagram (c).

(i) Berapakah jumlah panjang sistem spring itu?/What is the total length of the spring system?4 N → 1 cm Jumlah panjang/Total length∴ 12 N → 3 cm = 10 cm + 3 cm = 13 cm

(ii) Berapakah pemalar spring bagi sistem spring di (c)?/What is the spring constant of the spring system in (c)?

k = 12 N3 cm

= 4 N cm–1

Contoh/Example 2Rajah/Diagram(a)

(b)

(c)

F = 6.0 N

F = 12.0 N

F = 12.0 N

13.0 cm


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2

UNIT

Panjang asal spring ialah 12 cm. Dengan beban 20 g, panjang spring dipanjangkan kepada 15 cm. Berapakah tenaga keupayaan kenyal yang tersimpan di dalam spring?The original length of a spring is 12 cm. With a load of 20 g, the length of the spring is extended to 15 cm. What is the elastic potential energy stored in the spring?


E = 12

Fx

= 12

× (0.02 kg × 10 m s–2) × 0.03 m

= 0.003 J

Rajah menunjukkan graf daya, F, melawan pemanjangan, x, bagi spring. Berapakah tenaga keupayaan kenyal yang tersimpan apabila spring diregangkan sebanyak 0.4 m?The diagram shows a graph of force, F, against extension, x, for a spring. What is the elastic potential energy stored when the spring is extended by 0.4 m?


E = 12

Fx

= 12

(20 N) (0.4 m)

= 4.0 J

Rajah menunjukkan sebiji bebola keluli berjisim 10 g ditolak pada penghujung satu spring di sepanjang permukaan licin. Panjang asal spring ialah 14 cm dan pemalar spring ialah 200 N m-1.The diagram shows a steel ball of mass 10 g being pushed against one end of a spring along a smooth surface. The original length of the spring is 14 cm and its spring constant is 200 N m-1.

14 cm 10 cm

DayaForce

Tentukan/Determine(a) tenaga keupayaan kenyal tersimpan di dalam spring./the elastic potential energy stored in the spring.

E = 12

kx2

= 12

(200 N m–1) (0.04 m)2

= 0.16 J

(b) halaju maksimum bola itu selepas daya mampatan pada spring itu dialihkan.the maximum velocity of the ball after the force of compression on the spring is removed.

12

mv2 = 0.16 J

v2 = 2 × 0.16 J

0.01 kg = 32 m2 s–2

v = 5.66 m s–1

Contoh/Example 4

Contoh/Example 5

Contoh/Example 6

20

00.4

F/N

x/m

Perhatian/Note: J

kg =

Nmkg

= (kg m s–2)(m)

kg = m2 s–2


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2

UNIT

Latihan/Exercises

Rajah di bawah menunjukkan susunan radas dalam eksperimen untuk menentukan hubungan di antara pemanjangan spring, x, dengan berat, W. Hubungan di antara x dengan W ditunjukkan dalam graf di bawah.The diagram shows the arrangement of an apparatus in an experiment to determine the relationship between the extension, x, of the spring with weight W. The relationship between x and W is shown in the graph below.

0

1

2

3

4

5

0 42 6 8 14

6

7

1210

SpringSpring

Pemanjangan, x/cmExtension, x/cm

PemberatberslotSlotted mass

Berat, W/NWeight, W/N

(a) (i) Nyatakan hubungan antara x dan W.State the relationship between x and W.

x berkadar langsung dengan W, asalkan had kenyal tidak dilebihi.

x is directly proportional to W, provided the elastic limit is not exceeded.

(ii) Namakan hukum saintifik yang terlibat dalam hubungan yang dinyatakan di (a)(i).Name the scientific law involved in the relationship stated in (a)(i).

Hukum Hooke/Hooke’s Law

(b) Tandakan dengan tanda pangkah (×) had kenyal spring di dalam graf. Mark a cross (×) at the elastic limit of the spring on the graph.

(c) Berdasarkan graf, tentukan pemalar daya bagi spring, k (dalam unit N m–1). Based on the graph, determine the force constant of a spring, k (in N m–1).

k = 10 N

0.04 m = 250 N m–1

(d) Tenaga keupayaan kenyal tersimpan dalam spring apabila ia diregangkan. Kirakan tenaga ini dalam spring apabila ia memanjang sebanyak 4 cm.Elastic potential energy is stored in the spring when it is extended. Calculate this energy in the spring when it is extended by 4 cm.

E = 12

Fx

= 12

(10 N) (0.04 m)

= 0.2 J


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