C Satah condong/Inclined plane Fizik Tg4 B2C 2015(FSY4p) n.pdf · W –1= 3 kg × 10 N kg × 0.4 m = 12 J 25 N 30 N Contoh/Examples 1 Contoh/Examples 2 ... Tenaga bunyi/Sound energy

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© Nilam Publication Sdn. Bhd.113

MODUL • Fizik TINGKATAN 4

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 kerjaDefinition 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 Satahcondong/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

MEMAHAMIKERJA,TENAGA,KUASADANKECEKAPANUNDERSTANDING WORK, ENERGY, POWER AND EFFICIENCY2.10

Fizik Tg4 B2C 2015(FSY4p).indd 113 10/20/15 2:10 PM

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© Nilam Publication Sdn. Bhd. 114


PengiraankerjaCalculation 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 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

600

50 N

Contoh/Examples 3

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)

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

25 N

30 N

Contoh/Examples 1 Contoh/Examples 2


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Tiadakerjadilakukanapabila:/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.

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?

Penyelesaian/Solution

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 msin 30° )

= 512.0 J

Latihan/Exercises


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

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

Definisi tenaga keupayaan graviti/Define gravitational potential energy

Seorang budak perempuan membuat kerja apabila dia memanjat tangga sebuah papan gelongsor. Dia mempunyai tenaga keupayaan graviti 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 graviti 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)

Contoh/Examples

BeratWeight

PendulumBob

KotakBox

mm

m

hh

h

BeratWeight

PendulumBob

KotakBox

mm

m

hh

h

BeratWeight

PendulumBob

KotakBox

mm

m

hh

h


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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, ggravitational 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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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.

Penyelesaian/SolutionKerja = Tenaga kinetik = 1

2 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.

Contoh/Examples


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Aktiviti:KeabadianTenaga/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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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)


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 jatuhBefore 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)Contoh/Examples


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Menerangkan kecekapan peralatanExplain what efficiency of a device is

Kecekapan = Kuasa outputKuasa input × 100%

Efficiency = Output powerInput 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

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)Contoh/Examples

KBAT


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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

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 = 1

2 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 mvA

2 = mgh + 12 mB

2

12 (6 m s–1)2 = (10 m s–2)(1.5 m) + 1

2 vB2

12 vB

2 = (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)Latihan/Exercises

KBAT

KBAT

KBAT


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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?


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?



25% = Kuasa output

3 000 W × 100%

Kuasa output = 25%100% × 3 000 W

= 750 W

Efficiency = Output powerInput 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

S


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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?

Penyelesaian / Solution:


= 500 kg × 10 m s–2 × 12 m

8 s 45 000 W

× 100%

= 7 500 W45 000 W × 100%

= 16.67%

Efficiency = Output powerInput 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. Berapakah halaju pelajar itu sejurus sebelum dia menyentuh tanah? / 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. What is the velocity of the student just before he touches the ground?

Penyelesaian / Solution: 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

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. Diberi bahawa jisim Abu dengan basikalnya ialah 75 kg, cari / 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. Given 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.

Penyelesaian / Solution:

(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

12 m

2.5 m

u = 2 m s–1

3 m

v = 6 m s–1

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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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MEMAHAMIKEKENYALANUNDERSTANDING ELASTICITY2.11

Aktiviti1:Ideatentangkekenyalan/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 antarasatu 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 antarasatu 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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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, x

x = 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

Aktiviti2:HubunganantaradayadanpemanjanganspringActivity 2: Relationship between force and extension of a spring


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Menganalisis dataAnalysis data

Daya/Force, F/N

0 Pemanjangan/Extension, x/cm

PerbincanganDiscussion

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.

• Luasdibawahgrafadalahsamadengankerjayangdilakukanuntukmemanjangkanspring= 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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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.

• Springdenganpemalardaya,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.

• Springdenganpemalardaya,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 kekenyalanFactors 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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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

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

Contoh/Examples 1


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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

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 N

3 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 N

12 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

Rajah/Diagram(a)

(b)

(c)

F = 6.0 N

F = 12.0 N

F = 12.0 N

13.0 cm

Contoh/Examples 2

Contoh/Examples 3

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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

Contoh/Examples 4

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

20

00.4

F/N

x/m

Contoh/Examples 5

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 J0.01 kg

= 32 m2 s–2

v = 5.66 m s–1

Perhatian/Note: J

kg = Nmkg

= (kg m s–2)(m)

kg = m2 s–2

Contoh/Examples 6

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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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Latihan/Exercises


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1 Rajah 1 menunjukkan troli diletakkan di atas landasan terpampas geseran. Diagram 1 shows a trolley placed on a friction-compensated runway.

Rajah 1 / Diagram 1

Troli itu akan / The trolley willA bergerak dengan halaju bertambah. move with an increasing velocity.B bergerak dengan halaju berkurang. move with decreasing velocity.C bergerak dengan halaju malar/tetap. move with a constant velocity.D kekal tidak bergerak. remain stationary.

2 Tangki silinder minyak kereta En. Ali bocor. Tompokan minyak kelihatan sepanjang jalan dari A ke B seperti ditunjukkan dalam Rajah 2.

The cylinder oil tank of En. Ali’s car is leaking. Lubricant spots are seen along the road from A to B as shown in Diagram 2.

10 m

A B

15 m 20 m

Rajah 2 / Diagram 2

Ini menunjukkan bahawa kereta Ali bergerak denganIt can be concluded that En Ali’s car is moving withA pecutan malar. / constant acceleration.B nyahpecutan malar. / constant deceleration.C pecutan sifar. / zero acceleration.D pecutan berkurang dengan malar/tetap. acceleration that is decreasing constantly.

3 Antara pernyataan berikut, yang manakah menghuraikan perlanggaran kenyal antara dua objek dengan betul? / Which of following statements best describes an elastic collision between two objects?A Hanya mementum terabadi sahaja. Only momentum is conserved.B Hanya tenaga kinetik terabadi sahaja. Only kinetic energy is conserved.C Kedua-dua momentum dan tenaga kinetik

terabadi. Momentum and kinetic energy are conserved.D Jumlah tenaga, momentum dan tenaga kinetik

terabadi. / Total energy, momentum and kinetic energy are conserved.

4 Rajah 4 menunjukkan graf halaju-masa untuk Adnan, Hamid dan Lim sepanjang perjalanan di jalan lurus mengikut urutan.

Diagram 4 shows the velocity-time graphs of Adnan, Hamid and Lim respectively travelling along a straight road.

v/m s–1

t/s

Adnan6 6

6

4 4

10 2 3 4

v/m s–1

t/s

Hamid

10 2 3 4

v/m s–1

t/s

Lim

10 2 3 4

Rajah 2 / Diagram 2

Jika mereka bertiga memulakan perjalanan dari tempat yang sama dan masa yang sama, susun nama mereka dalam urutan menaik berdasarkan sesaran mereka dari titik permulaan. / If they started their journey from the same place at the same time, arrange their name in accordance to increasing order of their displacements from the starting point.A Adnan, Hamid, LimB Adnan, Lim, HamidC Hamid, Lim, AdnanD Lim, Adnan, Hamid

5 Rajah 5 menunjukkan sebiji durian jatuh dari sebatang pokok.Diagram 5 shows a durian falling from a tree.

Rajah 5 / Diagram 5

Apakahkuantitifizikyangditetapkansemasadurianitu jatuh? / What is the physical quantity that is constant when the durian falls?A Halaju / VelocityB Momentum / MomentumC Pecutan / AccelerationD Tenaga kinetik / Kinetic energy

Latihan Pengukuhan/Enrichment Exercises


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6 Sebiji bola dilontar ke atas pada ketinggian h m dari tanah. Antara berikut, yang manakah betul tentang gerakan bola itu?A ball is thrown upwards to a height of h m from the ground. Which of the following is true about the motion of the ball?A Tenaga kinetik adalah malar sepanjang gerakan. Its kinetic energy is constant throughout the motion.B Tarikan graviti pada ketinggian h adalah sifar. The gravitational attraction at the height h is zero.C Masa yang diambil untuk mencapai ketinggian

h meter adalah sama dengan masa yang diambil untuk jatuh dari ketinggian h meter ke tanah.

The time taken to reach a height of h metres is the same as the time taken to fall from a height of h metres to the ground.

7 Seorang budak lelaki berjisim 50 kg berada di dalam lif. Berat badannya mencapai 500 N apabilaA boy of mass 50 kg stands in a lift. His weight reaches 500 N when

(g = 10 N kg–1)A lif bergerak ke atas dengan halaju malar. the lift moves upwards with a constant velocity.B lif bergerak ke bawah dengan pecutan seragam. the lift moves downwards with a constant acceleration.C lif itu jatuh bebas. the lift falls freely.

8 Rajah 8 menunjukkan seekor kucing berjisim 2.5 kg menaiki tangga setinggi 3.0 m dalam masa 1.2 s.Diagram 8 shows a cat with mass 2.5 kg running upstairs to a height of 3.0 m in 1.2 s.

3.0 m

Rajah 8 / Diagram 8

Berapakah kuasa yang dijanakan oleh kucing itu?How much power is generated by the cat?

A 6.3 WB 7.5 WC 9.0 WD 62.5 W

9 Antara berikut, yang manakah bukan ciri-ciri keselamatan yang dipasang pada kenderaan?Which of the following is not a safety feature installed in a vehicle?A Kapasiti enjin. Engine capacity.

B Beg udara automatik. Automatic air bag.C Roda stereng boleh lipat. Collapsible steering wheels. D Cermin kaca tahan air. Shatterproof windscreen glass.

10 Rajah 10 menunjukkan alas bebola, P, yang berjisim 50 g dilepaskan atas satu landasan licin.Diagram 10 shows 50 g ball bearing P, being released on a smooth plane.

1.6 mX

Alas bebola PBall bearing P

Landasan licinSmooth plane

Rajah 10 Diagram 10

Berapakah halaju alas bebola itu di X?What is the velocity of the ball bearing at X?A 1.60 m s–1

B 4.00 m s–1 C 5.66 m s–1

D 8.00 m s–1

11 Suatu beban 800 g digantung pada penghujung spring seperti yang ditunjukkan dalam Rajah 11. Panjang asal spring itu ialah 6.0 cm.A load of 800 g hangs at one end of a spring as shown in Diagram 11. The original length of the spring is 6.0 cm.

10.0 cm

800 g

Rajah 11 Diagram 11

Tenaga keupayaan kenyal yang tersimpan dalam spring ialahThe elastic potential energy stored in the spring isA 0.16 JB 1.60 JC 16.00 JD 1 600.00 J


2

UNIT



1 Rajah 1.1 menunjukkan sebuah bas yang bergerak dengan halaju 70 km/j dan menunjukkan kedudukan penumpang di dalam bas apabila brek tiba-tiba digunakan. Rajah 1.2 menunjukkan tukang masak cuba untuk berjabat sos daripada botol.Diagram 1.1 shows a bus moving with a velocity of 70 km/h and shows the condition of a passenger standing in the bus when the brakes are suddenly applied. Diagram 1.2 shows a cook trying to shake the sauce out of a bottle.

Rajah 1.2 / Diagram 1.2Rajah 1.1 / Diagram 1.1

(a) Berdasarkan kedua-dua rajah di atas, nyatakan satu ciri yang dialami oleh penumpang dan sos. Based on both of the diagrams above, state one characteristic experienced by the passengers and the sauce.

Penumpang bergerak ke hadapan dan sos cili mengalir keluar daripada botol. The passengers move forward and the chilli sauce comes out from the bottle.

(b) Apakah halaju penumpang / What is the velocity of the passenger (i) sebelum pemandu bas membrek? / before the bus driver brakes?

70 km/j / 70 km/h

(ii) selepas pemandu bas membrek? / after the bus driver brakes?

Halaju yang sama iaitu 70 km/j. / The same velocity that is 70 km/h.

(c) Apakah halaju botol apabila sos mengalir keluar? What is the velocity of the bottle when the sauce comes out?

Halaju sifar. / Zero velocity.

12 Rajah 12 menunjukkan seorang lelaki menarik sebuah blok dengan daya 20.0 N.Diagram 12 shows a man pulling a block with a force of 20.0 N.

Blok30°

20.0 NG

Block

Rajah 12 / Diagram 12

Daya komponen mencancang, G, yang terhasil ialahThe vertical component force, G, that is produced is

A 6.7 NB 10.0 NC 17.3 ND 20.0 N

13 Rajah 13 menunjukkan dua beban yang disambung dengan tali tak kenyal melalui takal licin.Diagram 13 shows two loads connected by an inelastic string through a frictionless pulley.

Takal licinFrictionless pulley

Tali tak kenyalInelastic string

BebanLoad

4 kg

6 kg

BebanLoad

Rajah 13 / Diagram 13

Pecutan beban 6 kg itu ialahThe acceleration of the 6 kg load is

A 2.0 m s–2 C 6.7 m s–2

B 5.0 m s–2 D 20.0 m s–2

Soalan Struktur/Structure Questions


2

UNIT



(d) (i) Berdasarkan jawapan (b)(i), tulis satu kenyataan pergerakan bas dan penumpangnya. Based on the answer to (b)(i), write a statement on the motion of the bus and its passengers.

Apabila bas brek, halaju bas berkurangan dengan tiba-tiba, penumpang terus bergerak dengan halaju awalnya.

When the bus brakes, the velocity of the bus decreases suddenly, the passenger continues to move with its initial velocity. (ii) Namakankonsepfizikyangterlibat./Name the physics concept involved here.

Inersia / Inertia

(e) Jika anda dikejar oleh anak gajah, bagaimana anda berlari untuk menyelamatkan diri? Terangkan. If you are chased by a baby elephant, how can you run to save yourself? Explain.

Sayakenaberlari secarazigzag.Anakgajahakan terusbergerakdalamgaris lurusapabila sayamenukar

arah saya dengan tiba-tiba. / I have to run in a zig zag manner. The baby elephant will continue to move in a straight

line when I change my direction suddenly.

2 Satu objek A dengan jisim 40 kg ditarik oleh objek B dengan jisim 80 kg di atas sebuah landasan condong PQ licin seperti yang ditunjukkan dalam Rajah 2. Kedua-dua takal adalah licin.Object A of mass 40 kg is pulled by object B of mass 80 kg up a frictionless inclined plane, PQ, as shown in Diagram 2. The pulleys are frictionless.

8 m

10 m

40 kg

θA

80 kg

B

R Q

P

Rajah 2 / Diagram 2

(a) Hitung kerja yang dilakukan oleh objek B ke atas objek A apabila digerakkan 5 m di sepanjang satah condong PQ. / Calculate the work done by object B on object A when it moves 5 m along the inclined plane, PQ.

(b) Hitung daya paduan apabila objek B menarik objek A. Calculate the resultant force when object B pulls object A.

(c) Hitung pecutan objek B. Calculate the acceleration of object B.

Pecutan B = Pecutan AAcceleration of B = Acceleration of A

= 560 N———–120 kg = 4.67 m s–2

Kerja / work done = mgh = 40 kg × 10 m s–2 × 3 m = 1 200 J

Daya yang bertindak ke atas B / Force acting on B = 80 kg × 10 m s–2 = 800 NDaya yang bertindak ke atas A / Force acting on A = 40 kg × 10 m s–2 × sin θ = 400 N × 0.6 = 240 N Maka, daya paduan / Therefore, resultant force = 800 N – 240 N = 560 N


Documents

C Satah condong/Inclined plane Fizik Tg4 B2C 2015(FSY4p) n.pdf · W –1= 3 kg × 10 N kg × 0.4 m = 12 J 25 N 30 N Contoh/Examples 1 Contoh/Examples 2 ... Tenaga bunyi/Sound energy