Sains Fizik

SCIENCE APPLIED MODULE 1: RATE AND RESISTANCE 1.1 RATE & RESISTANCE IN MECHANICAL SYSTEM. LINEAR MOTION (Gerakan Linear) A linear motion describes movement of an object in a straight line. E.g : 1) A student running a 100m sprint.

2) An Aeroplane flying across the sky

Displacement (Sesaran) Suppose a student walks from his house (labelled H) to a shop (labelled S) in a straight line as shown below:

H

100m

S

The length between H and S is 100m, then we say that displacement of student is 100m from H to S : Displacement , SHS = 100m, in the HS direction Linear Velocity (Halaju Linear) Define as : Linear velocity = Rate of displacement = Displacement Time v = _s_ t where, v = linear velocity in meter per second (ms-1) s = displacement in meter (m) t = time taken in seconds (s) E.g. 1 : Student walking from his house to the shop. If the student takes 50 seconds to walk the distance.

Since, we know that his displacement is 100m. Then, Linear velocity = v

= Rate of displacement Displacement Time = _s_ t

= 100m 50s = 2.0 ms-1 in the HS direction

E.g. 2 :

An aeroplane travels in a straight line with a velocity of 750 km h-1. How far will it travel in 3 hours? Since

v

Then

s = =

=

t

_s_

= v x t 750 km h-1 x 3 h 2250 km

ROTATIONAL MOTION (Gerakan berputar) A rotational motion describes movement of an object in a rotation. E.g : 1) The rotating blades of a fan.

2) The rotating wheel of a car

3) The rotating blades of food blender.

Angular Displacement (Sesaran Berputar) In rotational motion, we use angle to describe the displacement of the rotating object. Refer to Handout 1. (Look for the diagram) The measurement of the angle that the rotating object has gone through is called angular displacement, that is: θ

=

angular displacement

Angular displacement can be measured in:

360◦ = 2π radian

=1 revolution =1 cycl

=1 turn

Angular Velocity (Halaju Berputar) Angular velocity of an object is the rate of angular displacement or how fast the object rotates. Define as : Angular velocity ω

= =

t

Angular Displacement Time θ

where,ω = angular velocity in radian per second (rad s-1) θ = angular displacement in radian (rad) t = time in seconds (s) E.g. 2 : Referring to Diagram 3, suppose it takes 0.1 s for the point on the disc to make one complete turn. What is the angular velocity of the disc? Angular velocity ω = = =

= =

t

Angular Displacement Time θ

2π rad 0.1 s 2 x 3.14 rad 0.1 s 62.8 rad s-1

E.g. 2 : Calculate the angular velocity of: a) a wheel turning 6500 revolutions in 4 s. b) a shaft on a motor rotating 5600 revolutions in 1.5 minutes. a)

b)

a wheel turning 6500 revolutions in 4 s. ω = θ t = (6500 x 2π) rad 4 s = 10210 rads-1 a shaft on a motor rotating 5600 revolutions in 1.5 minutes. ω = θ t = (5600 x 2π) rad (1.5 x 60) s = 391 rads-1

Resistance in Mechanical Systems (Rintangan dalam Sistem Mekanikal) Resistance opposes motion.

Resistance in mechanical systems is called friction. It’s happen when the surfaces of 2 solid objects slide against one another. Friction is absolutely necessary in life. Without friction, we may not be able to sit on a chair, write on papers or even wear our clothes. Importance : 1) enables to turn pulleys, 2) brakes to stop and tyres to grip the roads.

The Brake System (Sistem Brek) Most cars have disc brakes on the front wheels. Some have disc brakes on all four wheels. The Basic Parts of a Disc Brake The main parts of a disc brake are : (a) the brake pads (b) the calliper, which contains a piston. (c) the rotor, which is mounted to the hub (the part where the wheel is attached)

Reducing Resistance in Mechanical Systems (Pengurangan Rintangan dalam Sistem Mekanikal) When 2 surfaces of solids rub against each other, heat is produced. e.g : When the piston in the engine of a car moves up & down, heat is produced and the engine will get hot. The surface of the piston will wear out after some time.

To reduce the surface wear and the heat caused by friction, parts of engine

(bearings, gears, chains etc) require lubrication. Lubricant have been proven to provide:

Using bearings can also reduce friction in machines. Bearing is a ball or roller that is placed between surfaces. When correctly lubricated & not subjected to extreme operational conditions, bearing can last longer.

1.2

RATE AND RESISTANCE IN ELECTRICAL SYSTEM

FREQUENCY, PERIOD & PEAK VOLTAGE FOR ALTERNATING CURRENT (Frekuensi, Tempoh & Voltan Puncak bagi Arus Ulang-Alik) VOLTAN/V

Figure 1 : Alternating current signal

VOLTAN/V

Figure 2 : Positive voltage AC VOLTAN/V

VOLTAN/V

Figure 4 : Alternating current changes direction in one cycle VOLTAN/V

Figure 5 : Alternating current changes direction 100 times in 50 cycles 50 cycles a second or 50 Hertz is a measurement of a quantity called frequency. Frequency is the number of cycles in a second. To calculate frequency, we need to know the number of cycles & the time it takes to complete the cycles Frequency

=

Number of cycles time taken n t

f =

Where f = frequency in cycles per second or Hertz (Hz) n = number of cycles t = elapsed times in seconds .

Example 1: A commercial generator produces an electrical signal that has 240 cycles in 4 s. Find the frequency of the signal. Solution: n = 240 cycles f = n

and

t = 4 s

t 240 cycles 4 seconds = 60 cycles seconds = 60 Hz =

The time taken for one cycle is called the period, T VOLTAN/V

Figure 5 : Alternating current changes direction 100 times in 50 cycles From Figure 5, if 50 cycles are made in 1 s, then 1 cycle will take 1/50 of a second. Therefore, the period T

is 1/50 s or 0.02 s.

The relationship between frequency, f is given by the following equation : f therefore

T

=

1

and period T =

1 T

f Where f = frequency in Hertz (Hz) T = period in seconds (s) Example 2 : What is the period for the commercial generator describe in Example 1? Solution:

T

=

1 f

=

1

= 0.017 s

60 Hz

MODULE 2: ENERGY- Defined as the ability to do work 2.1 ENERGY IN MECHANICAL SYSTEM POTENTIAL ENERGY (Tenaga Keupayaan) In Mechanical Systems consist two potential energy: 1)

Gravitational Potential Energy (Tenaga Keupayaan Graviti)

The gravitational force causes things to fall down. The higher object is brought up from the ground, the greater is the potential energy. The bigger the mass of the object, the more the potential energy. An object on the floor does not have potential energy no matter how big it’s mass. It does not have the potential to fall.

Figure 1: Stone on a hill; an example of gravitational potential energy. The formula for g.p.e is : Potential Energy E

= =

Mass x Acceleration due to gravity x Vertical height m x g x h

where, E = g.p.e in joule (J) m = mass of the object in kilogram (kg) g = acceleration due to the earth gravitational attraction (g = 10 ms-2) h = the vertical height of the object from a references level in meter (m) * g.p.e = gravitational potential energy E.g. 1 : A pile driver shown in Figure 2 is dropped from a height to hit the piling rod, so that it gets buried to a certain depth. Calculate the potential energy of

the 50 kg pile driver if it is dropped from heights: a) 2 m b) 3 m

Figure 2: A pile driver is dropped to hit the piling rod Solution : a) E

= m g h = 50kg x 10ms-2 x 2m = 1000 J

b) E

= m g h = 50kg x 10ms-2 x 3m = 1500 J

E.g. 2 : 5000 kg mass of water is at the top level of a damn, which is 50 m high above the river below. Calculate the g.p.e of the water. Solution : E

= = = =

m g h (5000 kg) x ( 10 N kg-2) x (50 m) 2 500 000 J 2500 kJ

Importance of GPE: Helps piling to go very deep into the earth. Allows us to build strong & tall structures. Water falling from hills & mountains in hydroelectric power.

2)

Elastic Potential Energy (Tenaga Keupayaan Elastik / Kenyal)

Elastic means that it has a tendency to return to it’s original shape whenever it is compressed or stretched. Examples : springs, rubber bands etc. EPE is the energy that an elastic object has when it is compressed or stretched. The more we compress/stretch the material, the more the potential energy is. The spring constant = The force you need to stretch/compress the spring Not all spring have the same spring constant The use spring depends in the spring constant. * EPE = Elastic Potential Energy

Photograph 1: Gadgets using elastic potential energy; (a) a spring for exercise (b) a playground toy using spring (c) rubber band toy cars *A stiff spring has a large spring constant, whereas a soft spring has a low spring constant. The formula for e.p.e is : Elastic Potential Energy E

=

½ x (Spring constant) x (Length of spring stretched/compressed) = ½ x k x x2

where, E = e.p.e in joule (J) k = the spring constant (N m-1) x = the length of spring stretched/compressed in meter (m) * e.p.e = elastic potential energy

E.g. 3 : A rubber band wound around the shaft of a toy car is stretched 3 cm when the wheel is rolled backward. How much e.p.e exists in the spring if the rubber band stiffness constant is 40 N m-1? Solution : Note that x must be in meter, 1 m = 100 cm: e.p.e of the rubber band: E

= ½ k x2 = ½ x (40 N m-1) x ( 0.03 m)2 = 0.018 J

Photograph 2: A rubber band wound around the shaft of a toy car. E.g. 4 : The rear spring of a car has a spring constant of 100 000 N m-1. After being filled with some loads, the spring gets compressed to 10 cm. Calculate the potential energy stored in such a spring.

Figure 3: The suspension system of a car Solution : E

= ½ k x2 = ½ x (100 000 Nm-1) x (0.1m)2 = 500 J

Importance of EPE: Useful in valves, car part such as the brake, suspension & pressure regulator. KINETIC ENERGY (Tenaga Kinetik) Object is in motion, it has kinetic energy. This motion can either be linear or rotational. 1)

Translational Kinetic Energy (Tenaga Kinetik Linear)

Kinetic Energy object)2 E where E

= m v

= =

½ x Mass of moving object x (velocity of moving ½ x m x v2

kinetic energy in joule (J) = mass of the moving object in kilogram (kg) = velocity of the object in meter per second (m s-1)

* TKE = Translational Kinetic Energy E.g. 5 :

(Refer to e.g 1)

When a pile driver is released from a certain height, the velocity when it hits the piling rod is 8 ms-1. Calculate the kinetic energy of the pile driver when it hits the piling rod.

Solution : E

= ½ m v2 = ½ x (50 kg) x (8 ms-1)2 = 1600 J = 1.6 kJ If the pile driver hits the piling rod at a velocity twice the above, calculate its kinetic energy. Twice the velocity means, v = 16 ms-1 Solution : E

= ½ m v2 = ½ x (50 kg) x (16 ms-1)2 = 6400 J = 6.4 kJ

Note that the gravitational potential energy stored in the pile driver has been changed to kinetic energy as it falls onto the piling rod. Rotational Kinetic Energy (Tenaga Kinetik Berputar) Rotational Kinetic Energy E where E meter (rad s-1)

= I ω

= =

½ x Moment of inertia x (Angular velocity)2 ½ x I x ω2

kinetic energy in joule (J) = moment of inertia of the moving object in kilogram (kg m2) = angular velocity of the object in radian per second

* RKE = Rotational Kinetic Energy In a linear motion: The larger the mass, the bigger the translational kinetic energy. In a rotating motion: The larger the moment inertia, the larger the rotational kinetic energy.

Photograph 3: An exercise bicycle, a top and a fan blade Table 1: Comparison between translational and rotational quantities Translational Rotational Mass, m Moment of inertia, I Linear displacement, s Angular displacement, θ Linear velocity, v Angular velocity, ω

Photograph 4: A top, a rotating wheel and a yo-yo Moment of inertia depends on the size and shape of the object. A thin object (small radius) has a small moment of inertia compared to a wide object (big radius)

An object rotated about a different axis causes it to have a different moment of inertia

Answer the question below : 1)

A flywheel is simply a rotating disc. If the moment of inertia of the flywheel is 100 000 kgm-2. Calculate the kinetic energy of the flywheel rotating at 10 rad s-1.

2.2 ENERGY IN ELECTRICAL SYSTEM We will study how energy is in stored in two important electrical devices: 1) Capacitors

: Function : Stores electrical energy

2) Inductors Function : Stores electrical energy Function : Stores electrical energy by creating voltage

Energy Stored in a Capacitors (Tenaga disimpan dalam Kapasitor) A Capacitor consists of 2 parallel conductors separated by an insulator. Once the 2 plates of a capacitor are connected to a battery, charges will accumulate at the plates of the capacitor. This creates a voltage different (V) across the plates of the capacitor. When the voltage across the plates is equal to the voltage across the battery, the flow of current will stop. Connecting a capacitor to a battery is just like stretching a spring. If a spring stores Elastic PE when it stretched, a capacitor stores Electrical PE when the battery causes a voltage difference between its plates * PE = Potential Energy The formula for the energy stored in a capacitor is : Energy stored in a capacitor = ½ x Capacitance of the capacitor x (Voltage different

E

=

between the plates of the capacitor)2

½ x C x V2

where, E = energy stored in joule (J) C = capacitance of the capacitor in farad (F) V = voltage used to charge the 2 plates of the capacitor in volt (V) Capacitance Capacitance is the quantity that tell us the amount of charge that can be stored across the plates of the capacitor when it is connected to a certain voltage. The unit for capacitance = farad (F) A 1-farad capacitor is a very big. Normally capacitors are measured in microfarads or picofarads. 1 microfarad =

__ _1____ farad = 1 x 10-6 (written as µF) 1 000 000

1 picofarad = __ _ 1____ microfarad = 1 x 10-12 (written as pF) 1 000 000 E.g. 1 : A camera flash circuit boosts the voltage supplied by the battery to 200 V. If the energy needed to light up an electronic flash is 3 J, what is the capacitance of the capacitor that supplies the charge needed? Solution : Energy stored in capacitors is given by E = ½ CV2 For V

= 200 V, 3 J = ½ x C x (200)2 = 1.5 x 10-4 F = 150 µF

Energy Stored in an Inductor (Tenaga disimpan dalam Induktor) An inductor is simply a coil of wire. Symbol :

It is stores energy by creating a voltage. This voltage oppose any change in current in a circuit. The circuit voltage The circuit voltage

, the inductors develops in the opposite direction. , the inductor adds voltage to the circuit.

This helps in maintaining a constant voltage across the inductor

Energy stored by an inductor is : Energy stored in an inductor = ½ x Inductance x (Current flowing)2 E = ½ x L x I2 where, E = energy stored in joule (J) L = inductance of the inductor in henry (H) I = current flowing through the inductor in ampere (A) Inductance Inductance is how much voltage difference develops across an inductor when current charges in the circuit. The unit for inductance = henry (H) 1-henry (1 H) = the inductor will develop a voltage of 1 V whenever there is a change of current at the of 1 A in 1 S. E.g. 2 : An inductor with the inductance of 10 H is connected to a car battery. Calculate the energy stored by the inductor if the current flowing is 1 A. Solution : Energy stored in the inductor is given by E = ½ LI2 For L = 10 H,

and

I = 1A

E = ½ x L x I2 = ½ (10)(1)2 = 5 J Answer the question below : 1) The amount of energy stored in an inductor is 10 J. if the is 0.8 H, how big is the current flowing through it?

inductance

2.3 ENERGY IN THERMAL SYSTEM To heat up something means supplying heat energy so that is temperature increases. Heat moves from hot objects to cold objects in contact. The hotter object loses the energy, while the cooler object gains heat energy. As long as there is difference in temperature between the 2 objects, heat will transferred.

Photograph 1 : Transfer of heat from hot to cold objects 3 factors affecting the amount of heat transferred: (3 faktor mempengaruhi kadar pemindahan haba) (a) Mass of the object involved. The bigger the mass of an object, the longer it takes to heat up. (b) Types of the material that receive & lose the energy. Objects of different materials behave differently towards heat. 1 kg of H2O takes longer time to heat compared to 1 kg of iron. (c) The increase or decrease in temperature. A cup of boiling water takes a longer time to get cooled off compared to a cup of warm H2O. The amount of heat energy transferred is given by :

where, Q = heat energy transferred (lost or gained) in joule (J) m = mass of the object in kilogram (kg) c = specific heat capacity of the object in joule per kg per degree kg-1 0C-1) θ = change in temperature of the object in degree Celsius (oC)

Celsius (J

Specific Heat Capacity (Muatan Haba Tentu) Define as quantity of heats are need to increase heat of an objects.

Figure 1 : Humans and plants are made up of cells. The water content ranges between 70% to 80% within a single cell. Photograph 2 : Earth is also comprised mainly of water Material Specific Heat Capacity (J kg-1 0C-1) Aluminum Copper Glass Gold Water Ice Iron Mercury Silver Human Body Wood 900 400 840 130 4200 2100 450 140 230 3470 1700 Table 1 : List of common materials & their specific heat capacity. E.g. 1 : By using the specific heat capacity from Table 1, calculate the total heat needed to increase the temperature of 1 kg of water from 300C to 1000C. Solution : Initial water temperature Final water temperature =

=

300C 1000C

Change in temperature, θ = Specific heat capacity of water, c = Mass of water =

700C 4200 J kg-1 0C-1 1 kg

Q = m x c x θ Q = 1 kg x 4200 J kg-1 0C-1 x 70oC = 294 000 J = 294 kJ Heat Energy Transferred between 2 Objects at Different Temperature (Pemindahan Tenaga Haba antara 2 Bahan pada Berlainan Suhu) When 2 objects at different temperatures are contact with each other, heat will be transferred from the hot object to the cold object. The hot object will lose heat while the cool object will gain heat. Photograph 3 : Oil refinery Thermal Energy at the workplace

Photograph 4 : A cold Storage room, Thermometer & Pyrometer

MODULE 3: POWER 3.1 POWER IN MECHANICAL SYSTEM POWER IN LINEAR MECHANICAL SYSTEM (Kuasa dalam Sistem Mekanikal Linear) The work done to move an object in a straight line E.g : pushing a cart in a straight line.

Therefore, in linear mechanical systems, Power is the rate of moving something in a straight line.

Photograph 1 : A crane in a factory

Photograph 2 : Ploughing machine

Photograph 3 : A water pump Power (Kuasa) Define as : Power =

work done @ energy consumed per unit time = Work done Time P = _W_ t

where, P = power in watt (W) or joule per second ( Js-1 ) W = work done in joule ( J ) t = time taken to do work in seconds (s) Linear Mechanical Power (Kuasa Mekanikal Linear) Define as : Power

= P

=

t

Linear mechanical work Time _W_

since, Work = Force x Distance W = F x s where, F = force applied s = distance along which the force acts For linear mechanical work such as lifting things, work against the force of gravity. Given by : Work

=

Mass x Acceleration x Height W = m x g x h

where, m = mass of the object in kilogram (kg) g = acceleration due to gravity in metre per second square (m s-2) Power, P = m x g x h t where,

W

P = power produced by machine in watt (W) = work done in joule ( J ) t = time involved in doing the work in seconds ( s )

E.g. 1 : An airport luggage belt takes 5 s to move a bag along a 5 m distance. If the force exerted by the motor of the luggage belt is 10 000 N, calculate the power of the motor. Solution :

Power

=

Work Time =

_F x s_ t

= 10 000 N x 5 m 5s = 10 000 W = 10 kW E.g. 2 : The slide before, luggage belt has to exert 15 000 N to carry an extra load. How long does it take to move the bag in 5 m? Solution :

time

=

Work Power =

_F x s_ P

= 15 000 N x 5 m 10 000 W E.g. 3 :

= 75 s

The time taken for a lift to move to a height of 50 m is 4 min. (a) If the power of the lift is 10 kW, what is the mass of the lift? (b) What would be the power, if we want it to carry twice the above mass? POWER IN ROTATIONAL MECHANICAL SYSTEMS (Kuasa dalam Sistem Mekanikal Berputar) Involves the rate of doing this rotational work. E.g : A 1 horsepower boat engine could turn the propeller 300 rads-1, a 2 horsepower boat engine might be able to rotate it twice as fast. For rotating machine, the rotational mechanical work, is equivalent to the rotational kinetic energy. Rotational Mechanical Power (Kuasa Mekanikal Berputar) Define as : W

=

½ I ѡ2

since power is,

P

=

_W_ t

Therefore, power is given by : P

½ I ѡ2 t

=

where, P = power delivered by machine in watt (W) W = work done in joule ( J ) I = moment inertia in kilogram metre square (kg m2) t = time involved in doing the work in seconds ( s ) Ѡ = angular speed of the machine in radian per second (rad s-1).

E.g. 4 : A washing machine with moment of inertia of 2 kg m2 spins at angular speed of 350 rad s-1. If the brake system in the washing machine has an average power of 1000 W, how long does it take to stop? Solution : Power, P

=

½ I ⍵2 t

=

t ½ I ⍵2

P = ½ x ( 2 kg m2) x ( 350 rad s-1)2 1000 W = 122.5 s = 2.04 min

DIFFERENCE BETWEEN POWER IN LINEAR & ROTATIONAL SYSTEM (Perbezaan antara Kuasa dalam Sistem Linear & Berputar) Linear Mechanical System Rotational Mechanical System Involve linear motion. Also involve FORCE, LINEAR VELOCITY, DISPLACEMENT & TRANSLATIONAL KINETIC ENERGY. Involve rotational works.

Also involve TORQUE, ANGULAR VELOCITY, ANGULAR DISPLACEMENT & ROTATIONAL KINETIC ENERGY. In most situation, both linear & rotational powers are involved together. Examples : Boat engine – A boat engine provides rotational power to rotate the propeller. The rotation of the propeller produces the linear motion of the boat. Examples for Linear Mechanical System: Photograph 4 : A large crane lifting bricks

Photograph 5 : An aeroplane taking off Photograph 6 : Cars moving in a straight line Examples for Rotational Mechanical System:

Photograph 7 : A gas turbine Photograph 8 : An air ventilator

Photograph 9 : A grass cutter MEASUREMENT & CONTROL OF POWER IN MECHANICAL SYSTEMS (Kawalan & Pengukuran Kuasa dalam Sistem Mekanikal) All machines have their own mechanical power. Their engine & parts are designed to suit their functions. Example :

Photograph 10 : The engine of a forklift is designed to lift lighter loads Photograph 11 : The engine of a crane is designed to lift heavy loads 3.2 POWER IN ELECTRICAL SYSTEM

ELECTRICAL POWER (Kuasa Elektrik) Some electrical devices use energy faster than others. A more high powered device uses energy more quickly than a low powered device. (a) Batteries

(b) A wall clock controlled car

(c) A remote

An electrical machine works by transferring electrical energy to an output. This output could be mechanical, light, heat or sound. E.g : A radio – electrical energy produces sound as the output.

Photograph 2 : Radio Electrical power (Kuasa Elektrik) Define as : P

=

V I

where, P = power (kuasa) delivered by machine in watt (W) V = voltage (voltan) difference in the electrical system in volt (V) I = current (arus) in the electrical system in ampere (A) E.g. 1 : The power window of a car uses 40 A current & operates at a voltage of 12 V. What is the power? Solution : Power =

V I =

(12 V ) x (40 A)

= 480 Watt E.g. 2 : An electrical iron rated at 1000 W operates at a voltage of 240 V. How much current does it use? Solution : I

= = =

P = P V 1000 W 240 V

V I

4.17 Ampere

Efficiency (% Kecekapan) We calculate efficiency in order to know whether certain electrical devices are working well @ not.

When a machine sows a very low efficiency, it tells us that something is wrong. If the POWER OUTPUT = POWER INPUT, then the device is 100% efficient.

where, efficiency is a unitless quantity P out = power output in watt (W) P in = power input in watt (W) E.g. 3 : A certain motor produces a shaft power of 200 W. The operating voltage is 20 V and it draws current of 13 A. Calculate : (a) The power input. (b) The power output. (c) The efficiency of the motor. Solution : (a) Pin =

V I

(b)

Pout = 200 W

(c)

Efficiency =

W:

=

(20 V) x (13 A)

=

Pout Pin

x

200 W x 100 260 W

=

260 Watt

100 = 77 %

E.g. 4 : The efficiency of a 220 V electric motor is 85 %. If the power output is 700 (a) (b)

What is the power input? How much current does the motor use?

Solution : (a)Efficiency

= 85

Pin =

Pout x 100 Pin =

70000 ÷ 85 Pin =

MODULE 4: ENERGY CONVERTERS

700 W x 100 Pin

823 Watt

(b)

I

= Pin V

= 823 W 220 V = 3.7 Ampere

are devices that convert energy from one form to another. 4.1 ENERGY CONVERTERS IN MECHANICAL SYSTEM MEC can be either kinetic energy or potential energy. MEC changes kinetic energy & potential energy to other forms of energy. Example : A windmill & turbine power generator is an example of a MEC. changes mechanical energy to electrical energy. MEC = Mechanical Energy Converters