NOTES FOR ENGINEERING SCIENCE (SEM 1)

7/22/2019 NOTES FOR ENGINEERING SCIENCE (SEM 1)

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Engineering Science -TL

Topic 1: Physical Quantitiest4"4{

^y-

eu+{|M

and MeasurementDut"oh"nf)e#nr,srhauu*'|{

o't t'Yre *uau'#'i'

f,gr6r'2{ , deriv'ed ?*-sl

Physical Quantity € pru*tP*'s r

Physicat quantity is the numerical value of a measurable property that describes a physical

system's state at a moment in time.

It is a physical property that can be quantified, can be measured with a measuring instrument.

Examplei of physical quantities are rnass, volume, length, time, temperature,electric current-

Quantities can be divided into two types :

1. Base quantities2. Derived quantities

Basic quantitiesBasic qLJantittes are the fundamental quantities that are not related to each other and that are

use to derive all other quantities.

There are seven basic quantities. They are

1. length2. time

3. mass4. Thermodynamic temperature

.

- 5. electric current

Q "rnorntof substance

(7)luminous intensity

Derived quantitigs" Derived quantities are just quantities that are derived from one or more basic quantities.

For exarnple:

Area is a derived quantity because it is derived from the basic quantity length.Area=length*length

Volume is a derived quantities because it is derived from the basic quantity length.

volume = length " length * lengtlt

Density is a derived quantity because it is derived from length and mass, two basic quantities.

density = mass/(length * length * lengttt)

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88101 fngineering Science -T1

lnternational System of Units

All systems of weights and measures, metric and non-metric, are linked through a network ofinternational agreements suppofting the lnternational System of Units. The lnternational

System is called the Sl, using the first two initials of its French name Sysfeme lnternational

d'lJnitds.

It is a scientific method of expressing the magnitudes or quantities of important naturalphenomena. There are seven base units in the system, from which other units are derived,

This system was formerly called the meter-kilogram-second (MKS) system.

Sl Base Units

The Sl is founded on seven S/ base unifs for seven base quanfifr'es assumed to be mutuallyindependent, as table 1.

Table {. Sl base units

Ba.se quantity Name $5'mbol

Sf base unit

lfftgth

nra.ss kilo.greu kg

elech'ic crnreflt arupere A

t#fiuddj .'G"g41 ',,, ,,kei , , K ''

amourtofsrlrstatrce roole mol..r.aluminousiat€asitrr :,. ,:3.andgla cd ,,. .., t, : ..''r i :..:.'' i-ra:,,:r.:,' :.,.::', .,

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88101 Engineering Science -fr I

Definitions of the $l base units

Unlt of length meter The meler is the tength of ttre path travelled iry light in vaurum dunng a time interval ot lllgg i92 458 cf a . .qeseconc tltr

Unitofmass kilogram lheklogramrstheunloimass:ltiseqra totrem*ssoftlprr.tematinnalprotolypeofthekrfogram *ffi(adar

Unitof time second fte selond is the duration of g 192 031 770 peiiods of the raiilation conesponding to $ie transiiian ,.9rg

beltireenthelt;o hvperfine levels oftne ground state ofifte cesiilrTi 133 atom. Tflffi

Unit of arnpere The ampere is ihai constanf cunent s'hich, ii mair$ained in lwa straighi parallei concuctors of irifrrute [en$h, ..ete

electric cunentof negltgrbie circular cioss-sec&on. and phced 1 nreter apan in vacium, vrould produce bet*een

lhese'

Hconductors a foree equal to 2 x 10-i neulion per rneter of lerEti.

Unftof kelvin Tlrcltelvin,unilofiherrnoivnamictenperature,isthefractronl2i3.lti0fthethennoCynamictemperafurecf ..r"p

thermodynamic tlu triple grornt of nater. H*temperature

Unlt of mole 1 The fiole is fre anrftnt of $$slance of a sysiem which cffiiains as ma*y elemenian entlites as tl€re are ..St'

amoun{ of afoins in 0.012 kiloEam ol carbon 12; its wr*ol is 'mo[." Hsubstance

2. F/hen the moi,e is useC. the elemeniarr e$rtres must tre speciiled and may be atoms. mo{ecules. .ofis, elediots.

other padcles. cr speu{ied groups of su$ par{i&s.

Unit of oandela The candela is tlte klmtnous intensitv. in a given dredion oi a source thai emits rnonochronutic radralron d#*t"luminous frquency 540 x 1012 hert and that has a radlant intensrty in tha{ direstion of 1/683 waft per s{eraiian ffiJ

intensity

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88101 Engineering Science -T1 II

Sl Derived Units

Other quantities, called derived quantities, are defined in terms of the seven base quantities

via a system of quantity equations. The Sl derived unifs for these derived quantities are

obtained from these equations and the seven Sl base units.

Table 2. Examples of Slderived units

gea : ,,

rrriune

Sp€E4rdoq'

aceleration

ware$rrober . ..,

sas: dcnsi&'

qpeiific rohne

ctrrcdemir.*

Cagnelhfeld sr€oge

aoou{-of-slb$ancc coac.er8diou

ffi'r..nusstaction

::.'

-,im-

.a

tn5

u's

.n'

tEfi'

:':-:{ml

r. ',,

aobn'

:,-ad# :

kgfu= I

SI derirerl onit.---...-.-.-.---.:-j:> ..*---.-.-,*-----

l--:Sdel {q .,,

,.,:..

cubicmeter

,oetefpq second ",' ,'. . .

metupa second sEared

recipocdneter .\hlcgrdr[per cubhneter

rubico4qpqirlag ry,

allprr€ Pef s,quafr netef

SrygrePer:meter ' :i::'

oclcpcr nrbic rcter

t1 f] pq,lge*;i .,,,,,.'; ..,:,,,:; :' , - , ,",-,,

,, :,1

ldogaru pcr hlogu, wbich ar4' be represeced bl the qnbcr 1

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88101 Engineering Science -T1

For ease of understanding and convenience, 22 Sl derived units have been given special names andsymbols, as shown in Table 3.

Derived quantity Name

Expression Expressionin termr of in terms of

Symbol oiher Sl units Sl base units

Table 3. Sl derived units with special names and symbols

Sl derived snlt

plane angle

solid angle

fteqUegCy:.,', ::'i , r: r',:: rr.::',,, ;,: :.: ri ,,',, ,

force

pressure,.stress . '...

energy, work. quantity of frcat

power,radiantflux :''

electric charge" quantilv of electricity

electricpotential difference, : ' :

electromotiveforce . ' . ' '

capacitance

electricresistano: t i, ,,electric conductance

rna$iiri"*ut ,,.,1,,.,,,.ii,,,.,:,,, -,.',.i' , .,,,.,t.', -.,t,, ,. ,

magneiicflux densig

indudance

Celsius temperature

iu*itiOiii" .j,..',.,.,'' ,.=;irt.,.' ,:,,ri';,,:. ,,,;1,' ,; , ' .I

illuminance

ffiVtpr ..i ianuA*j"l',-.ti11,,1,.i,.,,,..::,=',,.',.',

catalytic activity

radian {ai

steradian {a}

.,riierE :, :, :

ne,uton

'rPascal : :, :

joule

.,*att,'," ",,coulomb

,i.',.VOlt;t, ., ',::: , ,

farad

ohm,.,::,t:,:,

siemens

.:aeber , .,

tesla

,1gfrv '-,,,,,:,;,

degree Celsiusl'.l*nen.i 't,,

lul

:,';fggguere , ,.,:'

gray

' ,a5jbvert .r :1: ,' '"'

katal

. mnrl= 1{bi

rp2-r1-2 = 1 {bi

,'s'j.m'kg 5'2

m-r*g s-r

m2.kg-s-2

mr*g s-r

sA

,;,,.,,nftg'-s{A1 '

m-2.kg-t.s4.42

' ,r,'n zXjj:{2 '

rn-2.kg-1.s3,A2

:- #tgs€.A'1'

kg's-2-4-1

tnz.kg s-e.4-z

K.rnr-m-.2-Cri=Cd .,

'n?*.4 66 = 6-216

,.s-1'*'"t

. ml:s-z . :,

s-1-mot

rad

51 ic)

'r1z

tl

,Pa

J

c

',V i,

F

',as

,wb

T

11

"c,,*::

lx

rBq

Gy

€Y,,

kat

'

N,'nf

N.m

Ji's ,

;tlr1A,

cA/

v A

Ai/V9'Wbim2

v,lb,'A

cd;5rftl

lmim?

J/kg

'idni "':','1.1.t''I

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E

88101 Lngineering Science -T1l

Sl prefixes

A prefix may be added to aunit to produce a multiple of the original unit. All multiples are

integer powers of ten.

The 20 Sl prefixes used to form decimal multiples and submultiples of Sl units are as below.

Factor Name Symbol

1024 yotta y

1021 zetta Z

101s exa E

1015 peta p

1012 tera' T

10e giga G

106 mega M

103 kilo k

102 ftecto h

101 deka.da

FactorName Symbol

10-1 ,deii 'd

fi2 centi c

10-3 rnilli 'm

10-6 micro l,

10 9 nano n

1A-i2 pico p

10-15 femlo f

10-18 atto a

o'21 zepto z

10_24 yocto y

Conversion of Units

Conversion of units involves comparison of different standard physical values, either of asingle physical quantity or of a physical quantity and a combination of other physicalquantities

Refer some conversion factors, and examples of unit conversion on page 7-10.

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Table 3: Convers[qqGonversion factqls-

L'IIdI ILILY

Length 1m={ lzm

1606:= 1-o'cm, tn3 tn - 0-621 mi

Mass 1 kg =.1O00 g1 metric ton = 1qq9- g

@sr yeir = 8J6 x 103 h

i trort = 6o minulqg-= J.999 sesgnq

Time

f raaian (rad) = 57.30o

ro - n A1-74\ radngle

t froriepower (hP) = 746 w

1kw=1000 wower \

Bro o+rncrnrtr cIjw SGTENCE

1.2 colwERSlON OF Ul'llrs

Unifs conversion

Example I

Thetalleststudentinclassiswithaheightof2.02m.

Solution:

Provide the conversion fa'dor

1m = 3'281 ft1 = 3.281 ft/m2'o2'"

= [3:3i)&1 ?' o,*,={r36'Tt

Example2 -= 6"#++

How tall is he in feet?

A bacterium has an area of 2.35 square inches. Give this area in square

millimeters'

SOIutlon:

Zamszuln' 1

Basic. tlnit conversion

Lenqth Time Weight

t hour ) 60 minutes

1 minute ) 6O seconds

1 kq ) 1000q

1 km ) 1000m

1 m + 100cm

1cm) 10mm



FI I OO4-TDCHnrrCI]TN SCIENCE

provide the conversion factor1 in.? = (O.OZ54)2,rnm2

Etample 3

Convert 3.5 kilometre to meter.

SoLution:-1km :101m =1000m

3.S km = 3.5 Lr * l9:0h= 3.5,,x 1000 m =tkm \

Example 4

Convert 3.27 mtligmrn to gram.

_S_qAtjon;_

"-1fng -1O-39 =0.001 g

Examole 5

Convert 0.087 kW to mW.

3500 m

"rr'-" /"

$:-

*"\ 3D' ,.

.' I,,(7

mW

: o.o87 n* * 1000w

lkw

: B7w x looomw

lwI:

I

,i:

r

'ftr::

E,;r.

i&:d:,.

5'&lDi-ll:

&,'€:.

F'-'L1

f'

670sl,. lkQ = o.67ke1000c)

3.27 ms : 3.27y-fu ;g.z7x 0.001 s = o,00327 e

5"s"fuUon.'

kW*W*

0.087 kw

W* mW

Example 6

Solution:

mf)* fl*67O O00 mfl

Convert 670 0OO me to ke.

ka.

:67Oo00 mfl * lQ=

l000mC)

= 0.087 x 1000 w : 87 Watt

= 87 000 mW

?amaaui.s I



60 kmlh:

.-- 76.67 mls3600s

Example 8

Conveft 9.8 m/s to krn/h.

s.B m/s

=

n,tl.

,1*. " l9?0t =35.28

km/h.ls 1000m th ---

Example 9 S

>{A speed-trap mission in Utara-Selatan highway has determined the speed - * '-*

:Olirnit as 110 kilometer/hour (km/h). Express the speed in meter/secoitAirf6f* (&*Solution: // -/.,Y .,''g"i

provide the conversion factor *o /;,

s1 km = 1000 m and t hour = 3800 s

rewrite the factor of 1

)Ff

9)

1 = 1000 m/km and t hour i

11o krn/h = (110 km) (1000 m ) (

U36oA hr/s

t hr) = 30.5 m/s

m km 3G00 s

Exarnrfie 10

'A capacitor is with capacitance 1.2 x 10-6 F. what is its value inpicofarad(pF)?

Solution:

provide the convers.ion factor1 pF = 10-12 F

rewrite the factor of 1'f

- -^r?= r pF =1012oFTdtr2 F F

1.2 x 10-6 p = (1.2 x 10-6FX 1O12 pF) = 7.2x 106 pF

F

4tl

n:

4IH

rE

Ez..t:r^-=rir.



BIOO4-TECI{ISI CIA]\I SCIDNCE

ACTIVITY

.

1,1r. State: '

a) base, quantity

b) .derive quantity

L.2. Convert

a) 56km/htom/sM

b) 40 N/m to kN/m

1.3. Convert

a) 0.37 meter'to.decimete, dt

b) 5oo miligram to gram t l*-^Q,

L.4. Convert t'

a) 700 mg/cm3 to g/cm3

b) 900 9/cm3 to kglm3

1,5. Convert 90 km/h to m/s.

,l-o\i t tn:

l0C6w4: \jFEEDBACK

1.1. a) Base quantities its length, mass,

b) Derived quantities its area, volume, velocity and pressure.

L.2. a) 56km/h

b) 40 N/m

iooo= 56x- m/S

60x60

= 4ok*/.n

1000

= 15.56 m/s

= 0.04 kN/m

1.3- î-3 0.37m = 9Orn:

0'37x10dm =3.7dm

\ 1- t0-,

\ y,{st 5oo mg = 5oo x o.ool s = 0.5

. 9 dN)) 700 ms/cm3 = J9L sTcm' = a.7 slcm3\r'- ^N

\00u'

, n,b) e00 s/cm3 :

[#U)*, [#) -' = e00 000 ks/m3

1.s eokm/h = [H#) '" = 25 mls

la



'ITEKOU}(IW. :

Measuring instruments are v-eryimportant in order to measure physicall'guantiiies-:Three equipm'ent for measuring length is: .''- -- -'-''-:--' rEe'rsrrrLJ'

' i; . migrometer screw gauge '

iii. ruler

Use correctly the following measuring equipment

- Micrometer Screw Gauge- Vernier Caliper

Meter ruler

a) Micrometer screw gauge/ Tolok skru miirometer

' Used to measure very small length such as the diameter of a wire. Has two scales:

lrit'Pr.

E"

:"{;,

*;ti:tEirt '-g:

a:

tf,i

-t'

'I:;1. '

f::

f.'{t*€';

d:

ij:ll

e'+..

q:2'i

9'

'F:,:

(a) Main scale on the sleeve(skala utama mengufuk)(b) Circular scale on the thimble(skata bidal membuiat). Smallest divisiorr is 0.50 mm

. Canmeasure lengUi acolrately up to 0.01 mm

RaJche(race0

b) Vernier Calipers

a

a

aa

a

Used in rneasuring lengthSmallest division is 0.01 cm (0.1 mm)

Can measure length accurateiy up to b.ot cmUseful in measuring the inner diameter of an objectDivided into main scale and vernier scale

II

Vernier scale : 35 x O.Oi mm

= 0.35so, actual reading is: (6.5 + 0.35)mm

= 6.85 mm

jaws (rahang

o stJi).- 3e7"_

et -f l*"'



c)

tabung uJl

Meter ruterUsed in measuring lengthSmallest division is 0.1 cm (1 mm)Can measure length accurately up to 0.1 cm

Measurement haveto be recorded accurately to 0.1 cm

* r.=Oo h5

r-- -

.-in'

,* .1

;ir'

.i

.i

'Length of object = 4.2 cm

ACTIVITY

1. Discuss

a. Scalar quantities and give 3 examplesb. Vector quantities and give 4 examples

c. Mornentd.. Weight

2. Change the following units:a. 1O0 kg - 50 g (in gram)b. 1O tan (in gram) . ( +4^c. 15 micron (in meter) -'d. looTtTj (in m/s)e. 10 9/cm3 (in unit kg/m3)

t2



,.1'.rE.r' '.

/....

,,": .'t - t,

Exercise l.l' i liot x:rol ,

loh^ rA't> n;-,9

t. Convert ,(a) 3 x 10sm/s to l@lh {d) 7.g glcnf tn kdr.3

'ro

*-t

*

=Li{(b) 30tu1h to mls i"i ZSfr2 1,o n"- : ^q'b'(c) 1.3 kg/m3 to gl*tt (D 45 d to mm3 = | o :

'@-'o o)" *

2. Cilculatethe area of a circle ofradius l0 cm in'til;;;""

*tifi#"^ -ii-;;"o':* rvuru

3. calculate the volume of a spher-e ofradius l0 cm in ' tV

:.1,

' ':

(a) run3 (b) cmt (rl -t , ' '

4. ConvertI

(a) 60 ms to h (d) 30 pA to mA u

(b) 20 pm to m (e) 1000 km to cmD) zv pm ro m (e) rwu km to cm(c) Is s tops (0 400 JkgrICr to JgrKr - :

,ffi ci,I3>r'ro-{gc*1

5'' Convert the unit of the following quantities and write the final answer in the standard

@.(a) 13600 kd*t to glctl3(b) 6400 lsn to m

(c) 6x l0-7m to ;rm.

,*

Measuring instrrments are very important in order to measurephysical quantities. Threeimportant things should be remembired when using measuring iorn r*rott

(a) All measuring insfiuments have tbe smallest scale that tliey could measur€.

Therefore, the decimal places given by these instuments cannot be s'inallerthan their smallest scale. .t(b) The zero elTor of the inshuments must be determined before measurements

can be made. Zero elTor is a non-zero reading shown by an instrument whileit is not measuring any object

' (c) To get a true reading, we need to subfact the zero error from the obsbrvedreading.

>o prr *t'r .s

)o1,m K to- \' lf".

: )* tO -t'-

\ l-l^

t tr' '-\cx

tsoO nt

i9O c ^^lO *:t..

.:

\

(1.1)

1;g MEASTIREMENT oF. LENGTII

Observed reading - Zero ewor l

e\= srnn

lgI

ir

; \ L

'r:--l^f,

( qd$**

{c:-f

It

1 g;c;]

.i..\

t:.. d,0



MEASUREMENT

Metre ruleta

Mefre rule is a measuring insfiument that is commonly av,ailable in the laboratory and its

smallest scale is 0.1 cm; Thereforen the reading from a nnene rule must not be more than

one decimal place in the unit of centimete. The zero efiot for a mefie rule is the end

elTor, wlich is the end portion of a met'er rule which does not give an accurate*0"

rcading due to wear and tear.

Vernier Calliper

Vernier calliper is a measuring insfiilment that is usually used to measure a length of

between 0.10 cm and 12.00 cm. Figure 1.1 shows a venrier calliper which consists of the

following components: ' '6y Outstde jn+,s to measure the outside diameter or the length of an object,

(b) inside jcws to measure the the inside diameter of an object such as apipe,

(c) tatl to measure the depth of a hole or a test tube,

(d) main scale and(e) verniey scale.

main scale

vcmicrscalc

outsidejaws

1.70 em0.07 cro

1.70+0.07 =l.77cm

Fig. 1.1 Vernier calliper and how to read it .'

the main scale, the length I crn is divided into 10 divisisns. Therefore, a length of I

at the main scale is 0.10 crn. At the vernier scale, a length of 0.9 cm is divided1O divisions. Hence, a length of 1 division at the vernier scale is 0.09 cm. Therefore,

length of I division at the main scale is longer than the length of I division at the

scale by 0.10 - 0.09 : 0,01 cm. Figure l-2 shows the 'length of the main scale

to that of the vernier scale-

?.j

?.

V**0

Main scalercading =Vernicr scale rcading =Rea<ting =

insidejaws

l4-



,10.tcm

0.1-0.09;0.01rn

0.09ca

Fig. t:.2 The Tain and vemier scale on a Vernier Calliper',

ZeroError for a Vernier Calliper

E

o vtzaicrsrb

Zcro qror - 0

t

(\+q\u-rs's'.

Fc>s i+.\ra ac6C.?t o ?

;r-c.96..\rr-.1-< ? <ro

,<-??o?

zero error, and (c)

v

ok

7*ro etror for a vernier calliper (a) nonegative zero error.

rtrnltr

T*roantr = -0,02crn

Fig. 1.3

r5

zero error, (b) positive

, ],



MEASUREWNT

When the jaws of a vemier calliper are closed without any object in between them, thereading shown by this vernier calliper is its zero error. There are only tbree possibilitiesof this znta efior, which are: (a) no zero eror (zero e:rpr = 0), when 0 o.n the vernier

scale is in line with the 0 on thi main scalg (b) plsitive d*o *ir,when 0 on the vernierscale is on the right of the 0 on the main scale, and (c) negative zaro enor,when 0 on thevemier scale is on the left of 0 on the main scale. Figl.3 shows howthe tbree zero errors

ateread.

True Reading for a Vernier Calliper

Table 5 gives tlree examples of tue readings from a vender calliper.

Table 5 Three examples of tue read'iggs from a vernier calliper

*.'

Z.erc enot Obtaineiiieading ITme reading

Ocm lcm

W5to

Zero enor =O

.X4cm' \l(r

5 crn

; tl;Obtained reading = 4.46 cm

a\

Tnrereading =4.46-0= 446 om

0cm

7*roerror :,*0.04 cm; i.

0..b

4 cm \l'

Obtained reading= 4.46 cm

Truereading =4.46-0.04. =4.42cm "

\\ lc:n 4crn 5cm

s5 t0

Obtained reading : 4.56 cm

True reading = 4.56 - (-0.03)

=4.59 cm

#re



I'EplE 2

FEtrllrnlk Prurdlr $iltm, sdCnffin .frd ,zir shS



88101 Engineering Science - T2

,€.@

TOPIC2:LINEARMOT|ON

Most of the physical quantities encountered in physics are elther scalaror vector quantities.

Scalar Quantities

lsytyquantity is-defined

as a quantity that has magnitude only.Typical examples of scalar quantities are time,6"-Tji"rperature, and volume. Forexample, the units for time (minutes, days, nouEEtdJ repiesent an amount of time onlyand tell nothing of direction. Additional examples of scalar quantities are density, mass, andenergy.

Vector Quantities

A vector quantity is defined as a quantity that has both magnitude and direction. Velocitv.acceleration, and weight are ve3tor quantities. --"::' '"'3'v'

Linear motionLinear motion is motion along a straight line, and can therefore be described mathematicallliusing only one spatial dimension.

It can be :

uniform, that is, with constant verocity (zero accereration), orNon-uniform,that is, with a variable veiocity lnon-rero acceleration).

The motion of a particle along the line can be described by its position x, which varies with f(ti*:| An example of linear motion is that of a bal tnrown'stlfi;; u[aio talling nactstraight down.

Distance and displacement are two quantities that are used to describe motion.

1'DISTANCE is the total path lenqth trave

quantity and the Sl unit is the Metre (m)

Example 1 :

Distance travelled is the length of the path taken by theperson from the initial position to the final position.

It is a scalar

fMSK/sllee



.\I

'ia+.i.,:{r,:;3

l-\ l-


^\ \\

I

l.'--,-/--I

"-\. IJ

l"./lil,/

*. DISPLACEMENT is a vector quantity and the Sl unit is the metre (m)

Example 6

Displacement = 15 m 150 clockwise from the horizontal

Example 2

Distance travelled is the length of the path from A to B

thrrrugh all the turns

Exannple 3

Distance travelled is the length of the path AB+ the length of

the path BC + the length of the path CD

The displacement is a vector representing the shortest path from the initial position to the

final position. lt is thus a vector quantity.

Example 4

Displacement = 10 m to the east

or Displacement = 10 m to the right

Example 5

Displacement = 10 m 15o clockwise from the horizontal,

eF<- \,r"-\

-H--}"\ t-'- r'/') i\----/

[--^1 I\/\-/

________l'':---

-1"

&

fr

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c




&.ffii

.g *.lrr**-cc."G;€-SPEED is the distance per unit time, or, of It is a scalar

quantity and the unit is the metre per second (m s-1)

A boy inoves from point A and walks a distance of 50 m to the point B in a time of 20 s.

Dista*lg-trwelled>n??r =' TimetakeFit

Speed

oftlze bay

- : Z.C ns s-l

VELoclTVi*ltisavectorquantityandtheunitisalsothe metre per second(m s-1)

A boy moves from point A to point B performing a displacement of 50 m towards the right asshown in the diagram below.

Irelocifr.- -D ispl ace?tlent

' Thnetakew

50V=lctiiy

:i f ne &ay = * = 2.5 n, : -1 io*'cr 55 f ft : ng ftI it}

Acceleration

It is defined as the rate of chanqe of verocity with time.

So, it means that in order for an object to accelerate you need to have a change of velocityover a period of time.

There are 2 ways in which an object can accelerate that is its velocity can change overtime, i.e- the velocity can increase or decrease with time.

JMSK/sllee'



88101 Engineering.science - T2

arcefej"cfisf* =c&eagefn $elorify

*rue fckenfor c*c$geE fnlreptare

arr:eJsraflsl. _ {flftei r:ei.acil.y - entfmj L'e}sciry,}

fimetaire$

?f-f;n--

E

[The acceleration is a vector quantity since it is a vector quantity divide by a scalar quantity.]Deceleration happens when the velocity of an object decreases. ln calculations, a will benegative.

'iulial^(U)" .f,rnq, (v)

5 , v(\or*,'r q

EQUATIO": oF MOTTON "$-J ' J

A.",J7\a6a,r^4.6 ^c(t{C-q+tv ^, ,,

*t f('{'l'dr-*'$-rer^

acceleration = (change in velocity)/time taken = (finat velocity - initiarlrJr"i,vl /time r"nJ;

*""'

a =(v-u)/t

v=u+at

v is the final verocity ; u is the initiar verocity of the object.

The area under a velocity-time graph is the distance travelled by the object.

Area under graph =1lz(v + u)*t vdultrtrytl

s=t/r(u+ult

i

Hence

v=u+at

s =t/r(

[u +atJ + u)*t

s=t/r(u+at+u)"ts=tlr(2u+at)*t

s =1/z(2ut +at2)

S=ut+1lzat2

Titnefs

\__--)a__J

G--tr=t

vdqftflrytl

fMSKlsllee



BBl0l EngineeringScience - T2

S=

ut +tlzafi

v=u+at ,thus

s=t/"(v+u)*t,thus

3as = Ttrs -tf *vz - wt

Zgs:tsB-st3

Multiplying on both side by za

lzcl *s: fanl -]e* +'l&;s]

Zc;c: {u +:a"}{v-u)

fr-u)t-' t

&

,=|t*, *"1* rt

f=u2+2as

,F.r

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,':

ii) Total displacement .,, ,. -,l

The area under the graph = displacement

V (ms r)

?

Soltttion:

A vetoilSH = <?4aie-rnA f,\^t c+

A.nth.

,@

a c((\(:-c-tlie o >

A

V -1,, :L

.r.Ir+t.,4a Onnj

r(s)

=10-0 ^rc- \""o'ck-aOPl-=zKy

aPQ=gms-2(vr,r16"-;

= 0-10 -2oQR--l- -=-)ns -

son--];g+to)0b1=7so,

I

I

l:

7s asjt.z"g

qcc

ExamPle 2

which sedion of the graph shows the vehicle moving with constant acceleration?

V(ms-r) roat,-a-t..rt^.iJ.. -, ,,-,^.-.tB (PÊ+A+"at"d + q= G i'^s

c6.'sfu1*

dCcctarrztr9 n

Solution:

(ms-l'l

L G

v



.

A bus initially at rest rolls down a hill with

long will it take the car to travel 150 m, the

Splgtjw.: ' I': ' :i'

Given: u=0mls (from rest)

Find: a)v(velocitY)

'b)l (time elaPsed)

Given:

Find:

of the hill?

a=5-0mls2s = 1502r

The distance to the bottom of the hill may be found

s=ttt+latz

t5om= o+*(5.0m tr'\'

,': @-

Examoie 4t'= J6 =7 '75s

A driver is driving his vehicle at a velocity of 30m/s when he suddenly accelerate his

vehicle to 50-sals-Itlakes 3 secontC6r tjr-re vdhiEte to reaclr 60m/s. Determine the

amount & ".rypne vehicle in condition mentioned

Solutian:q Initial velocity -= 30 m/s

V Final velocity = 50 m/s

g Time taken 3 seconds

60 -30Q=-

a = l0 ns-2'

Example 5

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts offthe ground. Determine the distance traveled before takeoff

Solution: ;

ry

a=3.2mls2- t=32.8s u=0m/s

s=?? s=ut+0-5at25 = (0 m/s) (32.8 s)+ 0.5(3.20 m/d)(32.8 s)2

Answer:s=1720m

Example6

A car starts from rest and accelerates uniformly over a time of 5,21 seconds for a

distance of 110 m. Determine the acceleration of the car:



pcliteltnil( Frenier sulf€il sslahuddiil .4bdul .*eia $rsh




TOPIC 3 : FORGE ':i

The Goncept of Force

Definition of ForceIt is a quantitative description of the interaction between two physicat bodies, such as an object

and its environment. lf is a push or pull that one object exerts on another which produce ortends to produce motionq sfops, or tends fo stop the motion

Force is a vector. The Sl unit for force is the newton (N). One neMon is equal to 1 kg * m/s2.

Force is proportional to acceleration. ln calculus terms, force is the derivative of momentum with

respect to time.

2 broad catego;'ies of force

Contactforce is defined as the force exerted when two physicalobjects come in direct contactwith each other, eg. frictionalforces, tension forces, normal forces, applied force..

Action-at-a-distance force,such as gravitation and electromagne$c forces, cqn exertthemselves even across the empty vaiuum of space, result

"u"-nt*tun theQrolntbracting objects

are not in physical contact with each other.

Weight and Mass

Mass is a fundamental property of an object, is a measure of the amount of material in an object,being directly related to the number and typeof atoms present in the object.

weight is the force created when a mass is acted upon by a gravitational field'uJ e

^/\g

Weioht Massdeoend on the acceleration due to oravitv is a constant qlanlitu

-\o eu'at,4t

a vector quantitv a scalar ouantitvmeasured in Newton measured in ko

w=mq m

Fp e r*^b.

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F is the net force in Newton (N)m is mass of the body in kitogram ([g]a is acceleration of the body in m/sz

The force applied is directly proportional to the rate of change of linear momentum of the object.An object will accelerate in proportional to the net force acting upon it and in inverse pr"fii;;'i"the object's mass.

Forces in EquilibriumAccording to Nevvton's First Law of Motion, an object remaining at rest even when r9es are acting upon it does so, because there is no net forie acting on tne oolect.This means that the resultant or vector sum of all the forces acting o-n tn" object is zero.An object in this state or condition is said to be',in equilibrium.',

Resultant forceA force which is the sum of the net force adgd o _lhe

objecIt is a single rorc two or more forces acting on an object.It is the result of two or more forces actiig conjoinily.

BB1 01 Engineering Scienc/fS

Newton's Second Law

-Newton's second law of motion states that the net force on an object is equal to the mass of the

object multiplied by its acceleration.

F= ma

-iieuE t: '2'eclers i: :is sr* C**iasl

,, fM=Fi+F? a" L'**

=:il

i.r;.*I-

ti

#Ern{

..l#."Effftf---:r

t:i.

irgure3jPsallelogramconstrlclio[ ;;2ior 4lding lectors

F;gu€ 2. \,*ic6 in thecpNsh€di€ition

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Filue j:\jectors rn dfierem Ji€alions

tk&

f:



,,...--: .


Resultant forces can be determined applying graphical method (the sum of vectors),parallelogram method (tor 2 forces only), and component method (for 2 and a ove forces).

Resolution Method (Gomponent Method)

Force can be resolved into 2 component vectors, which have magnitudes,

F,= Fcos 0 horizontalcomponent

FF Fsin g verticalconiponent

Each force is resoved into its x and y componentsrwith negatively directed components taken as

negative

The scalar x-component F, of the resultant force is the algebraic sum of all the scalar x-

component.

The scalar y-component, F, is found in similar way

With these csmponents known, the magnitude of the resultant is given by

ft= , h lleafun

ln two dimensions, the angle of the resultant force with the x-axis can be found from the relation

tan9: Zr,E4

r S 'h deTrcz C')

r,Y *(Ir,I

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. l..lIfjf, i

I

L))

fut rn &y"<lbr--uu

4n clLie c4 ,'r grnol ttt b h exuir;rr;s,v1 if resur/a46y v.edtf Euet+ "il( 4l* f*uo *kn * +i 6lird is 2tr<o r a t,e c*a say r,o*es no np*ftve.e.

"rAj on ;1. '

(ratuplzs:

/. bolctt*tt- 7 g6 tlv^* *,ycl'g ri ll. $sJuvt cqn 6e gq}J'a'

egt;/rLt;u4,t -

F>=taN Fs =-2

Fr= 4N F="F:F,tFz

= €+to=16N

F,= $NFr=TN

F>= BN

Fs =?N

0t{

@

6)

I

t,+F = F.+Fs

F = Frrfs -Fr=(2cor60')tl0-4= | +1o -+ =f'TN

d)

?s= l6N

Fl{-F++FF

F'+Fs

7+16-8-5/oN

h=toN '+0,

Yz=3ON

Fg*F, + F

F.-Fs -Tr = 30 -(g cos +o') - l0

F=?

F+F'=F-t--':

Fr+FsFr+ft -F t5+t- 3 =?N

<) F= 8N3

t-r.a

F_t--

F^r=fN

16o'

eO- 6.t7 -(O = l3.sT N



t' DztrQ.rtwr'r*

a)

.:

tl,r. rtnl fi,r'cL tw each uanglr-; ( nrf in eTTqlibsu- .

:r

L)

-ffiF,=3N-r=fN

rueffo.ct = F, -fz= 3-J= -qN

ar 4 N {o tt,e" k++

Fs= 6 cos 80o=S.tq€N

N{t ft}rc.o = F, -Fz +Fg

= tt-5+ E.ne= (/../?6 N C +vc un-enlt9 *o +t,.r ub[*

F, =tOS

Fr= 0N 0

20

Fs=l?fr:-oN

Fr= 3N

Nclf,vcc---Fs-Fr-F=- 2okq.+-3

=)= 7- GN C n"a,ouns ,.l,pwo.nd." )

c) t'=t'*

W

F''+tI P'=:N

Notftre:? :Pt'= -, c'vzhêArê J*''-ovJs | )

d) II

v

.Fr tO c6s rtoq.+



Resol,^hon n^Lfhsd / Csvrpnar,* ,t^"aJks4

Tv det+t wae rest^tk t* f,rrczs o{ f o,,td ehoW N<Is ud"'ci arl ad;1g an

,ffiqd::.

*io e;rel .

Q.xqru"Fk I: Cr.J

I

I

tIt

I

t1

cltf4f€ {Az resu fiqn'* *vcz rnSmt24o A6r^r'a I :

-. 3N

60'

2N

hgwe I

Rpsr.llar?* frvcr, Q =

= J€F,l'+6Qry/crffis?sr1r.15;;g l = 3.ssN

rane= €E = *-T = -o-tro?

F = lo,rt-{-o.t7o7)=-9.7. = ffi' lg0'_ 1-1. -_

tVO.g-

F(N) F" = Fcos $ Fv= Fsvro3

5

3cos 6O0 = l-S

F cos t8Q' = _S C-D = -g2 cos )}Ou = r(6) s- I

$rt 60' = )-s?8

}rb t&O' = S(o) = 6

$''n )|0' --'2Gt) = -l

3

_s

.).

5F^ = -s.sN E' F7 = o.sgg

6



BB1O1

Additional.....

Movement with balanced and unbalanced forcesA car or bicycle has a driving force pushing itforwards. There are always counter forces of airresistance and friction pushing backwards.

You need to know how these forces compare ifyou are to predict what will happen to the speed ofa moving object.

. lf the driving force is greater than thecounter forces, there is a resultant forceforurards. This will make the car speed up.

. lf the driving forde isless than the counter

forces, there is a resultant force backwards.This will make the car slow down.lf the driving force is the same as the

counter forces, there is no resultant force,and so no change in velocity.

ilf'f s'e#$

I redionbrce

F,i::$ffiJ driuirqlorae

I fidi,Bl

tr airres{stere lf the car is already moving, it will carry on ata steady speed in a straight line.

o lf the car is not moving, it will stay still.

Newton's First Law Of Motion

There are four changes that can take place if a force acts on an objects:

1 - The speed of the object will increase from zero if it was at rest and the object will move in thedirection in which the force is acting.

2- The speed of the object will increase if the force is acting in the same direction as the the directionof motion of the object until the force no longer acts on tie object.

3. The speed of the object will decrease if the force is acting in i direction opposite to the direction ofmotion of the object- The speed of the object might decrease to zero if the force acts for longenough.

4. lf the object is moving and the force is acting in a direction perpendicular to the direction of motionof the object then the object will move in a circular path. When'the application of the for"e stop" ineobject will leave the circular path and would move in a straight line in a direction tangential to ihecircular path.

These four changes can be combined together to produce the Newton's first law of motion as shown below:

Unless a force is applied, an object will continue in its state of rest or uniform motion in a straight

line.

Another way to write this law:

lf a force is applied

(i) an objects that was at rest will no longer be at rest or (ii)if moving with a uniform will no longer have a

uniform or constant speed or lastly (iii) if moving in a straight line will no longer move in the same direction

T MSK/sllee




This law is thus explaining that if a force is acting ol?3.

object then the object will experience an

acceleration. However we are officially learn this only when we see the Newton's second law of motion.Newton's Second Law Of Motion

lf a force acts on an object, then the velocity of the object changes. The more force is applied, the greater

the change in velocity. This must mean that if the mass remain constant then the momentum of the object

changes. And hence the rnore force is applied the more the linear momentum will change.

A force F acts on an object of mass m for a time t such that before the application of the forcethe object

has a velocity v1 hence a.-linear momentum of p1 and after the force is applied the velocity of the object is v2

hence the linear momentum is p2.

Before the force is applied velocity = Vr ond linear momentum = p1

After the force-is applied velocity=V?2

?rid linear momentum is p2.

Now the Newton's second law of motion states that the force applied is directly proportionat to the rate

of change-,of linear momentum of the object.

fsrnse s,ral* oy' cfrcnge .}f :i|n6arflrrilEsr?:iirnr

duFrr '

dt._ Fr.*,rl s*.crn**:fs'l*- easftcl *ns,$*ssfrg.frFs

drPl-Pr

f-

'l?llfa - t?ttrrlu-

tm{vr- vr,Fa '- ^'

f{}E-$r'

lf r =athen

F ocn"m

F = Anna

lfk=1

Then this will reduce the equation to

This where the famous equation F = ma come from. This why it is also considered as the Newton's secondlaw of motion.

d*b.

IMSK/sllee



8B101 Engineering Science - T3

Concept of Moment Of Force

Just as a force is a push or a pull, a moment can be thought of as atwist.

ln physics, measure of the turning effect, or torque, produced by a force acting on a body. lt isequal to stance from its line oi action to ihe pcor pivot, about which the will turn. The turning force a

The moment of a force can be workedout using the formula

moment = force applied x perpendicular distance from the pivot.

lf the magnitude of the force is F newtons and the perpendicular distance is d metres then:

moment,M=Fxd in newton metre (Nm)

It is easier t0 und0 a D0[ using a long spenner than a shoft spdnner. This is because more turningforce is produced at the bolt (pivot) with less effort. A long spanner is an example of a forcemultiplier.

ln a simple balanced see-saw, the forces acting on the left- and right-hand sides of the pivot arethe same' This is known as balancing moments. Moving the load lt one end will cause the see-saw to become unbalanced. To regain balance, the load on the opposite side must either beincreased or its position changed. This is known as the principle of moments.

Definition of modrent of force

Moment of force (often just momenf) is a measure of the tendency of a force to twist or rotatean objeg-t about a specific point or axis. lt is the product of the iniensity of the force into theperpendicular distance from the point to the line of direction of the force, moment, M = F x d

Turrr;r13 {wce , ol:noruic

{ho,ueî,/s. a tu.e-as.*rQ. o{ hisv,) ynu& o..fuc.e a&,N6 an Ah 6Ljec} cata.ad lArtfr*d {o rshre or *anrt . U.n+ in A1rrr .

I MSK/sllee




Principle of moment of force

The Principle of Moments, also known as Varignon's Theorem, states that the moment of anyforce is equal to the algebraic sum of the moments of the components of that force.

Total normalforce (applied force) = Total reaction force

IFil. = IF{i

lf a body is in equilibrium: the sum of the clockwise moments is equal to the sum of theanticlockwise moments

&lEH

Sum of moment rotating anticlochrvise = Sum of moment rotating clockwise

Irfut = Ia.r$\

The principles of moment are applied in solving related problems in context of determining thecenter of gravity for action forces system in equilibrium, the reaction forces , and to determinethe distance resulted in specified reaction force in a system.

If o bdV ls ,h <4,FL'bn^wntt, dnder {AtadNn -f o, nun,L*Q

fw*t , -14'rz Ngobyoic g,JM af {A,r Motn z'8 *ryd q W etrCls ^Lo*t anf potnt, }s eyAl fo"RO

"'>M(I = >M)l

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f\4or wvt's Pvtl.lzvt Soln]ict

f,+F,+ = R, tApghad '6rce = Rroc*o,. *rrcz

Dalovnro c-: )hp c+yt/r-r- o* 3";A 4t ma;nlainJlrq rrqclrcn fohq ,"Jho dig*anoe

,:

I

Root pa suppor.s o{ roec/ion p

a,nd &, foied v;+h o toqd o{ 30kA/*-;

d gOfV ruspec^r.,c/y. The. ,aao*on"i ; is SoKN. F,d ,,a drsla nee x.

SorN c)^ ?r,

513KN

Sd'lroa :

XF{ = zFt3o KN + SOKN

G = 6orN

l5M

a:

l.Srn 2.s r 1

S"lu.horr :

xFt _ xF+Ra+Rs= lo+trot40

Rq = 7o -Ru= 7o- 4t.e7= 29.33 KN

= toKA/ + O,

MOr,r,-crut al P

I

F"a +qql^<oc-f,''orr foyc.g

o{ A ond B.3 rve 2rvl

Xvr a = Zrq9

lo(rs) + 20 C+) + Ae 0) = g Ru,

15t.80+2go=gRuRr= 37s L.il ,1. e7 rN

@ z*J .= ztr?M= Fxd

\ tyt*J- ir, sC-aftc egcftb;u*tquvd earroL*iaw sfuof<ol .

Zrut} = 5Mt , ,

3OKN (ann) + SOkN 05-x)r'a = 60 KN Clsrra

9Ofrlnn + l2oo kNru - BO:c.Kt.lrrn = gOOrfw,

8ox t$lr"r = ( go + t)oo - ?oo)fNJv"

x = 3?o-^ - t'd = 9'87sm



3 :

{h,e Ccn*er * ?Ya.v+-2 FA +r\.s Vs{otr x .=sn eXu;Llo;uaa.

8N

B

_xAsswvc€ {A-t cm.ltr op ga;la }s 'l+ X cl-s*a,,rcs {on. pur'^* A

frro h.rotvro,".* rvra{lôd :

5M(r, = :Mp(ts*-n rs)Cx) = :o c6-x) t 8cg-x) ,

6.34x = /f,o-)ox +7)-gf6.3+x + fox- f g r. __ r?2

. 'i

'.,-i.

ij\

o a'\

'?'mw

9 Rosuilar.t marvlu.tvL nlqlhad:

= O+llo +713+.3+i

= g-39rrt

(t0+q4 + 5ir-?z- rr-S -_qoq -(|-2

l>O'9r} lao'9rF

Co)t9 +Q+) (r.9+(:sinu$Gs) - ECre{5b + l+ +(7s s1h Its)-E

X= (92. 39.3y

= s.S?nr

(tssnrs')Co) + ro(6)j 8C?)

6.3+ * to +8

Fg/\^* ,F: belerrv,^ne J+< eg^-her- + q,,o ;tu) so J4o,f -tl-e ngstet^ wXI, &. rn

e-11Wr(Ila*u*t.

<)-tvl )l

4\,--\

CD Fer*'c6vr,rad-ro,..dhod

.'

Yir\A) : En4J(m)cx->) +( l+)(x-3.9 +0s s'rn ns9(x-s-s) =1-sXle-s-x)

. S-o)L-ro-o t t+x-49+ 6t.++)c-337 -12-= -17.9t 9x5(tx+lq-x +6{.?9x -"sx = -_17-s+too+ *9 +3i1.q2l>0-4*x -- 4o9.42

f,=3.rf

rn,r

@ fus.rtlah* srasnêof T*JA'J ; )(:#-'

._i_.. : _ -c



I'EPIE 4

PElilu{nft F d r $ltm Sddrd*1, ftfd, Âi' Sfrh



t

BBl0L EngineeringScience -T5 iI

'ly:'ry*:* .t*i'3.w :r' ':r{'''

Topic 4 : WORK, ENERGY & POWER

WORK

Definition :

The product of an apptied force, and the displacement of an object in the direction o the applied

force.I

Work = Force x distance in the direction of force ; W=F x d

W= F x 6=(m x a) x d= 1m x 1xdt

Therefore, the unit is + kgx L xs

kg-'s

xm + + Nm orJoule

It is a scalar quantity, with unit in Joule.

Joule : Work done by i force of one Newton acting over a distance of one meter.

1 joule is approximately the energy required to lift a small apple L meter straight up.lf there is no motion (no displacement), then, there is no work is done

Mechanical work , I W = Fd cosO-l ; more general equation of work when the force is at an angle 0 tothe object's displacbnEnr

J

ENERGY

Definition :

The ability to do work, or, The capacity ol o system to enable it to do the work, or,The capdcity to dowork.

tt is a scalar quantity, Sl unit : Joule

The law of 'Conservation of energy'This law states that the total amount of energy remains the same for transformation of energy fromone form to another.

Kinetic Energy

The energy porsesseO by an object because of its motion.rr---T---]Kinetic energy, I Er. = |mv,

I

t2l

Potential EnergyThe energy possesse{ ry.tn $eracting object waiting to be let loose, due to its position.

Potential energy, F;*

Gravitational Potential EnergyThe energy which an object has because of the gravitational interaction.

sllee1,



Engineering Science -T5

OWER

: The rate at which work is done, or the time rote of doing work ;unit : watt (w) ; joule/second

watt is, the rate work is done, when an object moves 1 meter per second against a force of 1 Newton.term of power of machine, t horsepower is approximately 746 watts l

aw Fd-

-;_; fy

t

- Fd c.os o =|^@tL_

-work 1 forcl )( displaceme nt )=-time fime

force velocity

A and B were given a job to lift a parcel from point C to point D. A managed to have this donequickly compared to B, so, we say A has a greater power than B

energy lost because of friction etc, so, not all input energy goes into doing work

=P-,

,lO0 yoP,n

mgh

t_



The Lgw of Conselr-ation of f Delg\'

Erag.r in a a.saoar.talic mraiu bms (cg. kinaic" pot@ial. hu li.$r)- Th. lrtr of m;m-aiou cf arg'satcs dat aFrg-t ma:: neirhr bc crcacd nu dcsto-'cd Tbuatac $e m sf all $c

cargies in ttu sistm is a co;tat.

fle mo;t cooo..olv u;ed t:aqpb L Se pcaddru*

1 r fomrla to cdorldc dIe potanial or31- is:

Jlre mass ofdre bal = i0keThcLeiglt.h=0.3mThe acelentioa dne to gnr-itr'. g = -$ 6 5'11

Substitute 6r ulues irno 6e fonula ad yw gcc

= (to r.6) tc. 8' r*s-) Lo'zd

=lq.L 9fld 6JouleS2-v

PE = ngh

i -O, *. Or. bal is +prochins thc guplr ball posirim iie P= is dccruiq liic $c X bgca;iag: At ciacdi haFsa;' bttrm dc bht ad puplc ball positim de ?E - fiIIi...-..i:;ti t' ttr< rositio of *r pupl: ball is drae the Kirctic E:ilrg;..'is at its raariom lhlc the Pot:oial haar @ ) - |1

i -tt *is poic- tbcs:dcaXr. all Sc P ba; randornrn irao K > Iluefse mn rh: K-E = I9.6J n'hle *r Pi = 0-

I

i iln. p*irioo Adrcpinli ball is rtse&r Porrtial Eoag;,(PD is osce agah ariis oarimrn ad&c iibaic o<g'$iN = 0-

| 1\'e ca nos'''aaaod

und:r:raad tb:a

ibc sulr of Pi ad KE i tlc total re chaliel eoerg':

?I+Kf,=0

PE = -XE

Total Mcc[adcal Enera' = PE = IE

\OT1: Tlis is uith iLc abscm:r of outside foces ;uc as $gigq

fsrrs brro: of dr tn' .ri6.r65g17d6 qf dlirg'r-

CONVERNNG POTENTIAT ENERGY TO KINETIC ENERGY

Roller Coaster Marbles

Slow and clanking, the string of cars is pulled up to the crest of the r:allest point on the roller coaster. One b'' . . -, the cars start

downhlll on the oiher siOe, rintit gravity takes over and the full weight of the.train is careening down into cL:'ves, twists. and tums.

The roller coaster is a great exa;ple of converstons between potenlial energy (stored energy) and kinetic energy (the energy of

motion). As the cars are being puiled up to the top of the first hill, they areicquiring potential energy. The chain that pulls them up

the hill works against the forie'of gravi'ty. At the top of the hill, the cars' potential energy is at it's maximum. When the cars start

down the otherltde, this potential enerdy ts converted to kinetic energy. The cars pick up speed as they go doerfihill. As the cars go

through the n€xt upiill section, they slow down. Some of the l.inetic energy is now being converted to potential energy, which will

be be releBsed when the cars go down the other side.

3

a1\

------+

T BBtoI.lLSL



a

aa

PRINCIPLE OF CONSERVATION OF:

The principle states that energy can neither be created nor destoyed.

Energy can be converted from one form to another.Exarrple of energy conversion is electric energy to kinelig energy.

ENERGY

KE

The ram of a pile-driver possesses mechanical energy - theability to do work- Hhen held at a height, it possesses

mechanical energy in the form of potential ener.gy. As it fallsit possesses mechanical energy in the form of kinetic energy-

As it strikes the spike, it applies a force to displace thespike - i.€., it does work on the spike_

Energy transfer from potential energy to kinetic energy and the opposite

' Object of any mass falting verlically downwards from static , its loss in potential €nergyis equal to the amount of kinetic energy gained, assuming that no energy is tost during theprocess.

mgh %mf

o If object is rebound to any height after hitting the ground and left it, its maximum kineticenergy while movrng

upwards will be equal to its potential energy at its maximum height

lrmv2 mgh

' For an object that slides down a friction free slope, its potential energy is equal to itskinetic energy at the bottorn

mgh = Yrmv2

,+



Exanplel:

A steel ball of a rnass 2 kg is released from a height of 8 m from the ground' On hitting

the ground" the ball rebounds to a height of 3.2tb. Assuming there is no other energy

leavls or enters the system during the process, find: , .

êtl E<firyzreaol)4a) the kinetic euergy of the ball a+it+edre+the ground' Jbi rhekinetic

"""[email protected]^ê-vc

--ci the velocity of the ball onrebound. rânx hl^'(ht 3

solation nikn rt rebcrui"J

a) the kinetic energy of the bail as it reaches tbe ground.

t:*+zasI = (o)2 + (2) (e.81) (8)

f = l-56.96/, (2) 156.96

1s6.96 J

b) the kinetic energy ofthe ball as it leaves the ground on rebound.

Kinetic energy of the ball as it leaves the ground is equal to its maximum

potential energy on rebound

mgh

2 (e.81) (3.2)62.784 J

c) the velocity of the ball on rebound

E6

Es62.7841

62.7841

v

Yrmv2

%Q)it?

7.9236 mls

Example 2:

An object of 30 kg ru$s was lifted as high as 3 m from the ground and was let to fall

under the gravitational reaction, Calculate the gravitational potential energy aud the

kinetic ettergy possessed by the object under these situations:

Solution

a) before it was let to fallb){ meter under free fallc) right after its touched the ground

a) before it was let to fall

mgh

30x9.81x3882.9 J

%ml% (30)(o)

Ep

Eg



+.+ + . -- ,

6

b) I meter under fue fall

FDp mgh

30x9.81x2= 588.6 J -

Er. %(30)(v':)

ObJain v ffi-e the formulae I = u2 + 2as

:0-+2 (9.81) (l)

E6 Vrmvt

= %(30)(v2\

- ";f.l','nn' \--1

c) right after its touched the ground

Er mgh30 (e.81) (0)

:0

"'Er Yrmvz(30) (v,)

Obtain v from the formulae rl = u2 + 2as

tv- u2 + 2as

= 0+2(9.81)(3)58,86 n/s

E-- VrmvzK /2 Uta

%(3qfh

: 88219 J \--1

Example 3:A boy drops a 1.2 kg of vase from height of 6.4 rr.(a) What is the kinetic energy-of &e vase when it is at a height of 4"2 rrfl

(b) With what speed will the irase hit ihe groundt(Neglect air resistance)

*-SoIution

(a) First" find the total mechanical energy Since this quantity is conperved wliile it is

falling. Initially, with vo = 0, the vase's total mechanical energy is all poteutial.

Taking the groqnd as the zero reference poing we bave

c



i

't

E-K*U=0+mghop = (t.2kgfu st* t t'\u.o^)

E:75.34J

wehaveK =E-(J,

K:E-mgh

K :75.34J -(t.z*gft.swt s'?\+.Zm)

K:25.9J

Just before the vase strikes fte grouid,@ = O t : O), lhc total mchanicalenergy is all kinetic.

E = K =lmvz

Thus,

tb)

.'.itj::,...;

':.: ,' .{ i'

1



Topic 4 : Work, Energy and Pourer

a(l)EoC)P)o-uoC,

l-(o(l)J

At the end of this lesson, you should

able to

r'statelhe ltw of Cooservalion ol Energy

/calculate kinelic energy and polentbl enetgy

appling fornulas

Energy cannot be created or destroyed;therefore, the sum of all the energies in thesystem is a c6nstant-

The totalimount of energy remains thefor transformation of energy from one form

The Law of Conservation of Energystates that

Changing Forms of Energy

o Energy is most noticeable as ittransfbrms from one form to another.

o What are some appliances whereelectrical energy .undergoes

transformation?

Video Points of lnterest

....Efficiency,

... .Energy Conversion.

t"

3

Kinetic and Potential E

O ln many sitrlations., there is a conversionbetween potential and kinetic energy.

8



Pendulum SystemEp- PolentblEk- Kinetic

Ep max

Ek min

EpEK

7p min

Mechanical Energy

o The mechanicalenergy DOES NOTCHANGE because the loss in potentialenergy is simply transfened into kineticenergy.

o The energy in the system remains

Erample 1: Calculation of Energr

Epmax=?.Ek min = ?

Trythis out...

-\A coconut falls frorn a tee fom a height of 20 m.

'Whatis lhe'Velocity of cooonut just before hitting

q =.E*

mgh = 14 mv2

tn(9.81x20) =l mv2

vo = 4O0

v = 20 ms'l

Calculation ofEnergy and Velocity

ED- Polential

Ep=E16=2

Whatis thevelocitv ol

Action question.....

Amarkerpen of mass 100 gnamisfrom a height of 2 meter. Calculate theenergy possessed by the maiker right at themoment it is being released

[Assume that the acceleration dueto gravityis 9.81msl

12

I



.. i. :.

t

t

Summary

1. What isihe Law of Conservation ofEnersy ? tfl

23hatis tne foimuta ior.:::.1:. . r::

, ,:,.,..kin tic energy ?

. potential energy ?

2

b

,g

' ' Thank you

'Kl5

to



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t ulJlL 3

pfililcknit( Frcni$en $ultan Sglshuddin Adul eir shsfi



ri l\

BB101 Engineering Science-T5

TOPIC 5: SOLD & FLIIID

SOLIDSolid is a sample of matter (substance) that retains its shape and density when not confined, it isfirm or compact of definite shape and volume, tends to keep its form rather than to flow or spreadout likea liquid or gas.

Example : Rock, glass, wood , ice, most metals, etc.

LrQrrrDA state of matter, has a definite volume but not a definite shape; able to move and change shape

without separating when under pressure; particles easily move and change their relativJpositionswithout separation of mass.

Example : Water, mercury, oil and etc.

GASA state of matter, consists of particles, without definite shape or volume; conforms to the shape ofa container in which it is held.

Example : Oxygen, helium, carbon dioxide and etc.

Properties of solid, fluid & gas in term of characteristics of particle(Particles arangement; force between'particles; shape & volume)

Solid : Particles close together, arrange properly; bond with strong force, vibrate at a point;have certain shape & volume

Fluid: Particles loose and not arranged; bond with weak forces, glide freely; shape change &follow container but maintain its volume

Gas : Particles separate far from each other, distribute randomly; very weak bond withnegligible forces"move freely & randomly; shape and volume changes follow container.

MASS DENSITY

The measurement of,moss content in s sabstance, os mass per wit volume; unit is in kgm-3.

It is a characteristic of a body, usually used to determine whether an object or material floats orsinks if placed in a fluid.

'ilp-l+iil . Unit in kgm-3

m-masinkgp - the Greek letter 'rho'v - volume in m3

]MSK/Sllee



BB1O1 Engineering Science-T5

Relative Density

Sometimes known as specific gravity - older literature Usually means RD with respect to water.It is the ratio of lhe density of a subtance to the density of a given reference material.

For example, suppose an object has a density of 4000 kg/m3' To calculate the Relative Density

with respect to water which has a density of 1000 k1/m', the RD is 4000 kg/m3

1000 kglm3

PRESSURE

Define as the perpendicularforce per unit area acting on a surface. SI unit is Pa @ Nm-2

:4

Unit in Pascal

F - Force in Newton

A - Area in m2

It describes the influences upon fluid behavior.....

PASCAL'S IJAW_t'.

Pascal's Law states thatpressure applied to an enclosedfluid causesforce to be transmitted ::'<'equally in all directions;@orce acts at right angles to any surface in contuet with thefluid(the walls of the container )

Application

The Pascal's Law states that the liquid is spread to all directions and equal transmission in

confined. The weight is changed in between the elevations. The hydraulic press principle, water

towers, dams.and scuba divers are application of Pascal's Law. We can use the vessels for

verification of liquid. It uses the pressure as main aspect in application.

IMSK/Sllee



BB 1O 1 Engineering Science-T5

Explanation for Pascal's Principle Application

The hydraulic machines eg: hydraulic press hydraulic, hydraulic brakes and hydraulic lift are

based in Pascal's law principle.

Suppose there are two cylindrical vessels A and B fitted with pistons. The vessels are connectedby a horizontal tube C. Liquid is fitted in vessels. The cross-sectional areas of vessels are Ar andA2 respectively. Such that Ar < 42. Suppose a force F1 is applied on piston A by putting a weight on it.

Then pressure applied on piston

Accordin to-pascal's law principle the pressure exerted on piston A is transmitted to liquid in B.

Thus pressure exerted on piston B upwards: -. If Fz is the upward force exerted on piston B,

then pressure on piston B :-

Therefore, -As Az> Ar ,Fz> F1. This indicates that a smsll force Ftapplied on smaller piston A can be usedto exert a large force Ft on bigger piston B.

Hydraulic Press Application

Hydraulic press is based on Pascal's law and was inverted by an engineer, Bramah. So it is alsocalled Biamah press- It is used.for compressing bales of cotton, paper, cloth etc.

A,

Weight of load, F1

P=LAl

F2, Upward force

trD-t2

A2

Pump plunger Press plunger

p :L- F' . therefore n,=',4,, Az' " A, Fl

Design with Az>Ar will cause smaller force F applied can be used to exert a large force F 2

t:v

fMSK/Sllee



88101 Engineering Science-T5

It consists of two cylindrical vessels A and B, fitted with water tight smoothly moving pistons and

connected together by means of a pipe C. The water is filled in vessels. The cross-sectional areas

of pistons fitted in vessels A and B, are Al and A2 respectively such that Al<A2. The piston of

smaller cross-sectional area is called the pump plunger and that of larger cross-sectional area is

called the press plunger.

The hydraulic press is applied in our daily life in such as csr's hydraulic breaks and cur's

hydraulicjachs.

Other examples of applications

. The underlying principle ofthe hydraulic jack and hydraulic press

. Force amplification in the braking system of most motor vehicles.

. Used in artesian wells, water towers, and dams.. Scuba divers must understand this principle. At a depth of 10 meters under water, pressure

is twice the atmospheric pressure at sea level, and increases by about 100 kPa for each

increase of l0 m depth.r'r

Additional....

ln the physical sciences, Pascal's law or the Principle of iransmission of fluid-pressure states that

"pressure exerted anywhere in a confined incompressible fluid is transmitted equally in alldirections throughout the fluid such that the pressure ratio (initialdifference) rernains the same."[l]

The law was established by French mathematician Blaise Pascal.t'r

AP - Bstxrh)

where

AP is the hydrostaticlessure (given in pascals in the SI system), or the difference in pressure at

two points within a fluid column, due to the weight of the fluid;p is the fluid density (in kilograms per cubic meter in the SI system);

g is acceleration due to gravity (normally using the sea level acceleration due to Earth's gravity inmetres per second squared);

Aft is the height of nuiC uUou, the point of measurement, or the difference in elevation between

the two points within the fluid column (in metres in SI).

The intuitive explanation of this formula is that the change in pressure between two elevations is

due to the weight of the fluid between the elevations.

Note that the variation with height does not depend on any additional pressures. Therefore

Pascal's law can be interpreted as saying that any change in pressure applied at any given point ofthe fluid is transmittedundiminished throughozr the fluid.

Equation: (Pr)ff r) : (Pz)fVz)

t:1,:::.-1

jMSK/Sllee



B B f0t Engineering Science-T5

ARCHIMEDE'S PRINCIPLE

A body immersed in aJluid is buoyed up ful aforce equal to the weight of the disptaced jluid.

p - fluid density in kglm3 ;

g - 9.81 m/s2, acceleration due to gravityh - height in meter, depth of fluid.

The Principle

An iron needle sinks in water but a hug; ship floats on the surface of water.

All objects whether inside the liquid (or fluid) or partially inside, appear to be lighter than in air.This is because liquids (or fluids) exert upward force on the immersed object. This is called theupthrust, the Buoyant force or buoyancy. The upthrust due to liquid pressure, pushes the objectfrom all sides, but is greatest on the bottorn where the liquid is deepest.

Llp.thrrut

Statement of Archimedes' Principle

When a body is immersed wholly or partially in a liquid (or fluid) at rest, it appears to lose some

of its weight. The apparent loss in weight of the body is equal to the weight of liquid displaced.

B$d .

fMSK/SIIee



BB 101 Engineering Science-T5

Proof

Consider a solid cylinder of height 'h' and area of cross-section A, to be completely immersed in a

fluid of constant density r. The horizontal thrusts on the cylinder balance each other.

P = 9|l"

= A1lg[v)

& p=fF=aP

..=::a-:l:.:-c:::

Consider the vertical thrust on the cylinder.

Total downward thrust on the top face of the cylinder = (P + hrpg)A

Total upward thrust on the bottom face of the cylinder : (P + h2pg)A

Resu'[an'i'|hrus'[.

HI; il;;:r* = M3

= mg : Weight of the displaced ffuid

Mg:Mass of liquid x 5 Volume of liquid displaced x density x g)

upthrust: weight oftilE displaced fluid.

<- AP+ Qgl",) I

+ AP +qeU) t

[t#r]{.1-.

:>?

Applications of Archimedes's Princinple

r Floating ship displaces water equal to its own weight, icncluding that of cargo.o A submarine sinks by taking water into its buoyancy tanks. Once submerged, the upthurst

is unchanged, but the weight of the submarine increases with the inflow of water and itsinks faster. Cornpressed air is used to blow water out of the tanks when it has to resurface.

Balloons filled with hot air or hydrogen, weight of cold air that it displaces. Thereforece, the

upthrust is greater than its weight and the resultant upward force on the balloon, courses it to rise.

fMSK/Sllee



B B 1 O 1 Engineering Science-T5

TUTORIAL 5

l. State 3 differences between solid, liquid and gas.

2. Define (include with formula and unit)a) density b) pressure

3. Calculate the density of a rectangular block using dimensions as below:Length :27cm, width : 10cm, height : 30 cm, mass : 9509.

4.

5.

0-K/v

6. Calculate the pressure and force on an inspection hatch 0.75m diameter located in thebottom of a tank when it filled with oil of density 875 klm3 to a depth of 7m.

{= xts 'AF. *p= tr01^ =: ;4s:23_*u'

,/ ,/ -.600Ed. z5-

(t1. >t b5/u 7

Thedensity of airl.29kg/^t.Calculatethemassof airinaroomwithdimensionof 15mx 20m x4m.

'-,J3,,:-,r,

L'

Alan uses a knife with a sharp edge and a cross-sectional area of 0.025cm2to cut opena watermelon.

a) Ifthe force applied on the knife is l5N, what is the pressure exerted by theknife on the watermelon.

(' fiapa*

b) After that,tr?cuts open a pineapple using the same knife by exerted a pressure of3X105Pa. Calculate the migintude of foice applied to cut tire pineappll.

JMSK/Sllee

lr-- 4 rw

:g-fi=g=75-y^ : )dy3t v tv




350Nl00X20cm2

7. The figure I shows hydraulic press. The cross-seetional area of piston B iS 20 cm2. The

cross-sectional area of piston A is 100 times the cross-sectional of piston B.

Figure I

a) What is the pressure on piston A that is required to raise a load of 350N on piston B ?

B

8. The cynlindrical piston of a hydraulic jack has a cross-sectional area of 0.5m2 and theplunger has a cross-sectional area of 0.003m2.

. f* E{ls*N+I

b) If position A is pushed down a distance of 3 cm,

raised ?

c) Calculate the cross-sectional area of piston A.

i*, I.l'*

-:

Fignlc

a) The upward force for lifting a load placed in

the downward force on the plunger required.

top of the larger piston is 2000N. Calculate

ilN

how high would

3nr

0 2w\

piston B can beffsuoo ft

b) If the distance moved by the plunger is 50cm, what is the distance moved by the largerpiston ?

A= 0.003n0t I I< 1..- I A**'5nr1

fMSK/Sllee

0 bolwt - 0 -3 c-



I'EF]C E

ruilrlclt( Pfinlq: $liil Sdlhdffr frrhrl, fdr,$rfl



tt- '|'a.l+,

TOPIC 6 : TEMPERATURE & HEAT

WIIAT IS HEAT?

Heat is energy that can be defined as the sum total of the kinetic energies of all the constituent

molecules of abody, or as the intemal energy of a body. Heat, like other forms of energy, can

transform the matter it toucheq heat it and cause molecules to move, or cause a reaction likeburning.

The Universe is made of matter and energy. Matter is made up of atoms and molecules. These

particles and molecules me continuously moving around or vibrating back and forth. The

movement of the atoms and molecules create energy; this energy is also known as heat orthermal energy. If more energy is added to an object its molecules and atoms move faster

inueasing its energy of heat or motion. It does not matter if an object is cold, it still has some

heat energy because its atoms are still in movement.

In a simpler way, heat can be defined as the thermal enerry that transfers from a bodyto another due to temperature difference between the bodies.

The S.L unit of heat isjoule. A calorie ofheat is equal to 4.18 joules.

WHA'T IS TEMPERATI]RE?

The temperature of an object is referred to as the degree of internal energy of a body, which

determines the level ofwarmth or coldness we feel when contact with it. Thus temperature isa measure of the amount of heat present in a body. Or, in a simpler way, temperature is a

measure of the degree of hotness or coldness, and this physical quantity is measured byusing thermometer.

Temperature is measured in Celsius (C) or Kelvin (K). The Kelvin scale is equivalent toCelsius except that it begins at absolute zero andthe Celsius scale begins at the freezing pointof water. Absolute zero is the point at which an object contains absolutely no heat energy. At0 K or -273.15"C, the net internal energy of a body is defined as 0.

The S.I. unit of temperature is Kelvin (K).

One unit in Kelvin scale is equivalent to one unit in Celsius scale, therefore zero Kelvin orabsolute zero is equivalent to -273 "C.

Another example :27"C: (27+273)K : 300K, or xoC: (x+273)K

-lVhatis o Thermometer?

A thermometer is an instrument that measures the temperature of a system in a quantitativeway. The most simple way of doing it is by finding a substance that has a property thatchanges in a regular way with its temperature. One of these substances is mercury. Mercury is

BB 1O1 Engineering Science-T6

fMSK/sllee Page 1



',b

liquid in the temperature range of -38.9 C to 356.7 C. As a triquid mercury expands as it gets

wanner, this expansion is linear and can be accurately calibrated

HEAT TRANSFER

The second law of thermodynamics states that energy or heat transfer can only take place in

one direction, from a higher grade to a lower gade state. This means that heat will only flowfrom hot to cold, never the other way around. This tendency will always achieve an even

distribution of heat energy in an environment with no additional heat input. Heat frartsfer

between objects occurs in three ways: conduction, convection, and radiation.

Conduction

When one end of a rad is heoted. the atoms Iin the rod above the flome]

vibrate faster. bump into the neighbouring atoms and startthem vibrating. Inthis way, the atoms csnduct heat from the hot end to the cosl end. But duringthe process, the ntoms themselves do not move frsm one end of the rod to theother.

5imilorly. heot can be transfemed befween two bsdies e,g,. from o hot drinkinto and along a metal spoon.

{onduction is the tronsfer of heat as a result of the direct contsct of rapidSmoving atoms through a rnedium nr frsm one medium to onother. withoutmovement of the media.

Ittaterials that allow hent to travel through them in ihis way are colled

condu*t$ru, lfietals are good conductors of heat. Non-mefals such os plastic.clay, wood and paper ore poor conductsrs of heat they are also called

insulntorx,

BB 1O1 Engineering Science-T6

lMSKlsllee PageZ



-t.'

Convection

Convection occurs when a liquid 0r gas is in contactwith asolid body ot a differenttemperoture and is always

accompanied by the motion of the liquid or gos. In the kettle,

hotwater rises and cold water descends until ollthe water is

at the snme tempersture. In the atmosphere, convection results

in the movement of hot ond cold oir ,,. winds

f;cntmction is the transfer of heot by the physicol movement

of the heoted medium itself. Convection occurs in liquids ond

gases but not in solids.

Radiation

The sun connot transfer heot to eorth by conduction [becousethere is no physicol contactwith the flflrfh] or convection

fbecouse there is no liquid 0r gfls between them], The sun heats

sur earth by rndiation which does not require contoct or the

presence of any rnotter between them,

Rqdiution is the trsnsfer of hent in the form of waves fhrough

spdce fvocuum]. Dull block surfaces are betfer than white shining

ones at nbsorbing radiated heat,

4n electric heater trcnsfers heat by rndiatiorrfdirectlyrodiating its heat energyl as well as by convection. fn

convection, heoted air near the heater rises and is replaced bycooler air and the cycle repeats thus warming up the entire roam,

TIIER]VIAL E QUILIBRIT]M

Heat can be transferred from one body to another colder body if if they are placed in thermal

contact. The process continues for some times until both bodies achieve the same final

temperature. At this stage, these two bodies would not exchange energy by heat again, , and

there is no net heat transfer of flow This situation is called thermal equilibrium.

,"" " ""-

:[lii '

BB 1 01 Engineering Science-T6

f MSK/sllee PageS



)' ti

IIEAT CAPACITY (C)

The heat capacity of a substance is the amount of heat needed to raise the temperature of the

subtance byone degree Celcius (l"C or lK). The S.I. unit of heat capacrty iJJ'C-t or Jl(r

C_fsie{ ifenf Ete*r6y

'Fnillf f$nl: gre ln ?ra*rr.p*r*r?.r:rnl"F

{ Sf$.{ aI giia - }l+:' }' rl} U

c-

.-- Q

"-aoQ:C4O

SPECIFIC HEAT CAPACITY (c)

The specific heat capacity of a substance is the amount of heat needed to raise the

temperature of I kilogram of a subtance by one degree Celcius (1"C or lK). The S.I. unit ofspecific heat capacity is Jkg-r"C-r or JkgrK-l.

I*f*i *fnssI']

u--r,lAS

Q: -"6

"x F*fetf {trern** i il fr:.er.p* i-*.iare

*C water : 42OO J/kg"C

TUTORIAL 6

l. Define a) Temperature b) Heat

2. State and explain 3 ways of heat transfer.

3. Calculate the heat quantity needed to increase 2 kg of copper from 20oC to 85"C.Given: specific heat capacity of copper: 0.35 KJ/kg"C.

How much heat does 30g of aluminium gives offas it cools from 120oC to 60oC ?

Given: C aluminium : 880J/kg"C.

A l00g silver spoon at20.C is used to mix tea at 95oC. After a few moment, the spoonand tea achieved 88"C. If the mass of the tea is 3009,

a) Calculate the thermal absorbed by spoon.

b) Determine the specific thermal capacity for tea.

Given : specific heat capacity of silver spoon :0.23 KJlkg'C.

4.

5.

BB1O1

f MSK/sllee Page 4



---EI-.-T UFIL T

PEliFftiliK prcmicr Sulton Sslshuddin 4bdul'ziz

SfiEh




TOPIC 7 : BASIC ELECTRICITY

Charge(Q)Electric charge is made up of tiny particles calted atoms; unit in coulomb.

Q: It

Current (I)The rate of charge/elections flow pass a given point in a circuit. The direction of a currenl is

opposite to elections flow direction.

T-O

r- T

Voltage (V)An electrical force that causes current to flow in a circuig measured in volt.It is equal to the amount of work per electric charge that an electric source can do.

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Resistance (R)The opposition of a conductor to the flow of electric current.

It is the ratio of the voltage to the resulting current flow; SI unit is ohm (A).

p - resistivity of material in Qm; L - Length; A - Cross-sectional area

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Capacitance (C)A measure of the amount of charges stored on each plate for a given potential difference

orvoltage appears.between the plates; gI unit is Farad (F).

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Q\---V

Farad : I coulomb of charge stored due to lV applied across the plates.

The energy stored in a capacitor is equal to the amount of work need to cause the voltage

across the capacitor, given as :

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BBlO 1 Engineering Science-T7

Ohm's LawThe main basic electrical law, defines the relationship between voltage, current and resistance,

with one ohm is the resistance value through which I volt maintain I ampere at current.

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*irh-+ll--lC*

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.'. V: lR

Cbnncection of Circuit ElementsConnection of circuit can either in series or in parallel.

Series Connection

Rr R: R: R"

ffi---rAAF

RTerrl= Rr + Ra + R.S, - - Rn

Cl {bCi c*

AC

DC

tHHF--{-1L 1L+-+-+...-

Crotal Cr Cz Cz Cn

Parallel Connection

R1o61=R1 //XrttR://...R' Crotul: Ct + Cr*C3 + ... Co

11LTI

-:

-]---rRtotal Rr Rz R: Rn

Direct Current (DC)Unidirectional flow of electric charge, no change on the direction and the magnitude of the

flow of electric charge.

Alternating Currpnt (AC)Electric current which direction of the flow of elechical charge reverses cyclically (opposed

to direct current), and the magnitude changes in cycle.

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4 .'T,4ottal 7 I '

l- ., (i) Gives TWO (2) types of electric charge

ti

. (ii) Define the definition ofelectric charge, current, voltage and gives SI

units for each

2- l-:..^ firrn /.\ r--- ^3-r-^r:- -:-^--:r- - J J, v,. . Give TWO (2) t1@es of electic circuits and draw treir simple schematic

diagrams. (6mark)

tl: ' If 50 C of an electic charge flows through a wire in l-5 minuteg what is the

current in the wire? (4 marla)t

(2 marks)

(9 marks)

(3 marks)

ts:' A battery with a potential difference of t.Sv is connected to a bulb with a

resistance of 90 O. Calculate the guantity of current flowing through the bulb.

, fcl,o 5] (3 marks)

. 6.'. A 20V potential difference is used to charge a 4 pF capacitor. Calculate the total

charge. ICLO 5] qge7: Give TWO (2) differences between direct current (DC) and altemating current

(Ac). lcl,o s] (4 marks)

8-'. A 25 voltsuppry is used to chargea capacitor of3.5pF. Determinethe

charge stored in the capacitor plates. (4 marks)

f " ' ,. . A larnp X. is lighted'p

for a duration of 6 minutes using a dry.cell. Assume

that the dry cell provides a steady clurent during that duration-

(i) How much charge flows through the lamp X if the ammeter shows a

reading of 0.2 A? (3 marks).

(ii) Determine the number of electons that flows through the lamp X if

.the charge of one electon is l.g x l0rt? (4 ry+)

i:. i e/ic-A.s,, + /.'8 -:^.i:n 4 X tO," et '_' , ,t it --.3;, * 7J a.:, ."r_{'- ' (''

lA.. A torch bulb has a resistance of I O when cold. It drarvs a current of 0.2 A from a

sour.ceof 2 V and glows.{ClO 5l

r- Sketch the flow of curreniintne circuit. (2 marks)

ii- Calculate the resistance ofthe bulb when glowing. t3 marks)

iii. explain the reason of the difference in resistance. (2marks)



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In Diagram 6(c), three resistors 6f 22 gL, I 5 O and 10 O are connected in

parallel to a 6V battery. Calculatc

6V

I'1,Ir

22n

t+t

lz

150

13

IOJL

F

r@

a

. .t"T"t

The total resistance ofthe circuit

. The curreirt '11 , 12 and 13.

'Itre ioial curent flowing in the circuit,

the total resistance

tlre total current

(3 marks)

(6 marls)

(2 marts)

(4

a

."E

Ih.

i,tt

4tJ>oqs"\

'...... - ii.

. -lIl.

1.

ll.

A potential difference of r[ v is appried to u n"tn*t oiieristors asst o*n inthe figure below. ,-, Qr=3lz-

>73,ct,t, o .6 Aq; ii.

t'2+\

tt : 13.

12. A resistorof2 O.F c6'nn{dtao( paf.rcrrr-pornf 8*"-Ro,stors

3 Oand 7f) are in parallel be.tw-een A and B. If a Uattery of 4 \i is connected betwee,n A and.C as shorvn in Figure 6(e), calorlate [CLO 5]

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NOTES FOR ENGINEERING SCIENCE (SEM 1)