Physics for Health Sciences Slides - Copy.pdf

PHYSICS FOR HEALTH SCIENCES

1. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
INTRODUCTION

2. Physicsisthestudyofmatterandenergy,andtheir
interactions.
Measurementistheprocessofassigninganumericalvalueto
aphysicalquantity.
Themeasurementofanyphysicalquantityinvolves
comparisonoftheunknownquantitywithastandard.For
example,lengthsaremeasuredagainstastandardrule,and
timesagainstastandardclock.
Ingeneral,aphysicalquantitywillhavebothanumberanda
unitattachedtothenumber.Thenumberitselfisinsufficient.
Forexample,1meterand1footarequitedifferentquantities.
Thesolutiontoallphysicsproblemsmustincludeunits.

3. Adimensionisaphysicalquantitythatcanbemeasured.
Dimensiondenotesthephysicalnatureofquantity.Aunitisa
particularwaytoassignanumericalvaluetoadimension.A
quantitycanbemeasuredinseveralunits.
Forexample,length(L)isadimension,andmeter(m)isthe
SIunitoflength.
Allunitsforthesamedimensionarerelatedtoeachother
throughaconversionfactor.Theunitsofaquantitytellsus
thekindofquantityitis.
Thedimensionsofmass,lengthandtimearedenotedbyM,L
andTrespectively.
Thedimensionsofanyphysicalquantitycanbeexpressedin
termsofM,LandTonly.

4. Allthetermsofanyequationmusthavethesame
dimensions.
Onlyquantitieswiththesamedimensionscanbeaddedor
subtractedorplacedequaltooneanother.
Thesevenfundamentalorprimarydimensionsarelength,
mass,time,temperature,electriccurrent,amountoflightand
amountofmatter.
Massisthequantityorameasureofthequantityofmatterin
abody.Itisameasureoftheinertiaofabody.TheSIunitof
massisthekilogram(kg).
Volumeistheamountofspaceoccupiedbyabodyoran
object.TheSIunitofvolumeiscubicmeters(m
3
).

5. Avectorquantityisanyquantity,whichhasbothmagnitude
anddirection.
Avectorquantityconveysbothmagnitudeanddirection.The
simplestvectorquantityisdisplacement.
Displacementissimplythechangeinthepositionofan
object.
Ascalarquantityhasmagnitudeonly.Ascalarquantityisa
quantitythatcanbedescribedbyasinglenumber.The
numbertellsusthesizeofthequantity.
Commonexamplesofscalarquantitiesaretemperatureand
mass.Forexample,themassofapatientis70kg.The
number70tellsusthesizeofthemass.Thekgistheunitof
measurement.

6. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
MOTION IN A STRAIGHT LINE

7. Thepositionofanobjectreferstothelocationoftheobject
relativetoachosenreference.
Anobjectissaidtobeinmotionwhenitspositionchanges
withtime.
Mechanicsisthestudyoftherelationshipamongforce,
matterandmotion.Mechanicsdealswiththestudyof
stationaryandmovingobjectsundertheinfluenceofforces.
Matterisanythingthathasweightandoccupiesspace,anda
forceiseitherapushorapull.
Thebranchofmechanicsthatstudiesobjectsatrestiscalled
statics.

8. Dynamicsisthebranchofmechanicsthatdealswithobjects
inmotion.Dynamicsisconcernedwiththeeffectsofforceson
motion.
Thebranchofmechanicsthatisconcernedwiththe
descriptionofmotionwithoutregardtoitscausesisknownas
kinematics.Thus,kinematicsisthestudyofmotionwithout
consideringitscauses.
Distancereferstothelengthofapathfollowedbyanobject.
Thedisplacementofanobjectisdefinedasthechangein
positionoftheobjectinsometimeinterval.Itisanobject’s
distanceanddirectionfromitsstartingposition.

9. Displacementisavectorquantitythatpointsfromanobject’s
initialpositiontoitsfinalposition,anditsmagnitudeequals
theshortestdistancebetweenthetwopoints.
Thedisplacementofanobjectdependsonlyonthestarting
andendingpositions,andnotonthepathtakenbytheobject.
Distanceisanexampleofascalarquantity,whereas
displacementisanexampleofavectorquantity.
Ingeneral,avectorquantityrequiresthespecificationofboth
directionandquantity.Bycontrast,ascalarquantityhasa
numericalvalueandnodirection.
Themassofapatientisascalarquantity,whereasthe
weight,theforceofgravityonthepatient,isavectorquantity.

10. Theaveragespeedofanobject,ascalarquantity,isdefined
astotaldistancedividedbythetotaltimeintervalrequiredto
travelthatdistance.
Theaveragespeedofanobjectisnotthesameasitsspeed
atanyparticularinstantoftime.Speedatanyparticular
instantoftimeiscalledinstantaneousspeed.
Averagevelocityofanobjectisdefinedasdisplacement
dividedbythetimeintervalduringwhichthedisplacement
occurs.
Themagnitudeofaveragevelocityisnotaveragespeed.
tan

dis e
AverageSpeed
time


Displacement
AverageVelocity
Time


11. Thevelocityofanobjectataparticularinstantintimeis
calledinstantaneousvelocity.
Theinstantaneousvelocityofanobjectindicateshowfastthe
objectismovingandthedirectionofmotionateachinstantof
time.Themagnitudeofinstantaneousvelocityiscalled
instantaneousspeed.
Inphysics,thereisacleardistinctionbetweenspeedand
velocity.Speed,ascalarquantity,istherateatwhichan
objectchangesitsposition.Incontrast,velocity,avelocity,a
vectorquantitydenotesthespeedofanobjectwithits
associateddirection.
Theaccelerationofanobjectisthetimerateofchangeofits
velocitywithtime.

12. Theaverageaccelerationofanobjectisdefinedasthe
changeinvelocitydividedbythetimeintervalduringwhich
thatchangeoccurs.
Instantaneousaccelerationistheaccelerationataparticular
instantintime.


ChangeinVelocity
AverageAcceleration
Time


13. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
EQUATIONS OF UNIFORMLY
ACCELERATED MOTION

14. Uniformaccelerationmeansconstantacceleration.
2
2 2
(1)
1
( ) (2)
2
1
(3)
2
2 (4)
v u at
s u vt
s ut at
v u as
  
  
  
  

15. tan
u initialvelocity
v finalvelocity
s dis ce
t time
a acceleration






16. Equation(1)enablesustodeterminethevelocityatany
instantoftime,giventheinitialvelocityandtheconstant
acceleration.
Equation(2)simplysaysdistanceisequaltoaveragevelocity
timestime.Formotioninastraightline,distance,lengthof
travel,isthesameasmagnitudeofdisplacement.
Equation(3)givesthepositionordistanceoftheobjectatany
instantoftime,giventheinitialvelocity.
Themostimportantproblemtowhichtheequationsof
constantaccelerationapplyisthatofthefreefallofobjects
neartheearth.

17. Intheabsenceofair,allobjectsinthesamelocationfallto
earthwiththesameacceleration.Thisaccelerationiscalled
accelerationduetogravity.
Thevalueofthisaccelerationvariesslightlywithgeographical
location,butanywhereonearth,themagnitudeis
approximately9.8m/s
2
.
Thedirectionoftheaccelerationduetogravityisdownward
towardstheearth.
Forobjectsfallingverticallydownwards,itispositive.For
objectsmovingverticallyupwards,itisnegative.

18. 1.2 / 8.3 /
5 .
. min
.
Anobjectisobservedtoincreaseits
velocityfrom m sto m sina
timeof sCalculateitsaverage
accelerationDeter ehowfarit
travelswhileaccelerating

19. 1.2 /

8.3 /
5

u initialvelocity
ms
v finalvelocity
ms
t time
s
a averageacceleration








20. 2
8.3 / 1.2 /
5
1.42 /
v u
a
t
ms ms
s
ms





1
( )
2
1
(1.2 / 8.3 /)(5)
2
23.75
s u vt
ms ms s
m
 
 

2 2
2
,
2
1
2
alternatively
v u
s
a
OR
s ut at


 

21. 2

21 / .
,
int ?

10 / .
Abodyisthrownverticallyupwards
withaninitialvelocityof m sHow
highwillitgo andwhenwillitreturn
tothepo ofprojection Assumethe
valueof accelerationduetogravityis
m s

22. 2
.

21 /

0 / ( )
10 / ( )
Considertheupwardmotion
u initialvelocity
ms
v finalvelocity
ms momentaryrest
a acceleration
ms negativebecauseof upwardmotion







23. 2 2
2 2
2
2
tan ( )
2
(0 / ) (21 / )
2( 10 / )
22.05
( int)
0 / 21 /
10 /
2.10
s dis ce howhightheobjectgoes
v u
a
ms ms
ms
m
t time toreachthehighestpo
v u
a
ms ms
ms
s















24. 2
2 2
.

0 / ( )
tan ( int )
22.05
int
1
2
1
22.05 (0 / ) (10 / )
2
2
Consider thedownward motion
u initial velocity
ms momentaryrest
s dis ce fromhighest topo of projection
m
t timetoreturntoprojectionpo
s ut at
m ms t ms t
t





 
 
.1

2.1 2.1
4.2
s
T timetoreturn
s s
s

 


25. 2

2.5 .
- ?
10 / .
Kangarooscanjumpverticallyupwards
toaheightof cmWhatistheir
take offvelocity Assumethevalueof
accelerationduetogravityis ms

26. 2
2 2
2 2
10 /
tan ( )
2.5

0 / ( )
( - )
2
(0 /) 2( 10 / )(2.5 )
7.07 /
a acceleration
ms
s dis ce height
m
v finalvelocity
ms momentaryrest
u initialvelocity take off velocity
u v as
ms ms m
u ms







 
  


27. .
12 cov 100 ,
?
,
Anobjectmovesinastraightlinewith
uniformaccelerationIfitstartsfrom
restandtakes sto er m what
istheacceleration Ifitcontinueswith
thesameacceleration howlongwillit
ta cov 100
200 ?
keto erthenext mandwhat
willbeitsvelocityafterthe m

28. 2
2
2

0 / ( )
tan
100
cov 100
12
1
2
1
100 (0 / )(12 ) (12 )
2
1.4 /
u initialvelocity
ms startsfromrest
s dis ce
m
t timeto erfirst m
s
a acceleration
s ut at
m ms s s a
a ms







 
 


29. 2
tan
100 100
200
1.4 /
cov 200
s dis e
m m
m
a acceleration
ms
t timeto er mfromrest

 





30. 2
2 2
1
2
1
200 (0 /) (1.4 / )
2
16.9
cov 100
16.9 12
4.9
s ut at
m mst ms t
t s
T timerequiredto erthenext m
s s
s
 
 


 


31. 2
2 2
2 2
1.4 /
tan
200

0 /

2
(0 /) 2(1.4 / )(100 )
23.7 /
a acceleration
ms
s dis ce
m
u initialvelocity
ms
v finalvelocity
v u as
ms ms m
v ms







 
 


32. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
NEWTON’S LAWS OF MOTION

33. Incommonusage,aforceisapushorapull.Moreformally,a
forceisanythingthatcausesachangeinthevelocityofan
object.
Contactforcesareforcesthatarisefromthephysicalcontact
betweentwoobjects.Anexampleofacontactforceisthe
forceexertedbyyourfeetonthefloor.
Fieldforcesdonotinvolvephysicalcontactbetweenobjects,
andtheyactthroughemptyspace.Anexampleisthe
gravitationalforceofattractionbetweentwoobjects.Another
exampleiselectricforcethatelectricchargeexertson
another.
Newton’sfirstlawofmotionstatesthatanobjectcontinuesin
astateofrestorinastateofmotionatconstantvelocity,
unlesscompelledtochangethatstatebyanetaforce.

34. Newton’sfirstlawofmotionisalsocalledthelawofinertia.
FromNewton’sfirstlaw,whennoforceactsonanobject,the
accelerationoftheobjectiszero.
Newton’sfirstlawofmotionindicatesthatanobjectwill
remainatrestorinmotionwithconstantvelocityunlesssome
externalforceactsonit.
Constantvelocitymeansconstantspeedinaconstant
direction.Thus,accordingtoNewton’sfirstlawofmotion,the
objectcannotchangespeedanddirectionunlessanetforce
actsonit.
Quantitatively,theinertiaofanobjectismeasuredbyits
mass.Atrainhasmoreinertiathanabicycle.

35. Inertiaisthenaturaltendencyofanobjecttoremainatrestor
inmotionataconstantvelocity.Themassofanobjectisa
quantitativemeasureofinertia.Thus,atrainhasagreater
tendencytoremainatrestthanabicycle.
Pleasenotethatthemassofanobjectisindependentofits
location.
Inertiaistheinherenttendencyofanobjecttoresistany
attempttochangeitsvelocity.
FromNewton’sfirstlawofmotion,aforcecanbedefinedas
thatwhichcausesachangeinmotionofanobject.
ThenetforcementionedinNewton’sfirstlawofmotionisthe
vectorsumofallforcesactingontheobject.

36. Newton’ssecondlawofmotiondealswithwhathappens
whenanetforceactsonanobject.
Newton’ssecondlawofmotionstatesthatwhenanet
externalforceactsanobject,theaccelerationthatresultsis
directlyproportionaltothenetforceandhasamagnitude
inverselyproportionaltothemassoftheobject.
Thedirectionoftheaccelerationisthesameasthedirection
ofthenetexternalforce.
ThenetforceinNewton’ssecondlawofmotionincludesonly
theforcesthattheenvironmentexertsontheobject.Such
forcesarecalledexternalforces.
Massisthatpropertyofanobjectthatspecifieshowmuch
resistanceanobjectexhibitstochangesinitsvelocity.

37. Newton’ssecondlawofmotioncanbeexpressed
mathematicallyasfollows.
Thenetforceactingonanobjectisthevectorsumofall
forcesactingontheobject.Itissometimesreferredtoastotal
force,resultantforce,orunbalancedforce.
TheSIunitofforceistheNewton(N).


F ma
F netexternalforce
m massof object
a acceleration




2
1 1 /N kgms

38. Themassofanobjectisanintrinsicproperty,whichcanbe
takenasabasicmeasureoftheamountofmatteritcontains,
andasameasureofitsinertia.
Themassisafundamentalconstantforagivenobject,which
doesnotchangewithotherphysicalpropertiesandwitha
changeinitsstateofmotion.
Themomentumofanobject,avectorquantity,isdefinedas
theproductofitsmassandvelocity.
Newton’ssecondlawofmotioncanalsobestatedasthenet
externalforceactingonanobjectisequaltotherateof
changeofmomentumoftheobject.

39. Forceisavectorquantity,andthereareoftenseveralforces
actingindifferentdirectionsonasingleobject.
TheforceinNewton’ssecondlawofmotionisthenetforce
thatisthesumofallforcesactingontheobject.
Infindingthesumofforcesactingonanobject,itis
necessarytotakeintoaccounttheirdirections.Thisiscalled
vectorsum.
Newton’sthirdlawofmotionstatesthatwheneveroneobject
exertsaforceonasecondobject,thesecondobjectexertsa
forcebackonthefirstobjectthatisequalinmagnitudeand
oppositeindirection.
Newton’sthirdlawofmotionissometimesstatedas:forevery
action,thereisanequalandoppositereaction.

40. 2
1,100
0.15 / .
tan 150 ,

?
Acaravanof mass kgistowedby
acarwithanaccelerationof ms
If theresis cetomotionis N what
istheforceexertedbythecarthroughthe
towbar

41. 2
2

1,100
0.15 /
tan
150

(1,100 )(0.15 / ) 150
315
m massof caravan
kg
a acceleration
ms
R resis cetomotion
N
F forceexertedbycar
F R ma
F ma R
kg ms N
N







 
 
 


42. tan 50

( ).
2.0 1.5,
?
Acons thorizontalforceof Nacts
onanobjectonasmoothhorizontalplane
nofriction If theobjectstartsfromrest
andisobservedtomove min s what
isthemassof theobject

43. 2
2
50
0 / ( )
tan 2.0
1.5
1
2
1
2.0 (0 /)(1.5) (1.5)
2
F force N
u initialvelocity ms startsfromrest
s dis ce m
t time s
a acceleration
s ut at
m ms s s a
 
 
 
 

 
 

44. 2
2
1.78 /

50
1.78 /
28.10
a acceleration
a ms
m massofobject
F
a
N
ms
kg







45. Theearthattractsallobjectstoitself,andobjectsfall
downwardbecauseofgravity.
Everyobjectexertsanattractiveforceoneveryotherobject.
Newton’slawofuniversalgravitationstatesthatthe
magnitudeoftheattractiveforcebetweentwoobjectsis
directlyproportionaltotheproductoftheirmassesand
inverselyproportionaltothesquareofthedistancebetween
them.
Mathematically,themagnitudeoftheattractiveforcebetween
twoobjectsisgivenby:

46. 1 2
2
1
2
11 2 2



tan
tan
6.674 10 /
Gmm
F
R
F magnitudeof attractiveforce
m massof objectone
m massof objecttwo
R dis ceorseparationbetweenobjects
G universalgravitationalcons t
Nm kg







 

47. Theweightofanobjectonearthisthegravitationalforcethe
earthexertsontheobject.Theweightofanobjectexists
becauseofthegravitationalpulloftheearth.
Theweightalwaysactsdownward,towardthecenterofthe
earth.
Themagnitudeoftheweightofanobjectonthesurfaceof
theearthisgivenby:



W mg
W magnitudeof weight
m massof object
g accelerationduetogravity





48. Thevalueofaccelerationduetogravityonthesurfaceofthe
earthisgivenby:
Massisaquantitativemeasureofinertia.Therefore,massis
anintrinsicpropertyofmatteranddoesnotchangeasan
objectismovedfromonelocationtoanother.
Incontrast,weightisthegravitationalforceactingonan
objectandcanvarydependingonhowfartheobjectisabove
theearth’ssurface.
2
2
9.80 /
tan


E
E
E
E
GM
g ms
R
G universal gravitational cons t
M massof earth
R radiusof earth
 




49. Theweightofanobjectisproportionaltothemassofthe
object,butitisdistinctfromit.
Massisafundamentalmeasureofanobject’sinertiaandthe
amountofmatterpresentinit.Itdoesnotchangewiththe
positionoftheobject.
Incontrast,theweightofanobjectvariesfromlocationto
location.
However,thesevariationsareusuallyinsignificant.
Itisincorrecttosayapatientweighs70kg.Weightisaforce
andismeasuredinNewtons(N).

50. 2

100 ?
100
(100 )(9.80 / )
980
Whatistheweightofamanof
mass kg
m kg
W mg
kg ms
N





51. Theforceofgravityactsonallpartsofanextendedobject.
Eachpartexperiencesitsindividualforce.
Alltheseforcesmustbeovercomeiftheobjectistobelifted.
Thepointatwhichalltheforcesofgravitymaybeconsidered
tobeconcentratedforpurposesofbalancingandliftingis
calledcenterofgravityorcenterofmass.
Thecenterofgravityofasolidobjectisthesinglepointat
whichtheforceofgravitycanbeconsideredtoact.
Forauniform,symmetricobject,thecenterofgravityisatthe
geometriccenter.Forexample,thecenterofgravityofa
uniformbarisatitscenter.

52. Tensionisthetypeofforcewhichexistsinthebodyofa
flexiblecable,string,orrope.
Thecableorropeisassumedtobeinfinitelyflexible,and
hencenolateralforcesaresupported.
Anyforce,whichisnotdirectedalongcableresultsinbending
ofthecable,andhencetheonlyforces,whichexistinsidethe
cablearethosedirectedalongthecable.
Thecableisalsoassumedtobeinelastic:itdoesnotstretch,
anditisunbreakable.
Sincethecableisassumedtobeflexible,itcanonlypulland
onlyexertsaforcealongitslength.

54. Staticsisthestudyofobjectsinequilibriumsuchasbuildings,
bridgesorpatientsintraction.
Anobjectisinequilibriumwhenitsaccelerationiszero.An
objectinmotionwithaconstantvelocityisalsosaidtobein
equilibrium(thistypeofequilibriumiscalleddynamic
equilibrium).
Anobjectissaidtobeinequilibriumifthesumofforces
actingonitiszeroandthesumofthetorquesexerteduponit
iszero.
Torqueistheturningeffectofaforce.
Anyobject,whichisbalancedandremainsinafixedposition
issaidtobeinstaticequilibrium.

55. Therearetwoconditionsforstaticequilibrium:First,sumof
upwardforcesmustequalsumofdownwardforces,andsum
ofrightforcesmustequalsumofleftforces.Second,sumof
torquesaboutanypointmustequalzeroi.e.sumofclockwise
torquesmustequalsumofcounterclockwisetorques.
Therateatwhichanobjectinequilibriumisrotatingcannot
change.Tochangetherateofrotationrequirestheapplication
ofatorque.Torqueistheeffectivenessofaforceinproducing
rotation.Itistheturningeffectofaforce.
Threefactorsdeterminethetorqueproducedbyaforce.First,
themagnitudeoftheforce.Second,thedirectionoftheforce
(whichdeterminesthedirectionofrotation).Third,the
distancebetweenthepointofapplicationoftheforceandthe
pivotpoint.

56. Torqueistheproductoftheperpendicularcomponentofthe
forceappliedtoanobjectandthedistanceofthepointofthe
applicationoftheforcefromthepivot.
Foranobjecttobeinequilibrium,itmustsatisfytwo
conditions:thenetexternalforceactingontheobjectmust
equalzero,andthenetexternaltorqueappliedtotheobject
mustequalzero.

tan int

Torque F d
F perpendicularcomponentof appliedforce
d dis cefromthepivottothepo of
applicationof theforce
 



66. Aforceisavectorquantitysinceithasbothmagnitudeand
direction.
Vectoranalysisorvectorsumisusedtohandleforcesthat
arenotsimplyparallelorperpendiculartoeachother.
Itissometimesconvenienttosplitorbreakasingleforceinto
twoforceswhosecombinedorresultanteffectisequivalentto
thatofthesingleforce.Thatis,thesumofthetwoforces
equalstheoriginalforce.
Thistechniqueorprocessiscalledprojectionorresolutionof
aforceintoitscomponents.Thetwoforcesarecalled
componentsoftheoriginalforce.
Forconvenience,aforceisusuallyresolvedintohorizontal
andverticalcomponents.

67. Thecomponentofaforceinanydirectionistheproductof
themagnitudeoftheforceandthecosineoftheangle
betweentheforceandthedirection.


x
x
y
y
F horizontalcomponent
F FCos
F verticalcomponent
F FSin







68. Wheneverthereisrelativemotionbetweentwosurfaces,
whichareincontact,therewillbeaforcewhichresiststhis
motion.Thisforceiscalledfriction.
Frictionisaforcewhichresistsmotion.Thesizeofthe
frictionalforcedependsonthenatureofthetwosurfaces.
Besidesthenatureofthesurfaces,thefrictionalforce
dependsmainlyupontheamountofforcepressingthetwo
surfacestogether.
Thefrictionalforceisproportionaltotheforceexertedbyone
substanceonanotherperpendiculartothesurfacebetween
them.Thisforceiscalledthenormalforce.

69. Therearetwotypesoffriction.Thesearestaticfrictionand
kineticfriction.
Staticfrictionisthefrictionbetweentwoobjects,whichare
notmovingrelativetooneanother.
Incontrast,kineticfrictionisthefrictionbetweentwoobjects,
whicharemovingrelativetooneanother.
Frictionalforcesaregenerallylessforslidingobjectsthanfor
stationaryobjects.
Thefrictionalforcebetweentwosurfacesisproportionalto
theforcepressingthesurfacestogether,whichiscalledthe
normalforce.Thenormalforceisatrightanglestothe
surface.

70. Insummary,kineticfrictionisfrictionbetweenslidingor
slippingsurfaces.Staticfrictionisfrictionbetweenstationary
surfaces.





k k N
k
k
N
S
s
F F
F kineticfriction
Coefficientofkineticfriction
F normalreaction
F staticfriction
Coefficientofstaticfriction










71. Consideranobjectrestingonaninclinedsurface.

72. c
angleof inclination
W mg weightof object
mgSin Componentof weightparallelto
surface
mgCos Componentof weightperpendicular
tosurface
F reactionforceproducedbysurface
thatbalancesweightof o




 



bject

73. ,
c
c
N Componenntof F perpendicular
tosurface
normalforce
f Componentof F paralleltosurface
frictionalforce
If theobjectisstationaryontheinclined
surface then
N mgCos
f mgSin









76. 2
10 , ?
,
3 ?
If anobjectof mass kgisactedupon
byaforceof N whatistheacceleration
If itisinitiallyatrest whatwillbeits
velocityafter s

77. 2
2
10
10
2
5 /
m massof object kg
F force N
a acceleration
F
m
N
kg
ms
 
 




2
0 /
3

0 / (5 / )(3)
15 /
u initialvelocity ms
t time s
v finalvelocity
ms ms s
ms
 
 

 


78. 2

10
. .

10 / .
Achildandbasketwithatotalmassof
kgaresuspendedfromascalebya
cord Calculatethetensioninthecord
Assumeaccelerationduetogravityis
ms

79. 2
2


10

(10 )(10 / )
100

,
0 /
'
0
Let T tensioninthecord
m massof babyand basket
kg
W weight of babyand basket
mg
kg ms
N
Sincethebabyand basket are
not accelerating
a acceleration ms
fromNewton sSecond Law
T W







 
 
100T W N 

80. 8.0
. 30
45 .
.

Twochildrenpullinoppositedirections
anobjectofmass kgonasmooth
surfaceOneexertsaforceof Nand
theotheraforceof N Bothpull
horizontallyCalculatetheacceleration
oftheobj.ect

81. 1
2

. ,



30
45

N
Therearefourforcesactingonthe
object Theseare
W mg weightof object
F normalreactionof thesurface
ontheobject
F forceexerted bychild one N
F forceexerted bychild two N
Sincetheobj
 

 
 

, ' sec
,
. ,
N
ectisnotmovingupor
down thenfromNewton s ond
lawof motion thenetforceinthe
verticaldirectioniszero Therefore
F W

82. 2 1
2 1
2 1
2

' sec
45 30
8.0
1.875 /
H
H
LetF netforceinhorizontaldirection
F F
fromNewtons ondlaw
F ma
F F ma
F F
a
m
N N
kg
ms

 

 






83. 2
50.0
.
90.0 .
0.35 / ,

Anurseexertsahorizontalforceof N
onagurneywithapatientonit Thegurney
andpatienthaveatotalmassof kg If
thegurneyandpatientaccelerateat ms
whatisthemagnitudeof
?
thefrictionalforce
actingonthegurney

84. 2
2
50.0

90.0
0.35 /
' sec ,
50.0 (90.0 )(0.35 / )
18.5
R
R
R
F forceexertedbynurse N
F frictionalforce
m massgurneyandpatient kg
a acceleration ms
fromNewtons ondlaw
F F ma
F F ma
N kg ms
N
 

 
 
 
 
 


85. 2
70.0 lim .
lim ,
.
0.20 / ,
.
A kggymnastc bsaropeIf the
gymnastc bsatauniformspeed
calculatethetensionintheropeIf the
gymnastacceleratesupwardat ms
calculatethetensionintheropeAssume
accele
2
10 / .rationduetogravityis ms

86. 2


70.0


, 0.
' sec ,
0
(70.0 )(10 / )
700
LetT tensioninrope
m massof gymnast
kg
W weightof gymnast mg
If thegymnastcimbsatunifiom
speed thenaccelerationis
FromNewtons ondlaw
T W
T W mg
kg ms
N



 
 
 



87. 2
2 2
0.20 /
' sec ,
(70 )(0.20 / ) (70 )(10 / )
714
a acceleration ms
FromNewtons ondlaw
T W ma
T ma W
T ma mg
T kg ms kg ms
N
 
 
 
 
 


88. 2
1,500
0.50 / .
125 .
?

Acarpullinga kgtrailerproduces
anaccelerationof m s Thefrictional
forceinthewheelsof thetraileris N
Whatforceisthecarexertingonthetrailer
Whatforceisthetrailerex ?ertingonthecar

89. 2
2
1,500
0.50 /
125

' sec ,
(1,500 )(0.50 / ) 125
875

R
R
R
T
m massof trailer kg
a acceleration ms
F frictionalforce N
LetF forceexerted bycarontrailer
fromNewton s ond law
F F ma
F ma F
kg ms N
N
LetF forceexerted b
 
 
 

 
 
 


' ,
875
T
ytraileroncar
fromNewton sthird law
F N

90. ker ,
, 250
.
?
,
Twowor spushhorizontally inthesame
direction awoodencrateof mass kg
thatisonawoodenfloor Whattotalforce
musttheyexerttostartthecratemoving
Oncemoving oneexertsaforceo 500
600 .
.
.5
0.3.

f N
andtheotheraforceof N Calculate
theaccelerationof thecrate Assume
coefficientof staticfrictioniso and
coefficientof kineticfrictionis Take
accelerationduetogravitya
2
10 / .s ms

91. 1
2
250


ker 1
ker 2
0.5
N
s
k
m massof woodencrate kg
W weightof woodencrate
mg
F normalreactiononwoodencrate
W
mg
F forceexertedbywor
F forceexertedbywor
coefficientof staticfriction
coefficien


 







 
 0.3tof kineticfriction

92. ker
,

.
(0.5)(
s
k
s
R
s N
R
k N
T
T R s N s
F staticfriction
F
F kineticfriction
F
LetF totalforceexerted bywor s
Forthecratetostartmoving the
totalforcemustbeatleastequalto
thestaticfriction
F F F mg


 





  

2
250 )(10 / )
1,250
kg ms
N

93. 1
2
1 2
1 2
1 2
2
,
.
500
600
' sec ,
500 600 (0.3)(250 )(10 / )
250
k
k
net
R
R
k N
Oncethecratestartsmoving thefriction
changesfromstatictokinetic
F N
F N
fromNewton s ond lawof motion
F ma
F F F ma
F F F
a
m
F F F
m
N N kg ms
k




  
 

 

 

2
1.4 /
g
ms

94. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
WORK, ENERGY AND POWER

95. Ifapersonexertsaforceonaboxanditmovesadistance
acrossthefloor,thepersondoesworkonthebox.
Theworkdoneonanobjectbyaforceisdefinedasthe
productofthecomponentoftheforceinthedirectionof
motionandthedistancemovedbytheobject
Pleasenotethatonlythecomponentoftheforceinthe
directionofmotionisusedincalculatingtheworkdone.



tan
W f d
W workdone
f componentof forceinthedirection
of motion
d dis ce
 




96. Workisascalarquantity,andtheSIunitofworkistheJoule
(J).
Theworkdoneonanobjectbyanetexternalforceisequalto
thechangeinkineticenergyoftheobject.Thisisknownas
thework-energyprinciple.
Energyisdefinedastheabilityorcapacitytodowork.Itisa
scalarquantityandismeasuredinJoules.
Energycantakeseveralformssuchasheatenergy,electrical
energy,chemicalenergy,nuclearenergyandmechanical
energy.
Theprincipleofconservationofenergystatesenergycannot
becreatedordestroyed,butcanbeconvertedfromoneform
toanother.

97. Mechanicalenergyisusuallydividedintotwoclasses:
potentialenergyandkineticenergy.
Potentialenergyistheenergypossessedbyanobjectdueto
itsshapeorposition.
Potentialenergyduetopositioniscalledgravitational
potentialenergy.Thegravitationalpotentialenergyofan
objectistheenergypossessedbyanobjectbyvirtueofits
positionrelativetothesurfaceoftheearth.
Forexample,apolytankontopofabuildinghasgravitational
potentialenergy.

98. Gravitationalpotentialenergyisoftenreferredtosimplyas
potentialenergy.
Kineticenergyistheenergypossessedbyanobjectbyvirtue
ofmotion.
Forexample,amovingcarhaskineticenergy.Incontrast,a
stationarycarhasnokineticenergy.




GPE mgh
GPE gravitationalpotentialenergy
m massof object
g accelerationduetogravity
h heightof object






99. Thekineticenergyofanobjectdependsonitsmassand
speed.
21
2



KE mv
KE kineticenergy
m massofobject
v speedofobject





100. Powerisdefinedastheamountofworkdoneperunittime.It
istherateatwhichenergyisusedorexpended.
Powerisascalarquantity,anditismeasuredinwatts(W).
Thehumanbodycanbethoughtofasanenergyconverter.
Thesourceofenergyforthehumanbodyischemicalenergy
(foundinfood),whichthebodyconvertsintomechanicalwork
andthermalenergy,andbackintostoredchemicalenergy
(fat).
Energy
Power
Time
Energy Power Time

 

101. 0.025 tan 10.0
300 / ?
Whatforceisneededtostopanobjectof
mass kginadis ceof cmif
theobjecthasaspeedof ms

102. 2 2
0.025
300 /
0 /

1 1
2 2
tan 10 0.10

m massofobject kg
u initialspeed ms
v finalspeed ms
KE changeinkineticenergy
mu mv
s dis ce cm m
LetF requiredforce
 
 
 
 
 
  


103. 2 2
2 2
- ,
1 1
2 2
1 1
0.10 (0.025 )(300 / ) (0.025 )(0 / )
2 2
0.10 1,125
1,1250
fromthework energyprinciple
F s mu mv
F m kg ms kg ms
F m J
F N
  
  
 


104. 100 .
tan 5.0 .
1.0 /
80
Anambulanceattendantpusheshorizontally
onagurneywithaforceof N Hestarts
fromrestandpushesforadis ceof m
Thefinalspeedis msandthemassof the
gurneypluspatientis k.
.
.
g Calculatethe
amountof workdoneagainstfriction
Calculatetheforceof friction

105. 100
tan 5.0
0 /
1.0 /
80



R
f
R
f
F forceexertedbyattendant N
d dis ce m
u initialspeed ms
v finalspeed ms
m massof gurneypluspatient kg
W workdonebyF
F forcefriction
W workdoneagainstfriction
 
 
 
 
 




106. 2 2
2 2
=100 5.0
500

1 1
2 2
1 1
(80 )(1.0 / ) (80 )(0 / )
2 2
40

,
500 40
460
R
f
f
W F d
N m
J
KE changeinKEof gurneypluspatient
mv mu
kg ms kg ms
J
fromtheprincipleof conservationof
energy
W J J
J
 


 
 
 

 


107. 2 2 2 2
2
460
5.0
92

(1.0 / ) (0 / )
0.1 /
2 2(5.0 )
R
R
f R
f
R
W F d
W
F
d
J
m
N
thequestioncanalsobesolvedinthismanner
v u ms ms
a ms
d m
 



 
  

108. 2
' sec ,
100 (80 )(0.1 / )
92
92 5.0
460
R
R
R
f R
fromNewtons ondlaw
F F ma
F F ma
N kg ms
N
W F d
N m
J
 
 
 

 
 


109. 2

10 .
4 ?
10 / .
Acoconutfallstoearthfroma
heightof mHowfastisittravelling
whenitis mfromtheground Assume
accelerationduetogravityis ms

110. 1
2
1 2
2 2
10.0
4.0

0 /


1 1
2 2
h initialheight m
h finalheight m
PE changeinpotentialenergy
mgh mgh
u initialspeed ms
v finalvelocity
KE changeinkineticenergy
mv mu
 
 
 
 
 

 
 

111. 2 2
1 2

,
1 1
2 2
fromtheprincipleof
conservationof energy
PE KE
mgh mgh mv mu
 
  
 
2 2
1 2
2 2 2
1
( ) ( )
2
1
(10 / )(10.0 4.0 ) (0 /)
2
10.95 /
mgh h mv u
m ms m m m v ms
v ms
  
  


112. 2
10
30 .
tan 20 .
10 / .
1.
Ablockof mass kgslidesdownaplane
inclinedat tothehorizontal Theforce
of frictionalongtheplaneisacons t N
Assumeaccelerationduetogravityis ms
Computetheaccelerati


.
2. ,
2.0 ?
onof theblock
downtheinclinedplane
Startingfromrest howlongdoesittake
theblocktomove mdownthetheplane

113. 2
30
10


20
' sec ,
(10 )(10 / )
p
R
R
R
angleof inclination
m massof block kg
W mg weightof block
W componentof weightparalleltoplane
mgSin
F frictionalforce N
fromNewton s ond law
mgSin F ma
mgSin F
a
m
kg ms




  
 
 


 
 



2
30 20
10
3 /
Sin N
ms



114. 2
2
2 2
3 /
0 /
tan 2.0
1
2
1
2.0 (0 /) (3 / )
2
0.385
a acceleration ms
u initialvelocity ms
s dis ce m
s ut at
m mst ms t
t s
 
 
 
 
 


115. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
FLUIDS AND PRESSURE

116. Afluidisanymaterialthathastheabilityorcapacitytoflow.
Thus,afluidiseitheragasoraliquid.
Pressureisdefinedasforceperunitarea.
TheSIunitofpressureisPascal(Pa).
Commonunitsofpressureareatmospheres(atm),mmHg
andTorr.
Force
pressure
Area

2
1 1 /Pa Nm
1 760 101,325
1 1
atm mmHg Pa
mmHg Torr
 


117. Gaugepressureispressuremeasuredrelativetoatmospheric
pressure.
Absolutepressureortotalpressureismeasuredrelativeto
vacuum.
Bloodpressureisgaugepressure.Forexample,ifthemean
arterialpressureofapatientis75mmHg,itmeansitis75
mmHgmorethanatmosphericpressure.


1 760
Total gauge atm
Total
gauge
atm
P P P
P totalorabsolutepressure
P gaugepressure
P atmosphericpressure atm mmHg
 


  

118. 2
max , ,

max
140
25 .
Calculatethe imumforce inNewtons
exertedbythebloodonananeurysmin
theaortagiventhatthe imumblood
pressureis mmHgandtheareaof the
aneurysmis cm

119. 4
4 2
2
2 2
2
2 2
2
3 2
140
760 101,325
140
140 101,325
760
1.867 10
1.867 10 /
25.0
10,000 1.0
25.0
25.0 1.0
10,000
2.50 10
pressure mmHg
mmHg Pa
mmHg
mmHg Pa
mmHg
Pa
N m
area cm
cm m
cm
cm m
cm
m



 
 
 


 
 

120. 4 2 3 2
1.867 10 / 2.50 10
46.8
force
pressure
area
force pressure area
Nm m
N


 
   


121. Thepressureatadepthinafluidisduetotheweightofthe
columnoffluidabovethedepth.Thispressureiscalled
pressureduetothecolumnoffluid,andisagaugepressure.




depth
depth
P gh
P pressureatadepthdueto
columnof fluid
densityoffluid
g accelerationduetogravity
h depth








122. 3

2.50
.
? 1,000 / .
1
CalculatethepressureinPaatadepth
of mduetowaterinaphysiotherapy
bathWhatisthetotalpressureatthat
depth Thedensityof wateris kgm
Assumeaccelerationduetogravityis
2
0 / .ms

123. 3
2
3 2
2.50
1,000 /
10 /
(1,000 / )(10 / )(2.50 )
25,000
25,000 101,325
126,325
depth
Total depth atm
h depth m
densityof water kgm
g accelerationduetogravity ms
P gh
kgm ms m
Pa
P P P
Pa Pa
Pa


 
 
 



 
 


124. Pascal’sprinciplestatesthatanypressureappliedtoa
confinedfluidistransmittedequally(undiminished)toallparts
ofthefluid.
Archimedes’principlestatesthatanyobjectplacedinafluid,
eitherpartiallyorwholly(completelyorimmersed)will
experienceanupwardforceequaltotheweightofthefluidit
displaces.
Theupwardforceiscalledbuoyantforceorupthrust.
Thebuoyantforcemakesanobjectweighlesswhenitis
placedinafluid.
Theweightofanobjectinairiscalleditstrueweight.The
weightwhenplacedinafluidiscalledapparentweight.

125. B
B
B True App
True
App
F gV
F bouyantforceorupthrust
densityoffluid
V volumeofobjectinfluid
g accelerationduetogravity
F W W
W trueweight
W apparentweight







 



126. 3
log
80.0
2.0 .
1,000 / .

Aphysio istmeasuresthemassof a
personinairtobe kgandhisapparent
masswhensubmergedinwatertobe kg
Thedensityof wateris kg m Calculate
thevolumeanddensityof theper.son

127. 3
80.0
2.0
1,000 /

True
App
w
B w
B True App
m truemassof person kg
m apparentmassof person kg
densityof water kgm
LetV volumeof person
F gV
F W W


 
 
 


 

128. 3
3
( )
( )
80.0 2.0
1,000 /
0.0780
True App w
True App w
True App w
True App
w
W W gV
gm gm gV
g m m gV
g m m
V
g
kg kg
V
kgm
m




 
 
 





3 3
3
80.0
1.02 10 /
0.0780
p
mass kg
kgm
volume m
   

129. Flowrateisdefinedasvolumeflowingperunittime.Volume
flowingperunittimeisproperlycalledvolumetricflowrate.
Flowratebetweentwopointsorregionsinapipeortube
(suchasabloodvessel)isdirectlyproportionaltothe
pressuredifferencebetweenthetwopointsandinversely
proportionaltotheresistancetoflow.

Volume
Flowrate
Time



tan
P
Q
R
Q flowrate
P pressuredifference
R resis cetoflow



 


130. Pressuredifferencecausesfluidstoflow.Fluidsflowfroma
regionofhighpressuretoaregionoflowpressure.
Theresistancetoflowiscausedbyfrictionbetweenthefluid
andthetube(pipe),andwithinthefluiditself.Frictioninfluids
iscalledviscosity.
Theviscosityofafluidisameasureofthickness.
Constrictionsandobstructionsinatubecausesresistanceto
flowtoincrease.


-sec
Q Av
Q flowrate
v averagevelocity
A cross tional





131. Forlaminarflowofanincompressiblefluid,threefactors
determinetheresistancetoflow:thelengthofthetube,radius
oftube,andtheviscosityofthefluid.
Theexpressionforcalculatingresistancetoflowforalaminar
flowiscalledPoiseuille’slaw.
4
8
tan

cos

L
R
r
R resis ce
L lengthof tube
vis ityof fluid
r radiusof tube








4
8


r
Q P
L
Q flowrate
P pressuredifference


 

 

132. 3

50.0 / ,
1. , , .
2. cos , , .
3.
If theflowratethroughatubeisoriginally
cm s calculatethenewflowrate
if thepressuredifference P doubles
if thevis ityof thefluid doubles
if thelengthof thetube


, , .
4. , , .

.
L doubles
if theradiusof thetube r doubles
Assumethatineachcaseonlythefactor
mentioneddiffersfromtheoriginalconditions

133. 4
2 2
1 1
2
2 1
1
' ,
8
,

. ,
fromPoiseuilleslaw
R
Q P
L
If allotherfactorsremainthesame
flowrateisdirectlyproportionalto
pressuredifferenceTherefore
Q P
Q P
P
Q Q
P


 




 


134. 3
1
2
1
2 1
31
2
1
3
50.0 /


2
2
50.0 /
100 /
Q originalflowrate cm s
Q newflowrate
P originalpressuredifference
P newpressuredifference P
P
Q cm s
P
cm s
 

 
   

 



135. 2 1
1 2
1
2 1
2
1
2 1
31
2
1
3
,

cos . ,
cos
cos 2
50.0 /
2
25 /
If allotherfactorsremainthesame
flowrateisinverselyproportional
tovis ity Therefore
Q
Q
Q Q
originalvis ity
newvis ity
Q cm s
cm s





 



 

 
 


136. 2 1
1 2
1
2 1
2
1
2 1
31
2
1
3
,

. ,

2
50.0 /
2
25 /
If allotherfactorsremainthesame
flowrateisinverselyproportional
tolength Therefore
Q L
Q L
L
Q Q
L
L originallength
L newlength L
L
Q cm s
L
cm s

 

 
 


137. 4
2 2
4
1 1
1
2 1
4
2
2 1 4
1
4
31
4
1
3
,

. ,

2
(2 )
50.0 /
800 /
If allotherfactorsremainthesame
flowrateisdirectlyproportionalto
thefourthpowerof radius Therefore
Q R
Q R
R originalradius
R newradius R
R
Q Q
R
R
cm s
R
cm s


 
 
 


138. 3
2
20.0 /
100 0.50

25 O .

Aflowrateof cm sisobtainedthrough
atubeof length cmandradius cm
whenthereisapressuredifferenceof
cmH alongthetubeCalculatethenew
flowratewhenthefollowingchanges
, min
.
are
made assu gallotherfactorsremainthe
sameastheoriginalconditions

139. 2
2
1. 350
.
2. 0.40 .
3.
40.0 O.
4.
50.0 O
A cmlongtubereplacestheoriginal
one
Thetuberadiusisdecreasedto cm
Thepressuredifferenceisincreasedto
cmH
Thepressuredifferenceisincreasedto
cmH andthe
0.20 .
radiusisdecreased
to cm

140. 4
2 1
1 2
1
2 1
2
3
1
2
1
' ,
8
,

. ,
20.0 /

fromPoiseuille slaw
R
Q P
L
If all other factorsremainthesame
flowrateisinversely proportional
tolength Therefore
Q L
Q L
L
Q Q
L
Q original flowrate cm s
Q newflowrate
L origi


 

 
 


2
100
350
nal length cm
L newlength cm

 

141. 3
2
3
2
4
2 2
4
1 1
4
2
2 1 4
1
1
2
100
20.0 /
350
5.71 /
,

. ,
0.50
cm
Q cm s
cm
Q cm s
If allotherfactorsremainthesame
flowrateisdirectlyproportionalto
thefourthpowerof radius Therefore
Q R
Q R
R
Q Q
R
R originalradius cm
R ne
 


 
 
 0.40wradius cm

142. 4
3
2 4
3
2 2
1 1
2
2 1
1
1
(0.40 )
20.0 /
(0.50 )
8.19 /
,

. ,

cm
Q cm s
cm
cm s
If allotherfactorsremainthesame
flowrateisdirectlyproportionalto
pressuredifference Therefore
Q P
Q P
P
Q Q
P
P originalpressuredifferenc
 





 

 
2
2 2
25 O
40.0 O
e cmH
P newpressuredifference cmH

  

143. 32
2
2
3
4
2 2 2
4
1 1 1
2
40.0 O
20.0 /
25 O
32 /
,


. ,
cmH
Q cm s
cmH
cm s
If all other factorsremainthesame
flowrateisdirectlyproportional to
bothpressuredifferenceand the
fourthpower of radius Therefore
Q P R
Q P R
Q
 


 



4
2 2
14
1 1
4
3 32
2 4
2
50.0 O (0.20 )
20.0 / 1.024 /
25 O (0.50 )
P R
Q
P R
cmH cm
Q cm s cm s
cmH cm
 

   

144. Forflowofanincompressiblefluidinatube,flowratemustbe
thesameinallpartsofthetube.
Whatflowsthroughthelargerpartofthetubealsoflows
throughthenarrowpartofthetube.Therefore,thefluid
movesfasterthroughthenarrowpartthanitdoesinthelarger
part.
1
2
1
2
1 1 2 2
1 1 1
2
sec 1
sec 2
sec 1
sec 2
,
sec 1
A areaat tion of tube
A areaat tion of tube
v averagespeed at tion of tube
v averagespeed at tion of tube
fromtheContinuityEquation
Av Av
Av Q flowrateat tion
A





 
2 2
sec 2v Q flowrateat tion 

147. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
HEAT AND TEMPERATURE

148. Thetemperatureofanobjectisanindicatoroftheaverage
kineticenergyofitsmolecules.
Thehigherthetemperature,thegreatertheaveragespeedof
itsmolecules.
Thetemperatureofanobjectdetermineswhetherthermal
energyorheatflowsintooroutofitwhenitisincontactwith
anotherobject.
Heatisdefinedasenergythatflowsasaresultof
temperaturedifference.
Heatistheenergythatistransferredfromonelocationto
anotherlocationasaresultofthetemperaturedifference
betweenthem.

149. Heatalwaysflowsfromanobjectofhighertemperaturetoan
objectoflowertemperaturetillthermalequilibriumisreached.
Heattransfercancausetemperaturechangeinanobject.
However,sometimesheatcanflowintooroutofanobject
withoutatemperaturechange.
Heattransfercanalsocausephasechangessuchasmelting,
boiling,freezingandcondensation.
Theamountofheatneededtocauseatemperaturechange
inobjectdependsonthemassoftheobject,theamountof
temperaturechange,andthespecificheatcapacityofthe
material/substanceoftheobject.

150. Thespecificheatcapacityofasubstanceistheamountof
heatrequiredtochangethetemperatureof1kgofthe
substanceby1Kor1
o
C.Notethatatemperaturechangeof1
Kisequivalenttoatemperaturechangeof1
o
C.
TheunitofspecificheatcapacityisJ/kg/KorJ/kg/
o
C.





Q mc T
Q heatenergyneededtocausea
temperaturechangeof T
m massof object
T temperaturechange
c specificheatcapacity
 



 


151. Theheatcapacityofsubstanceistheamountofheatneeded
tochangeitstemperatureby1Kor1
o
C.
TheSIunitofheatcapacityisJ/KorJ/
o
C.


.
H
H
C mc
C heatcpacity
m massof object
c specificheatcapacity





152. 200
5 ?
4,200 / / .
Howmuchheatisneededtoincrease
thetemperatureof gramof water
by C Specificheatcapacityof
wateris Jkg C


200 0.2
5

4,200 / /
(0.20 )(4,200 / / )(5 )
4,200
m massof water gram kg
T temperaturechange C
c specificheatcapacityof water
Jkg C
Q mc T
kg Jkg C C
J
  
   

 
 
  


153. Energymustbeputintoasubstancetocausetomeltorboil.
Energymustbetakenoutofasubstancetocauseittofreeze
orcondense.
Energyusedtocauseaphasechangedoesnotcausea
temperaturechange.
Forexample,whenicemeltsat0
o
C,itbecomesliquidwater
at0
o
C.Whenwaterboilsat100
o
C,itbecomessteamat
100
o
C.Whenwaterfreezesat0
o
C,itbecomesiceat0
o
C.
Whensteamat100
o
Ccondenses,itbecomeswaterat100
o
C.
Energyassociatedwithphasechangeiscalledlatentheat
becauseitdoesnotproduceatemperaturechange.

155. Energymustberemovedtocausefreezingorcondensation.
Specificlatentheatoffusionisusedwhenthechangeof
phaseisfromsolidtoliquidandviceversa.
Specificlatentheatofvapourizationisusedwhenthechange
ofphaseisfromliquidtogasandviceversa.
Specificlatentheatoffusionistheamountofenergyrequired
tomeltorfreeze1kgofasolidorliquid.
Specificlatentheatofvapourizationistheamountofenergy
requiredtovapourizeorcondense1kgofasubstance.

156. tan
.
tan
tan
f
f
Q mL
Q amountof energyneededtomelt
orfreezeasubs cewithouta
changeintemperature
m massof subs ce
L specificlatentheatof fusionof subs ce





157. tan
.
tan

v
v
Q mL
Q amountof energyneededto
vapourizeorcondenseasubs ce
withoutachangeintemperature
m massof subs ce
L specificlatentheatof vapourization





158. 5.0
20 100 ?

4,190 / /
2,257,000 / .
Whatheatisrequiredtoconvert kgof
waterat Ctosteamat C The
specificheatcapacityof wateris
Jkg Candthespecificlatentheat
of vapourizationof wateris Jkg
 


159. 1 2
1


20 100
5.0

4,190 / /
1
Q amountof heatenergyneeded
Q Q Q
Q amountof energyneeededtochange
waterat Ctowaterat C
mc T
m massof water kg
c specificheatcapacityof water
J kg C
T temperaturechange

 

 
 
 

 
   00 20 80C C C    

160. 2

100 100 .


2,257,000 /
(5.0 )(4,190 / / )(80 ) (5.0 )(2,257,000 / )
12,961,000
v
v
Q amountof energyneeded tochange
waterat Ctosteamat C
mL
L specificlatentheatof vapourization
of water
J kg
Q kg J kg C C kg J kg
J

 



   


161. 0.50
20 ?

500/ /
270,000/ .
Whatheatisrequiredtomelta kg
blockofironifinitiallyitisat C
Thespecificheatcapacityofironis
Jkg Candthespecificlatentheat
offusionofironis JkgThe
melting


int 1,200 .po ofironis C 

162. 1 2
1


0.5 20
1,200 .
0.5
500 / /
Q Q Q
Q amountof heatrequired
Q amountof energyneededtoraisethe
temperatureof kgof ironfrom C
to C
mc T
m massof iron kg
c specificheatcapacityof iron J kg C
T temperatu
 




 
 
  
 
1
1,200 20 1,180
(0.5 )(500 / / )(1,180 )
=295,000
rechange C C C
Q kg J kg C C
J
     
  

163. 2
2

1,200

270,000 /
(0.5 )(270,000 / )
135,000
295,000 135,000
430,000
f
f
Q amountof energyneededtomelt
ironat C
mL
L specificlatentheatof fusionof iron
Jkg
Q kg Jkg
J
Q J J
J







 


164. .

70
Onemethodof reducingadangerouslyhigh
bodytemperatureveryquicklyistogivea
patientalcoholrubHowmanygramsof
alcoholmustbeevaporatedfromthesurface
of a kgpersontoreducehistemp
1.05 ?
3,419.6 / / .
854,760 / .
erature
by C Specificheatcapacityof tissueis
J kg C Specificlatentheatof
vapourizationof alcoholis Jkg



165. 1 2
1
1




854,760
854,760
a v
a
v
a
Q Q
Q amountof energyneededtoevaporate
thealcohol
mL
m massof alcohol
L specificlatentheatof vapourizationof
alcohol
J
Q m








166. 2
2

70

3,419.6 / /
1.05
(70 )(3,419.6 / / )(1.05 )
251,340.6
854,760 251,340
p p
p
p
a
Q amountof heatlostbyperson
mc T
m massof person kg
c specificheatcapacityof person
J kg C
T temperaturechange C
Q kg J kg C C
J
m

 
 

 
   
  

 .6
0.42
a
m kg

167. Heattransferoccurswhenthereisatemperaturedifference.
Therearethreemodesofheattransfer:conduction,
convectionandradiation.
Conductionisthetransferofheatthroughstationarymatter
byphysicalcontact.
Anexampleofconductionistheheattransferthatoccurs
betweenbarefeetandacoldfloor.
Therateofheattransferbyconductiondependsonthe
thermalconductivityofthematerial,thesurfaceareain
contact,thetemperaturedifferencebetweenthetwosurfaces,
andthethicknessofthematerialthroughwhichtheenergy
flows.

168. Convectionisthetransferofheatbythemovementofafluid.
Inconvection,gravitypullsthedenserportionofafluid
beneaththelessdenseportions
Energytransferbyconvectionrequiresafluid.
Convectioncanbenaturalorforced.Innaturalconvection,
theforceofgravitypullsthedenserportionofafluidbeneath
thelessdenseregionsofthefluid.
Heattransferbyradiationisthetransferofheatenergyby
electromagneticradiation.

169. PHYS 101 PHYSICS FOR
HEALTH SCIENCES
CURRENT, VOLTAGE,
RESISTANCE, OHM’S LAW, AND
POWER

170. Chargeisabasicphysicalpropertycarriedbycertain
subatomicparticles,suchaselectrons,anditisthebasisof
electromagneticforce.
Therearetwotypesofcharges:positiveandnegative.
Positivechargesdenotedeficiencyofelectronsandnegative
chargesdenoteexcessofelectrons.
Electronscarrynegativechargeandprotonscarrypositive
charge.
Electrostaticforcesresultfromtheseparationofpositiveand
negativecharges.

171. Theprincipleofconservationofchargestatesthatcharges
canbeseparated,buttheycanneitherbecreatednor
destroyed.
Likechargesrepel,andunlikechargesattract.
TheSIunitofchargeistheCoulomb(C).
Thechargesoftheelectronandprotonarethesmallest
amountsofchargethatareobservedinnature.Theyare
exactlyequalinmagnitudeandoppositeinsign.
19
19
1.6 10
1.6 10
arg
arg
e
p
e
p
Q C
Q C
Q ch eonanelectron
Q ch eonaproton


 
 



172. Coulomb’slawgivesaquantitativeexpressionforthe
electrostaticforcebetweentwostationarycharges.
1 2
2
1
2
9 2 2

arg 1
arg 2
tan arg
' tan
9.0 10 /
QQ
F K
d
F magnitudeof electrostaticforce
Q magnitudeof ch e
Q magnitudeof ch e
d dis ceorseparationbetweenthech es
K Coulombscons t
Nm C






 

173. 9
'
arg 1.0 10
1.0 .
? arg

CalculatetheforceinNewtonsbetweentwo
identicalch esof Cseparatedby
cmWhatistheforceif theseparationis
doubled Whatistheforceif thech es
areincreasedbyafactoro


9 2 2
10
1.0 . '
tan 9.0 10 / .
f andthe
separationiskeptat cmCoulombs
cons tis Nm C

174. 1 2
2
9
1
9
2
2
9 2 2
9 9
9 2 2
2 2
5
1.0 10
1.0 10
1.0 1.0 10
9.0 10 /
(1.0 10 )(1.0 10 )
9.0 10 /
(1.0 10 )
9.0 10
QQ
F K
d
Q C
Q C
d cm m
K Nm C
C C
F Nm C
m
N



 



 
 
  
 
 
  

 

175. Everychargecreatesanelectricfieldarounditself.Similarly,
allmagnetscreateamagneticfieldandallmassescreatea
gravitationalfield.
Anelectricfieldisaregionorspacearoundachargewithin
whichanelectricforceisexertedonothercharges.Thus,an
electricfieldisafieldofforce.
Voltageorpotentialdifferenceisenergyperunitcharge.
TheSIunitforvoltageisthevolt(V).
arg
Energy
Voltage
Ch e

1 1 /V JC

176. Currentisdefinedastheamountofchargethatflowsperunit
time.
TheSIunitofelectriccurrentistheampere(A).
Currentiscausedbyanelectricalpotentialdifference
betweentwolocationsinaconductor.
argCh e
Current
Time

1 1 /A Cs

177. 10
19
10
10
19
9

arg
2.0 10 ?
1.6 10 1
2.0 10
2.0 10 1
1.6 10
1.25 10
Howmanyelectronsmustberemoved from
anobjecttoleaveitwithach eof
C
C e
C
C e
C
e






 

  

 

178. Ohm’slawstatesthattheamountofcurrentthroughagiven
circuitisdirectlyproportionaltothepotentialdifferenceand
inverselyproportionaltotheresistanceofthecircuit.
Resistanceistheoppositiontocurrent.
TheSIunitofresistanceistheOhm(Ω).

tan
V
I
R
I current
V voltageorpotentialdifference
R resis ce





179. tan 15
3.0 ?
3.0
tan 15
3.0
15
0.20
Whatcurrentflowsthroughaflashlight
if itsresis ceis anditsbatteries
haveatotalpotentialdifferenceof V
V potentialdifference V
R resis ce
V
I
R
V
A

 
  





180. Essentially,anelectriccircuitconsistsofasourceofvoltage,
aload,andwiresthatconnecttheloadtothesourceof
voltage.
Aloadisanythingordevice,whichdrawscurrentfroma
sourceofvoltage.Conventionally,currentflowsfromthe
positiveterminalofthevoltagesource,throughtheloadand
backtothenegativeterminalofthevoltagesource.

183. Aresistorisamelectricaldevicethatoffersresistancetothe
flowofcurrent.Resistanceisapropertyofaresistor.
Twoormoreresistorsareconnectedinseriesifthesame
currentpassesthroughthem.
Forresistorsconnectedinseries,thetotalorcombinedor
effectiveresistanceisthesumoftheindividualresistances.
Thevoltagedropacrossanyresistoristheproductofits
resistanceandthecurrentpassingthroughit.
Foraseriescircuit,thesumofthevoltagedropsisequalto
theappliedvoltageorsourcevoltage.

184. 1 2
1 1
1
2 2
2
1 2
tan



T
T
R total or effectiveresis ce
R R R
V applied voltageor sourcevoltage
V voltagedrop across R
IR
V voltagedrop across R
IR
V V V

 





 

185. Twoormoreresistorsordevicesareconnectedinparallelif
thesamevoltageappearsacrosseachandtheircombined
resistor.
Forresistorsconnectedinparallel,thetotalorequivalent
resistanceisthereciprocalofthesumofthereciprocalsofthe
individualresistances.
Thevoltagedropacrosseachresistoristheproductofits
resistanceandthecurrentpassingthroughit.
Foraparallelcircuit,thetotalcurrentorsourcecurrentisthe
sumoftheindividualcurrents.

186. 1 2
1 2
tan
1 1 1
1
1 1
T
T
T
R totalresis ce
R R R
R
R R

 


1 1
2 2
1 2


sin
sin
V appliedorsourcevoltage
I totalorsourcecurrent
I currentpas gthroughR
I currentpas gthroughR
I I I




 

189. Thepowerrequired,consumedordissipatedbyanelectrical
deviceorcomponentisgivenby:
2
2

P IV
P I R
V
P
R
P electricalpower
I current
V voltage







190. 2
2
2
tan 100
120 .
100
120
(120 )
100
144
Calculatetheresis ceof a W
lightbulbthatuses Velectricity
P power W
V voltage V
V
P
R
V
R
P
V
W
 
 



 

191. Thecostofusinganelectricaldevicedependsontwothings:
thepowerratingofthedevice(i.e.howmuchpowerthe
deviceconsumes)andhowlongthedeviceisused.
Sinceenergyconsumedistheproductofpowerandtime,a
largeelectricaldevice(i.e.onewithalargepowerrating)uses
moreenergyinagivenamountoftimethanasmallone.
Ontheotherhand,smallelectricaldevices,suchaslight
bulbs,thatareusedformanyhoursconsumelargeamountof
electricalenergy.
Inelectricbills,energyconsumedismeasuredinkilowatt-
hour(kWh).
( ) ( ) ( )energykWh PowerkW Timeh 

192. cos
0.011 / . cos
100 16 .
100 0.1
16
0.1 16
1.6
1 0.011
1.6
Theaverage tof electricenergyisabout
CediskWh Calculatethe tof
operating W lightbulbfor hours
Power W kWh
Time h
Energy Power Time
kW h
kWh
kWh Cedis
kW
 

 
 


1.6
0.011
1
0.0176
kWh
h Cedis
kWh
Cedis
 


193. Circuitbreakersandfusesaredesignedtominimizedamage
ifexcessivecurrentflowsinacircuit.
Electricityhastwobasichazards:thermalhazardsfrom
overheatingofwires,andelectricalshockhazardsfrom
currentpassingthroughaperson.
Circuitbreakersandfusesareusedinelectricalcircuitsto
preventlargecurrentsfromcausinglarge.
Acircuitbreakerorafusewillinterrupttheflowofcurrentif
thecurrentexceedsitratingorcapacity.
Electricalshockisdefinedasunwantedphysiological
responsetoelectricityandcurrent,andtherearetwotypes:
microshockandmacroshock.