sph4u unit #2mrneave.weebly.com/.../sph4u2introwork_oct20_2015.pdf · 2 is the angle between the...
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![Page 1: SPH4U Unit #2mrneave.weebly.com/.../sph4u2introwork_oct20_2015.pdf · 2 is the angle between the force and the displacement October 8, 2012 4U2 - Work 4 = W F ∆dcosθ Work PRACTICE](https://reader030.vdocument.in/reader030/viewer/2022040414/5f1a903708b1214eeb1edc23/html5/thumbnails/1.jpg)
SPH4U Unit #2 Energy and Momentum
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Work y The energy transferred to an object
when a force acting on the object moves it through a distance
y Work is a scalar quantity
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Equation and Units for Work y Force x distance
y Work = Fisdcoso
fl T NonN m
/N=1kgYg1Jouk=INnI J=1N 'm
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Positive Work y Adds energy to an object y For example the work:
◦ Increases the speed of the object ◦ Increases the height of the object
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Negative Work y Removes energy from the object y For example the work:
◦ Decreases the speed of the object ◦ Decreases the height of the object
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Zero Work y Object experiences a force or a
displacement or both, yet no work is done on the object.
y Example: Carrying a bookacross the room
.
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Work and Multiple Forces
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Homework: y Pg. 170 #1-6
SO ?
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10/8/2012
2
Work
NOTE!In the previous example, the force needed to raise the backpack and thedisplacement of the backpack were in the same direction. However, this isoften not the case. Consider the situation where the force is at some angle2 to the displacement. The component of the force that is parallel to thedisplacement, Fcos2, causes the objet to undergo the displacement. Inthis case, W=(Fcos2))d.
October 8, 2012 4U2 - Work 3
Work
WORK (W)
where W is the work done (J) 7 1 J = 1 N@m = 1 kg@m2/s2
F is the applied force (N))d is the displacement (m)2 is the angle between the force and the displacement
October 8, 2012 4U2 - Work 4
dcosθF W ∆=
Work
PRACTICE1. An emergency worker applies a force of 16 N to push a patient
horizontally for 2.5 m on a gurney with nearly frictionless wheels.Determine the work done in pushing the gurney if the force is applied:(a) horizontally.
(a) W = 40 J 7 2 = 0E
October 8, 2012 4U2 - Work 5
W = Fddcoso
W=¢6N#5n)( cost)t4oNm)( 1 )
W = 40J
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10/8/2012
3
Work
PRACTICE1. An emergency worker applies a force of 16 N to push a patient
horizontally for 2.5 m on a gurney with nearly frictionless wheels.Determine the work done in pushing the gurney if the force is applied:(b) at an angle of 25E below the horizontal.
(b) W = 36 J
October 8, 2012 4U2 - Work 6
Work
PRACTICE2. A woman pushes a lawnmower with a force of 150 N at an angle of
35E down from the horizontal. The lawn is 10 m wide and requires 15complete trips across and back. How much work does she do?
W = 3.7 x 104 J
October 8, 2012 4U2 - Work 7
Positive Work
In the gurney problem, the work done was positive because the force anddisplacement were in the same direction. Positive work indicates anincrease in the energy of an object (i.e. the object speeds up, the heightincreases, ...)
POSITIVE WORK! the force and displacement are in the same direction! object’s energy increases – speeds up, height increases, ...
October 8, 2012 4U2 - Work 8
W = Fad co so
=@N) (2 . snkcos 2 5)W : 36
t,30*10mm300
W = (so N)( 300 m)( cos 35 )w = 36862% ,
3.7×104 J
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Pg 167
#Za) Fg.5.21 KN o
o=o= 5210 N
Dd =355mW= Fddcoso=§21oN)( 355 - )( cos of
= 1849 550J
=/ . 85×1065= 1
. 85×103 KT
Work ist ) because it
slows the plane down .
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b) E.5210NW= - 1.52×106 J
→ motion
f- Ff0=1800
W= Fsdcose-
1.52×106 Nm=§210N)Dd( cos 180'
)- 1.52×106 Nm
-
'- 5210N Ad
ad = - 1.52×106Nhk£µd=29Z€
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Pg 169 #l .
h?q•#7E=hn
FrictionsWitsdcoso
=¢23w)( 223kcal 8DW = -1.6×10
'
J
Hiker Wi ' Fidcoso=(122N)¢23m)¢os3¥
W= 2.2×10"
J
Wttd ' White,
+ Weiter= 2.11×10 'T - 1.61×1045
= 5.6×103 J
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Pg 170*5 .
m'
' 24kg
,0=30.0
°
,x
⇒ tt.
.IE#Fgxi.Fgcos6o'
=
Mg cos 60°
€24kg)(98⇒us60 .
= 117.6N
Fgx =124D Ent =o %9PF±n=E
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c) 0 -
angle between
motion and appliedForce .
W= Fisdcoso=H8NX23mXus§
=H8NXDDLDW= 2714J
W = 2.7×103 J
d) e is 0° Eaker
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FN
qgawideFg
E Fx = 0
0 = Fa . TI .
Fg cos 60
EF8.gs#y-Fgsin6fFn = Fg sin 60
'
Fk =µ .
ten : Esnktssidi
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0 -
. En . Fk - Fg cos 60°
0 = Fn AKFG sin 60 - Fgusbi
Fa = Mg (Mksinbo 't cos 64
Fn =(24kD( 9.8*(0.254%6) topFa = 44kg) (9.8M$ (0.7165)
Fa = 168.52 N
w=@9N)( 16 D( cos ofW = 2704 JW = 2700J
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Fun,.Fgsn6o
'
=@2D(24kD(98⇒fn6i)
Fti 50.9NW=§a9Nµ6n)⇐l8i)
W=i8_JW = . 8.1×10 'T
WT 's 2704J - 815J
Wt= 1889 Wil .9×o3J