sph4u unit #2mrneave.weebly.com/.../sph4u2introwork_oct20_2015.pdf · 2 is the angle between the...

SPH4U Unit #2 Energy and Momentum

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

Equation and Units for Work y Force x distance

y Work = Fisdcoso

fl T NonN m

/N=1kgYg1Jouk=INnI J=1N 'm

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

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

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

.

Work and Multiple Forces

Homework: y Pg. 170 #1-6

SO ?

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

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

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 .

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€

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

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

c) 0 -

angle between

motion and appliedForce .

W= Fisdcoso=H8NX23mXus§

=H8NXDDLDW= 2714J

W = 2.7×103 J

d) e is 0° Eaker

FN

qgawideFg

E Fx = 0

0 = Fa . TI .

Fg cos 60

EF8.gs#y-Fgsin6fFn = Fg sin 60

'

Fk =µ .

ten : Esnktssidi

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

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