Download Lecture 14 - UD Physics

Phys 207
Announcements
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Hwk 5 submission deadline = 1st thing Monday morn.
Quizzes in DSC next week
Today’s Agenda
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Work & Energy
(Chapter 7)
ÍDiscussion
ÍDefinition
Dot Product
Work of a constant force
ÍWork/kinetic energy theorem
Work of multiple constant forces
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Work & Energy
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One of the most important concepts in physics
ÍAlternative approach to mechanics
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Many applications beyond mechanics
ÍThermodynamics (movement of heat)
ÍQuantum mechanics...
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Very useful tools
ÍYou will learn new (sometimes much easier) ways to
solve problems
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Page 1
Forms of Energy
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Kinetic: Energy of motion.
ÍA car on the highway has kinetic energy.
ÍWe have to remove this energy to stop it.
ÍThe breaks of a car get HOT!
ÍThis is an example of turning one form of energy into
another (thermal energy).
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Mass = Energy (but not in Physics 207)
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Particle Physics:
E = 1010 eV
(a)
e+
e+ 5,000,000,000 V
(b)
- 5,000,000,000 V
M
E = MC2
( poof ! )
(c)
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Page 2
Energy Conservation
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Energy cannot be destroyed or created.
ÍJust changed from one form to another.
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We say energy is conserved!
ÍTrue for any isolated system.
Íi.e. when we put on the brakes, the kinetic energy of the
car is turned into heat using friction in the brakes. The total
energy of the “car-brakes-road-atmosphere” system is the
same.
ÍThe energy of the car “alone” is not conserved...
» It is reduced by the braking.
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Doing “work” on an isolated system will change its “energy”...
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Definition of Work:
Ingredients: Force (F), displacement (∆r)
Work, W, of a constant force F
acting through a displacement ∆r
is:
W = Fi ∆r = F ∆r cos θ = Fr ∆r
F
θ
dis
“Dot Product”
Fr
∆r
t
en
m
ce
pla
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Page 3
Definition of Work...
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Only the component of F along the displacement is doing
work.
ÍExample: Train on a track.
F
∆r
θ
F cos θ
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Aside: Dot Product (or Scalar Product)
a
Definition:
ba
.
a b = ab cos θ
θ
= a[b cos θ] = aba
a
= b[a cos θ] = bab
Some properties:
aib = bia
q(aib) = (qb)ia = bi(qa)
ai(b + c) = (aib) + (aic)
b
θ
ab
b
(q is a scalar)
(c is a vector)
The dot product of perpendicular vectors is 0 !!
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Aside: Examples of dot products
y
.i=j.j=k.k=1
i.j=j.k=k.i=0
i
j
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Suppose
k
x
i
Then
.
.
.
a b = 1x4 + 2x(-5) + 3x6 = 12
a a = 1x1 +
2x2 + 3x3 = 14
b b = 4x4 + (-5)x(-5) + 6x6 = 77
a=1i+2j+3k
b=4i -5j+6k
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Aside: Properties of dot products
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Magnitude:
a2 = |a|2
.
=a a
= (ax i + ay j) (ax i + ay j)
= ax 2(i i) + ay 2(j j) + 2ax ay (i
= a x 2 + ay 2
.
.
ÍPythagorean Theorem!!
.
a
. j)
ay
ax
j
i
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Aside: Properties of dot products
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Components:
a = ax i + ay j + az k = (ax , ay , az) = (a
z
Derivatives:
. i, a . j, a . k)
d
da
db
(a⋅b ) =
⋅b + a⋅
dt
dt
dt
ÍApply to velocity
d 2 d
dv
dv
⋅v + v ⋅
= 2v ⋅ a
v = (v ⋅v ) =
dt
dt
dt
dt
ÍSo if v is constant (like for UCM):
d 2
v = 2v ⋅ a = 0
dt
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Back to the definition of Work:
Work, W, of a force F acting
through a displacement ∆ r is:
W = Fi ∆ r
F
∆r
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Page 6
Lecture 14, Act 1
Work & Energy
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A box is pulled up a rough (µ > 0) incline by a rope-pulleyweight arrangement as shown below.
ÍHow many forces are doing work on the box?
(a) 2
(b) 3
(c) 4
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Lecture 14, Act 1
Solution
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Draw FBD of box:
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Consider direction of
motion of the box
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Any force not perpendicular
to the motion will do work:
T
v
N
f
N does no work (perp. to v)
T does positive work
f does negative work
3 forces
do work
mg
mg does negative work
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Work: 1-D Example
(constant force)
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A force F = 10 N pushes a box across a frictionless
floor for a distance ∆x = 5 m.
F
∆x
Work done by F on box :
WF = Fi∆x = F ∆x
(since F is parallel to ∆x)
WF = (10 N) x (5 m) = 50 Joules (J)
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Units:
Force x Distance = Work
Newton x Meter = Joule
[M][L]2 / [T]2
[M][L] / [T]2 [L]
mks
N-m (Joule)
cgs
Dyne-cm (erg)
= 10-7 J
other
BTU
calorie
foot-lb
eV
= 1054 J
= 4.184 J
= 1.356 J
= 1.6x10-19 J
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Page 8
Work & Kinetic Energy:
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A force F = 10 N pushes a box across a frictionless
floor for a distance ∆x = 5 m. The speed of the box is v1
before the push and v2 after the push.
v2
v1
F
m
i
∆x
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Work & Kinetic Energy...
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Since the force F is constant, acceleration a will be
constant. We have shown that for constant a:
Ív22 - v12 = 2a(x2-x1) = 2a∆x.
1/ mv 2 - 1/ mv 2 = ma∆x
Ímultiply by 1/2m:
2
2
2
1
1/ mv 2 - 1/ mv 2 = F∆x
ÍBut F = ma
2
2
2
1
v2
v1
F
m
a
i
∆x
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Page 9
Work & Kinetic Energy...
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So we find that
Í1/2mv22 - 1/2mv12 = F∆x = WF
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Define Kinetic Energy K:
K = 1/2mv2
ÍK2 - K1 = WF
ÍWF = ∆K (Work/kinetic energy theorem)
v2
v1
F
m
a
i
∆x
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Work/Kinetic Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
Wnet = ∆K
= K 2 − K1
=
z
1
1
2
2
mv 2 − mv1
2
2
We’ll prove this for a variable force later.
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Page 10
Lecture 14, Act 2
Work & Energy
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Two blocks have masses m1 and m2, where m1 > m2. They
are sliding on a frictionless floor and have the same kinetic
energy when they encounter a long rough stretch (i.e. µ > 0)
which slows them down to a stop.
Which one will go farther before stopping?
(a) m1 (b) m2
(c) they will go the same distance
m1
m2
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Lecture 14, Act 2
Solution
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The work-energy theorem says that for any object WNET = ∆K
In this example the only force that does work is friction (since
both N and mg are perpendicular to the block’s motion).
N
f
m
mg
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Lecture 14, Act 2
Solution
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The work-energy theorem says that for any object WNET = ∆K
In this example the only force that does work is friction (since
both N and mg are perpendicular to the blocks motion).
The net work done to stop the box is - fD = -µmgD.
Í This work “removes” the kinetic energy that the box had:
Í WNET = K2 - K1 = 0 - K1
m
D
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Lecture 14, Act 2
Solution
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The net work done to stop a box is - fD = -µmgD.
ÍThis work “removes” the kinetic energy that the box had:
ÍWNET = K2 - K1 = 0 - K1
This is the same for both boxes (same starting kinetic energy).
µm2gD2 = µm1gD1
Since m1 > m2 we can see that
m2D2 = m1D1
D2 > D1
m1
m2
D1
D2
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A simple application:
Work done by gravity on a falling object
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What is the speed of an object after falling a distance H,
assuming it starts at rest?
Wg = Fi ∆r = mg ∆r cos(0) = mgH
v0 = 0
∆r
Wg = mgH
mg
j
H
Work/Kinetic Energy Theorem:
Wg = mgH = 1/2mv2
v = 2 gH
v
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What about multiple forces?
Suppose FNET = F1 + F2 and the
displacement is ∆r.
The work done by each force is:
W1 = F1i ∆r
F1
W2 = F2 i ∆r
FNET
WTOT = W1 + W2
= F1i ∆r + F2i ∆r
= (F1 + F2 )i ∆r
WTOT = FTOTi ∆r
∆r
F2
It’s the total force that matters!!
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Page 13
Comments:
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Time interval not relevant
ÍRun up the stairs quickly or slowly...same W
Since W = Fi ∆r
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No work is done if:
ÍF = 0
or
Í∆r = 0
or
Íθ = 90o
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Comments...
W = Fi ∆r
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No work done if θ = 90o.
T
ÍNo work done by T.
v
v
N
No work done by N.
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Page 14
Lecture 14, Act 3
Work & Energy
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An inclined plane is accelerating with constant acceleration a.
A box resting on the plane is held in place by static friction.
How many forces are doing work on the block?
a
(a) 1
(b) 2
(c) 3
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Lecture 14, Act 3
Solution
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First, draw all the forces in the system:
FS
a
mg
N
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Page 15
Lecture 14, Act 3
Solution
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Recall that W = Fi ∆r so only forces that have a
component along the direction of the displacement are
doing work.
FS
a
mg
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N
The answer is (b) 2.
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