Transcript F - HEPG

Physics 111: Lecture 9
Today’s Agenda
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

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
Work & Energy
Discussion
Definition
Dot Product
Work of a constant force
Work/kinetic energy theorem
Work of multiple constant forces
Comments
Physics 111: Lecture 9, Pg 1
Work & Energy

One of the most important concepts in physics
Alternative approach to mechanics

Many applications beyond mechanics
Thermodynamics (movement of heat)
Quantum mechanics...
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Very useful tools
You will learn new (sometimes much easier) ways to
solve problems
Physics 111: Lecture 9, Pg 2
Forms of Energy
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Kinetic: Energy of motion.
A car on the highway has kinetic energy.
We have to remove this energy to stop it.
The breaks of a car get HOT!
This is an example of turning one form of energy into
another (thermal energy).
Physics 111: Lecture 9, Pg 3
Mass = Energy (but not in Physics 111)

Particle Physics:
E = 1010 eV
(a)
e+
e+ 5,000,000,000 V
(b)
(c)
- 5,000,000,000 V
M
E = MC2
( poof ! )
Physics 111: Lecture 9, Pg 4
Wilberforce
Energy Conservation
Returning
Can

Energy cannot be destroyed or created.
Just changed from one form to another.
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We say energy is conserved!
True for any isolated system.
i.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.
The energy of the car “alone” is not conserved...
» It is reduced by the braking.

Doing “work” on an isolated system will change its “energy”...
Physics 111: Lecture 9, Pg 5
Definition of Work:
Ingredients: Force (F), displacement (r)
Work, W, of a constant force F
acting through a displacement r
is:
W = F r = F r cos  = Fr r
F

Fr
r
“Dot Product”
Physics 111: Lecture 9, Pg 6
Definition of Work...
Hairdryer

Only the component of F along the displacement is doing
work.
Example: Train on a track.
F

r
F cos 
Physics 111: Lecture 9, Pg 7
Aside: Dot Product (or Scalar Product)
a
Definition:
.
a b = ab cos 
= a[b cos ] = aba
ba

b
a
= b[a cos ] = bab
Some properties:
ab = ba
q(ab) = (qb)a = b(qa)
a(b + c) = (ab) + (ac)

b
ab
(q is a scalar)
(c is a vector)
The dot product of perpendicular vectors is 0 !!
Physics 111: Lecture 9, Pg 8
Aside: Examples of dot products
y
.i=j.j=k.k=1
i.j=j.k=k.i=0
i
j
z
Suppose
a=1i+2j+3k
b=4i -5j+6k
k
i
x
Then
.
.
.
a b = 1x4 + 2x(-5) + 3x6 = 12
a a = 1x1 +
2x2 + 3x3 = 14
b b = 4x4 + (-5)x(-5) + 6x6 = 77
Physics 111: Lecture 9, Pg 9
Aside: Properties of dot products

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
= ax 2 + ay 2
.
Pythagorean Theorem!!
.
.
a
. j)
ay
ax
j
i
Physics 111: Lecture 9, Pg 10
Aside: Properties of dot products
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Components:
a = ax i + ay j + az k = (ax , ay , az) = (a

Derivatives:
. i, a . j, a . k)
d
da
db
(ab ) 
b  a
dt
dt
dt
Apply to velocity
d 2 d
dv
dv
v  (v v ) 
v  v 
 2v  a
dt
dt
dt
dt
So if v is constant (like for UCM):
d 2
v  2v  a  0
dt
Physics 111: Lecture 9, Pg 11
Back to the definition of Work:
Skateboard
Work, W, of a force F acting
through a displacement  r is:
W = F  r
F
r
Physics 111: Lecture 9, Pg 12
Lecture 9, Act 1
Work & Energy

A box is pulled up a rough (m > 0) incline by a rope-pulleyweight arrangement as shown below.
How many forces are doing work on the box?
(a) 2
(b) 3
(c) 4
Physics 111: Lecture 9, Pg 13
Lecture 9, 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
N
v
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
Physics 111: Lecture 9, Pg 14
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 = Fx = F x
(since F is parallel to x)
WF = (10 N) x (5 m) = 50 Joules (J)
Physics 111: Lecture 9, Pg 15
Units:
Force x Distance = Work
Newton x Meter = Joule
[M][L] / [T]2 [L]
[M][L]2 / [T]2
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
Physics 111: Lecture 9, Pg 16
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.
v1
v2
F
m
i
x
Physics 111: Lecture 9, Pg 17
Work & Kinetic Energy...

Since the force F is constant, acceleration a will be
constant. We have shown that for constant a:
v22 - v12 = 2a(x2-x1) = 2ax.
1/ mv 2 - 1/ mv 2 = max
multiply by 1/2m:
2
2
2
1
1/ mv 2 - 1/ mv 2 = Fx
But F = ma
2
2
2
1
v1
v2
F
m
a
i
x
Physics 111: Lecture 9, Pg 18
Work & Kinetic Energy...
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So we find that
1/2mv22 - 1/2mv12 = Fx = WF
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Define Kinetic Energy K:
K = 1/2mv2
K2 - K1 = WF
WF = K (Work/kinetic energy theorem)
v2
v1
F
m
a
i
x
Physics 111: Lecture 9, Pg 19
Work/Kinetic Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
Wnet  K
 K 2  K1


1
1
2
2
mv 2  mv1
2
2
We’ll prove this for a variable force later.
Physics 111: Lecture 9, Pg 20
Lecture 9, Act 2
Work & Energy

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. m > 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
Physics 111: Lecture 9, Pg 21
Lecture 9, 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
Physics 111: Lecture 9, Pg 22
Lecture 9, 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 = -mmgD.
 This work “removes” the kinetic energy that the box had:
 WNET = K2 - K1 = 0 - K1
m
D
Physics 111: Lecture 9, Pg 23
Lecture 9, Act 2
Solution
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The net work done to stop a box is - fD = -mmgD.
This work “removes” the kinetic energy that the box had:
WNET = K2 - K1 = 0 - K1
This is the same for both boxes (same starting kinetic energy).
mm2gD2  mm1gD1
Since m1 > m2 we can see that
m2D2  m1D1
D2 > D1
m1
D1
m2
D2
Physics 111: Lecture 9, Pg 24
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 = F 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
Physics 111: Lecture 9, Pg 25
What about multiple forces?
Suppose FNET = F1 + F2 and the
displacement is r.
The work done by each force is:
W1 = F1 r
W2 = F2  r
F1
FNET
WTOT = W1 + W2
= F1 r + F2 r
= (F1 + F2 ) r
WTOT = FTOT r
r
F2
It’s the total force that matters!!
Physics 111: Lecture 9, Pg 26
Comments:

Time interval not relevant
Run up the stairs quickly or slowly...same W
Since W = F r
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No work is done if:
F = 0
or
r = 0
or
 = 90o
Physics 111: Lecture 9, Pg 27
Comments...
W = F  r
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No work done if  = 90o.
T
No work done by T.
v
v
No work done by N.
N
Physics 111: Lecture 9, Pg 28
Lecture 9, 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
Physics 111: Lecture 9, Pg 29
Lecture 9, Act 3
Solution
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First, draw all the forces in the system:
FS
a
mg
N
Physics 111: Lecture 9, Pg 30
Lecture 9, Act 3
Solution

Recall that W = F Δr so only forces that have a
component along the direction of the displacement are
doing work.
FS
a
mg

N
The answer is (b) 2.
Physics 111: Lecture 9, Pg 31
Recap of today’s lecture

Work & Energy
(Text: 6-1 and 7-4)
Discussion
Definition
(Text: 6-1)
Dot Product
(Text: 6-2)
Work of a constant force
(Text: 7-1 and 7-2)
Work/kinetic energy theorem
(Text: 6-1)
Properties (units, time independence, etc.)
Work of a multiple forces
Comments
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Look at textbook problems Chapter 6: # 1, 50, 65
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Physics 111: Lecture 9, Pg 32