Transcript Chapter 7 - s3.amazonaws.com

Midterm 1: July 9
Will cover material from Chapters 1-6
 Go to the room where you usually have
recitation
 Practice exam available on-line and in the
library

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1
Chapter 7
Kinetic Energy and Work
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Energy
A property of the state of an object
 Scalar quantity – no direction
 Conserved – cannot be created or
destroyed, but it can change from one
form to another or be exchanged from one
object to another
 Units: Joule = kg m2/s2

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Kinds of Energy
Kinetic (movement)
 Potential

 Gravitational
 Spring
 Chemical
bonds
Mass (E=mc2…)
 Thermal
 And more…

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Kinetic Energy
Kinetic Energy  Energy of motion
For an object of mass m moving with speed v:
K  mv
1
2
2
Also kinetic energy associated with rotation,
vibration, etc.
More on that later…
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Kinetic Energy: Orders of Magnitude
K = ½mv2 for some common objects:
Earth orbiting sun: 2x1029 J
Car at 60 mph: 100,000 J
Baseball pitch: 300 J
A man walking: 40 J
Angry bee: 0.005 J
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Work
Work  Energy transferred by a force
Work done on an object is the
energy transferred to/from it
W > 0  energy added
W < 0  energy taken away
W = F • r  Work done on an object
by a constant force F
while moving through a
displacement r
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Dot or Scalar Product
   
A  B  A B cos f  Ax Bx  Ay By  Az Bz
f is the angle between the vectors if you put
their tails together
B
f
Measures “how much”
one vector lies along
another
A
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What Does It Mean Physically?
F
r
F
q
r
 
F  r  Fr cos q
 F||r
W > 0 if q < 90°
force is adding energy
to object
W < 0 if q > 90°
force is reducing
energy of object
W = 0 if
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Fr
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Work Examples

Push on a wall
 r = 0, so no work is done (W = 0)

Lift a weight against gravity at constant
speed…
d


Wgrav  Fgrav  d  mgd
 
Wlift  Flift  d  mgd
Ftot = 0  Wtot = 0
Note: Kinetic Energy is constant…
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Work Against Gravity
Wlift=mgh
A weightlifter does work when lifting a weight
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Work for Tension, Gravity, Friction
What is the work done
by gravity, tension,
and friction?
Guess that the weight of the pack is 250 lb. = 113 kg
Iguazu Falls: 269 ft = 82 m
Vertical ascent: Wgravity = -mgh
= -(113 kg)(9.8 m/s2)(82 m) = -90,800 J
Wtension = -Wgravity = 90,800 J
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Work for Tension, Gravity, Friction
What is the work done
by gravity, tension,
and friction?
Horizontal pull (at the end): use d~50 ft=15.2 m, μk = 1.0
Wfriction = Ff d = -μkmg d
= -(1.0)(113 kg)(9.8 m/s2)(15.2 m) = -16,800 J
Wtension = -Wfriction = 16,800 J
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Work Due to Friction
v
Ffr
The frictional force always opposes the motion:
Moving to the right:
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Moving to the left:
W   F fr x
W   F fr x
x  0
x  0
W negative in both cases
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Another Work Example
Consider a pig sliding down frictionless ramp:
N
d
Fg
q
Work done by the
normal force:
 
WNormal  N  d  N  d  0
Work from the gravitational force:
 
Wgrav  F  d
 mgd cos(90  q )
 mgd sin q
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 mgh
d
h
θ
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Work-Kinetic Energy Theorem
v f  vi  2a( x f  xi )
2
For constant acceleration:
v f  vi 
2
Since Ftot = ma:
Multiply by ½m:
But ½mv2 = K:
2
1
2
2
2Ftot, x
m
( x f  xi )
mv f  12 mvi  Ftot, x ( x f  xi )
2
2
K f  K i  Ftot, x x
Easy to extrapolate to three dimensions
 
K  F  d  Wtot
The total work done on an object is the change in its
kinetic energy!
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Total Work in One Dimension
F
For a small segment x,
W  F(x)x
x
From x1 to x2:
Wtot 
S F(x)x
= area under curve
x1
x
x2
x1 < x < x2
To be exact:
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x2
W   F ( x)dx
x1
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Work in Two Dimensions
In one dimension motion and force are always
in the same direction (or opposite directions)
This is not true in two dimensions
F
dr
How do we generalize work and kinetic energy
to motion in two dimensions?
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Work Along an Arbitrary Path
Looking at a small
patch of the path:
F
l
W  Fx x  Fy y
 
 F  l
Δy
Δx
Over the whole path: W  
finish
start
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 
F  dl
x2
y2
z2
x1
y1
z1
  Fx dx   Fy dy   Fz dz
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Restating the Work-Kinetic Energy Theorem
K  W
Constant force:
 
( 12 mv )  F  d
2
Variable force:


( 12 mv 2 )   F  dl
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A Formal Derivation
xf
W=
Newton’s
 F(x) dx
xi
2nd
Law
xf
W = m  a(x) dx
xi
Using the chain rule:
dv dv dx dv
a(x)= =
=v
dt dx dt
dx
xf
vf
dv
W=m  v dx=m  v dv
dx
xi
vi
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vf
vf
1 2
W = m  v dv = mv
2
vi
vi
1 2 1 2
= mv f  mvi
2
2
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An Example
vi=60 mph =26.8 m/s
μk=0.9
x
150 ft
45.7 m
Will the car be able to stop before hitting the moose?
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An Example (continued)
Friction from the tires
on the road will slow
the car:
F fr    k N    k mg
The work done by
friction will be:
From the work-energy
theorem:
K f  Ki  W
1
2
mv f  mvi    k mg x
2
1
2
2
2
W fr  F fr x
   k mg x
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1 vf
1 (26.8 m / s) 2
x

2  k g 2 (0.9)(9.8 m / s 2 )
Just miss the moose!
x  40.8 m
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Example: Pile Driver
Drop a big mass to drive a nail into a board
Big Mass
d
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The Pile Driver
How much force is exerted on the nail?
m
1) Work done by gravity on freely falling pile
W1 = (-mg)(-h-x) = mg(h+x)
h
2) Work done on the nail
W2 = F(-x) = -Fx
3) Total Work = 0
x
W = W1+W2 = mg(h+x) – Fx = 0
F=
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mg(h+x)
x
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Example: Slowing a Bus
5.45 x 103 kg
vi=75 mph=33.5 m/s
vf=65 mph=29.0 m/s
How much work is done slowing down a bus?
Ki =
1
2
K f = 12 m v2f
mvi2
= 12 ( 5.45 103 kg) ( 33.5 m/s)2
= 12 ( 5.45 x 103 kg) ( 29.0 m/s)2
= 3.0 106 J
= 2.3 x 106 J
W  K f  K i  2.3  106 J  3.0  106 J
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W  7  105 J
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Example: (continued)
How far does the bus travel while slowing?
Fbrake = 2.75 x 104 N
xf
W=
 F(x) dx
xi
F
xi
xf
x
Area under F(x) curve:
W = Fbrake (x f -xi )
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Fbrake
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Example: (continued)
W = Fbrake ( x f -xi )
W
( x f -xi ) 
Fbrake
-7 105 J
( x f -xi ) 
-2.75 10 4 N
7 105 kg m2 /s 2

2.75 10 4 kg m/s2
( x f -xi )  25 m.
Fbrake  2.75 104 N
2
a



5
.
0
m/s
m
5.45 103 kg
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The Spring: A Variable Force
Springs exert force
when stretched or
compressed
Hooke’s Law:
F = - kx
k = "spring constant"
big k  stiff spring
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positive x
x=0
Defines equilibrium position
x = 0  no force
x < 0  F > 0 spring pushes out
x > 0  F < 0 spring pulls in
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Work Done By a Spring
How much work by spring in moving from xi to xf?
Fspring
x=0
Fspring
xi
x=0
xf
xf
xi
xi
W   F ( x)dx  
xf
1 2 1
2
(kx)dx  kxi  kx f
2
2
If |xi| < |xf| then spring takes energy away
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Example:
10 cm
Fspring
Fpull
You stretch a spring 10 cm and must apply a 10 N
force to hold the spring in place. What is the spring
constant, and how much work did you do on the
spring to stretch it?
Fspring  kx
k 
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Fspring
x
(10 N )

 100 N / m
0.1 m
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Example:
10 cm
Fspring
Fpull
The work done by the spring is:
Wspring  kxi  kx f
1
2
2
  12 kx f
1
2
2
2
  12 (100 N / m)(0.1 m) 2
 0 . 5 J
So the work I do on the spring is
W pull  Wspring  0.5 J
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Power
Work doesn’t depend on the time interval
Work to climb a flight of stairs
10 s
1 min
1 hour
~3000 J
Power is work done per unit time
W
Pavg 
t
dW
dx
Instantaneous Power: P 
F
 Fv
dt
dt
Average Power:
Units
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Work
time
1J
= 1 Watt
1s
1 hp = 746 W
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Power in 3-D
For a constant force:
dW  Fx dx  Fy dy  Fz dz
So power will be:
dW
P
dt
dx
dy
dz
 Fx  Fy
 Fz
dt
dt
dt
 Fx v x  Fy v y  Fz v z
 
 F v
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Express Elevator
Say a 900 kg elevator travels
135 floors (400 m) in 40 s. It
accelerates at 2.4 m/s2 until it
reaches a velocity of 12 m/s.
Find the average power (going up)
W = mgh = (900 kg)(9.8 m/s2)(400 m)
m = 900 kg
a = 2.4 m/s2
vmax = 12 m/s
= 3.5 x 106 J
Pavg
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W 3.5 106 J


 8.8 10 4 W
t
40 s
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Express Elevator
(continued)
Maximum Instantaneous Power:
P = Fv
F
a
Maximum just when
elevator reaches
cruising speed going up
F – mg = ma
F = m(g+a)
m = 900 kg
a = 2.4 m/s2
vmax = 12 m/s
P = (900 kg)(9.8 m/s2+2.4 m/s2)(12 m/s)
= 130,000 W
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An Application
A car engine can supply some maximum amount
of power.
How will the car’s velocity change when it
accelerates at constant power?
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Acceleration at Constant Power
Work
Power 
time

Work  Power  time
Start from v = 0 and use work-energy theorem:
1 2
mv  Pt
2
2 Pt
2
v 
m
2 Pt
v
m
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Acceleration is
not constant!
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Why Cars Need Gears
P  Fv
There is some maximum power at which the
engine can operate. Shifting to a higher gear
reduces the force applied to the wheels, allowing
for a higher top speed.
F must be larger than the air resistance for the car
to continue accelerating.
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Accelerating from 0 to 60
How much average horsepower is needed to
accelerate a 1100 kg car from 0 to 60 mph in 3.1
seconds?
1
K f  Ki  W
1
2
mv  Pt
P
2
1
2
mv 2
t
2

1100
kg

(
60
mph
)
P 2
3.1 s
1100kg  (26.8m / s ) 2

3.1 s
1
2
 1.27 105 W
 171 horsepower
In reality, there are transmission losses, etc., so you need
much more horsepower to achieve this level of performance.
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Example: (Problem 7.37)
The force (but not the power) to tow a boat
at constant velocity is proportional to the
speed. If a speed of 4.0 km/h requires 7.5
kW, how much power does a speed of 12
km/hr require?
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