Download 6.3 Kinetic Energy - Purdue Physics

PHYSICS 149: Lecture 15
• Chapter 6: Conservation of Energy
– 6.3 Kinetic Energy
– 6.4 Gravitational Potential Energy
Lecture 15
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ILQ 1
Mimas orbits Saturn at a distance D. Enceladus
orbits Saturn at a distance 4D
4D. What is the ratio of
the periods of their orbits?
A)
B)
C)
D)
Tm/Te = 1/8
Tm/Te = 1/4
Tm/Te = 1/2
Tm/Te = 2
Lecture 15
T2 ∝ r 3
2
⎛ Tm ⎞ ⎛ D ⎞
⎜ ⎟ =⎜
⎟
T
4
D
⎝
⎠
⎝ E⎠
⎛ Tm ⎞
1 1
=
=
⎜ ⎟
64 8
⎝ TE ⎠
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2
ILQ 2
A pendulum bob swings back and forth along a circular
path. Does the tension in the string do any work on the
bob? Does gravity do work on the bob?
A) only tension does work
B) both do work
C) neither do work
D) only gravity does work
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Energy
• Energy is “conserved” meaning it can not be created nor
destroyed
– Can change form
– Can be transferred
• Total Energy of an isolated system does not change with
time
• Forms
–
–
–
–
Kinetic Energy
Potential Energy
Heat
Mass (E=mc2)
Motion
Stored
• Units: Joules = kg m2 / s2
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Definition of “Work” in Physics
• Work is a scalar quantity (not a vector quantity).
• Units: J (Joule), N⋅m, kg⋅m2/s2, etc. .
– Unit conversion: 1 J = 1 N⋅m
N m = 1 kg⋅m
kg m2/s2
• Work is denoted by W
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(not to be confused by weight W).
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Total Work
• When several forces act on an object
object, the “total”
total
work is the sum of the work done by each force
individually:
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ILQ 1
You are towing a car up a hill with constant velocity.
The work done on the car by the normal force is:
A) positive
B) negative
C) zero
FN
V
T
Normal force is perpendicular to direction of
di l
displacement,
t so work
k iis zero.
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ILQ 2
You are towing a car up a hill with constant velocity.
The work done on the car by the gravitational force is:
A) positive
B) negative
C) zero
FN
V
T
Gravity is pushing against the direction of
motion
ti so it iis negative.
ti
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ILQ 3
You are towing a car up a hill with constant velocity.
The work done on the car by the tension force is:
A) positive
B) negative
C) zero
FN
V
T
The force of tension is in the same direction as
th motion
the
ti of
f the
th car, making
ki the
th work
k positive.
iti
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ILQ 4
You are towing a car up a hill with constant velocity.
The total work done on the car by all forces is:
A) positive
B) negative
C) zero
FN
V
T
The total work done is positive because the car is
moving up the hill. (Not quite!)
W
W=KEf-KEi=(0.5mvf2) - (0.5mvi2). Because the final
and initial velocities are the same, there is no
change in kinetic energy, and therefore no total
work
k iis d
done.
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Problem
A box is pulled up a rough (μ > 0) incline by a ropepulley-weight arrangement as shown below. How
many forces are doing (non-zero) work on the box?
A) 0
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B) 1
C) 2
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D) 3
E) 4
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Solution
Draw FBD of box:
N
T
z
Consider direction of
motion of the box
z
Any force not perpendicular
to the motion will do work:
v
f
N does no work (perp. to v)
T does positive work
f does negative work
3 fforces
do work
mg
mg does negative work
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Example: Block with Friction
•
A block is sliding on a surface with an initial speed of 5 m/s. If the
coefficient of kinetic friction between the block and table is 0.4, how
far does the block travel before stopping?
N
yy-direction:
direction: F
F=ma
ma
N-mg = 0
N = mg
Work
WN = 0
Wmg = 0
Wf = f Δx cos(180)
= -μmg Δx
5 m/s
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f
y
x
mg
W=ΔK
-μmg Δx = ½ m (vf2 – v02)
-μg
μg Δx = ½ (0 – v02)
μg Δx = ½ v02
Δx = ½ v02 / μg
= 3.1
3 1 meters
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Kinetic Energy: Motion
• Apply constant force along x-direction to a
point particle m.
W = Fx Δx
= m ax Δx
= ½ m (vf2 – v02)
1 2 2
recall : ax Δx = (vx − vx 0 )
2
• Work changes ½ m v2
• Define Kinetic Energy K = ½ m v2
W=ΔK
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For Point Particles
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Work-Kinetic
W
k Ki ti
Energy
Th
Theorem
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Translational Kinetic Energy
• When an object of mass m is moving with speed v (the
g
of instantaneous velocity),
y), the object’s
j
magnitude
“translational kinetic energy” is defined as follows:
•
•
•
•
Kinetic energy is a scalar quantity.
Units: J, N
N⋅m,
m, kg
kg⋅m2/s2,
m2/s2, etc.
Kinetic energy is denoted by K.
Translational kinetic energy means the total work done on
the object to accelerate it to that speed starting from rest.
• Translational kinetic energy is often called the “kinetic
energy” if it is clearly distinguished from the rotational
energy
energy or internal energy.
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Work - Kinetic Energy Theorem
= K f − Ki =
1 2 1 2
mv f − mvi
2
2
• The work done on an object by the “net” force
(whether the net force is constant or variable) is
equal to the change in the kinetic energy.
• Or, the “total” work done on the object is equal to
the change in the kinetic energy
energy.
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ILQ
Compare the kinetic energy of two balls:
Ball 1: mass m thrown with speed 2v
Ball 2: mass 2m thrown with speed v
A)
B)
C)
D)
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K1 = 4K2
K1 = K2
2K1 = K2
K1 = 2K2
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Work Done by Gravity 1
• Example
p 1: Drop
p ball
Yi = h
Wg = (mg)(s)cosθ
s=h
Wg = mghcos(00) = mgh
mg
Δy = yf-y
yi = -h
h
Yf = 0
y
x
Wg = -mgΔy
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S
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Work Done by Gravity 2
• Example
p 2: Toss ball up
p
Yi = h
Wg = (mg)(s)cosθ
s=h
Wg =
mghcos(1800)
= -mgh
Δy = yf-y
yi = +h
h
Wg = -mgΔy
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mg
S
y
Yf = 0
x
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Work Done by Gravity 3
• Example
p 3: Slide block down incline
Wg = (mg)(s)cosθ
s = h/cosθ
Wg = mg(h/cosθ)cosθ
h
Wg = mgh
θ
mg
S
Δy = yf-yi = -h
Wg = -mgΔy
mgΔy
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Work Done by Gravity
• Depends only on initial and final height!
• Wg = -mg(y
mg(yf - yi) = -mgΔy
mgΔy
– Independent of path
– If you end up where you began
began, Wg = 0
– Note: can do work “against” gravity, then get gravity to
do work back.
– Define: Potential Energy……
We call this a “Conservative Force” because we can
define a “Potential
Potential Energy
Energy” to go with it
it.
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Work Done by Gravity
•
Question: Does the work done by gravity depend on the path taken?
θ
mg
Δr
Left Case: Wgrav = F⋅Δr⋅cosθ = mg⋅|Δy|⋅cos0° = mg|Δy|
Middle Case: Wgrav = F⋅Δr⋅cosθ = mg|Δy| (because Δr⋅cosθ = |Δy|)
Right Case: Wgrav = mg|Δy| (because each segment can be
treated like the middle case)
•
•
Answer: The work done by gravity is independent of path–that is, the
work depends only on the initial and final positions (|Δy|)
(|Δy|).
This kind of force is called “conservative force.”
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Potential (stored) Energy
• “Stored” gravitational energy can be converted to
kinetic energy
ƒ -m g Δy = ΔK
ƒ 0 = ΔK + m g Δyy
ƒ 0 = ΔK + ΔUg
W = ΔK
define Ug = mgy
gy
ΔU = -WC
• Works for any CONSERVATIVE force
ƒ Gravity Ug = m g y
ƒ Spring Us = 1/2 k x2
ƒ NOT friction
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Potential
Only change in potential energy is important
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What is Potential Energy?
•
An object is thrown up vertically, and it reaches top.
Assume no air resistance.
The work done by gravity (near the surface of Earth) is
Wgrav = F⋅Δ
Δr⋅cosθ = mg⋅Δ
Δy⋅cos180° = –mg
mgΔy
vf=0
In this problem, gravity is the only force. Thus,
Wtotal
mgΔy
t t l = Wgrav = –mg
According to work-energy theorem,
Wtotal = –mgΔy = –½mvi2 (because ½mvf2=0)
That is, the initial kinetic energy (Ki = ½mvi2) has been
“stored” in (or transformed into) the form of mgΔy at the
top. And, it has the “potential” to do work (or to become
kinetic energy).
•
vi
Stored energy due to the interaction of an object with
something else (in this case, gravity) that can easily
be recovered as kinetic energy is called potential
energy.
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Definition of Potential Energy
• The change in potential energy is equal to the negative of
the work done by the conservative forces.
– Potential energy can be defined only for the conservative forces
forces.
For the non-conservative forces, potential energy can not be
defined in the first place.
– There is no way
y to calculate the absolute value of the p
potential
energy. Only the change in potential energy is important.
– The choice of the zero point of potential energy is arbitrary.
• Potential energy is a scalar quantity.
• Units: J, N⋅m, kg⋅m2/s2, etc.
• Potential energy is denoted by U (so ΔU means the
change in potential energy).
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ILQ
A hiker descends from the South Rim of the
Grand Canyon to the Colorado River
River. During this
hike, the work done by gravity on the hiker is
A)
B)
C)
D)
E)
positive and depends on the path taken
positive and independent of the path taken
negative and depends on the path taken
negative and independent of the path taken
zero
W g = Fg ( Δy ) cos θ =
mgh cos(0) = mgh
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Work and Potential Energy
• Work done by gravity independent of path
– Wg = -mg
g (yf - yi)
• Define Ug = mgy
• Works for any CONSERVATIVE force
• Modify Work-Energy theorem
∑Wnc = ΔK + ΔU
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Conservative Force
• If the work done by a force is independent of
path (that is, depends only on the initial and final
positions),
iti
) th
the fforce iis called
ll d ““conservative.”
ti ”
– Example: gravitational force, spring force, and
electrical force
– Note that the work done by conservative forces for
any closed loop is zero.
• If the work done by a force depends on the path
taken, the force is called non-conservative.
– Example: frictional force and air resistance
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Work - Energy with Conservative Forces
Work Energy Theorem
∑Wi = ΔK
ΔU = -WC
ΣW = Wcons + Wnc = ΔK
Move work by conservative forces to other side
∑Wnc = ΔK + ΔU
W n c = Δ K − W co n s
If there are NO non-conservative forces
0 = ΔK + ΔU = ΔEmech
0 = K f − Ki + U f − U i
Ki + U i = K f + U f
Lecture 15
Ei = E f
Conservation of
mechanical energy
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Mechanical Energy
• The sum of the kinetic and potential energies is
called the (total) mechanical energy (Emechh).
)
Emech ≡ K + U
• Mechanical energy is a scalar quantity.
• Units: JJ, N⋅m
N⋅m, kg⋅m2/s2, etc.
etc
• Mechanical energy is denoted by Emech.
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Conservation of Mechanical Energy
• If the work is done by only conservative forces (this is, the
), the
work done byy non-conservative forces is zero),
mechanical energy is conserved.
Emech = Ki + Ui = Kf + Uf = const (if Wnc = 0)
• If the work is done by also non-conservative forces, the
mechanical energy is not conserved and the following
relations are satisfied.
or
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Falling Ball Example
• Ball falls a distance of 5 meters. What is
its final speed?
Emech = Ki + Ui = Kf + Uf = const (b/c Wnc = 0)
Ui = mgyi
Uf = 0
Ki = 0
Kf = ½mvf2
Only force/work done be gravity
Wg = m ½ (vf2 – vi2)
Fg h = ½m
½ vf2
mg
mgh = ½m vf2
Vf = sqrt( 2 g h ) = 10 m/s
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Example: Pendulum
Vi = 0
•
•
•
mg
In this case, there are two forces acting on the object.
But, the direction of tension is perpendicular to the displacement of the
object, so the work done by tension is zero.
Gravity (conservative force) is the only force which does work
work, so the
mechanical energy is conserved.
Emech = Ki + Ui = Kf + Uf = const (b/c Wnc = 0)
Ui = mgh
Uf = 0
Ki = 0
Kf = ½mvf2
Æ 0 + mgh = ½mvf2 + 0
Thus, vf = sqrt(2gh) Å the same result as two previous results
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ILQ
Imagine that you are comparing three different ways of having
a ball move down through the same height.
In which case does the ball get to the bottom first?
A)
B)
C)
D)
Dropping
Slide on ramp (no friction)
Swinging down
All the same
correct
A
B
C
The
h time
t m to g
gett to the
th bottom
ottom iss height
h ght / y-component
y compon nt of velocity
oc ty
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ILQ
Imagine that you are comparing three different ways of having
a ball move down through the same height.
In which case does the ball reach the bottom with the highest
speed?
A))
B)
C)
D)
Dropping
pp g
Slide on ramp (no friction)
Swinging down
correct
All the same
A
B
C
Conservation of Energy (Wnc=0)
Kinitial + Uinitial = Kfinal + Ufinal
0 + mgh = ½ m v2final + 0
vfinal = sqrt(2 g h)
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Pendulum ILQ
• As the pendulum falls, the work done by the string is
A) Positive
B) Zero
C) Negative
W = F d cos θ. But θ = 90 degrees so Work is zero.
• How fast is the ball moving at the bottom of the path?
Conservation of Energy (Wnc=0)
ΣWnc = ΔK + Δ U
0 = Kfinal - Kinitial + Ufinal - Uinitial
Kinitial + Uinitial = Kfinal + Ufinal
0 + mgh = ½ m v2final + 0
vfinal = sqrt(2 g h)
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Pendulum Demo
A pendulum is released from a height h above the
minimum. At the bottom of its swing, the string hits a
peg, reducing the length. What is the final height y
the ball reaches?
A)) h < y
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B)) h = y
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C)) h > y
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Galileo’s Pendulum
How high will the pendulum swing on the other side now?
A) h1 > h2
B) h1 = h2
C) h1 < h2
Conservation of Energy (Wnc=0)
ΣWnc = ΔK + Δ U
Kinitial + Uinitial = Kfinal + Ufinal
0 + mgh1 = 0 + mgh2
h1 = h2
m
h1
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h2
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Gravitational Potential Energy
• If the gravitational force is not constant or nearly constant,
we have to start from Newton’s law
m1m2
F =G 2
r
• The gravitational potential energy is:
m1m2
U = −G
r
if U = 0 for r = ∞
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