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Physics 231 Topic 4: Energy and Work Wade Fisher September 17-21 2012

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Physics 231 Topic 4: Energy and Work Wade Fisher September 17-21 2012
Physics 231
Topic 4: Energy and Work
Wade Fisher
September 17-21 2012
MSU Physics 231 Fall 2012
1
Recall from Last Week
Force and Mass
Newton’s Laws of Motion
 Inertia, F=ma, Equal/Opposite Reactions
Application of Newton’s Laws
 Force diagrams
 Component Forces
Friction and Drag
 Kinetic, static, rolling frictions
Uniform circular motion
MSU Physics 231 Fall 2012
2
Car Driving in a Circle
A car is driving along a circular, flat road with radius of 240m. The
coefficient of static friction between the car’s tires and the road is
0.84. Find the maximum speed the car can travel without sliding.
MSU Physics 231 Fall 2012
3
Key Concepts: Work and Energy
Work and Energy
Work and it’s association with forces
 Constant forces (driving up a hill)
 Variable forces (stretching a spring)
Kinetic Energy
 Work-Energy Theorem
 Relationship to velocity
Potential Energy
Conservation of Mechanical Energy
Power
Covers chapter 5 in Rex & Wolfson
MSU Physics 231 Fall 2012
4
Work and Energy
Work: ‘Transfer of energy’
Quantitatively: The work W done by a constant force on
an object is the product of the force along the direction
of displacement and the magnitude of displacement.
W = (Fcos)·x
Units: N·m = Joule
1 calorie = 4.184 J
F

Fcos
1 Calorie = 4184 J
= 1 kcal
x
MSU Physics 231 Fall 2012
5
Work and Energy
When work is done,
energy can be stored.
Example: Lifting a ball
a distance y creates
“Potential Energy”.
Where does the energy
go in this case?
Answer: the energy
goes into friction (the
ground/sled heat up!)
MSU Physics 231 Fall 2012
6
Non-constant force/angle
W=(Fcos)x: what if Fcos is not constant while covering x?
Example: what if  or F changes while
pulling the block?
F

Fcos
Fcos
x
Area=A=(Fcos)x
x
W=(A)=total
area
x
x
The work done is the area under the graph of Fcos vs x
MSU Physics 231 Fall 2012
7
Example
4m
A person drags a block over a floor
with a force parallel to the floor.
Force
4N
2N
After 4 meters, the floor turns
rough and instead of a force of
2N a force of 4N must be applied.
0
4
The force-distance diagram shows the situation.
8m
distance
How much work did the person do over 8 meter?
a) 0 J
b) 16 J c) 20 J d) 24 J e) 32 J
MSU Physics 231 Fall 2012
8
Clicker Question: Friction and Work
A box is being pulled
a) friction does no work at all
across a rough floor at a
b) friction does negative work
constant speed. What can
c) friction does positive work
you say about the work
done by friction?
MSU Physics 231 Fall 2012
9
Clicker Question: Tension and Work
A ball tied to a string is being
whirled around in a circle.
What can you say about the
work done by tension?
a) tension does no work at all
b) tension does negative work
c) tension does positive work
MSU Physics 231 Fall 2012
10
Power: The rate of energy transfer
Work (the amount of energy transfer) is independent of time.
W=(Fcos)x … no time in here!
To measure how fast we transfer the energy we define:
Power = P = W/t (J/s=Watt)
P = (Fcos)·x/t = (Fcos)·vaverage
1 Watt = 0.00134 horsepower
MSU Physics 231 Fall 2012
11
A Runner
While running, a person dissipates about 0.60 J of
mechanical energy per step per kg of body mass.
A) If a 60 kg person develops a power of 70 Watt
during a race (distance = L), how fast is she running
(1 step=1.5 m)?
B) What is the force the person exerts on the road?
W=Fx
P=W/t=Fv
MSU Physics 231 Fall 2012
12
Quiz Question
A worker pushes a wheelbarrow with a force of 50 N over a
distance of 10 m. A frictional force acts on the wheelbarrow in
the opposite direction, with a magnitude of 30 N.
What net work is done on the wheelbarrow?
Wnet = ∑Fx
a)
b)
c)
d)
e)
don’t know
100 J
200 J
300 J
500 J
MSU Physics 231 Fall 2012
13
Example
A toy-rocket of 5.0 kg, after the initial acceleration
stage, travels 100 m in 2 seconds.
A) What is the work done by the engine?
B) What is the power of the engine?
h=100m
MSU Physics 231 Fall 2012
14
Clicker Question: Force and Work
a) one force
A box is being pulled up a rough incline
b) two forces
by a rope connected to a pulley. How
c) three forces
many forces are doing work on the
d) four forces
box?
e) no forces are doing work
MSU Physics 231 Fall 2012
15
Potential Energy
Potential energy (PE): energy associated with the position
of an object within some system.
Gravitational potential energy: Consider the work done by
the gravity in case of the rocket: W = (Fcos)·x
Wgravity = Fg cos(180o)h = -mgh
= -( mghf – mghi ) = mghi - mghf
= PEi - PEf
The ‘system’ is the gravitational field of the earth.
PE = mgh
Since we are usually interested in the change in gravitational
potential energy, we can choose the ground level (h=0) in a
convenient way.
MSU Physics 231 Fall 2012
16
Kinetic energy:
Consider an object that changes speed only
x=100m
t=0
VO=0
t=2s
Vf > 0
a) W = Fcosx = (ma)x … used Newton’s second law
b) v(t) = v0+at so t=(vf-v0)/a
c) x(t) = x0+v0t+½at2 so x-x0 = x = v0t+½at2
Kinetic energy: KE=½mv2
Combine b) & c)
d) ax = (v2-v02)/2
Combine a) & d)
W=½m(v2-v02)
When work is done on an
object and the only change
is its speed: The work done
is equal to the change in KE:
MSU Physics 231 Fall 2012
W = KEfinal-KEinitial
17
Conservation of energy
Mechanical energy = Potential Energy + Kinetic energy
Mechanical energy is conserved if:
• the system is closed (no energy can enter or leave)
• the forces are ‘conservative’ (see soon)
We’re not talking about this!
Heat, chemical energy
(e.g battery or fuel in an engine)
Are sources or sinks of internal
energy.
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18
Conservation of mechanical energy
Mechanical energy = potential energy + kinetic energy
In a closed system, mechanical energy is conserved*
V=100 m/s
ME = PE+KE = mgh + ½mv2 = constant
What about the accelerating rocket?
h=100m
M=5 kg
VO=0
At launch:
ME = 5*9.81*0 + ½5*02 = 0 J
At 100 m height:
ME = 5*9.81*100 + ½5*1002 = 29905 J
We did not consider a closed system!
(Fuel burning)
* There is an additional condition, see next lecture
MSU Physics 231 Fall 2012
19
Clicker Question: Work and Energy
Two blocks of mass m1 and m2 (m1 > m2) slide on
a frictionless floor and have the same kinetic
energy when they hit a long rough stretch ( > 0),
which slows them down to a stop. Which one
a) m1
b) m2
c) they will go the
same distance
goes farther?
MSU Physics 231 Fall 2012
20
Example of closed system
An object (0.2 kg) is dropped from a height of 35 m. Assuming
no friction, what is the velocity when it reaches the ground?
t=0 s
h=35 m
v=0 m/s
t>0 s
v at h=0?
MSU Physics 231 Fall 2012
21
Where is the kinetic energy…
1) highest?
2) lowest ?
Assume height of catapult is
negligible to the maximum height
of the stone.
And what about potential energy?
Parabolic
Motion

t=0
A
t=1
B
t=2
C
t=3
D
MSU Physics 231 Fall 2012
E
t=5
22
MSU Physics 231 Fall 2012
23
Clicker Question: Free Fall 1
Two stones, one twice the mass of
the other, are dropped from a cliff.
Just before hitting the ground, what
is the KINETIC ENERGY of the heavy
stone compared to the light one?
a) quarter as much
b) half as much
c) the same
d) twice as much
e) four times as much
MSU Physics 231 Fall 2012
24
A Swing
If relieved from rest, what is
the velocity of the ball at the
lowest point?
30o
L=5m
h
MSU Physics 231 Fall 2012
25
Work and Energy
 Work: W=Fcos()x Energy transfer
The work done is the same as the area under the graph of Fcos versus x
 Power: P=W/t
Rate of energy transfer
 Potential energy (PE) Energy associated with position.
 Gravitational PE: mgh Energy associated with position in grav. field.
 Kinetic energy KE: ½mv2 Energy associated with motion
 Elastic PE: ½kx2 energy stored in stretched/compressed spring.
 Conservative force:
Work done does not depend on path
 Non-conservative force:
Work done does depend on path
 Mechanical energy ME:
ME=KE+PE
Conserved if only conservative forces are present KEi+PEi=KEf+PEf
Not conserved in the presence of non-conservative forces
(KEi+PEi) - (KEf+PEf) = Wnc
MSU Physics 231 Fall 2012
26
Clicker Question: Free Fall 2
Two stones, one twice the mass of
the other, are dropped from a cliff.
Just before hitting the ground, what
is the VELOCITY of the heavy stone
compared to the light one?
a) quarter as much
b) half as much
c) the same
d) twice as much
e) four times as much
MSU Physics 231 Fall 2012
27
MSU Physics 231 Fall 2012
28
Race track
KE
PE TME NC
KE
With friction
PE TME NC
KE PE TME NC
MSU Physics 231 Fall 2012
KE
PE TME NC
KE
PE TME NC
29
Practice Question
Old faithful geyser in Yellowstone park shoots water hourly
to a height of 40 m. With what velocity does the water
leave the ground?
a)
b)
c)
d)
e)
7.0 m/s
14 m/s
20 m/s
28 m/s
don’t know
MSU Physics 231 Fall 2012
30
True or False?
5 kg
M1
3 kg
4.0 m
In the absence of friction, when
m1 starts to move down:
1) potential energy is
transferred from m1 to m2
2) potential energy is
transformed into kinetic
energy
3) m1 and m2 have the same
kinetic energy
M2
A. 1) and 2) and 3) are true
B. only 2) and 3) are true
C. only 1) and 2) are true
D. only 2 is true.
MSU Physics 231 Fall 2012
31
Conservation of mechanical energy
What is the speed of m1 and m2 when
they pass each other?
(PE1+PE2+KE1+KE2)=constant
5 kg
M1
3 kg
4.0 m
M2
196.2 = 156.8 + 4.0v2
v = 3.13 m/s
At time of release:
PE1 = m1gh1 =5.00*9.81*4.00
PE2 = m2gh2 =3.00*9.81*0.00
KE1 = ½m1v2 =0.5*5.00*(0.)2
KE2 = ½m1v2 =0.5*3.00*(0.)2
=196.2 J
=0.00 J
=0.00 J
=0.00 J
Total
=196. J
At time of passing:
PE1 = m1gh1 = 5.00*9.81*2.00
PE2 = m2gh2 = 3.00*9.81*2.00
KE1 = ½m1v2 = 0.5*5.00*(v)2
KE2 = ½m1v2 = 0.5*3.00*(v)2
=98.0 J
=58.8 J
=2.5v2 J
=1.5v2 J
Total
MSU Physics 231 Fall 2012
=156.8+4.0v2 J
32
Work
How much work is done by the
gravitational force by the point when
the masses pass each other?
5 kg
M1
3 kg
4.0 m
M2
MSU Physics 231 Fall 2012
33
Clicker Question!
5 kg
M1
3 kg
4.0 m
In the absence of friction, when
m1 starts to move down:
1) potential energy is
transferred from m1 to m2
2) potential energy is
transformed into kinetic
energy
3) m1 and m2 have the same
kinetic energy
M2
A. 1) and 2) and 3) are true
B. only 2) and 3) are true
C. only 1) and 2) are true
D. only 2 is true.
MSU Physics 231 Fall 2012
34
Springs!
Hooke’s Law:
Fs = -kx
k: spring constant (N/m)
x: distance the spring is stretched from
equilibrium
Fs (x=0) = -k(0) = 0 N
Fs (x=a) = -ka N
Ws = Fsx = (ka/2)a = ½ka2
The energy stored in a spring depends
on how far the spring has been
stretched: elastic potential energy
PEspring = ½ k x2
MSU Physics 231 Fall 2012
35
PINBALL!
The ball-launcher spring has a
constant k = 120 N/m.
A player pulls the handle 0.05 m.
The mass of the ball is 0.1 kg.
What is the launching speed?
MSU Physics 231 Fall 2012
36
Clicker Question Kinetic Energy
Two cars are driving along a highway.
Car #1 has twice the mass of car #2,
but they both have the same kinetic
energy.
a)
2v1 = v2
b)
 2v1 = v2
c)
4v1 = v2
d)
v1 = v2
e)
8v1 = v2
How do their speeds compare?
MSU Physics 231 Fall 2012
37
Conservative Forces
A force is conservative if the work done by the force when
Moving an object from A to B does not depend on the path
taken from A to B.
Example: work done by gravitational force
Using the stairs:
Wg = mghf-mghi = mg(hf-hi)
h=10m
Using the elevator:
Wg = mghf-mghi = mg(hf-hi)
The path taken (longer or
shorter) does not matter: only
the displacement does!
MSU Physics 231 Fall 2012
38
Non-Conservative Forces
A force is non-conservative if the work done by the force
when moving an object from A to B depends on the path
taken from A to B.
object on rough surface
top view
Example: Friction
You have to perform more work
Against friction if you take the
long path, compared to the short
path. The friction force changes
kinetic energy into heat.
MSU Physics 231 Fall 2012
39
Conservation of mechanical energy only holds if the
system is closed AND all forces are conservative
MEi-MEf=(PE+KE)i-(PE+KE)f=0 if all forces
are conservative
Example: throwing a snowball from a
building neglecting air resistance
MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc if some
forces are nonconservative.
Wnc=work done by non-conservative forces.
Example: throwing a snowball from a
building taking into account air resistance
MSU Physics 231 Fall 2012
40
Friction (non-conservative)
The pulley is now not frictionless. The
friction force equals 5 N. What is the
speed of the objects when they pass?
5 kg
M1
3 kg
4.0 m
M2
MSU Physics 231 Fall 2012
41
Overview
Conservation of mechanical
energy
Wnc=0
Closed system
Newton’s second Law
F=ma
Work
W=(Fcos)x
Work-energy Theorem
Wnc=MEf-MEi
Equations of kinematics
X(t)=X(0)+V(0)t+½at2
V(t)=V(0)+at
MSU Physics 231 Fall 2012
42
A
energy
Clicker Quiz!
B
In the absence of friction,
which energy-time diagram
is correct?
total energy
time
C
energy
potential energy
energy
time
kinetic energy
MSU Physics 231 Fall 2012
time
43
Ball on a track
A
h
end
B
h
end
In which case has the ball the highest velocity at the end?
A) Case A
B) Case B
C) Same speed
In which case does it take the longest time to get to the end?
A) Case A
B) Case B
C) Same time
MSU Physics 231 Fall 2012
44
Question
A ball rolls down a slope as shown in the figure. The starting
velocity is 0 m/s. There is some friction between the ball and
the slope. Which of the following is true?
h
a) The kinetic energy of the ball at the bottom of the slope
equals the potential energy at the top of the slope
b) The kinetic energy of the ball at the bottom of the slope is
smaller than the potential energy at the top of the slope
c) The kinetic energy of the ball at the bottom of the slope is
larger than the potential energy at the top of the slope
MSU Physics 231 Fall 2012
45
Extra credit quiz
A ball rolls down a slope and back up a more shallow slope as
shown in the figure. The starting velocity is 0 m/s. There is
some friction between the ball and the slope. Which is true?
h1
h2
a) The maximum height reached on the right (h2) is the same as
the height the ball started at on the left (h1)
b) The maximum height reached on the right (h2) is smaller than
the height the ball started at on the left (h1)
c) The maximum height reached on the right (h2) is larger than
the height the ball started at on the left (h1)
MSU Physics 231 Fall 2012
46
Question
A ball of 1 kg rolls up a ramp, with initial velocity of 6 m/s.
It reaches a maximum height of 1 m (i.e., the velocity is 0 m/s
at that point). How much work is done by friction?
a)
b)
c)
d)
e)
0.
8.2 J
9.8 J
18 J
27.8 J
MSU Physics 231 Fall 2012
47
Question
An outfielder who is 2M tall throws a baseball of 0.15 kg at
a speed of 40 m/s and angle of 30 degrees with the field.
What is the kinetic energy of the baseball at the highest
point, ignoring friction?
a)
b)
c)
d)
e)
0J
30 J
90 J
120 J
don’t know
MSU Physics 231 Fall 2012
48
Question
An outfielder who is 2m tall throws a baseball of 0.15 kg at
a speed of 40 m/s and angle of 30 degrees with the field.
How high does the ball go at its highest point, ignoring
friction?
MSU Physics 231 Fall 2012
49
For Next Week
Rex & Wolfson Chapter 6.1-6.5:
Momentum and Collisions
Homework Set 4: Covers Chapters 5
Due Wednesday 11PM
MSU Physics 231 Fall 2012
50
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