“It is important that students bring a certain
ragamuffin, barefoot irreverence to their studies;
they are not here to worship what is known, but
to question it.”
Jacob Chanowski
Reading: Chapter 6.
Raw grades are on Blackboard .
Homework 3 is on-line now and due on
Friday at 5pm.
Chapter 6
Work and Energy
Work
W = Fs*cos q
What happens between
points A and B?
A
B
C
What happens between
points A and B?
Work is done by gravity.
A
B
C
What happens between
points A and B?
What happens to the
speed of the person?
A
B
C
What happens between
points A and B?
What happens to the
speed of the person?
It increases.
The work done by
gravity goes into
changing the speed of
the person.
A
B
C
Back to our skier
We found that gravity
did work on this skier.
Where did that work
go?
Back to our skier
We found that gravity
did work on this skier.
Where did that work
go?
It went into making the skier go faster!
Work – Energy equivalency
principle.
The total work done on an object is
equal to the change in its energy.
Wtotal = ∆E
Work – Energy equivalency
principle.
The total work done on an object is
equal to the change in its energy.
In our slide and skiing examples,
work done by gravity is converted
into motion (kinetic energy) of the
person.
So now we need
to define some
energies.
Kinetic Energy (KE) is the energy of motion.
KE is measured in Joules (just like work).
KE = ½
2
mv
NOTE that because we have v2, we are
directionally impaired and answers will NOT
give the direction of the velocity.
Work-Energy Theorem
The total work done on an object is
equal to the change in its energy –
usually kinetic but it can be other
forms.
Wtotal = ∆KE = KEf-KEo
2
2
Fs*cos q = ½ m(vf -vo )
Kinetic Energy Example
What is the kinetic energy of a 0.1kg
hockey puck traveling at 25m/s?
KE = ½ mv2
A) 1.25J
B) 27.1J
C) 31.3J
D) 37.4J
E) 0J
Kinetic Energy Example
What is the kinetic energy of a 0.1kg
hockey puck traveling at 25m/s?
KE = ½ mv2
KE= ½ (0.1)(252) = 31.3J
Kinetic Energy Example
A football is thrown across the field and
caught by a person standing still. Just
before it's caught, it's moving at 28m/s. If
the mass of the football is 0.6kg, how
much work was done on the football in
catching it?
Where do we start?
Kinetic Energy Example
A football is thrown across the field and
caught by a person standing still. Just
before it's caught, it's moving at 28m/s. If
the mass of the football is 0.6kg, how
much work was done on the football in
catching it?
Where do we start?
W=F.s cosq
We don't know how much force was
applied or over what distance.
Kinetic Energy Example
A football is thrown across the field and
caught by a person standing still. Just
before it's caught, it's moving at 28m/s. If
the mass of the football is 0.6kg, how
much work was done on the football in
catching it?
Where do we start?
We don't know how much force was
applied or over what distance.
But we do know that its final motion is 0,
so ALL of its kinetic energy has been
converted into work.
DKE → W
Kinetic Energy Example
A football is thrown across the field and
caught by a person standing still. Just
before it's caught, it's moving at 28m/s. If
the mass of the football is 0.6kg, how
much work was done on the football in
catching it?
KE → W
DKE = ½ mv2
Kinetic Energy Example
A football is thrown across the field and
caught by a person standing still. Just
before it's caught, it's moving at 28m/s. If
the mass of the football is 0.6kg, how
much work was done on the football in
catching it?
KE → W
KE = ½ mv2
=1/2 (0.6)(282) = 235.2J
All this energy was lost to the work of
catching the ball.
Two women push a car (m=870kg) with a force of
235N for 23m. How fast is the car going when
they're done? (Assuming no friction.)
Two women push a car (m=870kg) with a force of
235N for 23m. How fast is the car going when
they're done?
The work of the women is converted into kinetic
energy:
W → DKE
Two women push a car (m=870kg) with a force of
235N for 23m. How fast is the car going when they're
done?
W=F.scosq → KE=1/2 mv2
235(23)=1/2 (870) v2
Another type of energy:
Gravitational Potential Energy.
The skydiver's kinetic energy increases after he
leaves the plane. That energy must come from
somewhere.
Another type of energy:
Gravitational Potential Energy: PE.
The skydiver is converting PE into kinetic energy.
PE = mg(h -h ) Caution: g is positive for energy!!!!!
f
Most often
PE = mgh.
o
Potential energy is the stored
ability to do work: stored energy.
EXTREMELY
IMPORTANT:
ENERGY IS NEVER LOST.
IT IS ONLY CONVERTED
FROM ONE FORM TO
ANOTHER.
(It may be lost to “a system” like a cup of coffee
loses heat to a room, but it is still there.)
The top of this platform is 10m
above the water. How fast will a
diver be moving when he/she hits
the water? (Note that this problem
can be solved without knowing the
diver's mass!!!)
Where to start?
The top of this platform is 10m
above the water. How fast will a
diver be moving when he/she hits
the water? (Note that this problem
can be solved without knowing the
diver's mass!!!)
Where to start?
We could use velocity and
acceleration from Chapter 3:
vf2=vo2+2gy with vo=0 and y=10m.
This gives an answer vf= -14m/s
The top of this platform is 10m
above the water. How fast will a
diver be moving when he/she hits
the water? (Note that this problem
can be solved without knowing the
diver's mass!!!)
Where to start?
We could use vf2=vo2+2gy from
Chapter 3: vf= -14m/s
Or we could convert the change in PE (use the
surface of the water as h=0) to a change in KE (the
diver begins at rest): PEo= KEf
mgh = ½ mvf2 → vf2= 2gh (look familiar?)
Converting forms of energy
As the ball rolls
down the incline, it
is converting
gravitational
potential energy
into kinetic energy.
Neglecting wind
and friction, no
energy is lost, only
converted.
Converting forms of energy
If I know the
energy at one
point, I know it at
ALL points.
EA=EB=EC
Only its form could
change.
Converting forms of energy
mghA=mghB+1/2mvB2=mghC+1/2mvC2
If I know the
energy at one
point, I know it at
ALL points.
EA=EB=EC
Only its form could
change.
Converting forms of energy
mghA=1/2mvB2=mghC+1/2mvC2
EA=EB=EC
I can define the
axes to make it
even simpler.
Converting forms of energy
mghA=1/2mvB2=mghC+1/2mvC2
EA=EB=EC
If the ball is 2kg,
how much energy
does it have?
Where to begin?
Converting forms of energy
EA=EB=EC
If the ball is 2kg,
how much energy
does it have?
Where to begin?
We only have
enough information
to solve the energy
for Point A.
mghA=1/2mvB2=mghC+1/2mvC2
Converting forms of energy
EA=EB=EC
If the ball is 2kg,
how much energy
does it have?
We only have
enough information
to solve the energy
for Point A.
EA=mgh=2(9.8)7
mghA=1/2mvB2=mghC+1/2mvC2
Converting forms of energy
EA=EB=EC
If the ball is 2kg,
how much energy
does it have?
We only have
enough information
to solve the energy
for Point A.
EA=mgh=2(9.8)7
= 137.2 J
mghA=1/2mvB2=mghC+1/2mvC2
Converting forms of energy
mghA=1/2mvB2=mghC+1/2mvC2
2
2
At
B:
137.2J
=
½(2)v
At C: 137.2J=2(9.8)3+½(2)vC
B
EA=EB=EC
If the ball is 2kg,
how fast is it
moving at points B
and C?
EA= 137.2 J
Converting forms of energy
mghA=1/2mvB2=mghC+1/2mvC2
2
2
At
B:
137.2J
=
v
At C: 78.4 = vC
B
EA=EB=EC
If the ball is 2kg,
how fast is it
moving at points B
and C?
EA= 137.2 J
Converting forms of energy
mghA=1/2mvB2=mghC+1/2mvC2
At B: vB = 11.7 m/s
At C: vC = 8.85 m/s
EA=EB=EC
If the ball is 2kg,
how fast is it
moving at points B
and C?
EA= 137.2 J
Conservation of Energy
Would the speed
of the ball at points
B and/or C be
different if the ball
took the blue or
orange path?
(still neglecting
friction and other
forces)
Conservation of Energy
Would the speed
of the ball at points
B and/or C be
different if the ball
took the blue or
orange path?
(still neglecting
friction and other
forces)
No! The speed will be the
same. Energy is not lost,
only converted. So the
path doesn't matter!
•Law of Conservation of
Energy – energy can neither
be created nor destroyed –
it can only change forms.
Mechanical => Kinetic (motion) &
Potential (spring)
Other forms of Energy => Chemical,
Heat, Nuclear, Solar, Geothermal,
Sound, etc.
Electromagnetic Energy
Nonconservative interactions:
Where energy is lost from the system.
How could our
ball example be a
nonconservative
example?
Nonconservative interactions:
Where energy is lost from the system.
How could our
ball example be a
nonconservative
example?
If we allow friction
or wind
resistance.
Nonconservative interactions:
Where energy is lost from the system.
How could our
ball example be a
nonconservative
example?
If we allow friction
or wind
resistance.
What's the speed of the ball at
points B and C if mK=0.2?