Transcript Work and Energy

Work and Energy
©2009 by Goodman & Zavorotniy
Conservation Principles
The most powerful concepts in science are called "conservation
principles". These principles allow us to solve problems without worrying
much about the details of a process.
We just have to take a snapshot of a system initially and finally; by
comparing those two snapshots we can learn a lot.
Conservation Principles
A good example is a jar of candy.
If you know that there are fifty pieces of candy at the beginning. And you
know that no pieces have been taken out or added...you know that there
must be 50 pieces at the end.
Now, you could change the way you arrange them...move them around,
whatever...but you still will have 50 pieces.
50 pieces
still
50
pieces
still 50 pieces
Conservation Principles
In that case we would say that the number of pieces of candy is conserved.
That is, we should always get the same amount, regardless of how they are
arranged as long as we take into account whether any have been added or
taken away.
50 pieces
still
50
pieces
still 50 pieces
Conservation Principles
We also have to be clear about the system that we're talking about. If we're
talking about one candy jar...we can't suddenly start talking about a
different one and expect to get the same answers.
We must define the system whenever we use a conservation principle.
50 pieces
still
50
pieces
still 50 pieces
Conservation of Energy
Energy is a conserved property of nature. It is not created or
destroyed, so in a closed system we will always have the same
amount of energy.
The only way the energy of a system can change is if it is open to the
outside...if energy has been added or taken away.
You could ask "what is energy?"
It turns out that energy is so fundamental, like space and time, that
there is no good answer to that question.
Conservation of Energy
However, just like space and time, that doesn't stop us from doing
very useful calculations with energy.
We may not be able to define energy, but because it is a conserved
property of nature, it's a very useful idea.
Conservation of Energy
If we call the amount of energy that we start with as "Eo" and the amount we
end up with as "Ef" then we would say that if no energy is added to or taken
away from a system that
Eo = Ef
It turns out that there are only two ways to change the energy of a system.
One is with heat, which we won't deal with here; the other is with "Work",
"W".
If we define positive work as that work which increases the energy of a
system our equation becomes:
Eo + W = Ef
Work
Work can only be done to a system by an external force; a force from
something that is not a part of the system.
So if our system is a box sitting on a table...and I come along and push the
box, I can increase the energy of the box...I am doing work to the box.
Work
Previously, we learned that the amount of work done to a system, and
therefore the amount of energy increase that the system experiences is
given by:
Work = Force x Distanceparallel
OR
W = Fdparallel
This is still valid, but we have to bring a new interpretation to that equation
based on what we know about vector components. First, let's review some
more basic facts.
Work
W = Fdparallel
If the object that is experiencing the force does not move (if dparallel = 0)
then no work is done: W = 0. The energy of the system is unchanged.
If the object moves in the direction opposite the direction of the force (for
instance if F is positive and dparallel is negative) then the work is
negative: W < 0. The energy of the system is reduced.
If the object moves in the same direction as the direction of the force (for
instance if F is positive and dparallel is also positive) then the work is
positive: W > 0. The energy of the system is increased.
Units of Work and Energy
W = Fdparallel
This equation gives us the units of work. Since force is measured in
Newtons (N) and distance is measured in meters (m) the unit of work is
the Newton-meter (N-m).
And since N = kg-m/s2; a N-m also equals a kg-m2/s2.
However, in honor of James Joule, who made critical contributions in
developing the idea of energy, the unit of energy is also know as a Joule
(J).
1 Joule = 1 Newton-meter = 1 kilogram-meter2/second2
1 J = 1 N-m = 1 kg-m2/s2
Units of Work and Energy
Eo + W = Ef
Since work changed the energy of a system: the units of energy
must be the same as the units of work.
The units of both work and energy are the Joule.
Force and Work
v
Previously, the force was
either parallel, anti-parallel,
or perpendicular.
F
Let's look at those three
cases.
Δx
Force and Work
F
v
W = Fdparallel = 0
Δx
v
F
W = Fdparallel = FΔx
Δx
F
v
W = Fdparallel = -FΔx
Δx
Force and Work
How do we interpret:
W = Fdparallel
in this case.
v
Δx
Force and Work
After breaking F into
components that are parallel
and perpendicular to the
direction of motion, we can
see that no Work is done by
the perpendicular
component, only by the
parallel component.
v
θ
Δx
Force and Work
W = FΔxcosθ
The work done on an
object by a force is the
product of the magnitude
of the force and the
magnitude of the
displacement times the
cosine of the angle
between them.
v
θ
Δx
Force and Work
This is a similar case. We just
have to find the component of
force that is parallel to the
object's displacement.
Force and Work
The interpretation is the same
in this case, just determine the
angle between the force and
displacement and use:
θ
W = FΔxcosθ
Δx
A force F is at an angle θ above the horizontal is used to pull a heavy
suitcase of weight mg a distance d along a level floor at constant
velocity. The coefficient of friction between the floor and the suitcase is
μ. The work done by the force F is:
1
A
Fdcos θ - μ mgd
B
Fdcos θ
v
C
-μ mg
D
2Fdsin θ - μ mgd
E
Fdcos θ - 1
θ
A force F is at an angle θ above the horizontal and is used to pull a heavy
suitcase of weight mg a distance d along a level floor at constant
velocity. The coefficient of friction between the floor and the suitcase is
μ. The work done by the friction force is:
2
A
Fdcos θ - μ mgd
B
0
C
-μd(mg - Fsinθ)
D
2Fdsin θ - μ mgd
E
Fdcos θ - 1
v
θ
A force F is at an angle θ above the horizontal and is used to pull a
heavy suitcase of weight mg a distance d along a level floor at constant
velocity. The coefficient of friction between the floor and the suitcase is
μ. The work done by the normal force is:
3
A
Fdcos θ - μ mgd
B
0
C
-μ mgd
D
2Fdsin θ - μ mgd
E
Fdcos θ - 1
v
θ
A force F is at an angle θ above the horizontal and is used to pull a heavy
suitcase of weight mg a distance d along a level floor at constant
velocity. The coefficient of friction between the floor and the suitcase is
μ. The work done by the gravitational force is:
4
A
Fdcos θ - μ mgd
B
0
C
-μ mgd
D
2Fdsin θ - μ mgd
E
Fdcos θ - 1
v
θ
A force F is at an angle θ above the horizontal and is used to pull a
heavy suitcase of weight mg a distance d along a level floor at constant
velocity. The coefficient of friction between the floor and the suitcase
is μ. The work done by the net force is:
5
A
Fdcos θ - μ mgd
B
0
C
-μ mgd
D
2Fdsin θ - μ mgd
E
Fdcos θ - 1
v
θ
6
A 4 kg ball is attached to a 1.5 m long string and whirled in a horizontal
circle at a constant speed of 5 m/s. How much work is done on the ball
during one period?
A
9J
B
4.5 J
C
Zero
D
2J
E
8J
Gravitational Potential Energy
A book of mass "m" is lifted vertically upwards a distance "h" by an outside
force. How much work does that outside force do on the book?
Fapp
mg
W = Fdparallel
Since a = 0, Fapp = mg
W = (mg) dparallel
Since F and d are in the same direction ...and dparallel = h
W = (mg) h
W = mgh
Gravitational Potential Energy
But we know that in general, Eo + W = Ef.
If our book had no energy to begin with, Eo = 0, then
W = Ef
But we just showed that we did W = mgh to lift the book... so
mgh = Ef
Or
Ef = mgh
The energy of a mass is increased by an amount mgh when it is raised by a
height "h".
Gravitational Potential Energy
The name for this form of energy is Gravitational Potential Energy (GPE).
GPE = mgh
One important thing to note is that changes in gravitational potential energy
are important, but their absolute value is not.
You can define any height to be the zero for height...and therefore zero for
GPE. But whichever height you choose to call zero, changes in heights will
result in changes of GPE.
7
A 2 kg block is held at the top of an incline plane.
gravitational potential energy of the block?
A
80 J
B
60 J
C
50 J
D
40 J
E
30 J
What is the
A 2 kg block released from rest from the top of an incline plane. There is
no friction between the block and the surface. How much work is done
by the gravitational force on the block?
8
A
80 J
B
60 J
C
50 J
D
40 J
E
30 J
Kinetic Energy
Imagine an object of mass "m" at rest at a height "h".
If dropped, how fast will it be traveling just before striking the ground?
Use your kinematics equations to get a formula for v2.
v2 = vo2 + 2ad
Since vo = 0, d = h, and a = g
v2 = 2gh
And we can solve this for "gh"
gh = v2 / 2
We're going to use this result later.
Kinetic Energy
In this example, we dropped an object. While it was falling, its energy was
constant...but changing forms.
It had only gravitational potential energy, GPE, at beginning, because it had
height but no velocity.
It had only kinetic energy, KE, just before striking the ground, as it had velocity
but no height.
In between, it had some of both.
Kinetic Energy
Now let's look at this from an energy perspective.
No external force acted on the system so its energy is constant.
the form of GPE, which is "mgh".
Its original energy was in
Eo + W = Ef
W = 0 and Eo = mgh
mgh = Ef
Solving for gh yields
gh = Ef/m
Now let's use our result from kinematics
gh = v2 / 2
v2 / 2 = Ef/m
Ef = 1/2 mv2 This is the energy an object has by
virtue of its motion: its kinetic energy
Kinetic Energy
The energy an object has by virtue of its motion is called its kinetic energy.
The symbol we will be using for kinetic energy is KE.
Like all forms of energy, it is measured in Joules (J).
The amount of KE an object has is given by:
KE = 1/2 mv2
9
A
D
A stone is dropped from the edge of a cliff. Which of the following
graphs best represents the stone's kinetic energy KE as a function
of time t?
C
B
E
10
A ball swings from point 1 to point 3. Assuming the ball is in SHM and
point 3 is 2 m above the lowest point 2. Answer the following
questions. What happens to the kinetic energy of the ball when it
moves from point 1 to point 2?
A
Increases
B
Decreases
C
Remains the same
D
Zero
E
More information is required
11
A ball swings from point 1 to point 3. Assuming the ball is in SHM and
point 3 is 2 m above the lowest point 2. Answer the following
questions. What is the velocity of the ball at the lowest point 2?
A
2.2 m/s
B
3.5 m/s
C
4.7 m/s
D
5.1 m/s
E
6.3 m/s
12
A 2 kg block released from rest from the top of an incline plane.
There is no friction between the block and the surface. What is the
speed of the block when it reaches the horizontal surface?
A
3.2 m/s
B
4.3 m/s
C
5.8 m/s
D
7.7 m/s
E
6.6 m/s
13
A satellite with a mass m revolves around Earth in a circular orbit
with a constant radius R. What is the kinetic energy of the satellite if
Earth’s mass is M?
A
½ mv
B
mgh
C
½GMm/R2
D
½ GMm/R
E
2Mm/R
14
An apple of mass m is thrown in horizontal from the edge of a cliff
with a height of H. What is the total mechanical energy of the apple
with respect to the ground when it is at the edge of the cliff?
A
1/2mv02
B
mgH
C
½ mv02- mgH
D
mgH - ½ mv02
E
mgH + ½ mv02
15
An apple of mass m is thrown in horizontal from the edge of a cliff
with a height of H. What is the kinetic energy of the apple just before
it hits the ground?
A
½ mv02 + mgH
B
½ mv02 - mgH
C
mgH
D
½ mv02
E
mgh - 1/2 mv02
16
A 500 kg roller coaster car starts from rest at point A and moves down
the curved track. Ignore any energy loss due to friction. Find the
speed of the car at the lowest point B.
A
10 m/s
B
20 m/s
C
30 m/s
D
40 m/s
E
50 m/s
17
A 500 kg roller coaster car starts from rest at point A and moves
down the curved track. Ignore any energy loss due to friction. Find
the speed of the car when it reaches point C.
A
10 m/s
B
20 m/s
C
30 m/s
D
40 m/s
E
50 m/s
18
Two projectiles A and B are launched from the ground with velocities of 50
m/s at 60 ̊ and 50 m/s at 30 ̊ with respect to the horizontal. Assuming
there is no air resistance involved, which projectile has greater kinetic
energy when it reaches the highest point?
A
Projectile A
B
Projectile B
C
They both have the same none zero kinetic energy
D
They both have zero kinetic energy at the highest point
E
More information is required
19
An object with a mass of 2 kg is initially at rest at a position x = 0.
A non-constant force F is applied to the object over 6 meters.
What is the total work done on the object?
A
200 J
B
150 J
C
170 J
D
190 J
E
180 J
20
An object with a mass of 2 kg is initially at rest at a position x=0. A
non-constant force F is applied to the object over 6 meters. What is
the velocity of the object at 6 meters?
A
150 m/s
B
25 m/s
C
300 m/s
D
12.25 m/s
E
Not enough information
21
A
D
A metal ball is held stationary at a height h0 above the floor and then
thrown upward. Assuming the collision with the floor is elastic, which
graph best shows the relationship between the total energy E of the
metal ball and its height h with respect to the floor?
B
E
C
22
A toy car travels with speed vo at point x. Point Y is a height H below
point x. Assuming there is no frictional losses and no work is done by a
motor, what is the speed at point Y?
A
(2gH+1/2vo2)1/2
B
vo-2gH
C
(2gH + vo2)1/2
D
2gH+(1/2vo2)1/2
E
vo+2gH
Elastic Potential Energy
Energy can be stored in a spring, this energy is called Elastic Potential Energy.
Hooke observed the relationship between the force necessary to compress a
spring and how much the spring was compressed.
Fspring = -kx
k represents the spring constant and is measured in N/m.
x represents how much the spring is compressed.
The - sign tells us that this is a restorative force.
Elastic Potential Energy
The work needed to compress a spring is equal to the area under its force vs.
distance curve.
W = 1/2 (base) (height)
W = 1/2 (x) (F)
W = 1/2 (x) (kx)
W = 1/2 kx2
Elastic Potential Energy
The energy imparted to the spring by this work must be stored in the Elastic
Potential Energy (EPE) of the spring:
EPE = 1/2 kx2
Like all forms of energy, it is measured in Joules (J).
A force of 20 N compresses a spring with spring constant 50 N/m. How
much energy is stored in the spring?
23
A
2J
B
5J
C
4J
D
6J
E
8J
24
A block with a mass of m slides at a constant velocity V0 on a
horizontal frictionless surface. The block collides with a spring and
comes to rest when the spring is compressed to the maximum
value. If the spring constant is K, what is the maximum compression
in the spring?
A
V0 (m/k)1/2
B
KmV0
C
V0K/m
D
m V0/K
E
V0 (K/m)1/2
F
(V0m/K)1/2
25
A block of mass m is placed on a frictionless inclined plane with an
incline angle θ. The block is just in contact with a free end of an
unstretched spring with a spring constant k. If the block is released
from rest, what is the maximum compression in the spring?
A
kmgsinθ
B
kmgcosθ
C
(2mgsinθ)/k
D
(mg)/k
E
kmg
Recalling conservation of energy, we can now solve more complicated problems if
energy is conserved.
A roller coaster is at the top of a track that is 80 m high. How fast will it be going at the
bottom of the hill?
Eo + W = Ef
W=0
Eo = Ef
E0 = GPE, Ef = KE
GPE = KE
Substitute GPE and KE equations
mgh = 1/2 m v2
Solving for v yields
v2 = 2gh
Substitute the values and solve
v2 = 2 (9.8) 80
This method works with many energy
v =39.6 m/s problems
26
A student uses a spring (with a spring constant of 180 N/m) to
launch a marble vertically into the air. The mass of the marble is
0.004 kg and the spring is compressed 0.03 m. How high will the
marble go?
27
A student uses a spring gun (with a spring constant of 120 N/m) to
launch a marble vertically into the air. The mass of the marble is
0.002 kg and the spring is compressed 0.04 m.
How fast will it be going when it leaves the gun?
28
A student uses the lab apparatus shown below. A 5 kg block
compresses a spring 6 cm. The spring constant is 300 N/m. What
will the block's velocity be when released?
29
A mass is attached to an ideal spring on a smooth horizontal
surface. It is displaced an amount Δxo and released. Which of
the following is true?
I. The KE is largest when the mass passes through the
equilibrium point.
II. The PE is largest when the mass has a displacement of
±Δxo.
III. The PE and KE will never be equal
A
I only
B
II only
C
III only
D
I and II only
E
I and III only
GPE and Escape Velocity
The expression GPE = mgh only works near the surface of a planet. As an
object goes to a great height a more general expression is needed based
on Universal Gravitational Force: FGravity = GMm/r2 .
Deriving this accurately requires Calculus, but you can remember it by
thinking of it this way
W = FΔxcosθ
FGravity = GMm/r2; cosθ = -1; d = r
W = -(GMm/r2)r
UG = -GMm/r
GPE and Escape Velocity
To escape a planet, an object needs to have sufficient kinetic energy to
overcome its negative gravitaional potential energy.
Eo + W = Ef
UG + KE + W = 0
-GMm/r + 1/2 mv2 + 0 = 0
1/2 mv2 = GMm/r
v2 = 2GM/r
v = (2GM/r)1/2
Note that escape velocity is
independent of the mass
of the escaping object. It
only depends on the mass
and radius of the object
being escaped from.
30
A rocket of mass 5,000 kg will have _______ escape velocity as a
10,000 rocket?
A
larger
B
smaller
C
equal
31
Four objects are thrown with identical speeds in different directions from the
top of a building. Which will be moving fastest when it strikes the ground?
A
B
C
D
h
E
All will be the same
32
Four objects are thrown with identical speeds in different directions from the
top of a building. Which will strike the ground soonest?
A
B
C
D
h
E
All will be the same
33
Four objects are thrown with identical speeds in different directions from the
top of a building. Which will go the highest?
A
B
C
D
h
E
All will be the same
34
Four objects are thrown with identical speeds in different directions from the
top of a building. Which will land farthest from the base of the building?
A
B
C
D
h
E
All will be the same
35
Four objects are thrown with identical speeds in different directions from the
top of a building. Which will have the have the greatest horizontal component
of velocity at its maximum height?
A
B
C
D
h
E
All will be the same
36
Four objects are thrown with identical speeds in different directions from the
top of a building. Which will have the have the greatest kinetic energy at its
maximum height?
A
B
C
h
D
All will be the same
37
A rocket is launched from the surface of a planet with mass M and
radius R. What is the minimum velocity the rocket must be given to
completely escape from the planet's gravitational field?
A
2GM/16R2
B
(2GM/R)1/2
C
(GM/R)1/2
D
2GM/R
Power
It is often important to know not only if there is enough energy available to perform a
task but also how much time will be required.
Power is defined as the rate that work is done:
P=W/t
Since work is measured in Joules (J) and time is measured in seconds (s) the unit of
power is Joules per second (J/s).
However, in honor of James Watt, who made critical contributions in developing
efficient steam engines, the unit of power is also know as a Watt (W).
Power
P=W/t
Since W = Fd parallel
P = (Fd parallel) / t
Regrouping this becomes
P = F(d parallel / t)
Since v = d/t
P = Fv parallel
So power can be defined as the product of the force applied and the velocity of the
object parallel to that force.
Power
A third useful expression for power can be derived from our original statement
of the conservation of energy principle.
P=W/t
Since W = Ef - E0
P = (Ef - E0) / t
So the power absorbed by a system can be thought of as the rate at which the
energy in the system is changing.
38
A driver in a 2000 kg Porsche wishes to pass a slow moving school
bus on a 4 lane road. What is the average power in watts required to
accelerate the sports car from 30 m/s to 60 m/s in 9 seconds?
A
1,800 W
B
5,000 W
C
10,000 W
D
100,000 W
E
300,000 W
39
A force F is applied in horizontal to a 10 kg block. The block moves at a
constant speed 2 m/s across a horizontal surface. The coefficient of
kinetic friction between the block and the surface is 0.5. The work
done by the force F in 1.5 minutes is:
A
9000 J
B
5000 J
C
3000 J
D
2000 J
E
1000 J
40
A crane lifts a 300 kg load at a constant speed to the top of a
building 60 m high in 15 s. The average power expended by the
crane to overcome gravity is:
A
10,000 W
B
12,000 W
C
15,000 W
D
30,000 W
E
60,000 W
41
In a physics lab a student uses three light frictionless PASCO lab carts.
Each cart is loaded with some blocks, each having the same mass. The
same force F is applied to each cart and they move equal distances d.
In which one of these three cases is more work done by force F?
A
cart I
B
cart II
C
cart III
D
the same work is done on each
E
more information is required
42
In a physics lab a student uses three light frictionless PASCO lab carts.
Each cart is loaded with some blocks, each having the same mass. The
same force F is applied to each cart and they move equal distances d.
Which cart will have more kinetic energy at the end of distance d?
A
cart I
B
cart II
C
cart III
D
all three will have the same kinetic
energy
E
more information is required
In a physics lab a student uses three light frictionless PASCO lab carts.
Each cart is loaded with some blocks, each having the same mass. The
same force F is applied to each cart and they move equal distances d.
Which cart will move faster at the end of distance d?
43
A
cart I
B
cart II
C
cart III
D
all three will move with the same
speed
E
more information is required
44
A box of mass M begins at rest with point 1 at a height of 6R, where
2R is the radius of the circular part of the track. The box slides down
the frictionless track and around the loop. What is the ratio between
the normal force on the box at point 2 to the box's weight?
A
1
B
2
C
3
D
4
E
5
45
A ball of mass m is fastened to a string. The ball swings in a vertical
circle of radius r with the other end of the string held fixed. Neglecting air
resistance, the difference between the string's tension at the bottom of
the circle and at the top of the circle is:
A
mg
B
2mg
C
3mg
D
6mg
E
9mg
Work and Energy
Calculus Based
Work
We learned before that work can be defined as:
This equation comes from the scalar product of the force vector and
displacement vector.
If the force however is changing the previous equation for work does not
apply. To account for the varying force like for a varying velocity when you
are trying to find displacement requires the use of Calculus.
Work
F
x
By breaking the varying force into small segments we can precisely calculate
the work done by a varying force.
Gravitational Potential Energy
Work is defined as the change in energy, so to find the potential energy due to gravity
we integrate the gravitational force.
Power
For a constant force the power equation can be represented in several ways
The average power can be represented as:
To find the instantaneous power we use a limit expression in which the
interval is brought to essentially zero
Work along a curved path
As you move from point 1 to point 2 along a curved path, the segments of
the curve can be divided into an infinite amount of vector displacements
denoted by dl.
Φ
Force and Potential Energy
We already have learned that work is the change in energy.
The force is dependent on the components of the potential so the force can be
described as the negative gradient of the potential.
Energy Diagrams
An energy diagram shows the relationship between the objects position and the
amount of potential energy.
KE
U
-A
0
A
-A
0
A
This energy diagram shows the change of the potential for a spring. At a
maximum displacement (A or -A) the object on the spring has maximum potential
and at zero the object has maximum kinetic energy.
Energy Diagrams
Some energy diagrams are far more complex than that of a spring. To find the
magnitude of the force along the curve we have to take the derivative because
U
unstable equilibrium
stable
equilibrium
x
Energy Diagrams
In energy diagrams, stable equilibrium is anywhere the curve has a minimum and unstable
equilibrium occurs when the curve has a maximum. To find the critical points take the
derivative of the equation and set it equal to zero. The minimum is where the slope
changes from negative to positive, and the maximum is where the slope changes from
positive to negative.
Energy Diagrams to Force Diagrams
U
U
x
x
F
F
x
x
46
Given the equation U(x) = x5 - 3x2 + 8. Find F(x).
A
5x4 + 6x
B
-5x4 + 6x - 8
C
-5x4 + 6x
D
-5x4 - 5x - 8
E
5x4 - 5x
47
Given F(x) = -24x3 + 3x2 + 8x - 7. Find U(x) from x = 0 to x = 1.
A
0J
B
2J
C
4J
D
6J
E
8J
A particle moves along the x-axis while acted on by a force parallel to
the x-axis. The force corresponds to the potential energy function
graphed.
a) What is the direction of the force on the particle
when it is at point A?
U(J)
Negative
A
b) What is the direction of the force on the
particle when it is at point B?
B
C
0
Positive
(m)
c) What is the force on the particle when it is at
point C?
Zero