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AP-C Physics Unit 6
Work, Energy, & Power
Date
In Class
Reading
11/29/16
Work
11/30/16
Integrals
Reading 1:
Integrals
12/1/16
Work and Energy
7.1 – 7.5
12/2/16
Potential Energy,
Conservation of Energy
7.6, 8.2
12/5/16
Non-Conservative Forces
7.7, 8.3 – 8.4
12/6/16
Quiz 6.1
Assignment
U6 WS 1: The Dot Product
Reading 2:
Potential
Energy Curves
U6 WS 2: Integrals
PS 6-1
Ch. 7: 5, 15, 17, 29, 34, 38
PS 6-2
Ch. 8: 3, 5, 6, 9, 59, 64
PS 6-3
Ch. 8: 15, 22, 23, 63, 65
12/7/16
Potential Energy Curves
U6 WS 3: Potential Energy Curves
12/8/16
Problem Solving, Graphs
12/9/16
Power
12/12/16
Review
U6 Review WS
12/13/16
Unit 5 Test
Catch your breath
12/14/16
Grade Test and Discuss
12/15/16
Semester 1 Final Review
U6 WS 4
6.10
PS 6-4
Ch. 8: 29, 30, 31, 32, 39
4th Period – 12/19 2:00PM
5th Period – 12/20 2:00PM
AP-C Physics Unit 6
Work, Energy, & Power
Unit 6 Objectives: Upon completion of Unit 6, you should be able to:
1. Define the concept of work and state in terms of work done whether energy is being
transferred into or out of a system.
2. Relate the work done by a force to the area under a force vs. position graph, and use the
graph, or the integral of the function, to calculate the work done if initial and final positions are
known.
3. Define kinetic energy of an object and relate it to the speed and mass of the object.
4. State the work-energy theorem, and use it to determine the change in kinetic energy of an
object resulting from the net work done. Use it to determine the work done by the net force, or
by each of the forces comprising the net force, on an object given the change in speed or kinetic
energy of the object.
5. Calculate potential energy relative to a specific point of an object near the earth=s surface
(where gravitational force is fairly constant) and on a spring (Hooke’s Law).
6. Define and identify conservative and non-conservative forces.
7. Define total mechanical energy, and identify conditions where total mechanical energy is
conserved. Apply the relation between the work done by non-conservative forces and the
change in the total mechanical energy of a system.
8. Given a graph of the potential energy of an object as a function of position, determine the
magnitude and direction of the force on the object at a specified position. Determine the kinetic
energy of the object at a specified position, given the total mechanical energy.
9. Calculate the potential energy function U, given a conservative force, F which varies in one
dimension. For example, given F(x), calculate U(x). [And vice versa, given U(x), find F(x)]
10. Define power, and apply the appropriate relationships between power, work, time, force,
velocity, and kinetic energy to solve problems involving the motion of an object.
Given on the AP Test
Not Given on the AP Test
Fs = -kx
W = Fx (W = Fx cos θ) for constant force
W = ò F ·dr
Wf = µFn • x
K = ½ mv2
Ug = mgh
Us = ½ kx2
P=
dW
= F ·v
dt
F(r) = Pave =
W
t
dU(r)
dr
AP Physics Unit 6: Work, Energy & Power
Worksheet 1: The Dot Product
r
r
r
1. A floating ice block is pushed through a displacement of r = (15m)i - (12m) j along a straight
ur
r
r
embankment by rushing water, which exerts a force F = (210N)i - (150N) j on the block. How
much work does the force do on the block during the displacement?
2. Vector a has a magnitude of 2.83 N and vector b has a magnitude of 3.16
m, and vector a acts at an angle of 63.43 degrees with respect to vector b as
shown in the diagram. Find the dot product of the two vectors, a  b.
3. A force of F = (6N)i - (2N) j acts on a particle that undergoes a displacement of r = (3m)i - (2m) j
A) Find the work done by the force on the particle.
B) Find the angle between the force and the displacement.
4. A force of 100 N exerted at an angle of 30 degrees pulls a block of mass 5 kg a distance of 5
meters across the floor at a constant speed of 2 m/s.
A) How much work is done by the applied force?
B) How much work is done by gravity?
C) How much work is done by the normal force?
D) How much work is done by friction?
AP Physics Unit 6 Reading 1
Calculus in AP Physics: The Definite Integral
Definite Integrals
Before we used calculus to help us find the instantaneous slope of a curve and any point on the
curve in order to determine velocity from a position-time graph, or acceleration from a velocitytime graph. However, we also learned that the area under a velocity vs. time graph represents
displacement, and the area under an acceleration vs. time graph equals change in velocity. Now
that we are studying energy, we have also learned that the area under a force vs. displacement
graph equals work done by the force. Last year, we only used simple functions like a constant
velocity or force, or a linear relationship for velocity vs. time or for spring force vs. displacement.
For a constant force, the area is a simple rectangle A = bh, or
W = F Δx.
For a force that varies in a linear fashion, the area is a simple triangle, and can
be found by using W = ½ bh, which would be
W= ½ (Δx)(Fsp),
or replacing Fsp = k Δx, then
W = ½ (Δx)(k Δx) = ½ k (Δx)2.
Now that we are learning calculus, we will find that when the force varies as any
other function of x, there are ways to approximate the area under the curve, by
adding up small rectangles or trapezoids. As Δx approaches zero, we start
adding up small “differentials” of area. The process of adding up an infinite amount of
differentials (an infinite sum) between two limits is called integration.
A definite integral represents an area within two limits, and evaluating a definite integral
("integrating" in calculus language) is the inverse of finding a derivative - like subtraction is the
inverse of addition. In physics, the area under a velocity vs. time graph represents
displacement, so the definite integral of velocity gives displacement. The area under an
acceleration vs. time graph equals change in velocity, so the definite integral of acceleration
tells you the change in velocity. The area under a Force vs. displacement graph represents the
work done by that force.
The diagram at right shows the relationship between the definite integral
notation and the area it represents. The numbers 0 and xm ("a" and "b" on the
diagram above) - called "limits of integration" - go at the bottom and top of the
big "S" which represents the “sum” or integral. "F(x)" is the function being
integrated (the "integrand"), and "dx" says that "x" is the variable being used.
The notation is read "the definite integral from 0 to xm of f of x, dx".
ò F ·dx for an object moving in the x-direction. (Don’t forget
that the Dot product indicates that Work is a scalar product, or W = ò F cosq dx ).
Therefore, for variable forces, W =
Rules for Definite Integrals:
Just like you can subtract 5 from 12 by thinking, "What do I have to add to 5 to get 12?", you
can evaluate definite integrals by thinking "What function would I have to differentiate to get the
function in this integral?" That function is called the "integral" or "antiderivative." The
mathematical symbol for the antiderivative looks like the definite integral without the limits of
integration.
Antiderivative Rules for Particular Functions: If you are a calculus student, you will notice that
we are ignoring an important mathematical point in the following rules.
Rule in English
Rule in Math. Notation
Example
The antiderivative of a constant is the
constant times t.
The antiderivative of t to a power is t to
the power-plus-one, divided by the
power plus one.
(This is the inverse of the
derivative formula
)
The antiderivative of the sine of t is the
(This is the inverse of the derivative
negative of cosine t.
formula)
The antiderivative of the cosine of t is
the sine of t.
(This is the inverse of the derivative
formula)
Antiderivative Rules for Combinations of Functions: In the rules below, u and w represent
functions of time, t.
Rule in English
The antiderivative of a
constant times a function
equals the constant times the
antiderivative of the function.
The antiderivative of the sum
(or difference) of two
functions equals the sum (or
difference) of their
antiderivatives.
Rule in Math. Notation
Example
The Fundamental Theorem of Calculus - tells you how to evaluate definite integrals based on
what you know about antiderivatives.
Rule In English Rule in Math. Notation
Example
If F(t) is an
antiderivative of
f(t), then the
definite integral
from a to b of f
equals the
function F
evaluated when t
equals b minus F
evaluated when t
equals a.
When we know the limits of the integration, we can evaluate the integral at each limit and find
the difference
b
 f ( x)dx  F (b)  F (a)
a
However, when the limits of integration are not known, we must account for them anyway.
Remember that when we took derivatives of a constant (like v0), that value would disappear
(d/dx of a constant = 0). So when we are working backwards and taking an antiderivative (or
integral), we must include a constant of integration, C, to account for possible initial conditions
that may or may not be known.
For example, the area under a velocity vs. time graphs tells us the displacement of an object,
but it does not tell us the exact position. In order to know what an object’s position is at a given
time, we must know what its initial position was. Therefore, the constant of integration would
represent the object’s initial position. The constant of integration from an acceleration vs. time
graph would represent the initial velocity.
Now that we know how to integrate, we can go back to our kinematic equations to find another
way to derive them from the graphs. If we know an object undergoes constant acceleration, then
we can use an integral to find the change in velocity from the graph.
v = ò a dt =at + v0
and
x=
ò v dt = ò ( v
0
1
+ at )dt = v0t + at 2 + x0
2
AP Physics Unit 6: Work, Energy & Power
Worksheet 2: Integrals
1.
ò (x
2.
ò (5m
3.
  x
3
+ 2x 2 + 4) dx
4
2
+ 9m3 + 6)dm

3
3

 5dx
2
x

8
4.
r
3
dr
1
4
5.
ò (x
3
- 6x 2 + 9x +1)dx
1
6.
ò
2
0
(Ax 2 + 5x)dx = 34 ; Solve for ‘A’
7. A 100-g bullet is fired from a rifle having a barrel of 0.6 m long. The force exerted by the
expanding gas on the bullet is F = 15,000 + 10,000x – 25,000x2, where x is in meters and force
is in Newtons.
A) Determine the work done on the bullet as it travels the length of the barrel.
B) If the barrel were twice as long (1.2 m), how much work is done and how does this compare
to the work calculated in part A?
8. A dung beetle pushes his ball of dung with a variable force, represented by F(x) = 2x 2 - 3x +1
(with x in meters), moving the ball from x = 0 m to x = 1.5 m. Find the work done by the dung
beetle.
9. A force of F(x) = 3x – 2 in Newtons does 16 Joules of work moving a piñata in a straight line.
Find the displacement of the piñata in meters.
r
r
10. A variable force, represented by F(x) = (3x 2 N)i + (4N) j (with x in meters) acts on a
salamander, changing only the salamander’s kinetic energy. How much work is done on the
salamander as it moves from coordinates (2 m, 3 m) to (3 m, 0 m)? Does the salamander’s
speed increase, decrease, or remain the same?
Table of Integrals
1.
 kdx  kx  C
2.
x
3.
4.
5.
6.
n
dx 
1 n 1
x C
n 1
dx
 x  ln x  C
 sin xdx   cos x  C
 cos xdx  sin x  C
 e dx  e  C
x
x
AP Physics Unit 6: Reading 2
Potential Energy Curves
Change in Energy from Forces
Now that we know about derivatives and integrals (or antiderivatives), we can take our
understanding of work and energy one step further. In our previous reading, we learned that
work is the area under a Force vs. Displacement graph, and can therefore be found by taking an
integral:
W=
ò
xf
xi
F(x)dx
If a particle-like object is being acted upon by a conservative force, than the work done on that
object changes its potential energy. The change in potential energy of the particle is equal to the
negative of the work done (ΔU = -W). Therefore, substituting in the change in potential energy,
DU = - ò F(x)dx
xf
xi
Examples:
Gravitational Potential Energy: First consider an object moving vertically on the y-axis, being
acted upon by the force of gravity. To find the change in potential energy, we integrate along the
y-axis and substitute –mg for the force, being that it acts in the downward direction. We then
have
yf
yf
yf
yi
yi
yi
DUg = - ò (-mg)dy = mg ò dy = mg [ y] = mg(y f - yi ) = mgDy
And of course, if yi = 0 m (ground level), then ΔUg = mgy
Elastic Potential Energy: Now consider a block on a horizontal, frictionless surface, attached to
a spring with a spring constant k. If the block is pulled (and the spring is stretched) then
released, the spring exerts a conservative force of –kx on the block. The change in potential
energy then is
DUel = - ò (-kx)dx = k ò
xf
xf
xi
xi
x
é x2 ù f 1
1
x dx = k ê ú = k(x 2f - x i2) = kDx 2
2
ë 2 ûxi 2
Potential Energy Curves
We learned that when dealing with kinematics graphs, one can find the change in velocity from
a position vs. time graph by taking a derivative and the acceleration is the derivative of the
velocity-time graph. Once we learned integrals, we found that it is possible to go the other way;
you can find the change in velocity from an acceleration-time graph by taking the integral and
you can find the change in position from the integral of the velocity-time graph. The same basic
procedure can be used with work (change in energy) and forces.
If Work (or change in potential energy) is the integral of a force vs. displacement graph, then the
derivative of a potential energy vs. displacement graph can be used to find the force. If we take
the derivative of both sides of the equation:
DU = - ò F(x)dx
xf
xi
-
dU
= F(x)
dx
This shows that the derivative of a potential energy vs. displacement graph (or the slope)
represents the negative of the force acting on an object at that instant.
Potential energy curves can also be used to understand a lot about a particle’s motion. Consider
the graph below. If the object starts at position x3, then it begins with a total energy of E2. As the
object moves from x3 to xA, it loses potential energy. Because it is being acted upon by a
conservative force (no friction for example), than all that “lost” potential energy has been converted
into kinetic energy. (Notice that if you add up the potential energy at this point with the amount of
kinetic energy, it equals the total energy
of the system.) Once it reaches position
x4, it has regained its initial potential
energy (and therefore kinetic energy is
zero). That means the object cannot go
any higher. So what happens? If left to
its own, it will then “fall” back down
toward x3 again and oscillate back and
forth. If the object had instead started at
position x1, then it would have a total
energy of E1 and would therefore have
more kinetic energy upon reaching xA
and would thus be moving faster at this
point than it had before.
Using the graph you can also understand the force acting on the object. Remember that the
negative of the slope of the graph (or derivative) represents the force. At points x1 through xa,
the slope is negative. This means that the force is positive: or acting in the same direction of the
displacement. At points x4 and x2, the slope is positive (and less steep). This means that a
smaller force acts in the opposite direction as the displacement (which is why is slows down).
At any point, then the sum of the Kinetic and Potential energy equals the Total Energy of the
system, U(x) + K(x) = Emech. Therefore, to find the Kinetic energy at any point, K(x) = Emech –
U(x). Notice how this shows that Kinetic energy can never be negative (because v2 is always
positive).
Though this may all seem confusing at first, it can be very easy to understand and remember if
you visualize the curve as an actual valley. At the top (at points x1 and x2), the object feels a
force (like gravity) pulling it down into the valley. That means it will speed up on the way “down”.
At the lowest point on the graph, it will be moving the fastest before it slows down on its way
back up the other side of the valley.
Turning Points and Equilibrium Points
There are a few points of interest to note on potential energy curves. The first are called turning
points. A turning point is where the potential energy of the object equals the total amount of
energy. If all the energy is potential, that means that kinetic energy is zero and the object is
stopped. This is where it will turn around and start going in the opposite direction.
Equilibrium points are spots on the graph where the derivative is equal to zero, or the curve is
flat (also called an inflection point in calculus). At an equilibrium point then, the force acting on
the particle is zero momentarily. There are three kinds of equilibrium points:
Neutral equilibrium: The sum of the forces = 0. (ex. a marble on a table)
Unstable equilibrium: If a slight force in either direction will set it accelerating. (ex. a
marble on a bowling ball)
Stable equilibrium: (ex. a marble in a bowl)
Consider the graphs below. Graph (a) represents a potential energy curve in which the particle
starts with 5 J of potential energy. Graph (b) shows a plot of the force acting on the particle,
derived from the potential energy plot by taking its slope at various points. Graph (c) shows a
plot of three different possible values of Emechanical.
On graph (a): Notice there is only one turning point shown. Where are the equilibrium points?
Neutral equilibrium point:
__________
Unstable equilibrium point:
__________
Stable equilibrium point:
__________
On graph (c), if the object started with 1 J of potential energy, how would it behave?
And if it started with 3 J of potential energy?
AP Physics Unit 6: Work and Energy
Worksheet 3: Potential Energy Curves
1. A conservative force acts on a 2-kg squid that moves along the x-axis. The potential energy
curve associated with this force is shown below. When the squid is at 2 m, its velocity is 2 m/s.
A) What is the total mechanical energy of the
squid?
B) How much Kinetic energy does it have at
x = 3 m?
C) What is the squid’s velocity at x = 4 m?
D) Where are the turning points?
E) Label the equilibrium points on the graph as neutral, unstable, or stable.
2. The figure below shows a plot of potential energy U versus position x of a 0.240 kg particle that can
travel only along an x axis under the influence of a conservative force. The graph has these values:
UA = 9J
UC = 20J
UD = 24J.
The particle is released at the point where U
forms a “potential hill” of “height” UB = 12 J,
with kinetic energy 5.50 J.
What is the speed of the particle at
A) x = 3.5 m and
B) x = 6.5 m?
What is the position of the turning point on
C) the right side
and D) the left side?
E) What is the force acting on the particle at x = 2 m?
3. The graph below gives the potential energy function of a particle.
A) Rank regions AB, BC,CD, and DE
according to the magnitude of the
force on the particle, greatest first (use
the notation > or =,for example
BC=CD>AB>DE).
B) What value must the mechanical
energy Emec of the particle not exceed
if the particle is to be trapped in the
potential well at the left?
C) What value must the mechanical
energy Emec of the particle not exceed if the particle is to be trapped in the potential well at the
right?
D) What value must the mechanical energy Emec of the particle not exceed if the particle is to be
able to move between the two potential wells but not to the right of point H?
E) For the situation of D), in which of the regions BC, DE, and FG will the particle have the
greatest kinetic energy?
F) For the situation of D), in which of the regions will the particle have the least speed?
4. The potential energy of a diatomic molecule (like H2 or O2) is given by
U(r) =
A B
r12 r 6
where r is the separation of the two atoms of the molecule and A and B are constants. The
potential energy is associated with the force that binds the two atoms together.
Find the equilibrium separation (distance between the atoms at which the force on each atom is
zero).
AP Physics Unit 6: Work, Energy, & Power
Worksheet 4
1. A bullet is fired straight up with an initial velocity of 100 m/s.
A) How high does it go? (5E2)
B) What is the maximum height it can reach if it is fired from the same gun at an angle of 45°
with the horizontal rather than straight up? (2.5E2)
2. An ideal spring with a spring constant of 200 N/m is mounted at the base
of a frictionless plane inclined at an angle of 30° as shown in the diagram. A
20-kg box, released from rest, slides 6 meters down the plane before
coming in contact with the spring.
A) Determine the speed of the box just before it hits the spring. (7.8E0)
B) Determine the distance the box has compressed the spring when it comes to rest. (3E0)
3. The spring-loaded marble shooter shown is sitting on a table 1.0 meter
above the floor. The device contains an ideal spring with a constant of 100
N/m and shoots a 0.05 kg marble horizontally.
A) What is the launch speed of the marble if it lands in a cup on the floor
2.0 meters form the launch point? (4.5E0)
B) If the launching device is frictionless, what was the initial compression of the spring if the
marble lands in the cup? (1E-1)
C) Suppose the device is not frictionless and a constant friction force of 1.0 N acts on the
marble as it slides within the barrel of the shooter. In this situation, if the spring is compressed
0.1 m and fired, where must the cup be placed? (2E0)
4. The graph shows the potential energy of
a 2-kg object plotted as a function of its
position x.
A) What is the magnitude and direction of
the conservative force acting on the object
at
x = 4 m, x = 6 m, and x = 8 m?
B) In which region(s) is the force on the object in the direction of its velocity? …And opposite to
the direction of its velocity?
C) In which regions is the object in equilibrium?
D) What is the acceleration of the object at x = 2 m, x = 3.5 m, and x = 7.5 m?
E) In which region(s) is the work being done on the object positive? …negative? …zero?
Suppose the velocity of the object is 4.5 m/s when the object is at x = 6 m.
F) What is the total mechanical energy of the system?
G) What is the object’s kinetic energy at x = 4 m and x = 8 m?
H) What is the object’s speed at x = 4 m and x = 8 m?
I) What are the maximum and minimum values of x to which the object can move?
5. Graph A below represents the potential energy U(x) of an object plotted as a function of its
position x.
A) Identify all positions where the object is in equilibrium. State whether these are positions of
stable or unstable equilibrium.
Suppose the object has a total mechanical energy of 4 J as shown by the dashed line on the
graph.
B) Determine the kinetic energy of the object at x = 2 m and x = 4 m.
C) Can the object reach the position x = 0.5 m? Explain.
D) Can the object reach the position x = 5 m? Explain.
E) On the graph B above, carefully plot the graph of the force F(x) acting on the object as a
function of its position x, for 0 < x < 7 meters.
6. A weak-minded airhead that should never be allowed to handle a gun fires a 10-gram bullet
into a tree to test out his new rifle. The bullet enters the tree horizontally at a speed of 500 m/s,
travels across its 25 cm diameter, and emerges with a speed of 200 m/s.
A) Find the change in Kinetic energy of the bullet. (-1E3)
B) Find the average force exerted on the tree by the bullet. (4E3)