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PHYSICS SEMESTER ONE
UNIT 5
UNIT 5: WORK AND ENERGY
Relationship to Labs: This unit is supported by Lab 4: Conservation of Energy (RWSL)
ENERGY
Energy is the ability to do work. The energy in a moving hammer can drive a nail into a piece of wood
(create a bruise on your thumb). The potential energy of water is used to drive a water mill which grinds
grain into flour. The energy stored in a spring is used to accelerate a plunger dart when released. The
energy from fusing hydrogen nuclei in the sun heats the surrounding hydrogen and helium, which emits
light, which heats up the atmosphere to create wind, which drives a windmill that creates electricity that
powers my toaster, which creates toast and heats my kitchen.
Energy is a very important concept in physics. It helps us understand how fuel in a car makes the car move
forward. It helps us understand why a hot object will cool when it is connected to a cool object. It helps us
explain why a cyclist can coast from rest at the top of one hill, accelerate to a high speed in the valley
between hills, and slow in coasting up the next hill. It is also a very useful tool in problem solving that can
reduce a complex force problem to a simple comparison of the initial and final energies.
Using energy to solve problems usually requires the calculation of the initial and final energies. The most
important step in solving a problem using the energy is to take care to note the system that is being
considered. We also need to consider if energy is being added to (removed from) the system from outside,
energy is being transferred from one form to another within the system, or some combination of both.
WORK
In mechanical systems, energy added to a system using a force is known as work, W. The work is a product
of the force on an object and the displacement of the object while the force is applied. If the force and
displacement are in the same direction, the work done on an object by a constant force F is given by
W  Fd
where F is the magnitude of the force, and d is the magnitude of the displacement. If the force is not in the
same direction as the displacement, the work done is
W  Fd cos
where θ is the angle between the force and displacement vectors.
Work is a scalar.
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PHYSICS SEMESTER ONE
UNIT 5
2
SI units: joule (J) = newton·metre = kg·m /s
2
Quick Example 1
Bob drags a rock horizontally along the ground using a
rope around the rock. If Bob drags the rock 25 m
horizontally with a by pulling with a force of 450 N 33°
above the horizon, calculate the work done by Bob.
Define terms, state equation, substitute and evaluate.
The rock is dragged horizontally so the displacement is horizontal. The force is at an angle of 33° from
horizontal so   33 . The other variables are
F  450 N,
d  25 m,
W ?
W  Fd cos 
  450 N  25 m  cos 33
 9440 J
Bob did 9400 J of work in moving the rock.
Problem 1
A popular magazine slides across the desk, slowing due to friction. The friction provides a force of 0.25 N
over the 1.2 m where the magazine slows down.
What is the friction force angle relative to displacement?
Calculate the work done by the friction force on the magazine.
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PHYSICS SEMESTER ONE
UNIT 5
The Scalar Product of Two Vectors
We can derive the formula for the work using the scalar product of the force and displacement vectors. The
scalar or “dot” product of vectors is a way of multiplying vectors to obtain a scalar.
The dot product of vectors A and B is defined as
A  B  AB cos 
where θ is the angle between the vectors.
The dot product is commutative in that
B  A  BA cos 
 AB
The operations of multiplication and addition are also commutative.
47  7 4
47 74
The dot product obeys the distributive law of multiplication.


A B  C  AB  AC
This is similar to
3  4  5  3  4  3  5
If we write the vectors in terms of their components
A  Ax ˆi  Ay ˆj  Az kˆ
and
B  Bx ˆi  By ˆj  Bz kˆ ,
the dot product takes on the form
A  B  Ax Bx  Ay By  Az Bz
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PHYSICS SEMESTER ONE
UNIT 5
The easiest way to see this property is to imagine two vectors
A  Ax ˆi  Ay ˆj and
B  Bx ˆi  0ˆj ,
A
B
Bx = B
Ay = Asinθ
θ
Ax = Acosθ
A  B  Ax Bx  Ay By
  A cos   B    A sin   0 
 AB cos 
The scalar product can be used to define the magnitude of a vector
A  A  A  Ax2  Ay2
for a 2D vector
Using the dot product, the mechanical work becomes
W  Fd
Problem 2
Dot Product
a)
F  2.5 N ˆi  3.2 N ˆj, d  1.2 m ˆi  6.0 m ˆj,
b)
Find the angles (θF and θd) and magnitudes (F and d) for the vectors in part a), and show
Find W  F  d.
F  d  Fd cos  . Note that the angle between the vectors is the difference between the angles
   F  d or   d   F .
c)
Why doesn’t it matter whether we use    F  d or   d   F ?
d)
Assuming that the motion under the net force Fnet  F is in a straight line, use the magnitudes
of F and d, with vi = 11 m/s and m = 1.0 kg, to find vf.
W=0
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PHYSICS SEMESTER ONE
UNIT 5
Mechanical work formulas
W  Fd
when vectors are in same direction and magnitudes are known
W  Fd cos
when vector magnitudes and angle between vectors are known
W  F d
when know vectors in component form
 Fx d x  Fy d y
Quick Problem
Bob pushes on a car from the side with a force F  150 N ˆi while the car moves a displacement
d  1.5 m ˆj . How much work did Bob do? Terms are defined in question so state equation and run, or
simply state the answer. The answer is at the end of the line containing the link to the solution for problem
2.
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PHYSICS SEMESTER ONE
UNIT 5
Work Done by a Varying Force
Please note, it is recognized that you won’t see integration until the end of the semester if you are taking
Calculus. Integration is a very useful and often easily implemented tool, but don’t worry if this seems rather
confusing. You won’t get any integration problems on exams, quizzes or labs. Just keep this in mind as a
practical example when you cover integration.
Let’s consider the work done by a varying force on a object that displaces in the x-direction. In this case
W  Fx d x
 Fx x
If we have a force that varies with position between xi and xf, we can break the work into a set of works for
each subsection section of displacement Δx between xi and xf.
Each of these “little works” is the area of a rectangle between the force
and the x-axis on a graph of force versus position.
Fx
The total work done between xi and xf is the sum of all of the “little
works”
xf
W   Fx x
xi
This becomes the integral between xi and xf in the limit that the
displacements for the little works go to zero.
xf
 Fx x  x
x  0
W  lim
xf
i
xi
xi
xf
Δx
Fx dx
We can expand this to all directions
W 
xf
xi
Fx dx  
yf
yi
Fy dy  
zf
zi
Fz dz
If there is more than one force on the object we can replace the force in the integral with the net force to
find the net work
W 
xf
xi
Fnet , x dx
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PHYSICS SEMESTER ONE
UNIT 5
Example
A 2000 kg spaceship flies to an altitude of 100 km. Ignoring the changes in mass as the fuel is consumed, the
gravitational force between the spaceship and Earth is
Fy 

GM E m
y2
8  1011 N  km 2
y2
where y is the altitude above the centre of the earth in km. Calculate the work done in raising the spaceship
from sea level, yi = 6380 km, to an altitude 100 km, yf = 6480 km.
W 

yf
yi
Fy dy
6480 8  1011
6380
N  km 2
y2
8  1011 N  km 2

y
dy
6480
6380
 8  1011   8  1011  
  
   
  N  km

 6480   6380  
  1.2346  108  1.2539  108   103 Nm


=1.94  109 J
2×109 J of work is done in raising spaceship from sea level to an altitude of 100 km.
This is roughly equal to the food energy in 2 000 chocolate bars (imagine 2 stacks of chocolate bars 10 bars
long, 10 bars wide and 10 bars high). The actual launch required much more energy due to air friction and
the fact that, during launch, the spacecraft had to carry its fuel up with it so it would have the fuel at higher
altitudes.
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PHYSICS SEMESTER ONE
UNIT 5
KINETIC ENERGY AND THE WORK-KINETIC ENERGY THEOREM
Energy can take on several forms, and it can be converted from one form to another. We will look at several
forms of energy in this course.
The kinetic energy of an object is the energy the object possesses due to its motion. The kinetic energy of
an object of mass m moving with a speed v is
K  12 mv 2
KE  12 mv2
or
Kinetic energy has the same units as work, the SI unit of energy is the J = kg·m2/s2 = Nm. We will use the
symbol K to denote kinetic energy. KE is another common symbol for kinetic energy.
Let’s look at how kinetic energy relates to the work done on an object. The drawing below shows the net
force Fnet in the x-direction on an object of mass m. that is moving horizontally.
Fnet  Fx ˆi
The work done by this constant force over a displacement dx is
W  Fx d x
According to Newton’s Second Law, this can be linked to the acceleration as
W  Fx x  max d x
We also have the kinetics equation
v2fx
 vix2
 2ax d x
or
ax d x 
v 2fx  vix2
2
We can now insert this into the work equation
W  max d x
 v 2fx  vix2
 m

2





 12 mv 2fx  12 mvix2
 K f  Ki
 K
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PHYSICS SEMESTER ONE
UNIT 5
From this analysis, we can see that the work done on an object is equal to the change in kinetic energy of
the object. This is the work-energy theorem. This only works when there is no change in the potential
energy (more on that in next section).
The work is positive if there is an increase in speed and negative if there is a decrease in speed.
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PHYSICS SEMESTER ONE
UNIT 5
POTENTIAL ENERGY
Potential energy is a form of energy that can be stored for later use. Stored
energy can be used to do work at some later time.
A clown opens the closet door while looking for a shoelace. A bowling ball,
stored high in the closet, rolls forward off the shelf, accelerates downwards, and
stops after flattening the clown’s shoes. The clown gave the ball potential
energy when he stored the lifted the ball from ground level to the shelf height. The elevated bowling ball
had energy stored (the potentially do work) due to its elevated location on the shelf. The bowling ball
expended this energy in doing work in flattening the clown’s shoes.
The gravitational potential energy Ug of a bowling ball with mass m is given by
U g  mgh
or
U g  mgy
where g is the gravitational field strength (acceleration due to gravity), and h or y is the height of the
bowling ball above the surface of Earth or some other reference point, like the tops of the clown’s shoes.
The above equation is an approximation that is good in regions near the surface. We will not consider the
case in this class, but a more accurate expression for the gravitational potential energy is
Ug 
GMm GMm

yf
yi
where M is the mass of the earth, G is the universal gravitational constant, yi and yf are the initial and final
distances from the centre of the earth. Under the restriction that the distances are close to the surface of
the earth, this equation is very close to the first equation with y  y f  yi .
Consider the case where a ball of mass m is raised at a constant speed from the ground at the height yi to a
higher elevation yf.
F
The lifting force and gravitational force are shown in the FBD to the right.
The speed is constant so the net force is zero and the magnitude of the applied force F must
equal that of the gravitational force Fg.
Fg
F  Fg  mg
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PHYSICS SEMESTER ONE
UNIT 5
The applied force and direction of movement are in the same vertical direction so the amount of work done
in lifting the ball is
W  Fd y

 F y f  yi


 mg y f  yi

The change in gravitational potential energy in the lift is
U g  U gf  U gi

 mg y f  yi

W
So the work done in lifting the ball is the change in gravitational potential energy.
Example
Calculate the potential energy of a 8.0 kg bowling ball lifted 2.0 m off the ground (toe level).
answer below and to the right
Define terms, state equation, solve:
m = 8.0 kg, h = 2.0 m
Ug = mgh
= (8.0 kg)(9.8 m/s2)(2.0 m)
= 157 J
The potential energy is 160 J.
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PHYSICS SEMESTER ONE
UNIT 5
ELASTIC POTENTIAL ENERGY
The concept of potential energy involves the storage of the energy until it can be released. Springs and
other elastic objects store energy in being stretched compressed. To understand the potential energy of the
spring, we first have to look at the relationship between the spring force and the displacement.
An ideal spring is one that has zero mass and doesn’t lose any energy when stretched and compressed.
Most springs are reasonably close to ideal over a small range of motion. When compressing/stretching an
ideal spring the spring force is related to the distance
Fs  kx
where k is the spring constant or elastic constant and x is the distance from the equilibrium position of the
spring. The equilibrium position is the position of the end of the spring when there are no forces on it. The
sign on the spring force indicates that it is always opposite the direction of compression (elongation). The
spring always pushes to return to its equilibrium position.
x = 0, Fs = 0
+x
x
x < 0, Fs > 0
Fs
x
x > 0, Fs < 0
Fs
Using Newton’s third law, the force needed to compress a spring by a distance x is
Fx  kx
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PHYSICS SEMESTER ONE
UNIT 5
Consider the work done in compressing a spring from the equilibrium position at 0 to the position x. In this
case, we will consider the motion and forces are limited to the x direction. We will also assume that the
acceleration of the system is negligible during the compression so the spring force is equal and opposite the
force applied to the spring. The force varies as a function of position so we can find the work by integrating
the force over movement of the spring from xi = 0 to
xf = x.
W 
xf
xi
Fx dx
x
  kxdx
(note, you are not expected to know integration at this point)
0
 12 kx 2
We can get the same relationship from a graph of the force versus position. The graph of the force versus
position is a line that goes through the origin. The work or potential energy is the area of the triangular
region between the line, x-axis and the origin.
Us
F
Us = ½ kx2
area
= ½ kx2
kx
x
x
x
The elastic potential energy stored in the spring due to this compression is equal to the work done in
compressing the spring.
For a spring with spring constant k and compression/extension x, the elastic potential energy is
U s  12 kx2
Note that the elastic potential energy does not depend on the sign of x. Energy is stored within the spring
with both compression and extension.
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PHYSICS SEMESTER ONE
UNIT 5
CONSERVATION OF MECHANICAL ENERGY IN AN ISOLATED SYSTEM
We have seen that both gravitational and elastic systems have forces associated with them. We also saw
that the changes in gravitational and elastic potential energies of an object are equal to the work done in
moving the object. Forces where the work done in moving an object is equal to the change in potential
energy are known as conservative forces. Gravitational and electrical forces are conservative forces. Spring
forces, in the case of an ideal spring are also conservative.
Conservative forces have two equivalent properties:
1.
The work done by a conservative force on a particle moving between any two points is
independent of the path taken by the particle.
It doesn’t matter how I lift a ball up to place it on a shelf, the amount of work done in lifting ball
from point A on the floor to point B on the shelf is always the same.
d
B
B
d
A
A
d
2d
Wc = Fc·d
Wc = Fc·d
2.
The work done by a conservative force on a particle moving through any closed path is zero. (A
closed path is one where the beginning and end points are identical.)
A pendulum that swings out and comes back to its initial position returns to its initial energy
level (no work added, no work done).
Δr = 0
B
Wc = Fc·Δr = 0
A
Non-conservative forces often involve things like friction where the energy is transferred away from the
system through a mechanism that can’t be reversed. Kinetic friction always goes against the direction of
motion.
It is highly recommended that you look at the notes on non-conservative forces in the textbook.
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PHYSICS SEMESTER ONE
UNIT 5
In ideal mechanical systems, energy can be converted from one form to another between gravitational,
spring and kinetic energies without loss of energy. Any work done on the system will change the total
amount of energy in the system. That stated, truly technically conservative systems are isolated with no
energy entering/leaving the system from/to external sources/sinks (no work, no change in energy).
Let’s define the mechanical energy of a system as the sum of kinetic and potential energies
Emech  K  U
In a system where energy is conserved, the mechanical energy of the system remains constant
Emech  constant .
Under conservation of energy, an increase in potential energy must be countered with a decrease in kinetic
energy. This can be expressed in many equivalent ways.
K  U
K  U g  U s
K  U  0
K  U g  U s  0
K f  U f  Ki  Ui
K f  U gf  U sf  Ki  U gi  U si
or
The ideal systems obeying such conditions are isolated and friction-free. Simple springs and gravitational
interactions are regarded as systems where energy is conserved. There are several systems in the real world
that are close enough to ideal that can model accurately under the assumption that energy is conserved.
When doing problems, conservation of energy is a very important tool. Dynamics problems where we have
forces acting on an object with changes in height or spring compressions can often be solved quite simply by
looking at the initial and final mechanical energies.
A puck slides along the ice (generic frictionless surface) and goes up an ice covered ramp
(collides with a spring, compressing it). The initial kinetic energy of the puck is converted
into gravitational (elastic potential energy). By comparing the initial and final energies, we
can determine how high the puck slides up the ramp (how much the spring is compressed).
Conservation of energy is a very useful tool in problem solving. You are strongly advised to try several
energy problems within the textbook to help give you a feel for when you can use conservation of energy to
solve a problem.
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PHYSICS SEMESTER ONE
UNIT 5
Example
Use energy conservation to find the speed of an 8.0 kg bowling ball after falling 2.0 m with an initial speed of
0 m/s.
Define terms:
m  8.0 kg, h  2.0 m, vi  0
The change in gravitational potential energy is
U g  mg h
  8.0 kg  9.8 kg  2.0 m 
 157 J
No other forces are mentioned so we will assume energy is conserved
K  U  0
K  U
The reduction in gravitational potential energy is matched by the increase in kinetic energy.
K  U
1
2
mv 2f  12 mvi2  U
mv 2f  mvi2  2U
vf 

mvi2  2U
m
 8.0 kg  0 2  2  157 J 
8.0 kg
 6.26 m/s
The speed of the ball at the bottom of the fall is 6.3 m/s.
Problem 3
A spring launcher consists of a spring and an 8.0 kg bowling ball. The spring constant is 2.0×102 N/m. The
spring is compressed 2.0 m and then released, launching the ball. Find the speed of the ball when it passes
the equilibrium position.
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PHYSICS SEMESTER ONE
UNIT 5
Problem 4
A 8.0 kg bowling ball is positioned on a shelf 2.0 m above the equilibrium position of a spring, k = 2.0×102
N/m. The ball falls off the shelf, landing on the spring compressing it. Calculate the maximum compression
of the spring.
Example
A m = 8.0 kg bowling ball slides (no rotation) along the floor with a speed of v = 26 m/s before it runs into a
spring, k = 5100 N/m, coming to a momentary stop as it compresses the spring. Assuming friction is
negligible; calculate the maximum compression of the spring.
final compression x  ?,
xi  0,
v f  0, all other terms defined in question
The spring is initially in the equilibrium position so there is no elastic potential energy. All of
the initial energy is kinetic energy.
Emech,i  12 mv2
The speed is zero at maximum compression so all energy at maximum compression is elastic
potential energy.
Emech, f  12 kx2
With the absence of friction, energy is conserved and the mechanical energy is constant.
1
2
kx 2  12 mv 2
x

mv 2
k
The maximum compression is 1.0 m.
8.0 kg  26 m/s 
2
5100 N/m
 1.03 m
Problem 5
A m = 8.0 kg bowling ball slides (no rotation) along the floor with a speed of v = 26 m/s before it runs into
ramp with a slope of 32°. Find the speed of the ball when it has traveled r = 7.5 m along the ramp (in the
distance along the ramp, not horizontally or vertically).
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PHYSICS SEMESTER ONE
UNIT 5
NON-CONSERVATIVE FORCES
Conservative Forces
1.
The work done by a conservative force on a particle moving between any two points is
independent of the path taken by the particle.
2.
The work done by a conservative force on a particle moving through any closed path is zero.
Non-conservative forces are forces that do not satisfy the properties of conservative forces. The most
common example of a non-conservative force is the force of friction. Imagine sliding a book around a
horizontal desk between two points. The magnitude of the friction force is essentially constant and the
direction is opposite the direction of motion. The friction does negative work regardless of the direction.
The (negative) work done by friction is proportional to the distance traveled. This violates conservative
force property number 1.
B
B
A
A
2d
d
Wf = -Ffd
Wf = -2Ffd
In addition, because the friction opposes the direction of motion, the work done is always negative (in this
particular case). On closed path that includes points A and B, the book loses energy moving from point A to
B and then loses additional energy in moving back to A. The work done is less than zero. This violates
conservative force property number 2.
B
WfAtoB = -Ffd
WfBtoA = -Ffd
A
d
WfAtoBtoA = -2Ffd ≠ 0
Please note, we can also use friction to do positive work. Imagine a book placed on a moving conveyor belt.
Here, the friction between the book and conveyor belt can be used to increase the kinetic energy (speed) of
the book, doing positive work. Moving the book from A to B and then back to A with the friction doing
positive work all of the time, the next amount of work done will again be non-zero.
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PHYSICS SEMESTER ONE
UNIT 5
Energy is not conserved when work is done on the system. The work can be due to an applied force or
through friction. The amount of work done is related to the change in mechanical energy through
W  Emech
We have already seen that this works for the kinetic and potential energies separately. We can find
solutions to complicated dynamics problems by looking at the initial and final mechanical energies and the
work done. This is another very important problem solving tool.
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PHYSICS SEMESTER ONE
UNIT 5
Example
A m = 8.0 kg bowling ball slides (no rotation) along the floor with a speed of
vi = 26 m/s just before it runs into a spring, k = 5100 N/m, coming to a momentary stop as it compresses the
spring.
a)
Assuming friction is negligible; calculate the maximum compression of the spring.
Ki = ½ mvi2,
We did this in the example
about 2 pages back.
Usi = 0
The maximum compression
Kf = 0,
was 1.03 m or 1.0 m
Usf = ½ kx2
x
b)
Assuming the force of friction on the ball is Ff = 1500 N during the spring compression,
calculate the maximum compression.
Variables: m = 8.0 kg, vi = 26 m/s, vf = 0, k = 5100 N/m, Ff = 1500 N, x = ?
The spring is initially in the equilibrium position so there is no elastic potential energy. All of
the initial energy is kinetic energy.
Emech,i  12 mvi2
The speed is zero at maximum compression so all energy at maximum compression is elastic
potential energy.
Emech, f  12 kx2
Friction is present so we can’t assume energy is conserved. The change in mechanical energy is
equal to the work done by the friction. The friction work is
W  Ff x
 Emech, f  Emech.i
 F f x  12 kx 2  12 mvi2
We know all terms in this equation except for the compression distance. Rearranging
1 kx2
2
 Ff x  12 mvi2  0
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PHYSICS SEMESTER ONE
UNIT 5
This is a quadratic equation, so we use the quadratic formula to solve it.
x


 12 k    12 mvi2 
2  12 k 
Ff  Ff 2  4
 F f  F f 2  kmvi2
k
 1500 N  
1500 N 2   5100 N/m 8.0 kg  26 m/s 2
5100 N/m
 0.294  1.071 m
 1.365 m or 0.777 m
Toss out the negative compression because that is an extension. The negative value is also
unphysical because the work from friction would be positive which never happens.
6000
The graph shows the mechanical energy and elastic
potential energy as a function of the spring
mechanical 5000
energy
compression. The initial kinetic energy is the
elastic
potential
energy
4000
mechanical energy at the equilibrium location.
As the compression becomes more positive, the
3000
work done by friction removes more of the
2000
mechanical energy. When the
compression is negative, friction
initial kinetic energy
1000
adds to the mechanical energy (impossible).
0
-1.5
The maximum compression is 0.78 m.
-1
0
0.5
1
final
spring compression (m)
location
equilibrium
location
Friction removes some of the energy from the
-0.5
motion
system so we expect this compression to be less than the friction-free compression in part a).
After reaching the noted “final location”, the ball reverses direction and moves back past the
equilibrium, eventually stopping at a point where the friction force exceeds the spring force.
Problem 6
A m = 8.0 kg bowling pin slides along the floor with a speed of vi = 2.6 m/s before it runs into ramp with a
slope of θ = 32°. If the coefficient of kinetic friction between pin and ramp is μk = 0.55, find how high the pin
slides up the ramp before coming to a stop.
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1.5
PHYSICS SEMESTER ONE
UNIT 5
RELATIONSHIP BETWEEN CONSERVATIVE FORCES AND POTENTIAL ENERGY
When we considered the energy stored in the spring, we looked at the work done by an outside force to
compress the spring. When the spring expands from a compression it converts its potential energy into
work on the bowling ball, and ultimately into the kinetic energy of the bowling ball. Because energy is
conserved, the sum of the work done by the spring Wc (c for conservative) and the change in potential
energy is zero.
Wc  U  0
Wc  U
or
If we assume that the motion is in one direction (x-direction)
Wc  
xf
xi
Fx dx
The change in potential energy can be written as an integral


U   U f  U i   
Uf
Ui
dU  
xf
xi
 dU

 dx

 dx

so we get
xf
x
i
Fx dx  
xf
xi
 dU 

 dx
 dx 
This has to be true for all value of xf and xi so the terms inside the integrals must be equal. This gives us the
following relationship
Fx  
dU
dx
To check, set U = ½kx2, where x is the distance of end of the spring from the equilibrium (zero force)
x = 0, Fs = 0
position.

+x

d 12 kx 2
dU
Fx  

 kx
dx
dx
x < 0, Fs > 0
This is the same expression we had when we started
looking at spring forces. The “–” sign indicates that
the spring force is opposite the direction of
compression (or extension). The spring force is
x > 0, Fs < 0
towards the equilibrium position.
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PHYSICS SEMESTER ONE
UNIT 5
Let’s try U = mgy, where y is in the up direction (away from centre of Earth)
Fy  
d  mgy 
dU

 mg
dy
dy
This is the well known gravitational force near the surface of Earth. The sign indicates that the force due to
gravity is towards the centre of Earth.
Any conservative force Fx , in the generic x direction, is related to the potential energy U c due to that
force by
U c   
xf
xi
Fx dx
and
Fx  
dU c
dx
The PHeT applet http://phet.colorado.edu/simulations/sims.php?sim=Masses_and_Springs does a good job
of illustrating the transfer of energy from one form to another (you have to select the spring in the “show
energy of” box). Energy lost to friction is displayed as the heat. Note that the heat energy never decreases.
Once energy is lost to friction, we can’t reverse the process and get the energy back. With no friction, the
energy is cycled continuously between the kinetic energy (KE), gravitational potential energy (PEgrav) and the
elastic potential energy (PEelas) without loss.
Energy skate park, http://phet.colorado.edu/new/simulations/sims.php?sim=Energy_Skate_Park gives a
good view of what is happening to the kinetic and gravitational potential energies for a skate boarder on a
roller coaster (right click track, select roller coaster mode) or regular track (default). The skate boarder is
not attached to the regular track so be careful with curves that will send the boarder flying.
The ramp simulation, http://phet.colorado.edu/new/simulations/sims.php?sim=The_Ramp, also shows how
the energies (work, kinetic, gravitational potential, heat (friction)) as an object is moved around a ramp.
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PHYSICS SEMESTER ONE
UNIT 5
POWER
Power is the rate at which energy (work) is delivered. Light bulbs are categorized by their power, which is
the rate at which they use energy. Cars engines are ranked by their maximum power output compared to
horses.
The average power delivered over a time t is
P
W
t
The units of power are watts: W = J/s.
Example
Anne and Bob both do 1.5 × 106 J of work in hiking to the top of Mt. Arrowsmith. Anne’s climb took 1.7 ×
103 s, while Bob’s climb took 2.1 × 104 s. What were the average powers of Anne and Bob during their
climbs?
solution below and to the left
WA  1.5  106 J, WB  1.5  106 J, t A  1.7  104 s, tB  2.1 104 s
PA 

WA
tA
1.5  106 J
1.7  104 s
 88 W
PB 

WB
tB
1.5  106 J
2.1  104 s
 71 W
Anne’s average power was 88 W and Bob’s average power was 71 W.1.0
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PHYSICS SEMESTER ONE
UNIT 5
The instantaneous power is the limit of the average power as the time interval goes to zero
W dW

t 0 t
dt
P  lim
This is the derivative of the work with respect to time. The power can also be written as the derivative of
energy with respect to time
P
dE
dt
Power is commonly expressed in kilowatts (kW) and horsepower (hp).
1kW = 1000 W
1 hp = 746 W
The energy use that shows up on your electricity bill is usually expressed in kilowatt-hours (kWh).
1 kWh = 1000 W · 3600 s = 3.6 × 106 J
In Canada, we are usually charged around $0.10 per kilowatt-hour of electricity.
Example
The energy used by a cyclist in accelerating away from a stop light is given by the formula
E  320 t
where t is the time after she left the stop light and the energy is in joules. Calculate the instantaneous
power exerted by the cyclist at:
a)
1 s after leaving the stop light,
b)
100 s after leaving the stop light.
solutions below and to the left
The instantaneous power is
P
dE d 320t1/ 2 1
160

 320t 1/ 2 
dt
dt
2
t
160
 160 W
1
a)
For t = 1 s, P 
b)
For t = 100 s, P 
160
 16 W
100
Recognizing problems that can be solved using energy analysis can often save a great deal of work (on
exams, assignments, life…). Check out several different problems in the text.
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PHYSICS SEMESTER ONE
UNIT 5
SUMMARY
Work
W  Fd
when vectors are in same direction and magnitudes are known
W  Fd cos
when vector magnitudes and angle between vectors are known
W  F d
when know vectors in component form
 Fx d x  Fy d y
Kinetic Energy
K  12 mv 2
W  K f  Ki
work-kinetic energy theorem
Potential Energy
U g  mgh
gravitational potential energy
U s  12 kx 2
elastic potential energy
K  U  0
W  K  U
conservation of mechanical energy
F 
dU
dx
change in mechanical energy due to work from an external source
relationship between a conservative force and the associated potential energy
Power
P
W
t
average power
P
dW dE

dt
dt
instantaneous power
Energy analysis is very useful for solving problems.
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PHYSICS SEMESTER ONE
UNIT 5
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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27