# Work, Energy and Power

```Work, Energy, and Power
Lesson 1: Basic Terminology and Concepts
– Definition and Mathematics of Work
– Calculating the Amount of Work Done by Forces
– Power
– Potential Energy
– Kinetic Energy
– Internal Energy
– Mechanical Energy
Lesson 2 - The Work-Energy Relationship
– Work-Energy Principle
– Internal vs. External Forces
– Analysis of Situations Involving External Forces
– Analysis of Situations in Which Mechanical Energy is
Conserved
– Application and Practice Questions
Definition and Mathematics of Work
force
• In physics, work is defined as a _________
acting upon an object to ____________
a
cause
displacement
__________________.
• In order for a force to qualify as having done
work on an object, there must be a displacement
cause
and the force must ___________
the
displacement
Work done
Work not done
Let’s practice – work or no work
1. A student applies a force to a wall and
becomes exhausted. no work
2. A calculator falls off a table and free falls
to the ground.
work
3. A waiter carries a tray full of beverages
above his head by one arm across the
room
no work
4. A rocket accelerates through space.
work
Calculating the Amount of Work Done by Forces
W  Fd cos
F
θ
• F - is the force in Newton, which
causes the displacement of the
object.
• d - is the displacement in meters
• θ = angle between force and
displacement
• W - is work in N∙m or Joule (J). 1 J
= 1 N∙m = 1 kg∙m2/s2
d
Fy
θ
d
F
• Work is a scalar quantity
Fx
• Work is independent of time the
force acts on the object.
Only the horizontal component of
the force (Fcosθ) causes a
horizontal displacement.
Example
• How much work is done on a vacuum
cleaner pulled 3.0 m by a force of 50.0 N
at an angle of 30o above the horizontal?.
W  Fd cos
W  (50.0 N )(3.0m) cos 30
W  130J
o
Work for vertical motion
• How much work is done in lifting a 5.0 kg
box from the floor to a height of 1.2 m
above the floor?
Given:
d = h = 1.2 meters; m = 5.0 kg; θ = 0
Unknown:
W=?
W = F∙dcosθ
F = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 N
W = F∙d = (49 N) (1.2 m) = 59 J
Only the horizontal component of the force
(Fcosθ) causes a horizontal displacement.
• Matt pulls block along a horizontal surface at constant
velocity. The diagram show the components of the force
exerted on the block by Matt. Determine how much work is
done against friction.
Given:
unknown:
Fx = 8.0 N
Fy = 6.0 N
dx = 3.0 m
W=?J
6.0 N
F
W = Fxdx
8.0 N
W = (8.0 N)(3.0 m) = 24 J
3.0 m
Relationships between work,
force and displacement
• A neighbor pushes a lawnmower four
times as far as you do but exert only half
the force, which one of you does more
work and by how much?
Wyou = Fd
Wneighbor = (½ F)(4d) = 2 Fd = 2 Wyou
The neighbor, twice as much
The sign of work
W = F∙d∙cosθ
Work done – positive,
negative or zero work
Positive work
negative work - force acts in
the direction opposite the
objects motion in order to
slow it down.
no work
When F is ┴ to d, W = 0
To Do Work, Forces Must Cause Displacements
On the force parallel to the displacement
causes displacement
W = F∙d∙cos0o = 0
Force vs. displacement graph
Example: a block is pulled along
a table with 10. N over a distance
of 1.0 m.
W = Fd = (10. N)(1.0 m) = 10. J
height
base
area
Force (N)
• The area under a force versus
displacement graph is the work done by
the force.
work
Displacement (m)
Example
• A student produced various elongations of a spring by
applying a series of forces to the spring. The graph
below represents the relationship between the applied
force and the elongation of the spring. Determine the
work done between 0.0 m to 0.20 m.
example
Which graph best represents the greatest amount of
work?
A.
B.
C.
D.
Example
• A box is pushed to the right with a varying horizontal
force. The graph below represents the relationship
between the applied force and the distance the box
moves. What is the total work done in moving the box
6.0 meters?
The angle in work equation
• The angle in the equation is the angle between
the force and the displacement vectors.
If F & d are in the same direction, θ is 0o.
d
F
What is θ in each case?
θ is 0o in all cases
What have we learned?
1. What is definition of work?
2. What is the work equation? DAP to
3. Give examples of positive, negative and
zero work
4. How to find work in F vs. d graph?
Power
• Power is the rate at which work is done. It is the
work/time ratio. Mathematically, it is computed using the
following equation.
• The standard metric unit of power is the Watt.
All machines are typically described by a power rating. For
example, a 60 Watt light bulb indicates 60 J of electrical
energy is transferred to light energy every second. A high
power car indicates that it can be accelerated very rapidly.
Some people are more power-full than others because they
are capable of doing the same amount of work in less time
or more work in the same amount of time
example
• Ben Pumpiniron elevates his 80-kg body up the
2.0-meter stairwell in 1.8 seconds. What is his
power?
It can be assumed that Ben must apply an (80 kg x 9.81
m/s2) -Newton downward force upon the stairs to elevate his
body.
Another equation for power
example
•
Two physics students, Will N. Andable and Ben
Pumpiniron, are in the weightlifting room. Will lifts the
100-pound barbell over his head 10 times in one
minute; Ben lifts the 100-pound barbell over his head
10 times in 10 seconds. Which student does the most
work? ______________ Which student delivers the
example
• When doing a chin-up, a physics student lifts her 42.0-kg
body a distance of 0.25 meters in 2 seconds. What is the
power delivered by the student's biceps?
kilowatt-hour is unit for energy
• Your household's monthly electric bill is often
expressed in kilowatt-hours. One kilowatt-hour
is the amount of energy delivered by the flow of l
kilowatt of electricity for one hour. Use
conversion factors to show how many joules of
energy you get when you buy 1 kilowatt-hour of
electricity.
Work and Energy are related
• When work is done on a system, that system’s change in
energy equals to the amount of work done on it.
• Work and energy have the same unit: Joule
• For example, if you push a cart, you do work on the cart,
the cart is going to speed up and its temperature may
increase, its energy is increased. If you lift a rock, you do
work on the rock and you increase the rock’s energy.
• There are many forms of energy.
– Potential energy
– Kinetic energy
– Internal energy
Potential energy
• An object can store energy as the result of
its position. Potential energy is the stored
energy of position possessed by an object.
• Two form:
– Gravitational
– Elastic
Gravitational potential energy
1. When you lift an object, you apply a force equals to
gravity. As a result, its position is higher, and it has
more gravitational potential energy.
2. Gravitational potential energy is the energy stored in an
object as the result of its height
3. The energy is stored as the result of the gravitational
attraction of the Earth for the object.
4. Gravitational depends on
a. mass, in kg
b. acceleration due to gravity, 9.81 m/s2
c. height, in m
Gravitational potential energy Equation
Gravitation potential energy equals to work done against
gravity
.
Wgrav  F  d  m  g  h
PEgrav  m  g  h
Gravitational attraction between Earth
and the object:
m: mass, in kilograms
g: acceleration due to gravity = 9.81 m/s2
change in height,
in meters
Unit of energy
• The unit of energy is the same as work:
Joules
• 1 joule = 1 (kg)∙(m/s2)∙(m) = 1 Newton ∙
meter
• 1 joule = 1 (kg)∙(m2/s2)
Work and energy has the same unit
∆h - Gravitational Potential Energy is relative
To determine the
gravitational potential energy
of an object, a zero height
position must first be
assigned. Typically, the
ground is considered to be a
position of zero height.
But, it doesn’t have to be:
– It could be relative to the
height above the lab table.
– It could be relative to the
bottom of a mountain
– It could be the lowest
position on a roller coaster
Change in GPE only depends on
change in height, not path
As long as the object starts and ends at the same
height, the object has the same change in GPE
because gravity does the same amount of work
regardless of which path is taken.
Example
• The diagram shows points A, B, and C at or near Earth’s
surface. As a mass is moved from A to B, 100. joules of
work are done against gravity. What is the amount of
work done against gravity as an identical mass is moved
from A to C?
100 J
As long as the object starts and ends at the same height, the
object has the same change in GPE because gravity does the
same amount of work regardless of which path is taken.
Work done against gravity = ∆GPE
•
•
•
•
•
How much potential energy is gained by
an object with a mass of 2.00 kg that is
lifted from the floor to the top of 0.92 m
high table?
Known:
Solve:
m = 2.00 kg
∆PE = mg∆h
h = 0.92 m
∆PE = (2.00 kg)(9.81m/s2)(0.92 m)
= 18 J
g = 9.81 m/s2
• unknown:
• PE = ? J
GPE vs. Vertical Height Graph
The graph of
gravitational potential
energy vs. vertical
height for an object near
Earth's surface is a
straight line. The slope
is the weight of the
object.
Elastic potential energy
• Elastic potential energy is the energy
stored in elastic materials as the result of
their stretching or compressing when a
force is applied.
• Elastic potential energy can be stored in
– Rubber bands
– Bungee cores
– Springs
– trampolines
Elastic potential energy in a spring
•
•
1
2
PEs  k  x
2
equilibrium
x
elongation
k: spring constant
x: amount of compression or
elongation relative to equilibrium
(unstretched) position
Elastic potential energy
Elastic potential energy is directly
proportional to x2
elongation
Example
• As shown in the diagram, a 0.50-meter-long spring is
stretched from its equilibrium position to a length of 1.00
meter by a weight. If 15 joules of energy are stored in the
stretched spring, what is the value of the spring
constant?
x = 0.50 m
PE = ½ kx2
15 J = ½ k (0.50 m)2
k = 120 N/m
Example
• The unstretched spring in the diagram has a length of
0.40 meter and a spring constant k. A weight is hung
from the spring, causing it to stretch to a length of 0.60
meter. In terms of k, how many joules of elastic potential
energy are stored in this stretched spring?
x = 0.20 m
PEs = ½ kx2
PEs = ½ k(0.20 m)2
PEs = (0.020 k) J
Elongation ( or compression) depends
on force
• For certain springs, the amount of force (F) is
directly proportional to the amount of
elongation or compression (x); the constant of
proportionality is known as the spring constant
(k).
Fs  k  x
Hooke’s Law
Fs  k  x
• F in the force needed to displace (by stretching or
compressing) a spring x meters from the equilibrium
(relaxed) position. The SI unit of F is Newton.
• k is spring constant. It is a measure of stiffness of the
spring. The greater value of k means a stiffer spring
because more force is needed to stretch or compress it
that spring. The SI units of k are N/m.
• x the distance difference between the length of
stretched/compressed spring and its relaxed
(equilibrium) spring.
The slope of Fs vs. x
• Spring force is directly proportional to the
elongation of the spring (displacement)
force
The slope represents spring
constant: k = F / x (N/m)
elongation
caution
• Sometimes, we might see a graph such as
this:
elongation
The slope represents the inverse of
spring constant:
Slope = 1/k
force
Example
• Given the following data table and corresponding graph,
calculate the spring constant of this spring.
Example
• A 20.-newton weight is attached to a spring,
causing it to stretch, as shown in the diagram.
What is the spring constant of this spring?
Example
•
The graph below shows elongation as a
function of the applied force for two springs, A
and B. Compared to the spring constant for
spring A, the spring constant for spring B is
1. smaller
2. larger
3. the same
Work done stretching a spring
equals to Elastic Potential Energy
• Elastic potential energy stored in a spring
equals to the work done in stretching it.
Work = Area = ½ (base)(height)
Work = ½ (x)(F)
Work = ½ (x)(k∙x)
Work = ½ k∙x2
PEs = Work = ½ k∙x2
PEs  F  x
Example
• If a mass of 0.55 kg attached to a vertical spring
stretches the spring 2.0 cm from its original
equilibrium position, what is the spring constant?
Example
• Determine the potential energy stored in the spring with
a spring constant of 25.0 N/m when a force of 2.50 N is
applied to it.
Solve:
Given:
PEs = ½ k∙x2
Fs = 2.50 N
To find x, use Fs = kx,
k = 25.0 N/m
(2.50 N) = (25.0 N/m)(x)
Unknown:
x = 0.100 m
PEs = ? J
PEs = ½ (25.0 N/m)(0.100 m)2
PEs = 0.125 J
Example
• A 10.-newton force is required to hold a
stretched spring 0.20 meter from its rest
position. What is the potential energy
stored in the stretched spring?
Kinetic energy
• Kinetic energy is the energy of motion.
• An object which has motion - whether it be
vertical or horizontal motion - has kinetic energy.
• The equation for kinetic energy is:
KE = ½ mv2
– Where KE is kinetic energy, in joules
– v is the speed of the object, in m/s
– m is the mass of the object, in kg
• Like potential energy, kinetic energy is a
scalar quantity.
Questions
1. Which of the following has kinetic energy?
a. a falling sky diver
b. a parked car
c. a shark chasing a fish
d. a calculator sitting on a desk
2. If a bowling ball and a volleyball are traveling at the
same speed, do they have the same kinetic energy?
3. Car A and car B are identical and are traveling at the
same speed. Car A is going north while car B is going
east. Which car has greater kinetic energy?
Speed has more impact on kinetic energy
• KE is directly proportional to m, so doubling the mass doubles
kinetic energy, and tripling the mass makes it three times
greater.
Kinetic energy
Kinetic energy
• KE is proportional to v2, so doubling the speed quadruples
kinetic energy, and tripling the speed makes it nine times
greater.
speed
mass
Example
• A 7.00 kg bowling ball moves at 3.00 m/s.
How much kinetic energy does the bowling
ball heave? How fast must a 2.45 g tabletennis ball move in order to have the same
kinetic energy as the bowling ball? Is this
speed reasonable for a table-tennis ball?
Example
•
An object moving at a constant speed of 25
meters per second possesses 450 joules of
kinetic energy. What is the object's mass?
Known:
• KE = 450 J
• v = 25 m/s
Unknown:
• m = ? kg
Solve:
KE = ½ mv2
450 J = ½ (m)(25 m/s)2
m = 1.4 kg
Example
•
a.
b.
c.
d.
A cart of mass m traveling at a speed v has kinetic
energy KE. If the mass of the cart is doubled and its
speed is halved, the kinetic energy of the cart will be
half as great
twice as great
one-fourth as great
four times as great
Example
•
Which graph best represents the relationship
between the kinetic energy, KE, and the
velocity of an object accelerating in a straight
line?
a
b
c
d
Total Mechanical Energy
• Mechanical energy is the energy that is
possessed by an object due to its motion or
due to its position. Mechanical energy can be
either kinetic energy (energy of motion) or
potential energy (stored energy of position) or
both.
The total amount of mechanical energy is
merely the sum of the potential energy and the
kinetic energy
TME = KE + PE
TME = PEgrav + PEspring + KE
Note: PE includes both gravitational potential energy and elastic potential energy.
Mechanical Energy as the Ability to
Do Work
• Any object that possesses mechanical energy whether it is in the form of potential energy or
kinetic energy - is able to do work.
Internal Energy
• internal energy: TEMPERATURE/HEAT
Q
Resistance force, such as friction, air
resistance, or any force stopping motion,
produce internal energy. When resistance
force is zero, internal energy does not change.
Q  Wresistance   Fresisitance  d
Classification of Energy
Energy
Nonmechanical
Mechanical
Potential
Gravitational
potential
kinetic
Elastic
potential
Work done by forces other than gravity,
spring, or friction equals to change in total
energy of the system
Internal
energy
Chemical, nuclear,
electric
W other = ∆ET
ET = PE + KE + Q
What types of energy are changing?
Scenario #1
A car is moving along a flat surface with an increasing
speed (while external force is applied). What type(s) of
energy are changing?
PEg
X
PES
X
KE
↑
Q
?
What types of energy are changing?
Scenario #2
A car is moving up an incline with constant speed (while
external forces are applied.) What type(s) of energy are
changing?
PEg
↑
PES
X
KE
X
Q
?
What types of energy are changing?
Scenario #3
A car is applying its brakes (external force) and slowing
down as it moves down a slope. What type(s) of energy
are changing?
PEg
↓
PES
X
KE
↓
Q
↑
What types of energy are changing?
Scenario #4
A cart is moving on a flat surface toward a
wall and is stopped by a spring on its front.
What type(s) of energy are changing?
PEg
X
PES
↑
KE
↓
Q
?
What types of energy are changing?
Scenario #5
A rocket flies upward with an increasing velocity [ignore
air resistance]. What type(s) of energy are changing?
PEg
↑
PES
X
KE
↑
Q
X
Lesson 2: The Work-Energy Theorem
•
•
•
•
•
Work and Energy principle
Analysis of Situations Involving external forces
Analysis of Situations Involving a closed system
Analysis of Situations Involving an ideal system
Application and Practice Questions
Work and Energy Principle
• When an external force does work on a system, the
work done changes the system's energy.
Workexternal force = ∆ET= Ef - Ei
ET=PE + KE + Q
TME = PE + KE
Workexternal force = (TMEf + Qf) - (TMEi + Qi)
Workexternal force = ∆PE + ∆KE + ∆Q
What is a system? And what are
internal, external forces?
• Any sets of objects can be considered A SYSTEM.
The definition of system can be arbitrary. However,
according to New York State Regents Physics, examples
of a system are apple and Earth, two cars in a collision,
two cars connected by a spring, rifle and bullets, cannon
and cannon ball, spring and its launched toy, car and
• The forces within a system are called internal forces.
Example of internal forces include gravity (between apple
and Earth), force by the spring between two connected
cars, force of friction between car and road.
• Applied force to a system is an example external force
Problem solving strategies
• Identify the system in question.
• Identify external forces (applied force on the system) or
external work.
• Identify initial and final mechanical energy (KE & PE)
• Determine if there is any resistance force.
Q  Wresistance   Fresisitance  d
• Apply work-energy theorem
Workexternal force = ∆PE + ∆KE + ∆Q
Example 1
System: barbell
External force = 1000 N
Fresistance = 0
Q f  Qi
Wexternal force = ∆ET = Ef - Ei
Wext  (TME f  Q f )  (TMEi  Qi )
Wext  TME f  TMEi
(1000 N )(0.25m)  TME f  1500 J
TME f  1750 J
Example 2
System: cart and track
External force = 18 N
Fresistance = 0
Q f  Qi
Wexternal force = ∆ET = Ef - Ei
Wext  (TME f  Q f )  (TMEi  Qi )
Wext  TME f  TMEi
(18 N )(0.70m)  TME f  0 J
TME f  12.6 J
Example 3
• A person does 100 joules of work in pulling back the string
of a bow. What will be the initial speed of a 0.5-kilogram
arrow when it is fired from the bow?
System: bow
External work = 100 J
Fresistance = 0
Q f  Qi
Wext  (TME f  Q f )  (TMEi  Qi )
Wext  TME f  TMEi
100 J  KE f
1
100 J  (0.5kg)v 2
2
v  20m / s
ENERGY IS CONSERVED IN A CLOSED
SYSTEM
• A closed system is one in which there are no external
forces doing work on the system, no external work being
done by the system, an no transfer of energy into or out of
the system.
• Although the energy within a closed system may be
transformed from one type to anther, the total amount of
energy in a closed system must remain constant – energy is
never created or destroyed.
Wext = ∆ET = 0
PE  KE  Q  0
Example
Closed
system: car
and ground
Fresitance = 6000 N
∆Q = -Wresistance TMEbefore  Qbefore  TME after  Qafter
TMEbefore  TMEafter  Qafter  Qbefore
TMEbefore  TMEafter  Q
320000 J  TMEafter  [(8000 N )( 30m)]
TMEafter  8000 J
Example
• A shopping cart full of groceries is sitting at the top of a 2.0-m
hill. The cart begins to roll until it hits a stump at the bottom of
the hill. Upon impact, a 0.25-kg can of peaches flies
horizontally out of the shopping cart and hits a parked car with
an average force of 500 N. How deep a dent is made in the
car (i.e., over what distance does the 500 N force act upon the
can of peaches before bringing it to a stop)?
Closed system: can of
peaches and parked car
Fresitance = 500 N
∆Q = -Wresistance
Conservative vs. non-conservative Forces
1. When work done against a force is independent
of the path taken, the force is said to be a
CONSERVATIVE FORCE.
2. When work done against a force is dependent of
the path taken, the force is said to be a NONCONSERVATIVE FORCE.
Conservative
forces
Fgrav
Fspring
Non-conservative
forces
Fapp
Ffrict
Ideal Mechanical System
• An ideal mechanical system is a closed system in
which no friction or other non-conservative force acts.
The only forces doing work are gravity, or force of a
spring.
• In an ideal mechanical system total mechanical
energy remains the same:
Wext = ∆ET = 0
PEi  KEi  Qi  PE f  KE f  Q f Q  0
PEbefore  KEbefore  PEafter  KEafter
KE  PE
Ideal mechanical system
example #1 - Skiing
• In skiing, the only force doing work is gravity. Normal
force is perpendicular to the displacement. The system is
ideal, TME is constant. Energy only transforms from one
form to another.
Ideal mechanical system
example #2 - Pendulum
• In a pendulum, the only force doing work is gravity.
Tension is perpendicular to displacement. The system is
ideal, TME is constant. Energy only transforms from
one form to another,
Example
• As the 2.0-kg pendulum bob in the above diagram swings to
and fro, its height and speed change. Use energy equations
and the above data to determine the blanks in the above
diagram.
0.306
0.153
1.73
0
2.45
0.306
Example
•
1.
2.
3.
4.
As the pendulum swings from position A to position C
as shown in the diagram, what is the relationship of
kinetic energy to potential energy? [Neglect friction.]
The kinetic energy decreases more than the potential
energy increases.
The kinetic energy increases more than the potential
energy decreases.
The kinetic energy decrease is equal to the potential
energy increase.
The kinetic energy increase is equal to the potential
energy decrease.
example
• A pendulum is pulled to
the side and released
from rest. Sketch a
graph best represents
the relationship
between the
gravitational potential
energy of the pendulum
and its displacement
from its point of release.
PE
pos
Example
•
1.
2.
3.
4.
In the diagram, an ideal pendulum
released from point A swings freely
through point B. Compared to the
pendulum's kinetic energy at A, its
potential energy at B is
half as great
twice as great
the same
four times as great
Ideal mechanical system
example #3 – Roller Coaster
• In Roller Coster, the only force doing work is gravity.
Normal force is perpendicular to the displacement. The
system is ideal, TME is constant. Energy only transforms
from one form to another.
Ideal mechanical system
example #4 – Free Fall or projectile
• In Free Fall, the only force doing work is gravity. The
system is ideal, TME is constant. Energy only transforms
from one form to another.
Energy graph of a free falling object
The graph shows as a ball is dropped, how its energy is
transformed.
constant
• The total mechanical energy remains _____________.
• GPE decreases as KE increases
example
• A 3.0-kilogram object is placed on a frictionless track at
point A and released from rest. (Assume the gravitational
potential energy of the system to be zero at point C.)
Calculate the kinetic energy of the object at point B.
KEi + PEi = KEf + PEf Since only force is gravity, TME is constant
0 + mg(3.0m) = KEf + mg(1.0m)
KEf = mg∆h = (3.0 kg)(9.81 m/s2)(3.0 m – 1.0 m) = 59 J
(KE gained at B is potential lost from A to B
example
•
a.
b.
c.
d.
A 250.-kilogram car is initially at rest at point A on a
roller coaster track. The car carries a 75-kilogram
passenger and is 20. meters above the ground at point
A. [Neglect friction.] Compare the total mechanical
energy of the car and passenger at points A, B, and C.
The total mechanical energy is less at point C than it is
at points A or B.
The total mechanical energy is greatest at point A.
The total mechanical energy is the same at all three
points.
The total mechanical energy is greatest at point B.
example
• The diagram represents a 0.20-kilogram sphere moving
to the right along a section of a frictionless surface. The
speed of the sphere at point A is 3.0 meters per second.
• Approximately how much kinetic energy does the sphere
gain as it goes from point A to point B?
Since only force is gravity, TME is constant
KEi + PEi = KEf + PEf
KEf - KEi = PEi - PEf
The KE gained is PE lost
∆KE = ∆PE = mg∆h
KE = (0.20 kg)(9.81 m/s2)(1.00 m) = 2.0 J
example
• A 1.0 kg mass falls freely for 20. meters near the
surface of Earth. What is the total KE gained by
the object during its free fall?
Since only force is gravity, TME is constant
KEi + PEi = KEf + PEf
KEf - KEi = PEi - PEf
∆KE (gained) = GPE (lost)
∆KE = (1.0 kg)(9.81 m/s2)(20.m-0) = 2.0x102 J
Ideal mechanical system
example #5 – Pop Up toy
• In Pop Up toy, the only force doing work is gravity and
force on a spring. The system is ideal, TME is constant.
Energy only transforms from one form to another.
Initially, at the bottom: TMEi = ½ kx2
Finally, at the top: TMEf = mgh
TMEf = TMEi
½ kx2 = mgh
Work and Energy Thorem
1. When external force does work on a system, its total energy is
changed.
Wext  ET  PE  KE  Q
Q  Wresistance   Fresisitance  d
2. When external force is zero, the system is closed, its total
energy is conserved.
ET  0
PE f  KE f  Q f  PEi  KEi  Qi
3. In a closed system, if the only force doing work are
conservative forces (no friction), the total mechanical energy is
conserved.
Q  0
PE f  KE f  PEi  KEi
Example
•
A box with a mass of 0.04 kg starts from rest at point A and
travels 5.00 meters along a uniform track until coming to rest
at point B, as shown in the picture. Determine the magnitude
of the frictional force acting on the box. (assume the frictional
force is constant.)
A
0.80 m
B
0.50 m
example
diagram. A 250.-kilogram car is initially at rest at point A
on a roller coaster track. The car carries a 75-kilogram
passenger and is 20. meters above the ground at point
A. [Neglect friction.] Calculate the speed of the car and
passenger at point B.
20. m/s
example
• The diagram shows a 0.1-kilogram apple attached to a
branch of a tree 2 meters above a spring on the ground
below. The apple falls and hits the spring, compressing it
0.1 meter from its rest position. If all of the gravitational
potential energy of the apple on the tree is transferred to
the spring when it is compressed, what is the spring
constant of this spring?
Since only internal forces is doing work, TME
is constant
KEi + PEi = KEf + PEf
0 + mgh = 0 + ½ kx2
mgh = ½ kx2
(0.1 kg)(9.81m/s2)(2m) = ½∙k (0.1m)2
k = 400 N/m
Example
In the diagram below, 450. joules of work is
done raising a 72-newton weight a vertical
distance of 5.0 meters. How much work is
done to overcome friction as the weight is
raised?
Lab 15 – Power
Purpose: To determine my power requirement for climbing
a staircase - both by walking and by running. Compare
the two results.
Material: stop watch, ruler.
Data section: should contain colomns of measured and
calculated data for both walking and running up a flight
of stairs. The rows and columns should be labeled;
units should be identified. Work should be shown for
each calculation; the work should be labeled and easy
to follow.
Conclusion/discussion of results: The Conclusion should
(as always) answer the questions posed in the
Purpose.
Data table for individual
Name
Mass
(kg)
Weight
(N)
Total
work
(J)
Height of each
step
Average Power Average Power of
Time of
of
Time of
running
walking walking running
(watts)
(s)
(watts)
(s)
Number of
steps
Data table for each group
Name
Time for walking (s)
t1
t2
t3
t4
Height of each step
t5
Time for running (s)
t6 tavg t1 t2 t3 t4 t5 t6 tavg
Number of steps
Lab 16 - Energy of a Tossed Ball
OBJECTIVES
1. Measure the change in the kinetic and potential
energies as a ball moves in free fall.
2. See how the total energy of the ball changes during
free fall.
MATERIALS
computer
Vernier computer interface
Vernier Motion Detector
ball
Logger Pro
PRELIMINARY QUESTIONS
• For each question, consider the free-fall portion of the
motion of a ball tossed straight upward, starting just as the
ball is released to just before it is caught. Assume that there
is very little air resistance.
1. What form or forms of energy does the ball have while
momentarily at rest at the top of the path?
2. What form or forms of energy does the ball have while in
motion near the bottom of the path?
3. Sketch a graph of postion vs. time for the ball.
4. Sketch a graph of velocity vs. time for the ball.
5. Sketch a graph of kinetic energy vs. time for the ball.
6. Sketch a graph of potential energy vs. time for the ball.
7. Sketch a graph of total energy vs. time for the ball.
8. If there are no frictional forces acting on the ball, how is the
change in the ball’s potential energy related to the change in
kinetic energy?
DATA TABLE
Mass of the ball (kg)
Position
After
release
Top of
path
Before
catch
Time
(s)
Height
(m)
Velocity
(m/s)
PE
(J)
KE
(J)
TE
(J)
ANALYSIS
1. Inspect kinetic energy vs. time graph for
the toss of the ball.
2. Inspect potential energy vs. time graph for
the free-fall flight of the ball.
3. Inspect Total energy vs. time graph for
the free-fall flight of the ball.
4. Your conclusion from this lab
1. How does the kinetic and potential energy
change?
2. How does the total energy change?
Lab 14 – Hooke’s Law (1)
1.
2.
3.
4.
5.
Purpose: To determine the spring constant of a given spring.
Material: spring, masses, meter stick.
Procedure: Hook different masses on the spring, record the
force Fs (mg) and corresponding elongation x. Plot the graph of
Force vs. elongation
Data section: should contain colomns: force applied,
elongation.
– Data measured directly from the experiment. The units of
measurements in a data table should be specified in
Data analysis: Graph force vs. elongation on graph paper,
– What does the slope mean in Force vs. elongation graph?
– Determine the spring constant
Force (N)
Elongation (m)
Force (N)
Force vs. elongation
Elongation (m)
Do now
• A 1.2-kilogram block and a 1.8-kilogram block are initially at rest on a
frictionless, horizontal surface. When a compressed spring between the
blocks is released, the 1.8-kilogram block moves to the right at 2.0
meters per second, as shown.
• What is the speed of the 1.2-kilogram block after the spring is released?
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