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Energy
Physics 11
Page 217
 Multilab… Sources of Energy
 http://www.youtube.com/watch?v=iL
XDirj4JUA&feature=related
Try giving an example of…
 As many different types of energy
and an example that goes with each
What is energy?
 The measure of a system’s ability to
do work.
 http://www.youtube.com/watch?v=y
UpiV2I_IRI
Types of Energy
 There are 2 main classifications of energy:
 1) Potential Energy – The energy stored
in a body or system as a consequence of its
position, shape or form.
 Example: An object being held up has potential
energy because of its position (gravitational
potential energy).
 Example: A compressed spring has potential
energy (potential to spring open).
 2) Kinetic Energy – The energy of motion
 Example: When you walk across the classroom
you have kinetic energy.
 http://www.youtube.com/watch?v=0
ASLLiuejAo
Work and Energy Relationship
 There is not much difference between
work and energy.
 In order to do work, an object must
have energy.
 In order to have energy, an object
must have work done on it.
Energy Formulae
 Ek = KE = ½ mv2
 Eg = PE = mgh
 Units: Joules (J)
 NOTE: h = height of the object measured
from the reference level (measured in
metres).
What is the effect of doing work on
an object?
 You can give an object more kinetic
energy by doing more work on it.
 W = ΔKE
 W = KEf – Kei
 W = ½ mvf2 - ½ mvi2
Example 1
 A 145g tennis ball is thrown at a
speed of 25m/s.
 A) What is the ball’s kinetic energy?
 B) How much work was done to
reach this speed assuming the ball
started from rest.
Answers
 A) 45 J
 B) W = ΔKE = 45 J
Ex 2: Work on an moving object
A 2kg object is moving at 10 m/s
when a force is applied to it
accelerating it to 20m/s over a
distance of 5m. What is the work
done on the object?
Answers
1 2 1 2
W  mv f  mvi
2
2
1
1
2
2
W  2(20)  2(10)
2
2
W  400 J  100 J  300 J
Ex 3: Potential Energy and Work
What is the work done on a 12kg
object to raise it from the ground to
a height of 1.5m?
Potential Energy and Work
What is the work done on a 12kg
object to raise it from the ground to
a height of 1.5m?
W
W
W
W
 
 F  d
 
 mgd
 (12)(9.81)(1.5)
 176 J
Eg  W
E g  mgh
Ex 4:
 A 1000kg car moves from point A to point B
and then point C. The vertical distance
between A and B is 10.m and between A and
C is 15 m.
 A) What is the PE at B and C relative to A?
 B) What is the ΔPE (ΔPE = PEf – PEi) when it
goes from B to C?
 C) Repeat a) and b) but take the reference
level at C (switch all letters).
Answers
 A)
C:
 B)
 C)
B: 98100 J  98000J,
-147150J  -150 000J
a decrease of 245250 J 250000J
A: 150 000J
 B: 245250 J  250 000J
 Difference from A to B: increase of
100000J
Assignment on Energy – Kinetic
and Potential
Comprehension Check
 A truck pushes a car by exerting a
horizontal force of 500. N on it. A frictional
force of 300. N opposes the car’s motion as
it moves 4.0m.
 A) Calculate the work done on the car by
the truck.
 B) Calculate the work done on the car by
friction.
 C) Calculate the work done on the car
overall (net work).
Answers
 A) W = Fd = 500 x 4 = 2000 N
 = 2.00 x 103 J
 B) W = Fd = -300 x 4 = -1200. J
 C) Wnet = 2000 – 1200 = 800J
Comprehension Check
 Calculate the work done by a horse
that exerts an applied force of 100. N
on a sleigh, if the harness makes an
angle of 30’ with the ground and the
sleigh moves 30.m across a flat, level
ice surface (ie, no friction).
Answer
 W = Fd cosΘ = (100)(30)cos(30)
= 2.6 x 103 J
Comprehension Check
 A 50. kg crate is pulled 40. m along a
horizontal floor by a constant force
exerted by a person (100. N) which
acts at an angle of 37’. The floor is
rough and exerts a force of friction of
50.N. Determine the work done by
EACH FORCE acting on the crate, and
the net work done on the crate.
 DRAW A DIAGRAM!!!
 WFg = FdcosΘ  Work is 0J as the
force is perpendicular to gravitational
force.
 WFN = 0J (same reason as above)
 WFapp = Fdcos Θ =(100)(40)cos37’ =
3195J  S.F.  3200 J
 WFf = Fd = -50(40) = -2000 J
 -2.0 x 103J
 Wnet = 3200 – 2000 = 1200 J
Comprehension Check
 Mrs. Evans is holding a 2.4kg textbook at a height of
3.4m above the floor.
 a) What is the type of energy (potential or kinetic)?
How do you know?
 b) How much energy is there (use your equation)?
 c) What is the velocity of the book at this point (ie,
velocity initial)?
 d) If Mrs. Evans drops the book, what is the final
velocity assuming she doesn’t throw it (use your
kinematics equations!)?
 e) If Mrs. Evans drops the book as in d), what is the
type of energy when the book hits the floor?
 f) How much of this energy is there when it touches
the floor?
 g) Is there any time when there are both kinds of
energy? If so, when? Explain.
v 2f  vi2  2ad
Answers
A) Potential: It is not moving, it has the potential to
move (fall).
 B) PE = mgh = 2.4x9.81x3.4 = 80.J
 C) v = 0 (at rest, not thrown)
 D) vf2 = vi2 + 2ad = 2(9.81)(3.4) = 66.708
Vf = 8.2m/s [down]
E) KE = ½ mv2 = ½ (2.4)(8.2)2 = 80.J
F) 80.J
G) When the object is falling, there is both PE and KE.
When it falls 1.7m, there is equal PE and KE. Before
this point (higher than 1.7m) there is more PE. After
this (lower than 1.7m) there is more KE.

Work-Energy Theorem

“The net work done on an object is equal to its
change in energy"


If the object is experiencing KE:
if Wnet is +ve, KE increases (moves in direction of
force or speeds up)
if Wnet is -ve, KE decreases (moves in direction of
friction or slows down)




If the object is experiencing PE:
if Wnet is +ve, PE increases (is lifted)
if Wnet is -ve, PE decreases (is lowered)
Total Energy and Work-Energy
Theorem
 The total energy of an object is the kinetic
energy added to the potential energy.
 As an object is dropped, the kinetic
energy changes to potential energy until
there is 0 PE and only KE.
 ET = PE + KE
Law of Conservation of Energy
 Within a closed, isolated system, energy can change
form, but the total amount of energy is constant
 Closed - no objects enter or leave the system.
 Isolated - no net external force is exerted on it.
 Energy cannot be created or destroyed
 E1 = E2
 KE1 + PE1 = KE2 + PE2
Examples
KEi  PEi  EK f  PE f
 Book Drop
0  mghi  12 mv2f  0
h
v
EKi  Eei  EKf  Eef
 Collision into a spring
v
1
2
x
mvi2  0  0  12 kx2f
 Car coasting up a hill EKi  EGi  EKf  EGf
v
v
h
1
2
mvi2  0  12 mv 2f  mgh f
Example 1:
 A heavy object is dropped. If this
object reaches the floor at a speed of
3.2 m/s from what height was it
dropped?
Answer








Etop = Ebottom
KE + PE = KE + PE
0 + PE = KE + 0
mgh = ½ mv2
NOTE: The masses will cancel!
gh = ½ v2
9.81h = ½ (3.2)2
h = 0.52 m
Example 2:
 A heavy box slides down a frictionless
incline. The incline has a slope of 30°
and the length of the incline is 12m.
If the box starts from rest at the top
of the incline what is the speed at the
bottom?
Answer
 V = 10.8m/s
Example 3:
 A 4.0 x 104 kg roller coaster starts from
rest at point A. Neglecting friction, calculate
its potential energy relative to the ground,
its kinetic energy and its speed at points
B,C and D in the illustration above.
Answer
Page 287
 Questions 1, 2, 3, 4, 6, 8
 A 15.0 kg box slides down an incline. If the
box starts from rest at the top of the incline
and has a speed of 6.0m/s at the bottom,
how much work was done to overcome
friction?
 NOTE: The incline is 5.0m high (vertically)
and the incline that the box goes down is
8.0m long (hypotenuse).
 Remember: W = ΔE
Try This …
 A skier is gliding along with a speed of
2.00m/s at the top of a ski hill. The hill is
40.0m high. The skier slides down the icy
(frictionless) hill.
 A) What will the skier’s speed be at a
height of 25.0m?
B) At what height will the skier have a
speed of 10.0m/s?
 HINT: Use similar triangles!
 Known: vi = 2.00m/s
 h1 = 40.0m
 h2 = 25.0m
REMEMBER…
 W = ΔE
 That E can be potential or kinetic
energy!