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Conservation of Energy
Physics 11
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
 = 2.0 x 103 J
 B) W = Fd = -300 x 4 = -1200 J
 C) Wnet = 2000 – 1200 = 8.0 x 102 J
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 (copy)

“The net work done on an object is equal to its
change in energy"


If the object is experiencing EK:
if Wnet is +ve, EK increases (moves in direction of
force or speeds up)
if Wnet is -ve, EK decreases (moves in direction of
friction or slows down)




If the object is experiencing EP:
if Wnet is +ve, EP increases (is lifted)
if Wnet is -ve, EP decreases (is lowered)
Total Energy and Work-Energy
Theorem (Copy)
 The total energy of an object is the kinetic
energy added to the potential energy.
 ET = Ek + Eg + Ee
 As an object is dropped, the kinetic
energy changes to potential energy until
there is 0 EP and only EK.
Law of Conservation of Energy
(Copy)
 Within an isolated system, energy can change
form, but the total amount of energy is
constant
 Energy cannot be created or destroyed
 Ei = Ef
 So Far…
 Eki + Egi = Ekf + Egf
Mechanical Energy (copy)
 The sum of kinetic and potential
energy
 ***Not including light energy/sound
energy/etc.
 In the absence of friction, any
external work produces a change in
mechanical energy.
Conservation of Mechanical Energy
 In the absence of friction, any
external work produces a change in
mechanical energy.
 Why does there have to be an
absence of friction?
Conservation of Mechanical Energy
 If no work is done, then…
Conservation of Mechanical Energy
(copy)
 The total mechanical energy at any initial
point in an ISOLATED system is equal to
the total mechanical energy at any later
point (in the absence of friction)
 Energy can be transformed from one type
to another
 Example: A falling book starts with
potential energy. As it falls, the potential
energy gets transformed into kinetic
energy
Example 1 … Think and Discuss
Solution
 This is an example of a situation that is much easier to
analyze using conservation of energy, since a kinematics
analysis would involve calculating the accelerations using
vectors and using the distances to calculate final
velocities.
 So what do we know?
 Since both balls are released from rest, they have no
initial kinetic energy.
 Their initial potential energy is the same, since they
are released from the same height.
 When they reach the bottom, neither ball has any
potential energy.
 Same initial energy means same final energy (due to
conservation of energy).
 Since all of the energy is kinetic, having the same
kinetic energy (and same mass) means that they are
going the same speed.
Example 2
Solution
Example 3
http://video.mit.edu/watch/work-potential-energy-demo-lecture-11-2854/
Solution
SkatePark Assignment
 Let’s correct!
Check Your Understanding
1. If the car has a speed of 12.0 m/s at point A,
a)What will its speed be at point C?
b) What is the highest hill (above the ground)
that the car could
reach on this roller coaster?
Worksheets
 Conservation of Mechanical Energy
 Check Your Understanding
Check Your Understanding
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
Question 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 = 11 m/s
Question 3:
 A 4.0 x 104 kg roller coaster starts from
rest at point A. Assuming no 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
Conservation of Total Energy
Conservation of Mechanical Energy
Conservation of Mechanical Energy
What is Q?
 Heat energy!
Conservation of Total Energy
Summary Slide
 The total energy of a system is constant.
 HOWEVER… If work is done on a system,
the mechanical energy changes so…
 In the presence of friction, work (by
friction) causes a change in total energy.
 Equation: Eki + EPi = EKf + EPf + Q
 Where Q is heat
Example 1
 A 250 g car rolls down a ramp. The
car starts from a height of 32 cm, and
reaches a speed of 1.7 m/s at the
bottom of the ramp. Was mechanical
energy conserved?
Solution
 To determine if mechanical energy
was conserved, we must calculate the
total initial energy and the total final
energy. We will use the bottom of the
ramp as the reference level.
Example 2
Solution
Problem Solving Hints
Conservation of Energy Hints
Check Your Learning
Solution
Worksheet
 Conservation of Total Energy
Page 287
 Questions 1, 2, 3, 4, 6, 7, 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!
 http://www.youtube.com/watch?v=0
ASLLiuejAo
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