Transcript Energy - hrsbstaff.ednet.ns.ca

Energy
Physics 11
Think about… 5 min
1) Why is energy important?
2) Where does energy come from? Where
does it go?
3) How do we capture energy?
4) How does energy impact kinematics
(motion) and dynamics (forces)?
5) Name as many different types of energy
as you can. Think of an example to go
with each.
What is energy? (copy definition)
 Energy: The measure of a system’s
ability to do work.
 http://www.youtube.com/watch?v=y
UpiV2I_IRI
 Minutes 13 onward
 19:30 onward
Types of Energy (Summary)
 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 (elastic potential energy 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=iL
XDirj4JUA&feature=related
 Start at 1min
Work and Energy Relationship
(Copy)
 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 - Copy
 Kinetic Energy = Ek = KE = ½ mv2
 Gravitational Potential Energy= Eg= PE= mgh
 Units: Joules (J) – same as work!
 NOTE: h = height of the object measured
from the reference level (measured in
metres). It can be d from our kinematics
equations as well as long as it is vertical
distance.
What is the effect of doing work on
an object? (Copy)
 You can give an object more energy
(any type of energy) by doing more
work on it.
 W = ΔE = change in energy (kinetic,
gravitational, elastic)
 W = Ef – Ei
Work- Energy Theorem
 W = ΔE
 The work done is equal to the change
in total energy of an object.
Example 1
 A 145g tennis ball is thrown
horizontally 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 – 0J
 ** It was thrown from rest so kinetic
energy initial is zero!!!
Ex 2: Work on a moving object
A 2kg object is moving at 10 m/s
when a force is applied to it
accelerating it to 20 m/s over a
distance of 5m horizontally. What is
the work done on the object?
Answer
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
Worksheet – Kinetic Energy
Check your understanding
 A 1500. kg car is travelling at a speed of
50.0km/h when the driver applies the brakes.
The car comes to a stop after travelling 15 m.
 a) How much work was done by friction in
stopping the car?
 b) What was the magnitude of the average
frictional force applied to the car?
 c) If the same car were travelling at 100. km/h
when the driver applied the brakes and the car
experienced the same frictional force, how far
would the car travel before coming to a stop?
For your own viewing
 http://www.youtube.com/results?sear
ch_query=potential+and+kinetic+en
ergy&oq=potential+and+kinetic+ener
gy&aq=f&aqi=g4&aql=&gs_sm=e&gs
_upl=50335l56318l0l56512l40l36l4l2
0l21l4l307l2755l1.1.8.2l12l0
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
E g  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
Worksheet – potential energy
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 (copy)
 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)