Transcript Slide 1

P2.2.1 Forces and Energy
P2 Physics
Mr D Powell
Connection
•
•
•
Connect your learning to the
content of the lesson
Share the process by which the
learning will actually take place
Explore the outcomes of the
learning, emphasising why this will
be beneficial for the learner
Demonstration
• Use formative feedback – Assessment for
Learning
• Vary the groupings within the classroom
for the purpose of learning – individual;
pair; group/team; friendship; teacher
selected; single sex; mixed sex
• Offer different ways for the students to
demonstrate their understanding
• Allow the students to “show off” their
learning
Activation
Consolidation
• Construct problem-solving
challenges for the students
• Use a multi-sensory approach – VAK
• Promote a language of learning to
enable the students to talk about
their progress or obstacles to it
• Learning as an active process, so the
students aren’t passive receptors
• Structure active reflection on the lesson
content and the process of learning
• Seek transfer between “subjects”
• Review the learning from this lesson and
preview the learning for the next
• Promote ways in which the students will
remember
• A “news broadcast” approach to learning
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Index
P2.2.1 Forces and Energy
a)
When a force causes an object to move
through a distance work is done.
b) Work done, force and distance, are
related by the equation: W = F x d
c)
Energy is transferred when work is done.
(Candidates should be able to discuss
the transfer of KE i.e... shuttle re-entry
or meteorites burning up in the)
W=Fxd
P = E/t
d) Work done against frictional forces.
e) Power is the work done or energy
transferred in a given time. P = E/t
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Formulae help...
W=Fxd
P = E/t
Ep = mgh
Ek = ½ mv2
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When I move these objects I do what...
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a/b) What is work done?
C/D
TASK: Summarise this idea
then answer the
questions
1. When I move an apple (1N weight) a distance of 2m along a desk
what is the work done?
2. When I move 0.5kg a distance of 3m along a desk what is the
work done?
3. When a weight lifter lifts a weight of 80kg a height of 1.5m what
is the work done? (mgh)
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1. 2J
2. 15J
3. 1200J
Index
a/b) More on W.D.
TASK: Explaining / Devising
B/C
1. Using the formulae for reference can you explain
what happens to the W.D. If you have a bigger
mass or larger distance to move your object?
2. Can you think of a formulae which could relate
Power to Work Done. Write it out and explain how
it works....
3. Write out your own example of a “worked
problem” for another situation
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Friction at work
TASK: Explain several examples
where friction effects your life.
C
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Plenary...
1.
2.
3.
4.
96J
96J
90 & 4500J
WD= mgh or WD/mg= h = 270J/450N = 0.6m
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Power

This is a very easy thing to look at when we
consider work done.

Before we have looked at how energy is
transferred via work done (see example below).

If you want to think about the power you
simply divide the energy transferred by how
long it to transfer...
P = E/t
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Multichoice
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Summary Questions II
WD = Fd = 20N x 1.5m = 30Nm = 30J
WD = Fd = 600N x 100m = 60,000Nm = 60,000J = 60kJ
WD = Fd = 3000N x 30m = 90,000Nm = 90,000J = 90kJ
E = Pt so E / t = P = 60,000J / 20s = 3000J/s = 3000W
E = Pt so E / t = P = 90,000J / 20s = 4500J/s = 4500W
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P2.2.1 Forces and Energy
Ep = mgh
e) Gravitational potential energy is the
energy that an object has by virtue of
its position in a gravitational field.
Ep = mgh
f) The kinetic energy of an object depends
on its mass and its speed.
Ek = ½ mv2
Ek = ½ mv2
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What is happening...
What happens here....
Where has the energy gone?
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Pendulum...
PE  max
KE  0
PE  max
KE  0
PE  0
KE  max
1 2
KE  m v
2
PE  m gh
KE  PE
1. Label the
diagram with
the energy
changes...
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Where does it come from?
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Interactive
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Kinetic Energy of a horse..
A horse (with 3 legs) of mass 175kg
walks at a velocity of 3ms-1 its
kinetic energy is found as;
1 2
KE  m v
2
1
1 2
KE   175kg  3m s
2
1
KE   175kg  9m 2 s  2
2
KE  87.5kg  9m 2 s  2


KE  787.5kgm2 s  2
1 2
KE  m v
KE

787
.
5
J
2
1
Quick Version..... KE   175 32
2
KE  787.5 J
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Another worked example?
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Sorting....
A/B
2250 kgm2s-2 or
2250J
TASK: Firstly work in a pair and guess sort these
vehicles then work out the missing KE values and
then sort them into which has the largest and
which has the smallest?
1000 kgm2s-2 or
1000J
187.5 kgm2s-2
or 187.5J
100 kgm2s-2 or
100J
NB: take care with your units for KE
2000 kgm2s-2 or
2000J
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Practical Examination of Kinetic Energy
The kinetic energy of an object
is the energy it has due to its
motion. It depends on its mass
and its speed We can
investigate how the kinetic
energy of a ball depends on its
speed.

The ball is released on a slope from a measured height above the foot of the
slope. We can calculate the gravitational potential energy it loses from its
weight and its drop of height. The kinetic energy it gains is equal to its loss of
gravitational potential energy.

The ball is timed, using light gates, over a measured distance between X and
Y after the slope.
Q) How do light gates improve the quality of the data you can collect in this
investigation?
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Practical Examination of Kinetic Energy (HT)
Here are some results for a 0.5kg
ball which has been dropped.
Can you work out the missing
values if the formulas are;
EGPE = mgh
EKE = ½mv2
d/t = v
∆h (m)
0.05
0.1
0.16
0.20
PE of ball (J)
0.25
0.72
0.80
1.00
∆t to travel 1m from x -> y
0.98
0.71
0.57
0.50
Velocity ms-1
1.02
1.75
2.00
A/B
1.41
TASK: Complete the table by doing 4 calculations. There two
ways of thinking about it!
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Multichoice
PE = mgh
KE = ½mv2
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w = mg
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Homework Questions...
1. Work out the kinetic energy of a fighter aircraft with
a mass of 195kg travelling at a speed of 666ms-1.
2. What is the significance of the speed of the fighter
aircraft?
TASK: You have these 6
questions on a
worksheet to fill in
and then check
next lesson.
3. The designers of the Euro fighter Jet want to
improve the acceleration of the plane. How can they
achieve this without using larger engines? Justify you
answer?
orking
4. If the mass of the plane doubles by what factor does
the kinetic energy increase for the same speed?
nswer
5. If the velocity of the plane increases by a factor of x3
by what factor does the kinetic energy increase.
nit
6. A passenger plane burns up 70kg of fuel which
contains 1050J of energy to get to a height of 700m.
Taking account that the plane is 70% efficient, how
much potential energy does the plane now have at
this height?
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Answers....
1 2
KE  m v
2
1) See right.......
2
1
KE  195kg  666m s1
2
2) twice the speed of sound; mach 2
1
2 2
KE


195
kg

443556
m
s
3) F=ma, reduce the mass. Since the
2
acceleration will be increased therefore
KE  97.5kg  443556m 2 s  2
the force is reduced to accelerate the

plane. Same argument also applies to
streamlining
4) x2

KE  43246710kgm2 s  2
KE  43246710J
KE  43.2 MJ
5) x9 or 32
orking
6) 1050J x 0.7 or (1050J x 70%) /100%
gives; 735J. The 30% of wasted energy is
lost in heat to the surroundings.
nswer
nit
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W.D. In context
TASK:
1.
Logically reason out
what is happening in
the diagram.
2.
Draw out your own
copy and label what is
going on.
3.
Compare what is
happening here to the
formulae you have
already covered.
(Hint: W = mg)
Ep = mgh
A-B
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Formulae Warm Up......
You have 2 minutes to rearrange and sort out these formulae
2
1)
2)
Simplify?
Make the square the subject?
2
3)
Find the triangle?
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Formulae......
You have 2 minutes to rearrange and sort out this formulae
1 2
KE  mv
2
to get make the following the subject...
v=
m=
be prepared to come up to the board and demonstrate how
you got there!
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Back to waterfalls...

We can also do something clever when
thinking of this formula when looking at
Kinetic Energy as well.
EGPE = mgh
EKE = 0.5mv2
(eq1)
(eq2)
 If I think of a transfer involving the energy at
the top of a waterfall (GPE) and energy at
the bottom (mainly KE)
 I can work out the speed of the water falling
by equating the two equations and
assuming all the energy is transferred to
kinetic.
 In every case we find a value of “g” gravity
as 10 m/s2
1 2
m gh  m v
2
2m gh  m v2
2 gh  v 2
2 gh  v
20h  v
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What is the speed of the water...
 Using this idea can you apply the formula and
fill in the grid for water which drops from
different heights.
 You will need to use the Sqrt button on your
calculator...
h (m)
V (in m/s)
5
10.00
1 2
m gh  m v
2
2m gh  m v2
10
14.14
2 gh  v 2
20
20.00
2 gh  v
100
44.7
20h  v
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Physics of Space Battles - Advanced

We think of the EM spectrum from Year 10
as having more energy as  (wavelength)
is shorter. (think UV -> Gamma)

The actual reason for this is because the
energy for each photon of light is found
from E=hf.
 (h is a constant, f = frequency )

But what about simply using a matter
thrower.

Energy of particles is found from 0.5mv2

Just throw rocks fast enough and they
have lots of energy!
Task: Now you have watch the
movie make a comment in
your own words (you can use
my help) to explain what you
have seen. What do you
think? (1 paragraph)
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Plenary Questions
1.
Copy down the equation for Kinetic
Energy in your book.
2.
Use the formulae to explain why an
object with more mass would have more
KE. (explain in detail)
3.
Use the formulae to explain why an
object with double the velocity would
have four times the KE. (explain in detail
4.
What is the main point of the movie you
saw in relation to “space physics”
5.
Calculate the kinetic energy of a ball of
mass 0.5kg thrown at a speed of 5ms-1
Task: Do these questions /
activities on your own
without help from anyone!
1 2
KE  mv
2
WD  Fd
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Extension Quick Questions
PE = mgh
KE = ½mv2
w = mg
KE = ½mv2 = 0.5 x 80 x 102 = 4000J
KE = ½mv2 = 0.5 x 500 x 302 = 225kJ
KE = ½mv2 = 0.5 x 10,000 x 2002 = 9MJ
PE = mgh = 10m x 10 N/kg x 100kg = 10,000 Nm = 10,000J = 10KJ
PE = mgh = 3000m x 10 N/kg x 60kg = 1,800,000 Nm = 1.8MJ
PE = mgh = 50m x 10 N/kg x 0.100kg = 50 Nm = 50J
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Multichoice...
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P2.2.1 Forces and Energy
P2.2.1 Forces and Energy
P2.2.1 Forces and Energy
a)
When a force causes an object to
move through a distance work is
done.
a)
When a force causes an object to
move through a distance work is
done.
a)
When a force causes an object to
move through a distance work is
done.
b)
Work done, force and distance,
are related by the equation: W = F
xd
b)
Work done, force and distance,
are related by the equation: W = F
xd
b)
Work done, force and distance,
are related by the equation: W = F
xd
c)
Energy is transferred when work is
done. (Candidates should be able
to discuss the transfer of KE i.e...
shuttle re-entry or meteorites
burning up in the)
c)
Energy is transferred when work is
done. (Candidates should be able
to discuss the transfer of KE i.e...
shuttle re-entry or meteorites
burning up in the)
c)
Energy is transferred when work is
done. (Candidates should be able
to discuss the transfer of KE i.e...
shuttle re-entry or meteorites
burning up in the)
d)
Work done against frictional
forces.
d)
Work done against frictional
forces.
d)
Work done against frictional
forces.
e)
Power is the work done or energy
transferred in a given time. P =
E/t
e)
Power is the work done or energy
transferred in a given time. P =
E/t
e)
Power is the work done or energy
transferred in a given time. P =
E/t
f)
Gravitational potential energy is
the energy that an object has by
virtue of its position in a
gravitational field.
Ep = mgh
f)
Gravitational potential energy is
the energy that an object has by
virtue of its position in a
gravitational field.
Ep = mgh
f)
Gravitational potential energy is
the energy that an object has by
virtue of its position in a
gravitational field.
Ep = mgh
a)
The kinetic energy of an object
depends on its mass and its
speed.
a)
The kinetic energy of an object
depends on its mass and its
speed.
a)
The kinetic energy of an object
depends on its mass and its
speed.
Ek = ½ mv2
Ek = ½ mv2
Ek = ½ mv2
Formulae help...
W=Fxd
P = E/t
Ep = mgh
Ek = ½ mv2
Formulae help...
W=Fxd
P = E/t
Ep = mgh
Ek = ½ mv2
Work Done Revision
Work Done Revision
KE & PE Revision
KE & PE Revision