Transcript 7.3 Energy changes

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7.3 Energy Changes
7.3 Energy changes
Warm-up
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Jason is playing trampoline and he bounces
almost to the same height each time. Which
statement(s) seem(s) correct to you.

Trampoline gives him energy
to go up.
When he bounces up & down,
his KE is created & destroyed
continuously.

His PE is largest when he
reaches the top & his KE is
largest when he is just before
touching the trampoline.
7.3 Energy changes
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Warm-up
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When we slide down the chute from rest, our
speed may increase dramatically to 58 km h–1.
Which of the following is correct?

The increase in speed is
resulted from the work done
by gravity.
When we reach the bottom
of the chute, our PE is zero.
The increase in speed
depends on the slope of the
chute.
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Introduction
When energy changes from one form to
another, the amount of energy stays the
same.
Energy cannot
be created:
energy of an
explosion comes
from the chemical
energy in the
explosives.
7.3 Energy changes
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Introduction
Energy exists in many forms.
Energy cannot be created or destroyed.
Energy can change from one form to another.
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Conservation of energy
Energy changes in throwing a ball upwards.
chemical
energy
kinetic
energy
potential
energy
kinetic
energy
Principle of conservation of energy
sound &
internal
energy
Energy can be changed from one form to
another, but it can’t be created or destroyed.
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Conservation of energy
Simulation
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Example 6
KE and PE of a ball
A 0.1-kg ball falls from a height of 10 m
above the ground. (neglect air resistance)
(a)
Find KE, PE & total energy (E) of
the ball before it falls.
KE = 0
PE = mgh = 0.1  10  10
= 10 J
E = PE + KE = 0 + 10
= 10 J
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Example 6
KE and PE of a ball
A 0.1-kg ball falls from a height of 10 m
above the ground. (neglect air resistance)
(b) Find KE, PE & total energy (E) of
the ball when it falls 5 m.
Data:
PE = mgh = 0.1  10  5
E before it
=5J
falls = 10 J
E = KE + PE = 10 J (Conservation of energy)
 KE = E  PE = 10  5 J
=5J
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Example 6
KE and PE of a ball
A 0.1-kg ball falls from a height of 10 m
above the ground. (neglect air resistance)
(c)
Find KE, PE & total energy (E) of the
ball just before it hits the ground.
Data: E before it falls = 10 J
PE = 0
E = KE + PE = 10 J (Conservation of energy)
 KE = E  PE = 10  0 J
= 10 J
7.3 Energy changes
2 Example of conservation 11
of energy
a
Simple pendulum
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KE = 0
PE = mgh (max.)
E = KE + PE
Will the ball move
higher than h?
KE = 0
PE = mgh (max.)
E = KE + PE
PE = 0
KE = max.
E = KE + PE
h
h
7.3 Energy changes
2 Example of conservation 12
of energy
a
Simple pendulum
•
h
It will never move
higher than h if no
extra energy is added.
h
7.3 Energy changes
Interconversion of potential
energy and kinetic energy
Energy Conversion:
When an object falls down vertically it loses
P.E.. At the same time, the object is
accelerating and it gains K.E..
Loss in P.E. = Gain in K.E. (assuming no
energy loss)
P.E. loss = K.E. gained + energy loss
K.E. loss = P.E. gained + energy loss
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2 Example of conservation 14
of energy
Simulation
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Experiment 7a
Energy changes in a simple pendulum
Set up the following apparatus:
Video
7.3 Energy changes
Experiment 7a
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Energy changes in a simple pendulum
Pull the 0.5-kg mass to one
side, raising it to a height h
above the lowest position.
Start the ticker-tape timer
and release the mass.
h
7.3 Energy changes
Experiment 7a
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Energy changes in a simple pendulum
d
Measure the separation d of the 2 most widely
separated dots.
As ticker-tape timer puts 50 dots on the tape in 1 s,
time interval between 2 dots = 1/50 = 0.02 s
Max. speed of the mass as it passes the lowest
position,
d
v=
m s–1
0.02
7.3 Energy changes
Experiment 7a
Energy changes in a simple pendulum
Calculate the theoretical value of v as follows.
PE of the mass at the highest position = mgh
1
KE of the mass at the lowest position =
mv 2
2
1
mv 2 = mgh
2
v = 2gh
Compare the experimental value with the
theoretical value of v.
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Example 7
Speed of pendulum bob
The bob of a pendulum is pulled to one side
and raised through a vertical height of 0.2 m.
It is then released.
Find its speed at the
lowest position.
v
7.3 Energy changes
0.2 m
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Example 7
Speed of pendulum bob
At the highest position,
E = KE + PE = 0 + mgh
At the lowest position,
by conservation of energy:
E = KE + PE = mgh
1
mv2 + 0 = mgh
2
v
 v = 2gh = 2  10  0.2 = 2 m s1
7.3 Energy changes
0.2 m
b
Bungy jump
Bungy jump is a very
exciting game.
With an elastic string
attached to ankles,
participants jump
from a height of tens
of meters high.
Video
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b
Bungy jump
(i) when elastic
string starts to
stretch:
PE  elastic PE
+KE
(iii) in his rebound:
elastic PE  KE +PE
(ii) at the bottom:
PE  elastic PE
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c
Roller-coaster
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Does a roller-coaster
need an engine to
run?
Yes & No.
Engine pulls it up to
the top of the 1st hill.
But after that it rolls down the track on its own.
It moves because of interchange between PE & KE.
7.3 Energy changes
c
Roller-coaster
Simulation
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Example 8
Roller-coaster
A roller-coaster car is hauled to 50 m above
the ground & then released.
Total mass of car & passenger = 5000 kg
50 m
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Example 8
Roller-coaster
(a) Find its PE at the top. (Take PE at
ground level zero.)
PE = mgh
= 5000  10  50
mass m = 5000 kg
= 2 500 000 J
50 m
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Example 8
Roller-coaster
(b)
It travels 500 m to reach ground level.
Find the work done against friction in the
journey (average friction = 2750 N).
Work done against
friction:
W = Fs
= 2750 N  500 m
= 1 375 000 J
7.3 Energy changes
mass m = 5000 kg
50 m
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Example 8
Roller-coaster
(c)
Find the speed of it when it reaches the
bottom?
m = 5000 kg
E at the top = PE
= 2 500 000 J
Energy against friction
= 1 375 000 J
50 m
speed at the bottom = ?
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Example 8
Roller-coaster
By conservation of energy E = KE + PE +
energy against friction
m = 5000 kg
E at the top = PE
=2 500 000 J
At the bottom:
2 500 000 =
Energy against
friction =
1 375 000 J
1
mv 2 + 0 + 1 375 000
2
1
 v = 21.2 m s
speed = ?
7.3 Energy changes
3 Kinetic energy to
internal energy
Daily examples
A metal gets hot when it is hammered.
Brakes tyres & road get hot when
a car is braked to stop.
KE
change
to
internal energy
(& sound energy)
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Example 9
Temperature rise of a falling satellite
A disabled satellite falls back to the Earth. It
1
1
slows down from 900 m s to 100 m s on
passing through the atmosphere.
mass = 1000 kg
Assume the change in PE
of it is negligible compared
with the change in KE.
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Example 9
Temperature rise of a falling satellite
(a)
Data:
How much internal energy mass = 1000 kg
is produced as it falls?
900  100 m s1
Increase in
internal energy = loss in KE
of satellite
1
1
2
= mu 
mv2
2
2
1
 1000  (9002  1002)
=
2
8
= 4  10 J
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Example 9
Temperature rise of a falling satellite
(b)
The satellite has a net specific heat
1
1
capacity of 250 J kg C . Find its
temperature rise on passing through
the atmosphere.
Data:
mass = 1000 kg
900  100 m s1
increase in internal energy = 4  108 J
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Example 9
Temperature rise of a falling satellite
mass = 1000 kg
Increase in internal energy
= mcT
900  100 m s1
4  108 = 1000  250  T
Data:
increase in internal
energy = 4  108 J
 T = 1600 °C
net specific heat capacity
= 250 J kg1 C1
7.3 Energy changes
That’s why it
must be covered
with heat-resistant
materials.
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Q1 Balls X and Y slide down...
Balls X and Y slide down smooth inclined
planes from rest. If planes are of the same
height, which ball has a greater speed at P ?
A X
X
Y
B Y
C Both of them have the same
speed if they have the same mass.
P
D Both of them have the same speed even
if they do not have the same mass.
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Q2 A ball is thrown vertically...
A ball is thrown vertically upwards. Which
graph below shows the variation of the total
mechanical energy (E ) with the height (h)?
E
A
h
B
C
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h
Q2 A ball is thrown vertically...
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Because……
A
B
C
E
E
E
h
h
PE only!
PE = mgh
7.3 Energy changes
h
Total energy
is conserved.
h
KE only!
KE = E  PE
KE = E  mgh
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Q3 A 1-kg block slides down...
A 1-kg block slides down a smooth inclined
plane from rest at A. Find its KE & PE at B.
A
B
5m
3m
ground
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Q3 A 1-kg block slides down...
Take the ground as zero PE.
Total energy of the block at A,
E = PEA = mgh = ____
1 × 10 × ____
5 = ______
50 J
A
PE of the block at B (PEB)
30 J
= mgh = ______
B
5m
By the conservation of energy,
KE of the block at B
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= PEA – PEB = _______
J
7.3 Energy changes
1-kg block
3m
ground
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The End
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Return
7.3 Energy changes