Transcript Physics I Unit 9 work--energy - power - Simple - science-b

Work, Energy & Power
Chapter 10
1
Lesson 1 Feb 10
Specific Instructional Objectives
At the end of the lesson, students should be able to:
– Show understanding of the Physics concept of Work
– Correctly identify Work from given situations
– Recall and show understanding of the formula to
calculate work done
– Solve related problems involving work
–DO NOW:
– If you push on a 5.00 kg block with 20.0 N of force, how
fast (frictionless) will it be moving after being pushed
2.00 meters?
2
Physics concept of WORK
• WORK is done only when a constant
force applied on an object, causes the
object to move in the same direction as
the force applied.
3
Physics concept of WORK
•
What IS considered as work done in Physics:
– You push a heavy shopping trolley for 10 m
– You lift your school bags upwards by 1 m
•
What is NOT considered as work done:
– You push against a wall
– Jumping continuously on the same spot
– Holding a chair and walking around the classroom
4
Lesson 1 Feb 10
Physics concept of WORK
Homework:
Page 261 – 262 #’s 1-8 all
page 278 #’s 54, 61,63,65
WORK can be calculated by:
W =F x d
Units: [J]
[N]
[m]
SI Unit for Work is JOULE (J)
5
Lesson 1 Feb 10
Examples of WORK
• You are helping to push Romac’s heavy
shopping cart with a force of 50.0 N for 200.0 m.
What is amount of work done?
Work done,
W= F x d
= 50.0 x 200.0
= 10,000 J
or
10.0 kJ (kilo-Joules)
6
Lesson 1 Feb 10
Examples of WORK:
• Thomas put on his bag-pack of weight 120 N.
He then starts running on level ground for 100 m
before he started to climb up a ladder up a
height of 10 m. How much work was done?
From Physics point of view, no work is done on pack at
level ground. Reason: Lift is perpendicular to movement.
Work is done on pack only when Thomas climbs up the ladder.
Work done, W = F x s
= 120 x 10
= 1200 J or 1.2 kJ
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Lesson 2 Feb 13
Mechanical Advantage
Objectives
Lesson 2 MA Objectives:
–
–
–
–
Show understanding of the Physics concept of Mechanical Advantage
Show understanding of the Physics concept of IDEAL Mechanical Advantage
Correctly identify MA and IMA from given situations
Recall and show understanding of the formula to calculate Efficiency of a System
– Do NOW: If you push a 10.0 Kg object from rest along a
frictionless surface with a force of 2.0 N what is the work
done on the object in the first 5.00 seconds?
ANSWER DUE @ 10:27 SHARP
– Lesson 2 HOMEWORK:
– Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88)
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Lesson 2
What is a Simple
Machine?
• A simple machine has few or no
moving parts.
• Simple machines make “work”
easier
• Do NOW: If you Lift a 10.0Kg
mass 1.0 meter off the floor how
much work have you done?
Lesson 2 HOMEWORK:
Glencoe PAGE:280-281: #’s 79, 80,
81, 82, 85, 86, (87 – 88)
Glencoe Page 273 #29, 30, 32
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Write this DOWN
Apply the concept of mechanical advantage to
Lesson 2
everyday situations.
• Conservation of Energy
– Work in = Work out
• F in * d in = F out * d out
•
Ideal Mechanical Advantage {IMA}is the RATIO of the
–
Displacement exerted (in) to the Displacement load (out).
•
IMA = d in / d out
•
Mechanical Advantage {MA} is the RATIO of the
–
Force exerted (in) to the Force load (out).
•
MA = F out / F in
•
Efficiency (%) is the RATIO of the (MA) / (IMA) * 100
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How do they work?
• Conservation of Energy
Work and Energy are related
• Scientific Definition of Work
Work = Force x Distance
distance in direction of force
Lever
Incline Plane
Wedge
Screw
Wheel and Axle
Gears
Pulley
Change Force Direction
Same Force Size
Lesson 2
Practice
• If you lift a 10.0Kg mass 1.0 meter off the floor with a
lever. The Fulcrum is .50 meters from the mass and
3.50 meters from the other end. If it takes a 2.0 Kg
mass to balance the lever,
• What is the IMA of the Lever?
– IMA = Dist in / Dist out = 3.5 / 0.5 = 7:1
• What is the MA of the Lever?
– MA = Force out / Force in
– 10.0*9.8 / 2.0 *9.8
= 98 / 19.6
= 5:1
• What is the efficiency of the Lever?
•
Efficiency = MA / IMA * 100 = 5 / 7 *(100) = 71.4286 %
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Lesson 2 – A FEB 16
PE  KE
Grade Homework
• Glencoe Page 271 Example 4
– Lesson 2 HOMEWORK:
– Glencoe PAGE:280-281: #’s 79, 80, 81, 82, 85, 86, (87 – 88)
– Glencoe Page 272 #’s 24 – 28 Homework
IN CLASS
Page 273 #’s 29, 30, 32
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LESSON 3
KINETIC ENERGY Feb 16th
DO NOW: What is the acceleration of a 4,500 Kg
auto as it goes from 29.5 m/s to zero in 10.0
meters?
At the end of the lesson, students should be able to:
–
–
–
–
–
Understand the Physics concept of Kinetic Energy (KE) = ½ m v2
Recall and show understanding of the KE formula
Distinguish situations involving KE
Demonstrate knowledge of the Work Energy Theorem
Work = ΔKE
Lesson 3 HOMEWORK:
PAGE:287: #’s 1 – 3
Page : 291: #’s 4 – 8
Page: 297 #’s 15 – 18
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LESSON 3
Energy – Quick Re-cap
• Energy is the capacity to do work
• SI Unit: Joule (J)
• Many forms
• Common ones:
–
–
–
–
–
–
Kinetic
Potential
Electric
Chemical
Solar
Nuclear
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LESSON 3
Kinetic Energy (KE)
• Formula: KE = ½ mv2
• The amount of KE of a moving body depends on:
– Mass of body (kg)
– Velocity (m/s)
– When the Mass doubles : KE Doubles
– When the Velocity doubles: KE QUADRUPLES
SI Unit: Joule [ J ] … same as Work Done
Work = Δ KE = ½ m (Vf – Vi)2
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LESSON 3
Examples of KE
• Find the KE of an empty van of mass 1000kg moving at 2m/s.
KE of van at 2m/s = ½ x 1000 x (2)2
= 2000 J = 2 kJ
• Find the KE of van when it is loaded with goods to give a total
mass of 2000kg, and moving at 2m/s.
KE of van at 2m/s = ½ x 2000 x (2)2
= 4000 J = 4 kJ
• Find KE of unloaded van when it speeds up to 4m/s.
KE of van at 2m/s = ½ x 1000 x (4)2
= 8000 J = 8 kJ
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LESSON 3
Examples of KE
• A motorcycle accelerates at 2m/s2 from rest for
5s. Find the KE of motorcycle after 5s. Mass of
motorcycle is 200 kg.
Velocity of motorcycle after 5s,
a = (v-u)
t
v = 2(5) + 0 = 10m/s
KE of motorcycle at 10m/s = ½ x 200 x (10)2
= 10,000 J = 10 kJ
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Lesson 4 Feb 17
Potential Energy
DO NOW:
What is the work done on a 10.0 Kg book lifted to a table 1.5 meters
from the floor?
If Potential Energy is defined as m*g*h what is the Potential Energy
of that same 10.0 Kg book when it has been raised to the table?
At the end of the lesson, you should be able to:
– Show understanding of the Physics concept of Gravitational
Potential Energy
– Recall and understand the formula PE = mgh
– Distinguish situations involving GPE
– Solve related problems involving GPE
HOMEWORK: Page: 308 #’s 64 – 68 all
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Lesson 4 Feb 17
Potential Energy
• Potential energy is the energy possessed by an object
as a result of its POSITION or CONDITION.
• Two common kinds:
– Gravitational PE GPE
– Elastic PE (not in syllabus)
• In Physics, ground level is normally assumed to be at
ZERO GPE.
• Any object that is at ground level has ZERO GPE.
• If object is lifted a certain height above ground, its GPE
has increased.
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Lesson 4 Feb 17
Gravitational PE
• Can be calculated with:
GPE = mass  gravitational  height above
acceleration
=
m  g  h
g
earth
Units:
[J]
[kg]
[m/s2]
[m]
SI Units of GPE : Joule [J]
ground level
Object on top of
building, of mass, m
Distance from
ground, h
Ground,
0 GPE
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Lesson 4 Feb 17
Example of GPE
• You lifted your bags to the top of your table.
What can you say about the GPE of your bag?
– Zero, increase, decrease
• Lift the same bag on the Moon. What happens to
GPE?
– Zero, increase, decrease
• Will the GPE be the same on Earth and Moon?
– Same, less on Moon, more on Moon?
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Lesson 4 Feb 17
Examples of GPE
• You lifted a set of books of mass 3kg, for 2m. What
is the GPE gained by the books? Take g=10m/s2.
GPE = mgh
= 3  10  2
= 60 J
• Find the work done by you to lift the books.
Work done, W = F  d
= (m  g)  d
= 3 x 10 x 2
= 60 J
(F = weight of books)
(Note: same as GPE)
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EP – TL
ZB – PL
TP – BZ
WK - WS
BV - AL
CM - HH
Do Now:
Lesson 5 Feb 20
Conservation of
Energy
What is the PE of a 5.0 Kg mass raised 4.0 meters above the ground?
If all that energy is converted to Kinetic Energy after it fall 4.0 meters
what will the Velocity be when it hits the ground?
Use TWO Methods:
mgh = ½ mv2
Vf2 = Vi2 + 2ad
At the end of the lesson, students should be able to:
 Show understanding of conservation & conversion of energy
 Correctly distinguish situation involving energy conservation & conversion
 Solve related problems
Energy of an object can be thought of as the sands in an hourglass!
Note that energy CANNOT be created nor destroyed!
Energy always remain same or fixed in quantity!
But this sand can change position, from the top to bottom and bottom to
top! Likewise energy can change in form eg. From KE  PE
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Lesson 5 Feb 20
Conservation & Conversion
of Energy
http://hyperphysics.phyastr.gsu.edu/hbase/flobj.html#c1
http://hyperphysics.phyastr.gsu.edu/hbase/hframe.html
HOMEWORK LESSON 5
Page: 308 – 309
#’s 71, 73, 78, 80
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Lesson 6 Feb 21
More Conservation of Energy TE = PE + KE – Work {friction}
DO NOW:
A) Turn in last nights homework
B) What is the velocity of a 2000.0 Kg roller coaster
at the bottom of a 42.0 m hill?
C) What is the velocity ½ way down the hill?
HOMEWORK: Page 309 #’s 81 – 89 ODD
TEST


Conversion of energy is the term used to denote
change in energy from one form to another.
Eg.
 Burning candle: Chemical  Heat, Light
 Fuel: Chemical  Heat  KE  Electricity
 Nuclear explosion: Nuclear  Heat, light
 Spring: Elastic PE  KE
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Lesson 6 Feb 21
More of Conservation of Energy


A fresh Coconut of mass 5 kg is found growing at the end
of a tree branch 20 m above ground. When ripe, the
Coconut will by itself drops to the ground below.
Let gravity = 10m/s2.
Find the energy of the fresh coconut? What form is it?


GPE. GPE = mgh = 5 x 10 x 20 = 1000J
Find the GPE and KE of the coconut when it is 5m above
ground. Sum up both the GPE and KE and compare the
value with above. What can you infer from the results?



GPE = 5 x 10 x 5 = 250J.
s = ½ vt, v = gt
s = ½ gt2, t = √ 3 KE = ½
mv2 = ½ (5)(10 √ 3)2 = 750J
v = 10(√ 3)
Sum of energies = 250 + 750 = 1000J
Same as above => energy is conserved.
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Lesson 6 Feb 21
More Conversion of Energy

A car of 800 kg is moving at an average speed of 5 m/s. The
traffic light changed to red and so the driver stepped on the
brakes to bring the car to a quick, sudden and screeching
halt.
 Find
energy of moving car and what form of energy
is this?
 KE. KE = ½ mv2 = ½ x 800 x 52 = 10,000 J.
 What
energy does the car possesses when it stops?
 None.
 What
happened to the original energy of the
moving car?
 KE has changed to Sound and Heat Energy.
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Roller Coasters
- an ENERGY MACHINE
http://en.wikipedia.org/wiki/Euthanasia_Coaster
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Lesson 7 Feb 22
Work – Energy Review
 Work

W = F*d * COSƟ
 Simple


IMA MA EFF
Types
 PE

Machines:
and KE
mgh ½ mv2
 Conservation

TE = mgh + ½ mv2
Practice

38
Evan pulls Romac’s newly finished Summer Exploration Dory
along the Maritime Museum docks with a 4.5 meter long rope. The
rope forms a 45 angle from the horizontal. If he pulls the boat
with a force of 220.0 N for 15.5 m in 60.00 seconds;




What is the work he does on the boat? =_____________ ____
How much energy did he exert?
=__________________ ____
How much power did he generate? =__________________ ____


4 points:__________



What does he do work against?
=________________________________
Where did the energy go?
=__________________________________
Practice
39
Terry and Andy (both have a mass of 90.0 Kg) run up the Union
to Academic wing stairs (height 4.20 m) to get to the College
office to "talk" to Ms. Coleman. Terry, being slightly older takes
3.5 seconds and Andy, being oh so much younger, beats him in
3.2 seconds. Actually he elbowed him, but anyway…
Courtney has a mass of 55.00 Kg.
Who does the most work?
=________________________________
Who produces more power?
=________________________________
How much power do they each generate?
=___________________________
4 points:__________
How fast would Courtney have to run up the stairs to match
Andy’s power? =________________________________
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Practice
4 points:__________
Thomas and Zach are in a small rowboat with a
9.9 HP (horsepower) outboard motor. Thomas
convinces Zach that the boat (boat mass of 225
Kg) can move the boat through the "watah"
(check out that Mainer Accent!!) at a constant 5
knots (0.5128 m/s). The engine must produce
75% of its' 10 HP to maintain the applied force
of 10.906 X 103 N to keep it at this steady
speed.
What is the power that the engine is actually
producing? =__________________
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Practice
The tallest roller coaster in the world is the “Steel Dragon” in Japan.
The cars are raised from the loading platform to the top of the
initial hill using a series of hydraulic and pneumatic lifters.
It takes 35.00 seconds to reach the top of the hill.
The first hill has a vertical drop of 93.5 meters and
has an initial angle of 60 degrees from the base of the ride.
The length of the first hill is 107.965 meters.
The initial velocity of the cars at the top of the hill is 0.00 m/s.
The mass of the fully loaded six cars is 2.825 X 103 Kg.
As soon as the cars reach the bottom of the first hill,
they enter a vertical loop that has more of an “egg shape”
Rather than a perfect circle.
The top and bottom of the loop have a radius of 18.50 meters
with the top at 43.50 meters from the base of the ride.
Half way up the loop the radius is 22.50 m
and the height of the cars is 20.00 meters from the base of the ride.
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Practice
What is the total energy available for the six cars at the top of the
hill?
How much work was done to get the cars to the top of the hill?
What was the power expended to get the cars to the top of the
ride?
What is the velocity of the cars at the bottom of the first hill
assuming no friction ?
Now what is the velocity of the car at the top of the loop (No
Friction)?
How many NET “g’s” do the riders feel at the top of the
loop?________________
If the actual velocity at the bottom of the hill is 41.00 m/s what is the
force of friction on the cars while going down the first hill?