Transcript Uniform Motion

Science 10 Physics
Read pg. 465 - 477
Unit B
Conversions Review
How to Multiply Fractions
•
3
1
x
4
5
=
12
5
• 3 x
7 =
8
21
8
• 3 x
5
9
15
9
=
Unit Conversions (Distance)
• E.g. 3.75km  m
1 km = 1000 m
3.75 km x 1000 m =
3750 m
1 km
• Try: 5.85km  m
1 km = 1000 m
5.85 km x 1000 m = 5850 m
1 km
• E.g. 427cm  m
100 cm = 1 m
427cm x 1 m = 4.27 m
100cm
• Try: 865 cm  m
100 cm = 1 m
865 cm x 1m = 8.65 cm
100cm
• E.g. 67 mm  m
1000 mm = 1 m
67 mm x 1 m = 0.067 m
1000 mm
• Try: 765 mm  m
1000 mm = 1 m
765 mm x 1m = 0.765 m
1000 mm
• E.g. 580 m  km
1000 m = 1 km
580m x 1 km =
1000 m
0.580 km
Try Unit Conversions (Time)
• E.g. 2.75 h  min
1 hr = 60 min
2.75 h x 60 min = 165 min
1h
• Try:
42 min  h
42 min x
1h
60 min
= 0.70 h
• E.g. 2.10 h  s
1 hr = 3600 s
2.10 h x 3600 s = 7560 s
1h
•Try:
3hs
3h x
3600s
1h
= 10800 s
• Units to know for this unit:
•
•
•
•
•
•
•
•
Distance, height = meters (m)
Time = seconds (s)
Speed, velocity = meters per second
(m/s)
Acceleration = meters per second2
(m/s2)
Work, energy = Joules (J)
Force = Newtons (N)
Mass = kilograms (kg)
Efficiency = percent (%)
• Formulas to know for this unit:
v=d
t
vave = vi + vf
2
a = vf - vi
t
F = ma
W = Fd
Ep = mgh
Ek = 1/2 mv2
% Efficiency = useful output x 100%
total input
Rearranging formulas
• You need to ISOLATE the variable you are
trying to solve for
• What ever mathematical operation you do to
one side of the = you need to also do to the
other side
• Ex. v = d
Solve for d and solve for t
t
Drawing Graphs
Parts of a Graph
• All graphs should have:
– A horizontal axis (or x axis)
– A vertical axis (or y axis)
– A title (Y vs X)
– Labels on each axis
– Units for each axis
– Appropriate scale (numbering on both axis)
Example:
Distance-Time Graph
Label
(Units)
Scale
Title
*Note: In
Physics, time
will always be
the horizontal
axis
Label
(Units)
Read pg.126-131
Energy
• Causes changes in the motion to occur
to an object
• It can speed objects up, slow them down
or change their direction
Uniform Motion
• Describes a type of movement
• It occurs when an object travels in a
straight line at a constant speed
• is difficult to maintain so ….we use
AVERAGE SPEED
v = d
t
 = change
d = distance in m or km
t = time in s or h
v = speed in m/s or km/h
Average Speed = distance traveled
time
v = d
t
v = dfinal – dinitial
tfinal – tinitial
v = speed (m/s or km/h)
d = change in distance (m or km)
t = change in time (s or h)
Example 1
A baseball travels________
200 m in ________
1.50
seconds. What is the average speed
of the baseball?
Δd
v =
d
v
t
= 200 m
=?
= 1.50 s
Δt
v = 200m
1.50 s
v = 133 m/s
Try:
a.)
A baseball travels 20.0 m in 1.50
seconds. Calculate the average
speed.
d = 20.0 m
t = 1.50 s
v=?
v =
Δd
Δt
v = 20.0 m
1.50 s
v = 13.3 m/s
b.)
If Lance Armstrong bikes 200.0 m in 10.0 s,
what is the cyclist’s average speed?
v=
d = 200.0 m
t = 10.0 s
v=?
v=
v=
Δd
Δt
200.0 m
10.0 s
20.0 m/s
c.) If a train traveled 100 km in 0.500 hours
what is its speed in km/h and in m/s?
d = 100 km
t = 0.500 h
v=?
v=
v=
v=
200 km x 1000 m/km
1 h x 60 min/h x 60 sec/min
=
200 km/h
3.6
Δd
Δt
100 km
0.500 h
200 km/h
= 55.6 m/s
Example 2
A car travels________
1.00 km at a constant speed of
________
15 m/s. What time is required to cover this
distance?
v =
d
v
t
= 1.00 km
= 1000 m
= 15 m/s
=?
Δd
Δt
t=d
v
t = 1000m
15 m/s
t = 67 s
Try:
a.)
How long would it take a car to travel
4000 m if its speed was 40.0 m/s?
d = 4000 m
v = 40.0 m/s
t=?
Δd
v =
Δt
t=d
v
= 4000 m
40.0 m/s
= 100 s
b.)
How long would it take a car to travel
2000 m if its speed was 10.0 m/s?
d = 2000 m
v = 10.0 m/s
t=?
Δd
v =
Δt
t=d
v
= 2000 m
10.0 m/s
= 200 s
c.)
How long would it take a car to travel
8000 m if its speed was 30.0 m/s?
d = 8000 m
v = 30.0 m/s
t=?
Δd
v =
Δt
t=d
v
= 8000 m
30.0 m/s
= 267 s
Example 3
A motorist travels ________
406 km in 4________
hours and 15
minutes. What is the average speed in km/h
and m/s?
Δd
d = 406 km
v=
Δt
v = 406 km
4.25 h
v = 95.5 km/h
= /3.6
v = 26.5 m/s
v= ?
t = 4 hour
+ 15min
60 min/h
= 4.25 h
Example 4. How far of a distance will a car
cover if it travels 2.00 m/s for 1.00 min?
d=?
v = 2.00 m/s
t = 1.00 min
= 60.0 s
Δd
v =
Δt
d=vt
=2.00 x 60.0
= 120 m
Distance Time Graphs
• distance varies directly with time when
speed is constant
• Have the following components:
– time is the (x-axis)
– distance is the (y-axis)
– the slope of the line is the speed of an
object
• speed describes the rate of motion an
object has
t(s)
0
1.0
2.0
3.0
4.0
5.0
d(m)
0
20
40
60
80
100
Distance-Time Graph
d
t
0
1
2
3
time (s)
4
5
• The steepness of the graph is the slope
slope
= rise
run
= y2 – y1
x2 – x1
• Example:
slope = y2 – y1
x2 – x1
= 80m – 20m
4.0 s- 1.0 s
= 20 m/s
• The steeper the slope the higher the speed.
A
Distance (m)
B
C
time (s)
Which line shows the greatest speed? The slowest speed?
Try the Following:
• Make a Distance
time graph for the
following
• Calculate the speed
of the boat
v=
Δd
Δt
=
30 m – 10 m
6.0 s – 2.0 s
V = 5.0 m/s
Speed- Time Graphs
•The area under the graph is the distance an
object travels
•The slope of the line gives you information
about the speed
•E.g.
A slope of zero (flat line) = uniform motion
Upward slope = speed is increasing
Downward slope = speed is decreasing
6
5
4
3
1
2
Uniform Motion
0
Speed
v (m/s)
5.00
5.00
5.00
5.00
5.00
5.00
Speed (m/s)
Time
t (s)
0.0
2.0
4.0
6.0
8.0
10.0
0
2
4
6
8
Time (s)
10
• Can be used to determine the distance an
object travels…… calculate the area under
the line
Example 1
Calculate the area under the following
speed-time graph up to 10.0 s.
Speed (m/s)
10.0
5.0
0.0
5.0
Time (s)
10.0
Speed (m/s)
Solution
Calculate the area under the following
speed-time graph up to 10.0 s.
10.0
area =   w
A = (10.0 s)(5.0 m/s)
A = 50.0 m
5.0
0.0
10.0
Time (s)
Speed (m/s)
Example 2
Calculate distance travelled by an object in
20.0s.
10.0
0.0
20.0
Time (s)
Speed (m/s)
Solution
Calculate distance travelled by an object in
20.0s.
10.0
A= 1 b x h
2
A= 1 20 x 10
2
0.0
20.0
Time (s)
A = 100 m
Try the Following
Calculate distance travelled by an object in
40.0s.
Speed (m/s)
30.0
20.0
10.0
0.0
20.0
Time (s)
40.0
Solution
Calculate distance travelled by an object in
40.0s.
Speed (m/s)
30.0
20.0
10.0
0.0
20.0
Time (s)
40.0
A= 1 b x h
2
A= 1 40 x 25
2
A = 500 m
Vectors Verses Scalar
• Scalar quantities:
– involve only magnitude (amount)
E.g.) Speed = 20 m/s
• Vector quantities:
– involve both magnitude and direction
– Are drawn using arrows
E.g.) Displacement = 20 m [N]
Velocity = 20 m/s [N]
Distance Verses Displacement
• Distance (d)
– is how far an object travels
– It is a scalar quantity (magnitude)
• Displacement (d)
– Is change in both distance and direction
– It is a vector quantity (magnitude and
direction)
Speed verses Velocity
•Speed (v):
•is the rate of an objects motion
•It is a scalar quantity
•The formula is:
v= d
t
v = speed
 d = distance (dfinal – dinitial)
 t = time
• Velocity (v )
• Describes the rate of motion and the
direction of the object’s motion
• It is a vector quantity
• the formula is:
vave = d
t
vave = dfinal – dinitial
tfinal - tinitial
• when describing a vector, we have two
quantities that indicate the direction:
-Degrees from the x or y axis
-Degrees using compass directions
(N, S, E, or W)
Cartesian Method
up (90°)
left
(180°)
E.g. 6 m [30°]
E.g. 10 m [right]
2 m 40
°
6m
30
°
8m
down
(270°)
10
m
right
(0°)
Navigator Method
(+)
N
(+)
(+)
It uses N-S-EW .. the angle
is relative to E
(east) or W
(west), or a
direction at N,
S, E or W
W
(-)
angle (°)
angle (°)
N of W
N of E
angle (°)
angle (°)
S of W
S of E
(-)
(+)
E
(-)
S
(-)
N
+ y axis
W
 x axis 65
1.0 m/s
B
80
10 km
A
E
+ x axis
S
 y axis
vector A: 10 km, 80 E of the y axis
10 km, 80 E of S
vector B: 1.0 m/s, 26 S of the x axis
1.0 m/s, 26 S of W
+ y axis
N
C
15°N of W
 x axis W
55°W of S
15º
E + x axis
30º
55º
B
A
S
S
 y axis
30° S of E
Vectors Direction
• Vectors in the Same Direction:
– To find the distance:
Add them together
– To find the displacement:
 Add them together and include the
direction
10 m
5m
= 15 m
Example1
A person runs 25 m south and then
another 15 m south.
(a) What is the distance travelled?
(b) What is the displacement?
Solution
A person runs 25 m south and then
another 15 m south.
(a) What is the distance travelled?
40 m
25 m
15 m
(b) What is the displacement?
40 m South
• Vectors in Opposite Directions:
– To find the distance: Add them together
– To find the displacement: take the difference
between the two
numbers and
include the
direction
20 m
Distance = 25 m
5m
Displacement = 15 m E
Example 2
A plane flies 200 km north and then turns
around and comes back 150 km.
a) What is the distance travelled?
b) What is the displacement?
Solution
A plane flies 200 km north and then turns
around and comes back 150 km.
a) What is the distance travelled?
350 km
200km
b) What is the displacement?
50 km N
150 km
50 km
Example 3
A student throws a boomerang north. It
travels 35.0 m before it turns around and
travels 33.0 m back to him. If the total flight
of the boomerang took 5.00s, determine the
following:
a) distance
b) displacement
c) speed
d) velocity
Solution
A student throws a boomerang north. It
travels 35.0 m before it turns around and
travels 33.0 m back to him. If the total flight
of the boomerang took 5.00s, determine the
following:
a) distance d = 35.0 m + 33.0 m =
68.0 m
b) displacement
d = 35.0 m [N] - 33.0 m [S] =
c) speed
v = d/t
= 68.0 m/5.00 s
= 13.6 m/s
2.0 m [N]
d) velocity
v = d/t
= 2.0 m [N]/5.00 s
= 0.40 m/s [N]
Example 4
A plane flies south to Edmonton International Airport,
which is 465 km from the Fort McMurray Airport. If the
flight takes 50.0 minutes what is the average velocity
of the plane in km/h and m/s?
vave = d
t
= 465 km [S] – 0 km
0.8333… h
= 558 km/h [S]
t = 50.0 min
60min/h
= 0.8333…. h
558 km/h x 1000 = 155 m/s [S]
3600
Example 5
A train travels at 12.0 m/s [E] for 15.0 minutes.
What is the displacement of the train?
= d
t
12.0 m/s [E] = d
900 s
= 10800 m [E]
= 10.8 km [E]
vave
t = 15.0 min x 60 s/min
= 900 s
•When we graph to demonstrate velocity
we use a position time graph
Position- Time Graph
time t(s)
0.0
2.0
4.0
6.0
8.0
10.0
Position
d(m) [E]
0.0
10.0
20.0
30.0
40.0
50.0
0
2
4
6
8
time (s)
10
Try the Following
According to the data below,
what is the velocity of the car?
Solution
calculating the average velocity:



dfinal – dinitial
Δd
rise
slope =
=
=
tfinal – tinitial
Δt
run
=
40.0 m – 10.0 m
8.0 s – 2.0 s
= +5.0 m/s
= 5.0 m/s [E]
Try the Following
Time interval
(s)
0.0 to 2.0 s
2.0 to 4.0 s
4.0 to 6.0 s
6.0 to 8.0 s
8.0 to 10.0 s
Plot the following Data.
What type of motion is this?
Solution
Uniform Motion!
Acceleration
• Is a change in velocity (speeding up or
slowing down)
• The unites = m/s2
•Positive (+) Acceleration=
velocity
•Negative(-) Acceleration =
velocity
(deceleration)
Positive Acceleration
• Positive (+) acceleration occurs two ways:
1. If direction is positive (+) and velocity is
increasing
direction
increasing velocity
+
direction
2) If direction is negative (-) and velocity is
decreasing
decreasing velocity
+
direction
direction
Position - Time Graphs
Time t (s)
•Positive acceleration the slope is increasing
Velocity – Time
Time (s)
• positive acceleration the slope is increasing
•the slope gives the acceleration
Negative Acceleration
• Negative” (-) acceleration occurs in two ways:
1. If direction is positive (+) and velocity is
decreasing
direction
decreasing velocity
+
direction
2) If direction is negative (-) and velocity is
increasing
increasing velocity
+
direction
direction
Time t (s)
• negative
acceleration because the slope
is decreasing
Time (s)
•Negative acceleration since the slope is
decreasing
•The area under gives the distance traveled.
http://videos.howstuffworks.com/hsw/9610-physics-of-motion-acceleration-anddeceleration-video.htm
Uniform Motion
Accelerated Motion
d-t Graph
Distance
(m, km,
etc)
d-t Graph
Distance
(m, km,
etc)
Time
(s, h, etc)
slope = speed
Time
(s, h, etc)
slope of tangent = instantaneous speed
Uniform Motion
Accelerated Motion
v-t Graph
v-t Graph
Velocity
(m/s or
km/h)
Velocity
(m/s or
km/h)
Time
(s, h, etc)
area =distance
Time
(s, h, etc)
slope = acceleration
area = distance
Uniform Motion
Accelerated Motion
no a-t Graph
a-t Graph
Acceleration
(m/s2)
Time
(s, h, etc)
area = velocity
acceleration = change in velocity
change in time
a = v
t
where:
a = vf – vi
t
t = change in time in s or h
v = velocity in m/s or km/h
vi = initial velocity in m/s or km/h
vf = final velocity in m/s or km/h
a = acceleration in m/s2
Example 1
Rudy falls out of an airplane and after 8.0 s he is
travelling at 78.48 m/s. What is his acceleration?
t = 8.0 s
vf = 78.48 m/s
vi = 0 m/s
a = vf - vi
t
= 78.48 m/s – 0 m/s
8.0 s
= 9.8 m/s2
Example 2
The initial speed of a bicycle is 8.0 m/s and it is
moving for 6.0 s. If the final speed is 10.0 m/s,
what is the acceleration of the bicycle?
vi = 8.0 m/s
t = 6.0 s
vf = 10.0 m/s
a=?
a = vf – vi
t
a = 10.0 m/s – 8.0 m/s
6.0 s
a = 0.33 m/s2
Try the Following
What is the acceleration of a car if its speed
is increased uniformly from 40 m/s to 70 m/s
in 3.0 s?
Solution
What is the acceleration of a car if its speed
is increased uniformly from 40 m/s to 70 m/s
in 3.0 s?
vf – vi
t
a = (70m/s – 40m/s)
3.0s
a = 10 m/s2
a=
Example 3
Mike is traveling down Franklin Ave at 50 km/h.
He sees Joslin standing at the bus stop and hits
the brakes so he can pick her up. How long will it
take for him to come to a stop if his acceleration
is –5.0 m/s2?
vf – vi
t
-5.0 m/s2 = (0m/s – 13.8888 m/s)
t
t = - 13.888888 m/s
-5.0m/s2
t = 2.8 s
a=
vi = 50km/h x 1000
3600
= 13.88888 m/s
vf = 0 m/s
t=?
a = -5.0 m/s2
Try the Following
Jack is traveling down Thickwood at 10 km/h. He
sees Amber walking at the side of the road and
hits the brakes so he can pick her up. How long
will it take for him to come to a stop if his
acceleration is –2.0 m/s2?
Solution
Jack is traveling down Thickwood at 10 km/h. He
sees Amber walking at the side of the road and
hits the brakes so he can pick her up. How long
will it take for him to come to a stop if his
acceleration is –2.0 m/s2?
vf – vi
t
-2.0 m/s2 = (0m/s – 2.7777 m/s)
t
t = - 2.7777 m/s
-2.0m/s2
t = 1.38 s
a=
vi = 10km/h x 1000
3600
= 2.7777 m/s
vf = 0 m/s
t=?
a = -2.0 m/s2
Try the Following
Dan is driving down the highway and comes up behind
a logging truck traveling 22 m/s. He hits the brakes,
and accelerates uniformly at a rate of - 2.5 m/s2 for 5.0
seconds until he reaches the same speed as the truck.
What was his initial speed in m/s and km/h?
Solution
Dan is driving down the highway and comes up behind
a logging truck traveling 22 m/s. He hits the brakes,
and accelerates uniformly at a rate of - 2.5 m/s2 for 5.0
seconds until he reaches the same speed as the truck.
What was his initial speed in m/s and km/h?
a = vf – vi
t
-2.5 m/s2 = (22 m/s – x)
5.0 s
(-2.5 m/s2)(5.0 s) = 22 m/s - x
-12.5 m/s = 22 m/s - x
x = 22m/s + 12.5 m/s
x = 34.5 m/s
vi= 35 m/s
vi = ? m/s
vf = 22 m/s
t = 5.0 s
a = -2.5 m/s2
34.5m/s x 3600
1000
= 124 km/h = 1.2 x102 km/h
Acceleration Due to Gravity
• Is 9.8 m/s2… (applies to all objects)
• It is greater near sea level
• It is less on the top of a
mountain
• Larger masses have more
• There is always drag/air resistance…
ignore it
When Doing Calculations:
• if an object is falling:
a = 9.81 m/s2
• if an object is going up:
a = 9.81 m/s2
Example 1
How fast are you falling after 2.5 s of free fall.
Remember a = 9.81 m/s2
vi = 0 m/s
a = 9.81 m/s2
t = 2.5 s
vf = ?
vf = vi + at
vf = 0 m/s + (9.81 m/s2)(2.5 s)
vf = 24.525 m/s
vf = 25 m/s
Calculating average speed/velocity for
constant acceleration
vave = vi + vf
2
vave= average speed
vi = initial speed
vf= final speed
Example 1
A train traveling through the Rocky Mountains,
enters the Kicking Horse traveling at 35 m/s.
When it reaches the top of the pass 65 minutes
later it has slowed down to 15 m/s. What is the
average speed of the train?
vave = vi + vf
2
= 35 m/s + 15 m/s
2
Vave = 25 m/s
Vave
Example 2
A car traveling is travelling up Thickwood at 40
m/s. When it reaches the top of the hill 3 minutes
later it has slowed down to 10 m/s. What is the
average speed of the car?
vave = vi + vf
2
= 40 m/s + 10 m/s
2
Vave = 25 m/s
Vave
Try
A car traveling is travelling up highway 63 at 80
m/s. When it reaches the Fort McMurray 30
minutes later it has slowed down to 6 m/s. What
is the average speed of the car?
Solution
A car traveling is travelling up highway 63 at 80
m/s. When it reaches the Fort McMurray 30
minutes later it has slowed down to 6 m/s. What
is the average speed of the car?
vave = vi + vf
2
= 80 m/s + 6 m/s
2
Vave = 43 m/s
Vave
Work and Energy
Force
• Is any push or pull on an object
• It is measured in Newtons (N)
• Objects remain at rest unless unbalanced
forces act upon it
Balanced & Unbalanced Forces
• Balanced force:
– forces are the same size
but in the opposite direction
– cancel each other out.
• Unbalanced force
– Forces are in the opposite
direction
– one force is larger than the
other
• Deceleration (slowing down)
• the force is in the opposite
direction of the movement
• energy is transferred from the
source of the force to the object that
the force is acting upon
• Accelerating (speeding up)
• the force is in the same
direction as the moving
object
To Change the Motion of Objects
• A force is needed
F = ma
F = force (Kg • m/s2) or 1 Newton (N)
m = mass (kg)
a = acceleration (m/s2)
• Note: weight is the force due to gravity ( 9.81 m/s2)
Example1
A 1000 kg car is accelerated at 2.5
m/s2. What is the force acting on it?
Solution
Example1
A 1000 kg car is accelerated at 2.5
m/s2. What is the force acting on it?
F = ma
F = (1000kg)(2.5 m/s2 )
F = 2500 N
F = 2.5 x 103 N
Try the Following
A 500 kg car is accelerated at 4.5
m/s2. What is the force acting on it?
Solution
A 500 kg car is accelerated at 4.5
m/s2. What is the force acting on it?
F = ma
F = (500kg)(4.5 m/s2 )
F = 2250 N
F = 2.2 x 103 N
Example 2
What is the mass of a crate with a weight
of 450 N?
F = ma
450 N = m 9.81 m/s2
m = 450 N/9.81 m/s2
m = 45.9 kg
F = 450 N
m=?
a = 9.81 m/s2
Example 3
What force is needed to accelerate a 500
kg car from rest to 20 m/s in 5.0 s?
F = ma
F = (500 kg)(4.0 m/s2 )
F = 2000 N
F = 2.0 x 103 N
F=?N
m = 500 kg
a=?
a = vf – vi
t
a = (20m/s - 0 m/s)
5.0 s
a = 4.0 m/s2
Work
• Occurs when a person lifts a weight, shovels
snow or pushes a car
• Occurs when a force acts through a distance
• Work input (energy input) = the work done
on the object
• work output (energy output) = the result
of this work done on the object
• in the absence of any outside forces, such
as friction
• WORK INPUT = WORK OUTPUT
• For work to be done on an object…….
THREE CONDITIONS must apply
3 Conditions for Work
1. There must be
movement
2. There must be a
force
3. The force and the
distance the object
travels must be in
the same direction
Is this work?
Is this work?
Formula:
for calculating work done
W = Fd
Note: since F = ma then
W = mad
W = work (N•m) or J (Joule)
d = distance (m)
m = mass (kg)
a = acceleration (m/s2)
Example 1
A force of 800 N is needed to push a car
across a lot. If the car moves 20 m, how
much work is done?
W = Fd
W = (800 N)(20 m)
W = 16000 J
= 16 kJ
Example 2
How much work is done in lifting a 60 kg
crate a vertical distance of 10 m?
W = mad
W = (60kg)(9.81 m/s2)(10m)
W = 5886 J
W = 5.9 kJ
W=?J
m = 60 kg
d = 10 m
a = 9.81 m/s2
Energy
• Is the ability to do work
• When a body does work on an object it
loses energy while the object gains it is
∆E=W
∆ E = change in energy (J)
W = work (J)
Example 1
How much energy is gained by a 750 kg
weight that is raised to the roof of a 3
story building if each story is 3.2 m?
Solution
How much energy is gained by a 750 kg
weight that is raised to the roof of a 3
story building if each story is 3.20 m?
W = mad
W = (750 kg)(9.81m/s2)(9.60 m)
W = 70632 J
∆E=W
∆ E = 70.6 kJ
Try the Following
How much energy is gained by a 500 kg
weight that is raised to the roof of a 1
story building if each story is 5.50 m?
Solution
How much energy is gained by a 500 kg
weight that is raised to the roof of a 1
story building if each story is 5.50 m?
W = mad
W = (500 kg)(9.81m/s2)(5.50 m)
W = 26977.5 J
∆E=W
∆ E = 27.0 kJ
2.1
Forms of
Energy
Forms of Energy
1. Chemical energy
2. Electrical energy
3. Magnetic energy
4. Nuclear Energy
5. Solar Energy
6. Thermal Energy
7. Potential Energy
8. Kinetic Energy
9. Mechanical Energy
1.Chemical Energy
• is the potential energy stored in the chemical
bonds of compounds.
•
• it is released when chemicals react
• E.g. Energy stored in the bonds of glucose that
the mitochondria release to do work
2. Electrical Energy
• is the work done by
moving electrons
• can be converted to heat
and light
• E. g. light bulb
3. Magnetism
• Hans Oersted (1820) proved that electricity can
produce magnetism
• Michael Faraday (1831) proved that a magnet
can cause electrical current to flow through a
wire
4. Nuclear Energy
• is a type of energy stored in the nucleus of an
atom
• when the nucleus of an atom is split (fission), or
when the nuclei of two atoms combine (fusion)
this energy is released
5. Solar Energy
• Is energy released through fusion
• it reaches the Earth as electromagnetic
radiation
• it is converted into other forms of energy like
electrical energy
6. Thermal Energy
• Is heat energy due to the movement of atoms
• there is more of it ….the faster the particles
move
• the more of it there is…… the hotter the object
2.2
Potential
Energy
7. Potential Energy
•Occurs when an object has stored
energy because of its position
•There is no motion
• Types:
A.
gravitational EP
B.
elastic EP
C.
chemical EP
D.
electrical EP
E.
nuclear EP
A. Gravitational Potential Energy
•The energy stored due to the height of an
object
•Is the energy (Work) required to lift an object to a
certain height
•Falling will release the stored “potential”
energy
Weight verses Mass
• Weight
– is a measure of the Earth’s gravitational
pull on an object
– it is a vector quantity
– measured in N (force)
• Mass
– is the amount of matter there is
– it is a scalar measurement
– measured in kg
Ep(grav) = Work
W = Fd
Ep = max
h
Ep = 0
F = mg
Ep(grav) = mgh
EP = mgh
where:
m = mass in kg
g = acceleration due to gravity
= 9.81 m/s2
h = height in m
EP = gravitational potential
energy in J
Example 1
What is the gravitational potential energy of
a 63 kg base jumper on the edge of a cliff
450 m high?
m = 63 kg
a = 9.81 m/s2
h = 450 m
EP = ?
EP = mgh
= (63 kg)(9.81 m/s2)(450 m)
= 278113.5 J
= 2.8 x 105 J
Example 2
How much work is done to lift a 5.0 kg
hammer from the floor to a height of 2.0 m?
m = 5.0 kg
a = 9.81 m/s2
h = 2.0 m
W = EP = ?
W = EP = mgh
= (5.0 kg)(9.81 m/s2)(2.0 m)
= 98.1 J
= 98 J
Example 3
A student has a weight of 600 N. Calculate
his potential energy if he climbs a ladder to
a height of 3.5 m.
F = 600 N
h = 3.5 m
EP = ?
EP = Fh
= (600 N)(3.5 m)
= 2100 J
= 2.1  103 J
Example 4
A crane uses 8.0 kJ of energy to lift a crate
onto the roof of a building. If the crate has
a mass of 125 kg, how high is the building?
m = 125 kg
g = 9.81 m/s2
EP = 8.0 kJ
= 8000 J
h=?
EP = mgh
8000 J = (125 kg)( 9.81 m/s2)h
h = 6.523…m
= 6.5 m
B. Elastic Potential Energy
• stretching an elastic out has stored
potential energy (Ep(elas) )
• converted to gravitational potential energy
when the elastic is released
Ep(elas) = Ep(grav)
2.3
Kinetic Energy
and Motion
8. Kinetic Energy
• Energy of a moving object
• It depends on both mass and velocity
• When potential energy is released it is
converted to kinetic energy
• Types:
1.
mechanical EK
2.
thermal EK
3.
sound EK
4.
electrical EK
EK = ½ mv2
where:m = mass in kg
v = speed in m/s
EK = kinetic energy in J
Example 1
A 8.0 kg mass moves at 20 m/s.
What is its kinetic energy?
Ek = ½ mv2
= (8.0 kg)(20m/s)2
2
= 1600 J
= 1.6 kJ
Ek = ? J
m = 8.0 kg
v = 20 m/s
Example 2
A
0.00m
10.0m
B
20.0m
30.0m
A.) A ball rolls from position A to position B in 15.0 s
What is the distance travelled by the ball in 15.0s?
B.) What is the average speed of the ball?
c.) What is the kinetic energy of the 2.00 kg ball?
A
B
0.00m
10.0m
20.0m
30.0m
A ball rolls from position A to position B in 15.0 s
• What is the distance travelled by the ball in 15.0s?
25.0 m
2. What is the average speed of the ball?
v = d/t
v= 25.0 m/15.0s
v = 1.67 m/s
3. What is the kinetic energy of the 2.00 kg ball?
Ek = ½ mv2
Ek = ½ (2.00 kg) (1.67m/s)2
Ek = 2.78 J
Example 3
An electron with a mass of 9.11 x 10-31 kg moves at
a speed of 2.5 x 108 m/s. Find the electron’s kinetic
energy?
Ek = 1/2 mv2
= (9.11 x 10-31 kg) (2.5 x 108 m/s)2
2
= 2.8 x10-14 J
Ek = ? J
m = 9.11 x 10-31 kg
v = 2.5 x 108 m/s
Example 4
A bullet travelling at 3.0 x102 m/s has 2.0 kJ
of energy. What is the mass of the bullet?
Ek = ½ mv2
2000 J = (m) (3.0 x 102 m/s)2
2
4000 J = m 90000
4000 = m
90000
m = 0.0444 kg
m = 44 g
Ek = 2.0 kJ
= 2000 J
m = ? kg
v = 3.0 x 102 m/s
Example 5
A moving rocket of mass 2.00 x 102 kg has a
kinetic energy of 1.0 x 103 J. What is the rocket’s
speed?
Ek = 1/2 mv2
1.0 x 103 J = (2.00 x 102 kg) (v)2
2
(2)(1.0 x 103 J) = (2.00 x 102 kg) (v)2
2.0 x103 J
= v2
2.00 x102 kg
v = 10
v = 3.2 m/s
Ek = 1.0 x 103 J
m = 2.00 x 102 kg
v = ? m/s
2.4
Mechanical
Energy
9. Mechanical Energy
•Is energy due to the motion and the
position of an object
•Mechanical Energy = Kinetic Energy +
Potential Energy
Em = Ek + Ep
Em = ½ mv2 + mgh
Example 1
A 0.300 kg baseball is thrown in a straight line
through the air. At a height of 2.50 m above the
ground, it has a speed of 20.0 m/s. What is the total
mechanical energy of the baseball?
m = 0.300 kg
v = 20.0 m/s
h = 2.50 m
g = 9.81 m/s2
Em = ½ mv2 + mgh
Em = ½ (0.300 kg)(20.0 m/s)2 + (0.300 kg)(9.81 m/s2)(2.50m)
= 60.0 J + 7.36 J
= 67.4 J
Pendulum Energy
•
A moving pendulum is at maximum kinetic
energy at the bottom:
EK = highest (fastest)
EP = lowest
• At maximum height on either side,
maximum potential energy is achieved:
EK = 0
EP = maximum
because h = 0
Example 1
A 600 gram bob on a pendulum is lifted 50 cm from the
bottom of the swing.
(a) How much work is done to raise the pendulum?
(b) What is the max speed the pendulum reaches?
W = Ep = mgh
W = (0.600kg) (9.81 m/s2) (0.50 m)
W = 2.943J
Ek = Ep
Ek = ½ mv2
2.943 J = ½ (0.600 kg) v2
v = 3.1 m/s
50 cm
2.5
Energy
Conversions
Law of Conservation of Energy
• Energy cannot be created or
destroyed, but can be converted
from one form to another.
• The total amount of energy never
changes. (input = output)
• The amounts of potential energy a
can be converted to kinetic and
vice versa
Ep = Ek
Energy Conversions
• Potential energy is not useful until it is converted
into another form of energy E.g. Kinetic
Free Fall Example
Start
EP = max
EK = 0
***all EP at the top is
converted into EK at the
bottom
EP = decreasing
EK = increasing
End
EP = 0
EK = max
Pendulum
***all EK at the bottom is
converted back into EP at
the top
EP = max
EK = 0
EP = max
EK = 0
EP = EK
EP = EK
EP = 0
EK = max
***all EP at the top is
converted into EK at the
bottom
Example 1
A cliff diver has a mass of 75 kg. He is
standing 30 m above the surface of the water.
(a) What is his potential energy at the top of the cliff?
(b) What is his kinetic energy as he hits the water?
(c) What speed is he traveling when he hits the water?
(a) Ep = mgh
= (75 kg)(9.81 m/s2)(30 m)
= 22072.5 J
= 22 kJ
(b) Ek = Ep
Ek = 22 kJ
(c) Ek = ½ mv2
(87 km/h) 22072.5 J = ½ (75 kg)v2
v = 24 m/s
Example 2
A 70.0 g bullet is shot out of a rifle straight up
into the air with 2.00 kJ of energy.
(a) What was the initial velocity of the bullet?
(b) If all the kinetic energy is converted into potential
energy, how high did the bullet go?
(a) Ek = ½ mv2
2000J = ½ (0.0700 kg) v2
v = 239 m/s
(b) Ep = mgh
2000J = (0.0700 kg)(9.81 m/s2) h
h = 2.91 x 103 m
or 2.91 km
Evidence of Energy Conversions
1. MOTION - energy from one sources causes
something to move
2. CHANGE IN POSITION - whenever something
is raised above the surface of Earth, there is
evidence of gravitational potential energy
3. CHANGE IN SHAPE - ex. an archer’s bow, an
elastic band
4. CHANGE IN TEMPERATURE - heat is the
transfer of kinetic energy
Energy Conversions in Technological
Systems
• energy conversions take place in many
systems:
– Natural Systems
– Technological Systems
– Nuclear Systems
– Solar Cells
– Fuel Cells
1. Natural Systems
• the fusion of hydrogen atoms on the sun
releases solar energy ……it travels to the
Earth as electromagnetic radiation
• chlorophyll in plants converts solar energy
into chemical energy in the chemical bonds
of glucose…. (during photosynthesis)
• plants and animals release chemical
energy stored in glucose during cellular
respiration
2. Technological Systems
• a hydroelectric power station converts the
gravitational potential energy of water into
electrical energy
EP(grav) water → EK water → EK turbines → Eelec
• a coal burning power station converts chemical
potential energy stored in coal into electrical
energy
EP(chem) coal → heat water to steam → EK turbines → Eelec
3. Nuclear Systems
• after WWII nuclear energy was used to
generate electricity
• CANDU (Canadian Deuterium
Uranium) reactors cause
uranium to disintegrate during
nuclear fission releasing energy
as radiation
EP(nuclear) uranium → heat water to steam → EK turbines → Eelec
4. Solar Cells
• convert solar energy into electricity
• are composed of two layers of silicon…
one layer has phosphorus added and the
other layer has boron
How they work
•
•
•
•
The sun’s rays hit the layers
electrons are released
the two layers become charged
electrons move carrying the electric
current
• the electricity can be used directly or stored in
batteries for later use
Fuel Cells
• operate like a battery
• they convert chemical energy (in hydrogen) into
electrical energy
• They do not require recharging
• Produce are water and heat
• Are used in spacecraft, buses in Vancouver
3.1
Laws of Thermodynamics
System
• is a set of interconnected parts
• It can be:
– Closed
– Open
– Isolated
Closed System
• cannot exchange matter but can
exchange energy with its surroundings
• E.g. A sealed container
Open System
• exchanges both matter and energy with
its surroundings
• E.g. An open container
Isolated System
• cannot exchange either matter or
energy with its surroundings
• E.g. sealed vault
Thermodynamics
• Is the study of the relationship between
heat, work and energy
• It helps describe how energy behaves in
systems
• There are two laws that govern how
energy behaves
1st Law of Thermodynamics
• Energy can not be created or destroyed
it can only change form
Energy input = energy output
• Means that the total energy in the system
remains constant
2nd Law of Thermodynamics
• During every energy conversion some
energy is lost as heat
• Occurs because of the friction between
moving parts
The Perfect Machine
• It would have no friction between
moving parts… therefore no loss of
energy
• it would have perpetual motion
• all the input energy would be converted to
mechanical energy…
Engines
• Turn thermal energy into mechanical
energy… (allowing work to be done)
• operate on the principle that heat flows
from a hot substance to a colder one
• Types:
– Heat engine
– Heat pumps
HEAT ENGINE
•turns heat into mechanical energy
•Heat is always lost
HEAT PUMPS
• uses mechanical
energy to pump heat
from a cooler space
into a warmer space
•E.g. refrigerator
Refrigerant is pumped
through copper piping to
absorb the heat
from inside
Gunpowder Engine
• Inventor: Huygens
• Technology: explosion
from lit gunpowder
drives a piston to do
work on an object
• Problem: piston does
not come back, and
you could blow up!
Papin’s Heat Engine
• Inventor: Papin
• Technology: hot steam
drives a piston up,
cool water drives the
piston down; work is
done
• Problem: Papin couldn’t develop it; the
casing also eventually broke down!
Savery’s Water Pump
• Inventor: Savery
• Technology: a Papin
engine is used to
pump water out of a
mineshaft
• Problem: for deep
mines or taller
heights, boiler is so
hot, it could explode!
Newcomen Engine
Newcomen Engine
• Inventor: Newcomen
• Technology: a “see-saw” type pump uses a
Papin engine to pump water out of much
deeper mineshafts
• Problem: lots of heat was lost!
Watt Engine
•
Inventor: Watt
•
Technology: a see-saw pump and wheel uses
a Papin engine to run ships, trains, and mills;
much more efficient heating and cooling
•
Problem: really big, and still needed a lot of
fuel and water to operate properly!
Watt Engine
Internal Combustion
• Inventor: Lebon (and Steele)
• Technology: a spark ignites coal gas, with
the resulting explosion driving a piston
(inspired by Huygen’s explosive idea)
• Problem: good idea, but not enough force
in either gentlemen’s designs!
Internal Combustion
•
Inventor: Otto and Lebon
•
Technology: Four-stroke engine: air-gas
mixture is first compressed by a piston; spark
ignites it; high temperature and pressure push
the piston’s “power stroke”
•
Problem: good idea, but coal gas was used,
which is not very powerful!
Efficiency
• Is the measurement of how efficiently a
machine converts energy input to energy
output
• Is reduced by friction
• Remember:
– machines have moving parts
– as the parts rub together friction is
produced ……. causing energy loss in the
form of heat
Magnetic Technology
• allows us to minimize contact between moving
parts, thereby creating more efficient machines
• no system is 100 % efficient!
• efficiency is the measurement of the amount of
useful work output from the energy conversion process
% efficiency = useful output  100
total input
Example 1
A system requires a work (energy) input of
125 J and it gives 25.8 J of useful work
output. Calculate the efficiency of the
system.
Solution
A system requires a work (energy) input of
125 J and it gives 25.8 J of useful work
output. Calculate the efficiency of the
system.
% efficiency = useful output  100
total input
= 25.8 J  100
125 J
= 20.64 %
= 20.6 %
Example 2
An engine with an efficiency of 80.0 % is used
to do 55 kJ of useful work. Calculate the
amount of energy that must be supplied by the
engine to do the work.
Solution
An engine with an efficiency of 80.0 % is used
to do 55 kJ of useful work. Calculate the
amount of energy that must be supplied by the
engine to do the work.
% efficiency = useful output  100
total input
80.0 % = 55 kJ  100
x
x = 55 kJ  100
80.0 %
= 68.75 kJ
= 69 kJ
Example 3
A crane does 6.0  104J of work to lift a 200
kg crate to the top of a 30 m building.
Calculate the efficiency of the crane.
Solution
A crane uses 6.0  104J of work to lift a 200
kg crate to the top of a 30 m building.
Calculate the efficiency of the crane.
input = 6.0  104 J
m = 200 kg
h = 30 m
W = EP = mgh
g = 9.81 m/s2
= (200 kg)(9.81 m/s2)(30 m)
= 58860 J
% efficiency = useful output  100
total input
= 58860 J  100
6.0  104 J
= 98.1 %
= 98 %
3.4
Energy Applications
Energy Applications
• All primary energy sources are classified
into two main categories:
(a) solar
(b) non-solar
1. Solar Energy Sources
• Are those that are derived
directly or indirectly from
the sun
• Solar radiation- is energy
emitted by the hydrogen
fusion reaction
•Wind energy – is the result of the uneven
heating of the surface of
the Earth
•Water Energy- the result of the heating of the
surface water by the sun
•Biomass- is the result of chemical
bonds found in organic matter
E.g. wood, wastes
•Fossil Fuels- all were formed from
plants and animals that
have decayed millions of
years ago
2. Non-Solar Energy Sources
• energy sources having no relationship to
the sun
Nuclear energy – uses fission of
uranium to create
energy
Geothermal energy –is heat energy that is
generated in the earth
Tidal Energy- is the result of
gravitational pull of the
moon
Renewable Verses Non-renewable Energy
• Renewable:
– Are continuously available
– E.g. solar energy
• Non-renewable:
– There is a limited supply
– E.g. Fossil fuels
Factor Affecting Energy Demands
1. The amount of energy
people use
2. The population of the
world is growing
3. Many societies are using
renewable energy instead
of non-renewable
Ways to Conserve Energy
1. Reduce waste usage
2. Use more renewable sources of energy
3. Increase our amount of cogeneration:
using waste to generate more energy
We must come up with a sustainable
solution!