Ground Rules
PC1221 Fundamentals of
Physics I
„
Lectures 13 and 14
„
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Energy and Energy Transfer
„
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Dr Tay Seng Chuan
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Switch off your handphone and pager
Switch off your laptop computer and keep it
No talking while lecture is going on
No gossiping while the lecture is going on
Raise your hand if you have question to ask
Be on time for lecture
Be on time to come back from the recess break to
continue the lecture
Bring your lecturenotes to lecture
1
1.
If vectors A and B (collinear) pointing in the same
direction are added, the resultant has a magnitude equal to
4.0. If B is subtracted from A, the resultant has a magnitude
equal to 8.0. What is the magnitude of A?
A.
2.0
B.
3.0
C.
4.0
D.
5.0
E.
6.0
Answer:
A and B are parallel.
|A| + |B| = 4 …(1)
|A| - |B| = 8 …(2)
(1) + (2)
2|A| = 12, so |A|= 6.
2
2. The site from which an airplane takes off is the origin. The
x-axis points east; the y-axis points straight up. The
position and velocity vectors of the plane at a later time
are given by
ˆ+ 9.14× 103 ˆ
ˆm
r = (1.61× 106 i
j)m and v = +224i
.
s
The plane is most likely
a. just touching down.
b. in level flight in the air.
c. ascending.
d. descending.
e. taking off.
Answer:
(If A and B are opposite, |A| can be |-2|, and
|B| can be |6|.)
3
The y value is not equal to 0 meaning the
airplane is above the ground. The airplane is
cruising toward east at an altitude of about 9
km.
4
3. If P = 6.0 N, what is the magnitude of the force exerted
on block 1 by block 2?
3
A. 6.4 N
B. 5.6 N
2
1
C. 4.8 N
6
5.0 kg
D. 7.2 N
P
3.0 kg
2.0 kg
F2 F2
E. 8.4 N
F1 F1
Answer:
Let F1 be the force in between block 1 and block 2, and F2 be the
force in between Block 2 and Block 3. Let a be the acceleration.
6 - F1 = 2a …(1)
F1 - F2= 3a …(2)
F2 = 5a
…(3)
Substitute (3) in (2), F1-5a= 3a, i.e.,
F1= 8a
…(4)
Substitute (4) in (1), 6 - 8a=2a, i.e., 6 = 10a,
so a= 0.6 m/s2.
Using (1), F1 = 6 - 2a = 6 - 2x0.6 = 6 - 1.2 = 4.8N
4. A roller-coaster car has a mass of 500 kg when fully
loaded with passengers. The car passes over a hill of
radius 15 m, as shown. At the top of the hill, the car has a
speed of 8.0 m/s. What is the force of the track on the car
at the top of the hill?
8.0 m/s
A. 7.0 kN up
R
B. 7.0 kN down
C. 2.8 kN down
mg
D. 2.8 kN up
15 m
E. 5.6 kN down
Answer:
2
mg – R = m v /r, so
R = mg - m v2/r
2
= 500x9.8 – 500 x 8 /15 = 2766.66 N
= 2.8 kN up
5
6
Ground Rules
SM2 Students are to attend a
seminar this Thursday at 2pm,
LT32. Attendance is
compulsory.
„
„
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„
Class monitor is to take
attendance at 1:40pm.
„
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7
Switch off your handphone and pager
Switch off your laptop computer and keep it
No talking while lecture is going on
No gossiping while the lecture is going on
Raise your hand if you have question to ask
Be on time for lecture
Be on time to come back from the recess break to
continue the lecture
Bring your lecturenotes to lecture
8
Introduction to Energy
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„
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Energy Approach to Problems
The concept of energy is one of the most
important topics in science while it is not the
only important topic
Every physical process that occurs in the
Universe involves energy and energy transfers
or energy transformations
Giving a lecture now is an example of the
transfer and transform of energy but it is more
complicated (chemical energy from food I ate
this morning to the sound energy in my voice
now)
„
„
The energy approach to describing
motion is particularly useful when the
force is not constant
An approach will involve Conservation
of Energy
„
This could be extended to biological
organisms, technological systems and
engineering situations
9
10
Systems
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„
A system is a small portion of the
Universe
A valid system may
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Work
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be a single object or particle
be a collection of objects or particles
be a region of space
vary in size and shape
Energy in a system is conserved.
11
The work, W, done on a
system by an agent exerting a
constant force on the system
is the product of the
magnitude, F, of the force, the
magnitude Δr of the
displacement of the point of
application of the force, and
cos θ, where θ is the angle
between the force and the
displacement vectors.
W = F Δr cos θ
F
θ
F cosθ
Δr
12
Work, cont.
F
Did you do any work if you were
repelling along a rope down a
helicopter?
θ
F cosθ
„
W = F Δr cos θ
„
„
„
Δr
The displacement is that of the point of
application of the force
A force does no work on the object if the
force does not move through a
displacement
The work done by a force on a moving
object is zero when the force applied is
perpendicular to the displacement of its
point of application
13
Did you do any work if you
were running on a level
ground?
F
θ
F cosθ
Δr
14
Example. Batman, whose mass is 80.0 kg, is dangling on the
free end of a 12.0-m rope, the other end of which is fixed to a
tree limb above. He is able to get the rope in motion as only
Batman knows how, eventually getting it to swing enough that
he can reach a ledge when the rope makes a 60.0° angle with
the vertical. How much work was done by the gravitational
force on Batman in this maneuver?
Answer:
Why did you feel tired after the run?
Take up Physics major – biophysics. This
course will give you interesting knowledge. 15
The force of gravity on Batman is down.
Only his vertical displacement contributes to
the work gravity does. His original ycoordinate below the tree limb is -12 m. His
change in elevation (the distance traveled) is
Δy = -12 cos (60°) m – (-12 m) = 6 m.
The work done by gravity is
80 kg x 9.8 m/s2 x 6 m = 4704 J
16
Work in Pushing a Block
„
The normal force, n,
and the gravitational
force, m g, do no
work on the object
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More About Work
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The system and the environment must be
determined when dealing with work
„
The environment does work on the system
„
„
cos θ = cos 90° = 0
The force F does do
work on the object
Work by the environment on the system
The sign of the work depends on the
direction of F relative to Δr
„
„
Work is positive when projection of F onto Δr is in
the same direction as the displacement
Work is negative when the projection is in the
opposite direction
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18
Work Is An Energy Transfer
Units of Work
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Work is a scalar quantity
The unit of work is a joule (J)
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„
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1 joule = 1 newton 1 meter
J=N·m
.
„
If the work is done on a system and it is
positive, energy is transferred to the system
If the work done on the system is negative,
energy is transferred from the system
If a system interacts with its environment, this
interaction can be described as a transfer of
energy across the system boundary (e.g., heat
generated due to friction)
„
19
This will result in a change in the amount of energy
stored in the system
20
Scalar Product of Two Vectors
„
The scalar product
of two vectors is
written as A . B
(read as A dot B)
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Scalar Product, cont
A
„
„
„
„
It is also called the
dot product
.B
The scalar product is commutative
„
= A B cos θ
A .B = B .A
The scalar product obeys the
distributive law of multiplication
A . (B + C) = A . B + A . C
θ is the angle
between A and B
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22
Work Done by a Varying Force
Dot Products of Unit Vectors
„
î ⋅ î = ĵ ⋅ ĵ = k̂ ⋅ k̂ = 1
„
î ⋅ ĵ = î ⋅ k̂ = ĵ ⋅ k̂ = 0
„
Using component form with A and B:
„
A = A x î + A y ĵ + A zk̂
„
Assume that during a very
small displacement, Δx, F
is constant
For that displacement, W
~ F Δx
For all of the intervals,
xf
W ≈ ∑ Fx Δx
B = B x î + B y ĵ + B zk̂
Force
Displacement
xi
A ⋅ B = A xB x + A yB y + A zB z
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24
Work Done by a Varying
Force, cont
„
xf
lim
Δx →0
∑ F Δx = ∫
x
xi
„
xf
xi
Fx dx
Work Done By Multiple Forces
Force
„
Therefore, W = f Fx dx
∫
x
xi
„
The work done is
equal to the area
under the curve of
force versus
displacement
If more than one force acts on a system
and the system can be modeled as a
particle, the total work done on the
system is the work done by the net
force
∑W = Wnet = ∫
Displacement
xf
xi
( ∑ F )dx
25
Work Done by Multiple Forces,
cont.
„
x
26
Hooke’s Law
If the system cannot be modeled as a
particle, then the total work is equal to
the algebraic sum of the work done by
the individual forces
Wnet = ∑Wby individual forces
„
The force exerted by the spring is
Fs = - kx
„
„
27
„
x is the position of the block with respect to the equilibrium
position (x = 0)
k is called the spring constant or force constant and
measures the stiffness of the spring
This is called Hooke’
Hooke’s Law
28
Hooke’s Law, cont.
„
„
„
Hooke’s Law, final
When x is positive
(spring is stretched),
F is negative
When x is 0 (at the
equilibrium position),
F is 0
When x is negative
(spring is
compressed), F is
positive
„
„
„
The force exerted by the spring is
always directed opposite to the
displacement from equilibrium
F is called the restoring force
If the block is released it will oscillate
back and forth between x and –x
29
Work Done by a Spring
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„
„
30
Spring with an Applied Force
Identify the block as
the system
Calculate the work as
the block moves from
xi = - xmax to xf = 0, or
„
„
xi = xmax to xf = 0
x
0
1 2
Ws = ∫ Fx dx = ∫
−kx ) dx = kxmax
(
−x
x
f
i
max
„
2
„
„
The total work done as
the block moves from
–xmax to xmax is zero
31
Suppose an external
agent, Fapp, stretches
the spring
The applied force is
equal and opposite to
the spring force
Fapp = -Fs = -(-kx) = kx
Work done by Fapp is
equal to ½ kx2max
32
Example. If it takes 4.00 J of work to stretch a
Hooke's-law spring 10.0 cm from its unstressed
length, determine the extra work required to
stretch it an additional 10.0 cm.
Answer:
4.00 J=
1
k( 0.100 m
2
Kinetic Energy (F x)
„
)2
„
„
„
„
To stretch the spring to 0.200 m requires
„
1
( 800)( 0.200) 2 − 4.00 J= 12.0 J
2
33
Kinetic Energy, cont
„
W =∫
xi
∑ F dx = ∫
xi
„
ma dx
vf
W = ∫ mv dv
vi
1
∑W = 2mv
2
f
1
− mvi2
2
ma dx = m
A change in kinetic energy is one possible
result of doing work to transfer energy into a
system
34
„
xf
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
the expression ½ mv2 can be derived using F = ma, and
Work-Kinetic Energy Theorem
Calculating the work
by integration:
xf
K = ½ mv2
„
∴ k = 800 N m
ΔW =
Kinetic Energy is the energy of a particle due
to its motion
dv
dx = mv dv
dt
„
35
The Work-Kinetic Energy Principle
states ΣW = Kf – Ki = ΔK
In the case in which work is done on
a system and the only change in the
system is in its speed, the work done
by the net force equals the change
in kinetic energy of the system.
Kinetic energy possessed by an
object of mass m and velocity v is
K = ½ mv2
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Work-Kinetic Energy Theorem
„
The normal and
gravitational forces do no
work since they are
perpendicular to the
direction of the
displacement
W = F Δx = ΔK
= ½ mv 2 – 0
(the vertical normal force
has no effect)
Nonisolated System
„
„
„
37
Internal Energy
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„
„
A nonisolated system is one that interacts with
or is influenced by its environment, e.g., push a
block on a rough surface
An isolated system would not interact with its
environment (e.g., push a block on a friction-less
surface.)
The Work-Kinetic Energy Theorem can also be
applied to nonisolated systems. In that case
there will be a transfer of energy across the
boundary of an object (e.g., from the block to
the surface where heat is generated.)
38
Potential Energy
„
The energy associated
with an object’s
temperature is called
its internal energy, Eint
In this example, the
surface is the system
The friction does work
and increases the
internal energy of the
surface
„
Potential energy is energy related to the
configuration of a system in which the
components of the system interact by
forces
Examples include:
„
„
„
39
elastic potential energy – stored in a spring
gravitational potential energy, e.g., you
stand on top of Bukit Timah Hill
electrical potential energy, e.g., you switch
on electricity to iron your shirt
40
Example. A 2.00-kg block is attached to a spring of force
constant 500 N/m as shown in figure. The block is pulled 5.00
cm to the right of equilibrium and released from rest. Find the
speed of the block as it passes through equilibrium if (a) the
horizontal surface is frictionless and (b) the coefficient of friction
between block and surface is 0.350.
Answer:
(a) When the block is pulled back to
equilibrium position, the potential energy
in spring will be fully converted to the
kinetic energy in the block when the
spring returns to original length, i.e.,
In terms of initial sum and final sum
of energies, you can also treat it as
½ k x2 = μmg x + ½ m v2
k x
x = 500
2
m
x
2
½ 500 0.05 = 0.35 2 9.8 ½ 2 v
½ k x2 = ½ m v2.
So, v =
(b) If there is fiction on the surface, the
potential energy on the spring is first
wasted on the work done by friction.
The remaining energy is converted to
the kinetic energy in the block, i.e.,
½ k x2 - μmg x = ½ m v2.
x
x
x
x
+
x
x
2
v = 0.53 m/s
0.05 = 0.79 m/s
41
Example. The ball launcher in a pinball machine has a spring
that has a force constant of 1.20 N/cm. The surface on which the
ball moves is inclined 10.0° with respect to the horizontal. If the
spring is initially compressed 5.00 cm, find the launching speed of
a 100-g ball when the plunger is released. Friction and the mass
of the plunger are negligible.
Answer:
42
Example. You need to move a heavy crate by sliding it across a flat floor with a
coefficient of sliding friction of 0.2. You can either push the crate horizontally or pull the
crate using an attached rope. When you pull on the rope, it makes 30° angle with the
floor. Which way should you choose to move the crate so that you will do the least
amount of work? How can you answer this question without knowing the weight of the
crate or the displacement of the crate?
Normal
force
Normal
force
Answer:
Initial energy stored in the spring
= ½ k x2 = ½ x 1.20 N/(10-2m) x (5 cm x 10-2)2 = 0.15 J
This 0.15 J is used to (i) move the ball up the slope of 10°
for a vertical distance of 5 cm x sin (10 °) = 0.87 cm, and (ii)
provide the ball with a muzzle speed of v. So we have
0.1 kg x 9.8 m/s2 x 0.0087 m + ½ x 0.1 kg x v2 = 0.15 J
v = 1.68 m/s
43
When pushing the crate with a force parallel to the ground, the force of friction
acting to impede its motion is proportional to the normal force acting on the
crate—in this situation, the normal force is equal to the crate’s weight. When
pulling the crate with a rope angled above the horizontal, the normal force on the
crate is less than its weight—the force of friction is therefore reduced. To keep
the crate moving across the floor, the applied force in the parallel direction must
be greater than or equal to the force of friction—pulling on the rope therefore
requires a smaller parallel applied force. The work done in moving an object is
equal to the product of the displacement through which it has been moved and
the force component parallel to the direction of motion. The applied force
component parallel to the ground is smaller when pulling the crate with the
rope—thus, the work done to move the crate with the rope must be less,
44
regardless of the weight of the crate or the displacement.
s1
What if you pull at 0 °, i.e., in
parallel with the surface?
Push
F1
µ=1
Normal
force
Normal
force
Slide 45
s1
scitaysc, 10/2/2007
Pull
F2
µ=1
mg
mg
F1 = µmg
Work done = F1 x d
F2 cos 30° = µ (mg – F2 sin 30°)
F2 cos 30° + µ F2 sin 30° = µ mg
F2 (0.866 + 0.5) = µ mg
F2 = 0.732 µmg = 0.732 F1
Work done = F2 x d
= 0.732 F1 x d
45
Ways to Transfer Energy Into
or Out of A System
„
„
„
Work – transfer by applying
a force and causing a
displacement of the point of
application of the force, e.g.,
you push a book
Mechanical Waves – allow
a disturbance to propagate
through a medium, e.g.,
sound, earthquake, tsunami
Heat – is driven by a
temperature difference
between two regions in space,
e.g., hot tea
More Ways to Transfer Energy
Into or Out of A System
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„
„
46
Matter Transfer – matter physically
crosses the boundary of the system,
carrying energy with it, e.g., use of
gasoline in car
Electrical Transmission – transfer
is by electric current, e.g., use of
electrical hair dryer
Electromagnetic Radiation –
energy is transferred by
electromagnetic waves, e.g., use of
light bulb.
47
Conservation of Energy
„
Power
Energy is conserved
„
„
„
This means that energy cannot be
created or destroyed
If the total amount of energy in a
system changes, it can only be due
to the fact that energy has crossed
the boundary of the system by some
method of energy transfer
„
The time rate of energy transfer is
called power
The average power is given by
W
Δt
when the method of energy transfer is
work
P=
48
Instantaneous Power
„
„
Units of Power
The instantaneous power is the
limiting value of the average power as
Δt approaches zero
P = Δtlim
→0
„
The SI unit of power is called the watt
„
„
W dW
=
Δt
dt
„
dW
dr
= F ⋅ = F ⋅v
dt
dt
Power = force x velocity
1 hp = 746 W
Units of power can also be used to express
units of work or energy
1 kWh = (1000 W)(3600 s) = 3.6 x106 J
Energy transferred in
Not 1 kilo-watt
1 hr at a constant rate
per hour
51
„
50
1 watt = 1 joule / second
= ((kg x m/s2) x m ) / s = 1 kg . m2 / s3
A unit of power in the US Customary system
is horsepower
„
This can also be written as
P=
49
Example. An energy-efficient light bulb, taking in 28.0 W of power, can produce
the same level of brightness as a conventional bulb operating at power 100 W.
The lifetime of the energy efficient bulb is 10 000 h and its purchase price is
$17.0, whereas the conventional bulb has lifetime 750 h and costs $0.420 per
bulb. Determine the total savings obtained by using one energy-efficient bulb
over its lifetime, as opposed to using conventional bulbs over the same time
period. Assume an energy cost of $0.080 per kilowatt-hour.
Answer:
Energy and the Automobile
„
kWh
„
„
kWh
The concepts of energy, power, and friction
help to analyze automobile fuel consumption
About 67% of the energy available from the
fuel is lost in the engine
About 10% is lost due to friction in the
transmission, drive shaft, bearings, etc.
„
„
About 6% goes to internal energy and 4% to
operate the fuel and oil pumps and accessories
This leaves about 13% to actually propel the
car
52
Friction in a Car
„
53
Friction in a Car, cont
The magnitude of the total friction force
(ƒt ) is the sum of the rolling friction (ƒr )
force and air resistance (ƒa)
„
„
„
ƒt = ƒr + ƒa
At low speeds, rolling friction predominates
At high speeds, air drag predominates
54
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