Download 4. Transport/pdf (DR)

Speed
Speed is a measure of how quickly distance is being covered.
Speed=
Distance
time
v= d
t
v = speed or velocity in metres per second, m/s or ms-1,
d = distance in metres, m,
t = time in seconds, s.
Question
Express 18 km/h in m/s.
Question
Express 18 km/h in m/s.
Answer
v= d
t
v=
18,000
3,600
v = 5.0 m/s
v=?
d = 18 km = 18,000 m
t = 1 hour = 3,600 s
Average Speed
Experiment to find the average speed of a trolley running down a
slope.
QED
• The distance between the light gates is measured using a metre stick.
• The time between the light gates is measured using a QED timer.
• The average speed is then calculated using the equation:
Average Speed =
Distance between light gates
Time between light gates
Instantaneous Speed
Experiment to find the instantaneous speed of a trolley at the
bottom of a slope.
QED
• The length of the card is
measured using a metre stick.
• The time for the card to pass through the light gate is measured
using a QED timer.
• The instantaneous speed is then calculated using the equation:
Instantaneous Speed =
Length of card
Time to pass through light gate
Warning the words length and time on their own are too vague.
Acceleration
Acceleration is a measure of how quickly velocity is changing.
Example 1
Time
in seconds
0
1
2
3
Velocity
in m/s
0
4
8
12
The velocity is increasing
at 4 m/s every second.
Example 2
Time
in seconds
0
1
2
3
Velocity
in m/s
21
25
29
33
The velocity is increasing
at 4 m/s per second.
Velocity
in m/s
4
8
12
16
It is accelerating at 2 m/s per
second.
Example 3
Time
in seconds
0
2
4
6
Example 4
Time
in seconds
0
2
4
6
Velocity
in m/s
5
15
25
35
It is accelerating at
5 meters per second squared, (general)
5 m/s2, (credit)
5 ms-2, (higher).
Equation
Acceleration can be found using the equation:
v-u
a=
t
a = acceleration in metres per second squared, m/s2.
v = final velocity in metres per second, m/s,
u = initial velocity in metres per second, m/s,
t = time between initial and final velocities in seconds, s.
• On a few occasions the units of acceleration may be in mph per second.
• The acceleration equation can be rearranged into:
v = u + at
• Another way of writing the acceleration equation is:
a=
∆v
∆t
a = acceleration in metres per second squared, m/s2.
∆v = change in velocity in metres per second, m/s,
∆t = change in time in seconds, s.
Acceleration
Experiment to find the acceleration of a trolley running down a
slope.
Method 1: Single card with a double beam.
QED
The QED timer is used to record:
• The time for the card to pass through the first light gate (t1).
• The time for the card to pass through the second light gate (t2).
• The time for the card to move between the light gates (t3).
The length of the card is measured using a metre stick.
initial velocity (u) =
length of card
t1
final velocity (v) =
length of card
t2
acceleration (a) =
final velocity (v) - initial velocity (u)
time between velocities (t3)
Acceleration
Experiment to find the acceleration of a trolley running down a
slope.
Method 2: Double card with a single beam.
QED
The QED timer is used to record:
• The time for the first card to pass through the light gate (t1).
• The time for the second card to pass through the light gate (t2).
• The time between the first and second cards passing through the
light gate (t3).
The lengths of the cards is measured using a metre stick.
initial velocity (u) =
final velocity (v) =
acceleration (a) =
length of card
t1
length of card
t2
final velocity (v) - initial velocity (u)
time between velocities (t3)
Speed-time Graphs
v
Constant speed.
Acceleration = 0 m/s2.
t
v
Speed increasing at a
constant rate.
Constant acceleration.
t
Speed-time Graphs
v
Speed decreasing at a
constant rate.
Constant deceleration.
t
v
Speed increasing at a
increasing rate.
Increasing acceleration.
t
Speed-time Graphs
Consider the following speed-time graph.
velocity
in m/s
10
4
2
1
6
time in seconds
Distance travelled = average speed x time
d=vxt
d=
d=
Area under the graph = area 1 + Area 2
Area = rectangle + triangle
(u + v)t
2
Area = (l x b) + (1/2 x l x b)
(4 + 10) x 6
d = 42 m
2
Area = (4 x 6) + (1/2 x 6 x 6)
Area = 42 m
The area under any speed-time graph is equal to the distance
travelled.
Speed-time Graphs
Example:
velocity
in m/s
10
2
2
1
3
5
Questions:
• describe the motion.
• calculate both accelerations.
• calculate the distance travelled.
12
time in seconds
Answers:
For the first 5 seconds the speed is increasing at a constant rate. For the next 7
seconds the speed is decreasing at a constant rate.
a=v-u
t
v = 10 m/s
u = 0 m/s
t=5s
a=?
v = 2 m/s
u = 10 m/s
t=7s
a=?
10 - 0
5
a = 2 m/s2
2 - 10
7
a = -1.14 m/s2
acceleration = 2 m/s2
deceleration = 1.14 m/s2
a=
a=
Speed-time Graphs
Distance travelled = area under a speed-time graph.
Distance travelled = Area 1 + Area 2 + Area 3
Distance travelled = (1/2 x 5 x 10) + (1/2 x 7 x 8) + (7 x 2)
Distance travelled = 25 + 28 + 14
Distance travelled = 67 m.
Forces
Forces can:
• Change the speed of an object,
• Change the shape of an object,
• Change the direction of a moving object.
Forces can be measured using a newton balance.
Force
The unit of force is the newton (N). Use a small ‘n’ when writing
the word, but a large ‘N’ when writing the letter.
Weight and Mass
Mass is a measure of the quantity of matter that a body has.
Mass is measured in kilograms.
Weight is a measure of the force of gravity pulling on a mass.
Weight is measured in newtons.
Mass in kg
Weight in N
1
10
2
20
3
30
10
100
From the table we see that the weight of a body is ten times
its mass. The ‘ten’ is called the gravitational field strength.
Weight = mass x gravitational field strength
W = mg
W = weight in newtons, N
m = mass in kilograms, kg
g = gravitational field strength in newtons per kilogram, N/kg.
The gravitational field strength on the surface of the Earth is
approximately 10 N/kg.
Gravitational field strength is different on different planets and
decreases as you move away from the surface of a planet.
There is nowhere in the universe where the gravitational field
strength is 0 N/kg. Weightlessness is a myth.
Weight
Problem:
A girl has a mass of 50 kg on the Earth. Find her mass and weight
on the Earth and Moon. The gravitational field strength on the
Moon is 1.6 N/kg.
Answer:
Mass = 50 kg on the Earth and on the Moon.
Weight on Earth = mg = 50 x 10 = 500 N
Weight on Moon = mg = 50 x 1.6 = 80 N
Crash Test Dummy
What happens to the passenger when the driver suddenly applies the
brakes?
With a seat belt.
The seat belt applies a force to the passenger and this causes her to
slow down with the car (Newton’s 2nd law, F=ma).
Crash Test Dummy
Without a seat belt.
The passenger will obey Newton’s 1st law and will continue to
move in a straight line at a constant speed until she hits the
windscreen.
Forces and Motion
Consider a mass being dragged over a frictionless surface.
Pulling Force
Mass
Smooth surface
velocity
Time
The mass will accelerate at a constant rate.
Forces and Motion
Consider a mass being dragged over a rough surface.
Frictional Force
Pulling Force
Mass
Rough surface
velocity
Constant velocity
Increasing velocity
Decreasing acceleration
Time
• At first the mass accelerates because the frictional force is less than
the pulling force.
• As the mass moves faster the frictional force increases.
• Eventually the frictional force will equal the pulling force. The mass
will then move at a constant velocity because there is no unbalanced
force.
Friction
Friction can be reduced by:
• Polishing the surfaces,
• Using oil (in a car engine),
• Using rollers (wheels),
• Using a cushion of air (hovercraft)
• Streamlining the body (car).
Sometimes we need friction:
• Brakes,
• Shoes on road,
• Nails in wood.
Stopping distance
Speed
Stopping distance equals
the area under the graph
Reaction
time
Time to
stop
Time
Stopping distance
Speed
Wet roads, worn tyres or
icy roads will increase the
stopping distance.
Reaction
time
Time to
stop
Time
Stopping distance
Speed
A faster car has a larger
stopping distance.
Reaction
time
Time to
stop
Time
Stopping distance
Speed
A tired or drunk driver
will take longer to react.
They will have a larger
stopping distance
Reaction
time
Time to
stop
Time
Acceleration and Force
Aim: To find relationship between acceleration of a fixed mass and the
unbalanced force acting on it.
Apparatus:
QED
Double mask
Light gate
Vehicle
To air
blower
Linear air track
Method:
• The acceleration of the vehicle was measured using a light
gate connected to a QED computer.
• The unbalanced force was taken to be the weight of the
hanging masses, and was calculated using W=mg.
• The mass of the vehicle was kept constant throughout the
experiment.
hanging
masses
Table of Results:
Mass in
grams
Weight in
newtons
Acceleration
10
0.1
0.29
20
0.2
0.57
30
0.3
0.86
40
0.4
1.14
50
0.5
1.43
60
0.6
1.71
70
0.7
2.00
in m/s2
Acceleration and Force
Graph of Results:
acceleration
in m/s2
X
X
X
X
X
X
X
Force in newtons
Conclusion: Since we have a straight line graph through the origin we can
conclude that the acceleration of a fixed mass is directly
proportional to the unbalanced force acting on the mass.
aαF
-(1)
Acceleration and Mass
Aim: To find the relationship between the acceleration, produced by a
fixed force, and the mass of the object being accelerated.
Apparatus:
QED
Double mask
Light gate
Vehicle
To air
blower
Linear air track
Method:
• The acceleration of the vehicle was measured using a light
gate connected to a QED computer.
• The mass of the vehicle was measured using scales..
• The hanging mass and hence the unbalanced force was kept
constant throughout the experiment.
hanging
masses
Table of Results:
Mass in
kilograms
1/Mass in
1/kg
Acceleration
0.3
3.33
2.33
0.7
1.43
1.00
1.0
1.00
0.70
1.4
0.71
0.50
1.7
0.59
0.41
in m/s2
Acceleration and mass
Graphs of Results:
acceleration
in m/s2
acceleration
in m/s2
X
X
X
X
X
X
X
X
X
Mass in kilograms
X
1/Mass in 1/kg
Conclusion: Since the second graph is a straight line graph through the origin
we can conclude that the acceleration of a mass by a fixed force is
inversely proportional to the mass.
aα 1
m
-(2)
Newton’s Second Law
From the first experiment we discovered that
aαF
-(1), mass constant
From the second experiment we discovered that:
aα 1
m
-(2), unbalanced force constant
Combining equations (1) & (2) we get:
aα F
m
Rearranging this becomes:
Fαmxa
Combining equations (1) & (2) we get:
aα F
m
Rearranging this becomes:
Fαmxa
A proportion sign (α) can be replaced with a constant (k) and an
equals sign.
F = kma
By defining one newton (N) as the force which will accelerate a
mass of one kilogram (kg) at a rate of one meter per second
squared (m/s2) the constant will become equal to one and the
equation will become:
F = ma
This equation is known as Newton’s second law.
Energy
Definition of energy.
A body possesses energy if it has the ability to do work.
Conservation of energy.
Energy cannot be created or destroyed it can only change from
one form to another.
Forms of energy.
Energy
Heat, Eh = cm∆T, Eh = lm
Sound
Light, E = hf
Kinetic, Ek =1/2mv2
Potential, Ep = mgh
Chemical
Electrical, E = ItV
Nuclear, E = mc2
Units of energy.
Energy is measured in units of:
• joules (J),
• Kilowatt hours (kWh).
Work (Energy)
Energy is transferred (work is done) when a force is applied
through a distance.
Mass (m)
Force (F)
Distance (d)
Obviously the energy transferred to mass m will be directly
proportional to:
• the size of the force applied,
• the distance the mass is moved.
Energy Transferred α force - (1)
Energy Transferred α distance - (2)
Combining equations (1) and (2) we get:
Energy Transferred or Work done α force x distance
Work done (Ew) α force x distance
Ew = k x F x d
By defining one joule of energy as the energy transferred when
a force of one newton is applied through a distance of one
metre the constant becomes equal to one and the equation
becomes:
Ew = F x d
Ew = work done in joules, J
F = force in newtons, N
d = distance in metres, m
Work Done (EW)
Example:A man drags a 5kg box a distance of 20m using a
30N force.
(i) Calculate the work done by the force.
(ii) Suggest what is likely to happen to the work
transferred.
Answers
(i)
EW = F x d
EW = 20 x 30
EW = 600 joules
(ii) Some energy will change to heat energy
(the box, air and ground will heat up).
Some energy will be changed to sound.
Some energy will change to kinetic energy
(the box will move faster).
Potential Energy
A body can be lifted at a constant velocity if an upward force
equal to the body’s weight is applied.
Mass (m)
F
h
Mass (m)
W = mg
The potential energy gained by the mass will be equal to the
work done by the lifting force.
Ew = F x d
Ew = Ep
F = W = mg
d=h
Ep = mgh
Ep = potential energy in joules, J.
m = mass in kilograms, kg.
g = gravitational field strength = 10 N/kg on earth.
h = height in metres, m.
Kinetic Energy
time = 0 s
Initial velocity = 0 m/s
Mass (m)
time = t s
final velocity = v
Force (F)
Mass (m)
Distance (d)
If there are no frictional forces then all the work done will be
transferred into kinetic energy.
Ew = F x d
E w = Ek
m(v - u)
mv
=
t
t
vt
(v + u)t
d = vt =
=
2
2
F = ma =
mv
Ek =
t
2
Ek = mv
2
x
vt
2
Ek = kinetic energy in joules, J
m = mass in kilograms, kg
v = velocity in meters per second, m/s.
Relationship between Kinetic Energy and Velocity
Aim: To confirm that Kinetic Energy is directly proportional
to velocity squared.
Apparatus:
QED
Slotted weights
Vehicle
Single mask
Light gate
h
To air
blower
Linear air track
hanging
masses
Method:
• The velocity of the vehicle was measured using a light
gate connected to a QED computer.
• The kinetic energy transferred to the vehicle and masses
is equal to the potential energy lost by the hanging
masses. (Ek = Ep = mgh).
10g slotted masses were used.
g was taken to be 10N/kg.
h was measured using a metre stick.
• The total mass was kept constant by transferring slotted
masses from the vehicle to the hanging masses.
Relationship between Kinetic Energy and Velocity
Table of Results:
Mass in
grams
Energy in
joules
Velocity in
m/s
Velocity2 in
(m/s)2
10
0.08
0.48
0.23
20
0.16
0.68
0.46
30
0.24
0.83
0.69
40
0.32
0.96
0.92
50
0.40
1.07
1.15
60
0.48
1.18
1.38
70
0.56
1.27
1.61
80
0.64
1.36
1.84
Graph of Results:
velocity2
in
(m/s)2
x
x
x
x
x
x
x
x
EK in joules
Conclusion:
Since we have a straight line through the origin we can
conclude that kinetic energy is directly proportional to the
square of its velocity.
Ek α v2
Conservation of energy
Example
A pendulum swings as shown.
2 kg
0.4 m
v
Find:
• the kinetic energy at the bottom of the swing,
• the speed at the bottom of the swing.
The kinetic at the bottom will equal the potential energy at the top.
Ek = 8 J
Ek = Ep
Ek = mgh
Ek = 2
x
Ek = 8 J
10
x
0. 4
mv2 = 8
2
v2 = 8
x
2
2
v = 2.83 m/s
Conservation of energy
A shortcut
2 kg
0.4 m
v
Ek = Ep
mv2 = mgh
2
v2 = 2gh
Ek = Ep
mv2 = mgh
2
v2 = 2gh
v=
√2gh
The velocity does not
depend on the mass.
This equation is true when all the potential energy changes to
kinetic energy. The velocity at the bottom of the swing in the
previous question could have been calculated:
√2gh
v = √2 x 10 x 0.4
v = √8
v=
v = 2.83 m/s
Power
Power is a measure of how quickly energy is transferred.
Example:
If 20 joules of energy is transferred in 5 seconds then the
power will be 4 joules per second or 4 watts (W).
Power can be found using the equation:
Power =
Energy
Time
E
P=
t
P = power in watts, W
E = energy in joules, J
t = time in seconds, s.
1 watt = 1 joule per second.
Power of a pupil
Title: Power of a pupil.
Aim: To find the power of a pupil running up the stairs.
Apparatus:
Stopclock
Pupil
Scales
h
Method:
• The height of the stairs was measured using a metre stick.
• The time to run up the stairs was measured using a stop clock.
• The mass of the pupil was measured using a set of scales.
• ‘g’ for the Earth was found in data tables to 10 N/kg.
• (If the pupil is running at the same speed at both the bottom
and top of the stairs then there will be no gain in kinetic
energy.)
• The power was calculated using the equation.
P=
mgh
t
Results:
• height of stairs = 2.5 m.
• time to run up the stairs = 2.91 s.
• mass of pupil = 60 kg.
• gravitational field strength = 10 N/kg (tables).
P=
mgh
t
60 x 10 x 2.4
P=
2.91
P = 495 W
Power of a Water Pump
Title: Power of a Water Pump.
Aim: To find the power output of a water pump.
Apparatus:
Ignore - siphon effect
V
Labpack
A
1 litre mark
h
Stopclock
Water pump
2.54s
Procedure:
Measure
• The time it takes to pump one litre of water into the bottle.
• The height the water is being pumped.
• The labpack voltage (should be set at 12 V).
• The current drawn from the labpack.
Calculate:
• Input power to the pump.
• Output power of the pump (1 litre of water has a mass of 1kg).
• Efficiency of the pump.
Some equations
Output power =
Pout =
Potential energy
Time
mgh
t
m = 1kg
g = 10N/kg
Input power = current x voltage
Pin = IV
Efficiency
Efficiency =
Power out
x 100%
Power in
P
Eff = out x 100%
Pin