Download Year 11 Science

Year 11 Science
Science 1.1
Demonstrate an understanding of
aspects of mechanics in one dimension.
SLO’s
You will be able to give reasons why
phenomena, concepts or principles relate to
given situations in the context of;
• Distance, speed, interpretation of distance
and speed time graphs, average acceleration
and deceleration in the context of everyday
experiences such as journeys, sport, getting
going, etc. Using v =Δd/Δt and a= Δv/ Δt
• Mass, weight and the acceleration due to gravity,
balanced and unbalanced forces, in the context of
everyday experiences such as being stationary, moving at
constant speed, accelerating, etc. The relationship F net =
ma.
• Force and pressure in the context of everyday
experiences. The relationship P =F/A
• Work and power, gravitational potential energy, kinetic
energy, and the conservation of mechanical energy in
free fall situations in the context of everyday experiences
such as sports performance, dropping things, tossing
balls, etc. The relationships Δ EP = mgΔh, EK = ½ mv2,
W=Fd and P= W/t.
Units you will need to know
Name
Description
Symbol
Unit
Unit symbol
Distance
How far it
moves
d
Metre
M
Speed
How fast
v
Metres per
second
ms-1
Acceleration
How speed
changes
a
Metres per
second per
second
ms-2
Time
How long it
takes
t
Seconds
S
Mass
How much
matter
m
Kilogram
Kg
Force
How much
push/pull
F
Newton
N
Kinetic
Energy
Energy of moving
objects
Ek
Joule
J
Potential
energy
Energy stored in
objects
Ep
Joule
J
Power
How fast it works
P
Watt
W
g
10N kg-1
Gravitational The force which
acceleration pulls objects to
the centre of
Earth
Height
Distance off the
ground of an h
object
h
meters
M
Work
What is done
when a force
moves an object
W
Joule
J
Area
Space an object
takes up
A
Centimetres
squared.
cm2
Pressure
The amount of
force applied to
an area
p
Newton’s
per metre
squared
Nm-2
Weight
Mass with the
force of gravity
acting upon it
w
Newton’s
N
Height
Key formula’s.
Distance and speed.
• Velocity = change in distance
change in time
v= Δd
Δt
• Acceleration= change in speed
change in time
a= Δv
Δt
Mass and weight
• Net force = mass x acceleration
Fnet = ma
Force and pressure
• Pressure = force
P= F
area
A
Work and power
• Potential energy = mass x gravity x height
ΔEP = mgΔh
• Kinetic energy= ½ mass x speed squared
Ek = ½ mv2
• Work (energy)= force x distance
W=Fd
Ew = Fd
• Power = work
time
P = W/t
Using a formula.
• How do we know which formula to choose?
• Read the question; look for the data given,
find the equation which fits the data and the
question.
• Sometimes you will need to use 2 or 3
formulae.
• What if we need to rearrange the formula?
• Put the equation into a ∆ and look for the data
point you want to calculate.
Writing it out.
• When you answer any exam question;
1. Write out the formula.
2. Substitute the data.
3. Write out the answer with units.
W=mg
W= 50 x 10
W= 500N
Review
%
%
X
The car starts at rest and travels a distance of 6m in 3s, what is its average
speed?
Formula; _____________________
Answer;
____________________________________________________________
The cars mass is 400g. The car then travels a further 28m at a constant speed
of 4ms-1 for 7s, how much kinetic energy does the car have?
Formula; _____________________
Answer;
____________________________________________________________
An athlete runs for 8s at a speed of 14ms-1, calculate their acceleration.
Formula; _____________________
Answer;
____________________________________________________________
The athlete then does some weight training; he lifts a 100kg dumbbell, what is
the weight force of the dumbbell?
Formula; _____________________
Answer;
____________________________________________________________
The athlete lifts the dumbbell above his head 1.5m, what is the work being
done?
Formula; _____________________
Answer;
____________________________________________________________
While the dumbbell is in mid-air it has gravitational potential energy, calculate
how much energy is created.
Formula; _____________________
Answer;
____________________________________________________________
While he is holding the dumbbell, he is standing on both of his feet. His feet have
an area of 459cm2 then combined weight of the athlete and the dumbbells is
160kg, what pressure is being exerted?
Formula; _____________________
Answer;
____________________________________________________________
The athlete finishes off his workout by cycling; he accelerates at 1.5ms-2 over a
distance of 25m, he and the bike have a mass of 70kg. Calculate the net force
acting upon the bike when it is accelerating.
Formula; _____________________
Answer;
____________________________________________________________
If it takes 5s to cover the distance of 25m, calculate the power output of the
bike during this time.
Formula; _____________________
Answer;
____________________________________________________________
Formula; _____________________
Answer;
____________________________________________________________
Formula; _____________________
Answer;
____________________________________________________________
Distance & Time
• Distance is a measurement of how far apart two points are.
• The unit for distance is the metre (m), or kilometre (km).
• Displacement is a measurement of how far an object has moved
from its starting point.
• The equation is:
displacement = final – initial
distance distance
• Time is a measurement of the duration of an event.
• The unit for time is the second (s), or
hour (h).
d/t
Motion
Time (s)
Walk
22.15
Skip
7.73
Run
6.00
Distance (m)
D/t graph.
Time (s)
Motion of the mini.
Distance (cm)
Time (s)
20
1.6
30
1.8
40
2.1
50
2.2
60
2.5
70
2.6
80
2.8
90
3.5
100
4.0
Speed (cms-1)
d/t cars
Distance
(cm)
10
15
20
25
30
35
40
45
50
55
60
Time 1 (s)
Time 2 (s)
Time 3 (s)
Av time
Speed (ms-1)
Speed & Velocity
• An object has speed when it travels a distance in a time interval.
• At any moment in time, a moving object has instantaneous speed.
Since this is difficult to calculate, we usually use average speed.
• Since velocity is speed in a given direction, the equation for average
velocity is:
final distance – initial distance
velocity =
final time – initial time
∆d
vav =
∆t
• The unit for speed or velocity is metres per second (ms-1) or
kilometres per hour (kmh-1).
• Pg 8
Distance/time Graphs
• Time always goes along the horizontal (x) axis
• Distance always goes up the vertical (y) axis
• A flat line (slope = 0) means
the object is stationary
• A slope means the object is
moving. The slope gives the
speed of the object
A straight line indicates
a constant speed.
• A curved line means the
object is accelerating
(speeding up or slowing
down)
• Pg11
d
t
d
t
d
t
What are these objects doing?
d
d
t
t
d
d
t
t
Calculating distance.
Time (s)
1
Distance (m) 2.5
2
3
4
5
7.5
10
25m
Time (s)
1
2
3
4
5
Distance
(m)
5
10
15
20
25
75m
Speed/time Graphs
• Time always goes along the horizontal (x) axis
• Speed or velocity always goes up the vertical (y) axis
• A flat line (slope = 0) means
the object is travelling at a
v
constant speed
(no acceleration)
• A slope means the object is
accelerating. The slope gives
the acceleration of the object
v
• A curved line means the
rate of acceleration is
increasing or decreasing
• Pg14
v
t
t
t
2cm
3cm
Each is 0.1s
2/0.1 = 20ms-1
3/0.1 = 30ms-1
4/0.1 = 40ms-1
4cm
s/t
Time (min)
Speed (ms-2)
0
0
25
1
5
20
2
10
15
3
15
4
20
5
20
6
20
7
20
8
20
9
5
10
0
Series1
10
5
0
1
2
3
4
5
6
7
8
9
10
11
Area under graphs
• The area under a velocity/time graph can be used to calculate the
total displacement of an object
• For a simple straight line, simple calculate the area of the rectangle
or triangle under the line
• For more complex shapes, split them into rectangles and triangles
and add together the areas for each shape to calculate the total
displacement
eg. total displacement = ½ base x height
v
t
Area under the graph- 1 shape
V
½ base x height
t
Area under the graph- 2 shapes
V
Then add them
together
½ base x height
Base x height
t
Area under the graph- more than 2
shapes.
V
t
Acceleration
• An object changing its speed is said to be accelerating. If the
acceleration is:
– positive (eg. 2ms-2) = object speeding up
– negative (eg. -2ms-2) = object slowing down
• The equation is:
final speed – initial speed
acceleration =
time taken
∆v
a =
∆t
• The unit for acceleration is metres per second squared (or ms-2).
• Pg 9
Deceleration
• So deceleration is simple an object slowing down.
• The same formula is used, but the result will be a negative number.
• Pg 13, 14
Forces
• A force is a push, pull, or twist.
• Forces can change the:
– speed
– direction, and/or the
– shape of an object.
• The unit for force is the Newton (N)
• When a force acts on an object (the action force), an opposing force
appears as a reaction force (eg. gravity or friction).
eg.
driving force
(from engine)
friction force
(from road/air)
Pg 16
Thrust.
• The force that moves objects in the direction
of the force.
• The greater the thrust, the faster the
movement.
Thrust
Friction
• Friction is a force that opposes motion.
• Friction always works in the opposite direction
to thrust.
• Friction between an object and air or water is
called drag.
Friction
Pg 17
Friction
• Friction is a force that opposes motion.
• It is created when objects rub against each other, releasing energy
as heat.
• Friction between an object and air or water is called drag.
• Friction can be:
– useful – eg. brakes, tyres
– undesirable – eg. engine wear
• Friction can be reduced by using:
– lubricants
– bearings
Pg 17
Support
• The force that pushes upwards, so objects do
not fall to the centre of the Earth.
Support
Gravity
• The force that pulls objects toward the centre
of the Earth.
• It is a constant force at 10ms-2.
Gravity
Balanced Forces
• When the action and reaction forces are equal in size and opposite
in direction, they are balanced.
• If the forces on an object are balanced it will either remain
stationary, or continue moving at a constant speed.
•
eg.
450N
450N
Net force = 0N
Pg 18
Unbalanced forces.
• When the action and reaction forces are different in size or
direction, they are unbalanced.
• If the forces on an object are unbalanced it will accelerate in the
direction of the resultant or net force.
500N
450N
Net force = 50N
Force, Mass & Acceleration
• When an unbalanced force acts on an object, it accelerates in the
direction of the net force.
• The equation is:
force = mass x acceleration
Fnet= ma
• The same equation also applies to the effect of gravity on mass (ie.
weight).
weight force = mass x acceleration due
to gravity
Fw = mg
= m x10N (if mass is in kg)
• The unit for weight force is the Newton (N).
Pg 19
F= ma
Mass (kg)
Acceleration (cms-1) Stopping distance
(cm)
1
0.6
670
1.2
5.2
700
1.4
16.3
690
1.6
21.3
700
1.8
23.4
723
2
26.2
726
Force (N)
Gravity, Mass & Weight
• Since weight is the result of
the force of gravity acting on
mass, weight can change
depending on the force of
gravity
• On Earth, a person with mass
60kg would weigh 600N
• On the moon, a person with
the same mass would only
weigh 100N
• F=mg
600N
100N
Force and lifting.
• Force is needed to lift an object.
• This force equals the weight of the object.
• The formula is; Force = mass x gravitational acceleration
F = mg
• E.g. A mass of 5kg is lifted 2m off the ground
F = 5 x 10
F = 50 N
• Work is also done to lift the object
W=Fxd
W = 50 x 2
W = 100J
5kg
Forces and the shot-puts.
Experiment 1; lifting the shot-put (F=mg)
Experiment 2; work and the shot-put (W=Fd)
Experiment 3; dropping the shot-put (F=ma)
Mass (kg)
Acceleration
(ms-2)
Force (N)
1.5
3
6.25
We also need to measure the area
of the impact craters.
Gravity (ms-2)
Distance (m)
Force and pressure
• Pressure is created when a force is applied to an area.
• The formula is; Pressure = force / area
P=F
A
• The greater the force and the smaller the area, the greater the
pressure.
10N
8cm 2
P = 10/8
P=1.25Nm-2
2
cm2
P= 10/2 P= 5Nm-2
Pg 20
High heels.
• A 50kg woman stands in a pair of high heels.
• The area of each heel is 2cm x 2cm and she
stands with the front of her foot off the
ground.
• How much pressure is going into the ground?
Steps to solve;
8cm2 is the area of the shoes.
Convert to m2; 0.0008 m2
Calculate her force; F=mg
50 x 10 = 500N
P = F/A
P = 500/ 0.0008
P = 625,000Nm-2
Work
• Work (W) is done when a force moves an object.
• Work is measured in Joules (J).
• The equation is; work done = force x distance
W= F x d
W
F
d
• Work done on an object equals the energy gained by the object.
• E.g. Pushing a pedal car 3m using a force of 200N
• 200N x 3m = 600J
• The car gains 600J of kinetic energy.
Pg 26
Power
• Power is the rate at which work is done or
energy is transferred.
• Power (P) is measured in watts (W).
• The equation is; Power = Energy / time
• P = E/t
My
pet
has
power
E
P
t
Energy
• Energy- the capacity to do work; the property
of a system that diminishes when the system
does work on any other system, by an amount
equal to the work so done; potential energy.
Symbol: E
• Sorry what now?
Pg 23
Energy
Potential
• Gravitational objects that
are lifted can fall.
• Chemical- energy stored in
chemical bonds.
• Elastic- energy stored in
stretched objects.
• Nuclear- energy stored in
the nucleus of an atom
Active
• Electrical- currents of
electrons
• Light- waves of photons
• Heat/Thermal- movement
of particles
• Sound- vibrations
• Kinetic- movement energy
Vinegar bottle.
• We will now complete the vinegar bottle
rocket experiment and observe the energies &
forces involved.
Gravitational potential energy
• Gravitational potential energy is stored in objects placed above
ground level, energy is released when objects fall to the ground.
• The formula is; potential energy = mass x gravity x change in height
ΔEp = mgΔh
Pg 26
Kinetic energy
• Kinetic energy is found in moving objects.
• The formula is;
• Kinetic energy = ½ mass x velocity squared.
Ek = ½ mv2
• Use mass in kg and velocity in ms-1.
Calculating energy.
• Complete the experiment to calculate the
energy of a toy car.
Ramp height
(cm)
10
20
30
40
50
Observations
Ep (J)
V (ms-1)
Ek (J)
• If the car weighed 0.0288kg, gravity is 10ms-2
and it was set off from 10cm the equation
would be;
•
•
•
•
Ep = 0.0288 x 10 x 10 = 2.88J
V2 = 2.88 / 0.0144 = 200
V = √200 = 14.1 ms-1
Ek = 0.5 x 0.0288 x 14.12 = 2.9J (rounded to 1
d.p.)
What do you notice?
Chemical potential energy.
• Energy stored in chemical bonds.
• When you break a bond between 2
atoms, you release the energy.
• E.g. digestion of food; breaks the
bonds between food molecules,
releasing energy for your body to
use.
Elastic potential energy.
• Energy stored in compressed or stretched
objects, the energy is released when the
object returns to its original shape.
• E.g. Spray cans, rubber bands.
Nuclear potential energy.
• Energy stored in the nucleus of atoms;
released when the atom is destroyed.
• E.g. Nuclear power plant
Electrical energy
• The flow of electrons along a conductor.
Light energy
• Waves of photons moving in one direction.
Heat/Thermal energy
• Movement of particles towards a gas state.
Sound energy
• Vibrating particles which in turn vibrate the
bones of the middle ear, which is turn causes
cells in the cochlea to move.
Energy can change.
• Energy can be transformed or transferred; but
no energy is lost as it does this, this is called
THE CONSERVATION OF ENERGY.
Energy transformations.
• When one type of energy is transformed into
another, without any loss to the net amount
of energy.
80J
100J
20J
Energy transformation.
• Think back to the toy car; why was the
amount of gravitational potential energy equal
to the amount of kinetic energy?
• The gravitational potential energy was
transformed into kinetic energy as the car
moved, as the amount was conserved.
Push a toy car along a surface
Turning on a light bulb
Push a slinky down stairs
Whisking a batter
Frying pancakes
Bounce a tennis ball
Snow board down a half pipe
Riding a rollercoaster
Listening to your ipod
Eating an apple
Using a magnet to pick up metal
Law of Energy Conservation
1. Energy must come from somewhere.
2. Energy may change form but the total
amount is constant.
3. Energy cannot be destroyed, it can only be
transformed.
Conservation of mechanical energy in
free fall.
• Every object in a state of uniform motion tends to remain in that
state of motion unless an external force is applied to it.
• E.g objects in motion stay in motion until acted upon by another
force.
• For every action there is an equal and opposite reaction.
Pg 28
Energy transference.
• When energy is shifted without changing its
form.
• E.g. heat moving along a piece of metal.
To remember
•
•
•
•
Power= pet (my pet has power)
Work= fwd (forward)
Kinetic energy= mev (my mate mev)
Pressure= pfa (pfffaaa the sound of a can
opening)
• Force= mfg (my force goes)
• Force= fma
Q2
• Ep= mgh
v2 = Ek / ½ m
• 83x10x1.2= 996J
• 996/41.5=24 √ = 4.9ms-1
Q1
• EK = ½ m v2
• 75 x 625 = 46875
• 46875/1500 = 31.25m
h= Ep / mg
Revision questions.
Exam revision.
Toy Car
Eugene has a toy car which moves along a concrete path
when he winds up the spring and lets the car go. The car
accelerates constantly from rest to reach a top speed of 6
m s-1 in 0.5 s.
a. Calculate the toy car’s acceleration.
Show all working.
b. From its top speed, the car slows
steadily and takes 4 s to stop. Sketch
a graph of the toy car’s motion from its
release until it stops.
c. Describe one energy change which
occurs during the toy cars motion.
a= ∆v / ∆t
a= 6/0.5 a= 12ms-2
c. Elastic potential  kinetic energy
The Water Park Slide
Ben and Ranui were on a slide at a water park.
20 m
10 m
a. It took Ben 10 s to reach the
bottom of the slide. Calculate his
average speed.
s=d/t s= 20/10 s= 2ms-1
Diving Down
Ranui has a mass of 70 kg and Ben has a mass of 90 kg. Both
of them stepped off a 10 m diving board at the same time.
a. Explain why Ranui and Ben hit the
water at the same time.
Mass does not effect the rate of acceleration of an object.
b. Ranui and Ben both accelerate at 10 m s-2. Calculate the
force on Ranui as she falls through the air. Show all working.
State the unit.
F = ma
F= 70 x 10
F = 700N
c. Ben started from rest and hit the water with a speed of 8ms-1.
Calculate the time it took him to reach the water.
∆ t= ∆v / a
8/10 = 0.8s
t= d/s t= 10/8 t= 1.25s
Warbirds over Wanaka
The Spitfire LF Mk XVI is a spectacular aircraft that takes
part in the biennial “Warbirds over Wanaka” airshow.
a. During the show, the spitfire accelerates in a low-level
horizontal pass over the crowd.
(i) Draw a diagram of the spitfire showing the relative
size and direction of the FOUR forces acting on the plane
during this acceleration.
(ii) If it takes 3 s to accelerate from150 m s-1 to 180 m s-1,
calculate the acceleration of the spitfire
Warbirds over Wanaka
Gravity
Thrust
Drag/friction
Support
(ii) If it takes 3 s to accelerate from150 m s-1 to 180 m s-1,
calculate the acceleration of the spitfire
a= ∆v / ∆t a= 180-150 / 3 a= 30/3 a= 10ms-2
b. A Hurricane aircraft, mass 4 000 kg, repeats the pass
over the crowd with an acceleration of 8 m s-2 and covers a
distance of 500 m.
(i) Calculate the size of the unbalanced force acting on
the Hurricane during its acceleration using the equation F =
ma.
F=ma F=4000x8
F=32,000N
(ii) Calculate the work done accelerating the Hurricane
using the equation W = Fd.
W=Fd W=32000x500 w = 16,000,000 watts
Bungy Jump
Ranui tried the bungy jump at a local water park.
At the top and middle of the jump, the bungy cord was slack. At
the bottom of the jump, the bungy cord was fully stretched.
a. Complete the table below to show the main forms of energy
at each point.
Position
Main form of energy
Top of jump
Gravitational potential energy
Middle of jump
Kinetic energy
Bottom of jump
Elastic potential energy
b. Ranui did not bounce back up to the platform. Explain why.
Some of the kinetic energy was transformed to elastic energy, so
the total amount of kinetic energy was not available for her return
journey.
Monica’s Morrie
The starter motor of Monica’s Morrie would not work. To start
the car, Monica got three of her friends to push it down a gentle
slope. The effect of the slope is to balance the friction forces on
the car. Each of her friends pushed the car with a force of 450
N. The car with Monica inside it had a mass of 900 kg.
a. i) Using the information above, calculate the car’s
acceleration. Show your working.
a= ∆v / ∆t
a= 18/14
a=1.28ms-2
a= f/m a= 450/900 a= 0.5ms-2
ii) The table below shows how the car’s speed changed with
time. Draw a speed-time graph of the motion.
Time (s)
0 2 4 6 8 10 12 14
Speed (m s-1)
0 3 6 6 6 10 14 18
0
2
4
6
8
10
12
14 Time (s)
Monica’s Morrie Moves On
b. While recovering from pushing the car, her three friends
watched a dog chase after a stick. The distance-time
graph for the dog is shown below.
d (m)
72
36
t (s)
12
18
36
(i) Calculate the dog’s speed for the first 12 s.36/12=3ms-1
(ii) Find the dog’s speed during the time 12 s to 18 s.0
(iii) Describe the dog’s motion from the time 18 s to 36s.
He was accelerating.
Just Cruising
The diagram shows
the four main forces
Thrust
acting on a car as it
travels to the left at
a steady speed.
a. Use labels to describe the forces
acting on the car.
b. The force to the right is
measured as 550 N. What is the
size of the force to the left?
Explain your answer. 550N
Gravity
Drag/friction
Support
The Sand Yacht
The wind pushes a sand yacht along on dry sand. When
the unbalanced force on the yacht is 280 N its acceleration
is 2 m s-2.
a. Calculate the mass of the yacht. Give the unit with your
answer. 140kg
b. The yacht is blown onto wet sand where the friction
force has the same size as the wind force. Describe what
happens to the yacht. It will either stop or move at a
constant speed.
c. When the yacht moves onto soft sand the friction force is
larger than the wind force. Explain what happens to the
speed of the yacht. It will slow down.
What force is required to move a 15kg mass at 3ms-2?
_______________________________________________________________
A force if 20N causes an object to move at 7ms-2, what is its mass?
_______________________________________________________________
If a 10N unbalanced force is applied to a 10kg object; what will be its
acceleration?
_______________________________________________________________
Calculate the force acting upon an 18kg object here on Earth with a gravity
force if 10ms-2.
_______________________________________________________________
Calculate the size of the force acting upon a 50kg object on the moon.
_____________________________________________________________________
If a 80kg man stands on a platform which is 50cm2, how much pressure is he applying?
_____________________________________________________________________
If an object exerts 10pascals of pressure over 1m2, what force is being created?
_____________________________________________________________________
If 50N of force and 6 pascals of pressure are being exerted, what area are the acting
upon?
_____________________________________________________________________