Download AS Unit 10 Work Energy Power

AS Physics Unit 10
Work, Energy, Power
Mr D Powell
Chapter Map
Spec Link: 3.2.1 Principal of cons of energy,
applied to examples involving GPE, KE, and Work
Done against resistive forces
Mr Powell 2009
Index
10.1 Work and energy
Specification link-up 3.2.1: Work, energy
and power
What is energy and how do we measure it?
Does energy ever disappear?
What is meant by work (in the scientific
sense)?
Mr Powell 2009
Index
What is work?
Work is the energy transfer that takes place when a force causes an object to
move.
work done = force applied × distance moved in direction of force
What is crucially important is to realise that no energy transfer takes place
when the motion is not in the direction of the force!



work done is measured in joules (J)
force is measured in Newtons (N)
distance is measured in metres (m)
100 N
W = Fs
= 100N x 5m
= 500Nm
= 500J
5m
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Index
Work done by a force at an angle
The same box is now dragged by a rope, which is raised at an angle θ to the
horizontal.
F
θ
This time, the box moves in a different direction to the direction of the applied force.
How does this affect the work done? Can you think of any suggestions?
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Index
Calculating work done at an angle
F
When calculating the work done by a force
acting at an angle, it is useful to break the
force down into components.
θ
The tension in the rope can be broken down
into a horizontal and a vertical component.
The vertical component does no work
because the box does not move in that
direction.
F
Fsinθ
θ
So to calculate work done by a force at an
angle:
Fcosθ
work done = force in direction of movement × distance moved
W = Fscosθ
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Index
Practical
F
Using a force sensor can you confirm
this idea by conducting a simple
practical?
θ
Take care to control your experiment
and all variables apart from the
angle.
You may require…
•
•
•
•
Force sensor
String
Block
Weights
F
Fsinθ
θ
Fcosθ
work done = force in direction of movement × distance moved
W = Fscosθ
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Index
Quick Question....
10 N
A toy car is pulled along by a piece of
string which is at 30° to the
horizontal.
1) Calculate the work done in
pulling the toy if the tension in
the string is 10 N, and it is pulled
along 5 m.
2) Comment on why the actual
energy input will need to be
more than this?
3) Comment on why power delivery
to the car to make it move at a
steady speed will not be
uniform?
30°
5m
W = Fscosθ
= 10 × 5 × cos30°
= 43.3 J
2) We will need to overcome frictional
forces as we move so energy will be
wasted
3) The friction forces will be higher initially
until the car is moving when this static
friction is overcome the dynamic friction is
lower. Hence, power will need to change.
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Index
Force Distance Graphs Questions…
Using…
work done = force applied × distance moved in direction of force
&
WD = Fscos
We can now relate the energy transfer for a box to an area under the graph.
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Index
Force Distance Graphs Questions…
Using…
work done = force applied × distance moved in direction
of force
What is the energy transfer in Joules in Each of these
cases if the force was applied at 45 in the left picture
and 60  in the right case?
27J x cos 60 =13.5J
100J x cos 45 = 71J
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Index
10.2 Kinetic energy and potential energy
Specification link-up 3.2.1: Work, energy and
power
When an object is lifted, what happens to the
work done on it?
If it is then allowed to fall, what energy
change takes place?
What is the effect on the kinetic energy of a
car if its speed is doubled?
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Index
What is kinetic energy?
Kinetic energy (KE or Ek) is the energy of an
object due to its speed....
kinetic energy = ½ × mass × speed2
Ek = ½mv2
You should try and get in the habit of using the
subscript “k” to show kinetic every time you
quote energy from now on.
Also EP for dealing with potential energy or it
will get very confusing.



kinetic energy is measured in joules (J)
mass is measured in kilograms (kg)
speed is measured in metres per
second (ms-1).
Eureka Video
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Index
Deriving Ek = ½mv2
Consider a force F acting on an object of mass m, initially at rest, moving it a distance s in
time t. Using the SUVAT equations & Newton's 2nd law can you derive kinetic energy?

From ‘suvat’ equations:
s = ½ (u + v)t
a = (v – u) / t

Because u = 0 ms-1:
s = ½vt
a=v/t

Newton’s 2nd law:
F = ma

Substituting a = v / t:
F = mv / t

Work done by force:
W = Fs
W = (mv / t) × ½vt
W = ½mv2

Work done = energy transferred:
Ek = ½mv2
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Index
What is gravitational potential energy?
Gravitational potential energy (GPE, Ep or Egrav)
is the energy of an object due to its position in
a gravitational field.
The Ep gained by a mass is proportional to the
force used to lift it, and the distance it is lifted.
It is often talked about in terms of a change in
an object’s Ep due to a change in its height:
ΔEp = mgΔh
ΔEp = mgΔh
= 0.250kg × 9.81Nkg-1 × 2m
Example...
A supermarket employee lifts a baked bean tin,
weighing 250 g, from the floor, to a shelf 2 m
high.
How much gravitational potential energy does
it gain? (g = 9.81 N kg-1)
= 4.9Nm
= 4.9 J
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Index
Ek and Ep
If resistive forces, such as friction and air
resistance, are ignored, Ek and Ep are
related as follows:
loss of Ek = gain in Ep
or
lose of Ep = gain in Ek
For example, if an object of mass m is
released above the ground at height h, it
will gain speed, v, as it falls.
Due to the conservation of energy, and
assuming air resistance is negligible, after
falling a height of Δh:
½mv2 = mgΔh
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Index
Exam Question - Conservation of energy
A ball of mass 400 g is thrown upwards at a speed of 5 ms-1. (g = 9.81 N kg-1).
a) What is the ball’s Ek as it is released?
Ek = ½mv2
= ½ × 0.4 × 52
=5J
b) What is the ball’s maximum gain of Ep?
ΔEp = Ek
=5J
c) What is the ball’s maximum height?
Ep = mgh
h = Ep / mg
= 5 / (0.4 × 9.81)
= 1.27 m
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Index
Pendulum – Gains & Loses KE / GPE
Pendulums gain and lose PE & KE as the move.
As you change height from equilibrium position
(blue line) you will gain or lose PE energy
according to the formulae..
ΔEp = mgΔh
If we lose energy on the journey
ΔEp = mg(ho –h)
Where h0 is the original height and h is the height
h is the height the pendulum swings to on the
other side. (lower than original)
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Index
10.3 Power
Specification link-up 3.2.1: Work, energy and
power
Which physical quantities are involved in
power? (AO1a)
How could you develop more power as you go
up a flight of stairs? (AO2b)
Why is a 100 W light bulb more powerful than
a 40 W light bulb when each works at the
same mains voltage? (AO2a)
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Index
Making things Easy?
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Index
What is power?
Power is the rate at which work is done, or the rate at
which energy is transferred.
power = work done / time taken
P = W/t
power is measured in watts (W)
 work done or energy transferred is measured
in joules (J)
 time is measured in seconds (s).

Question...
A crane lifts a load of 1500 kg a
height of 25 m at a steady rate,
in a time of 2 min.
What is the power of the
crane?
W = energy transferred = ΔEp
ΔEp = mgΔh
= 1500 × 9.81 × 25
P=W/t
= 367 875 / 120
= 3070 W
= 367 875 J
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Index
Key Points….
If we think of a machine at work that lifts an
object…..
Work Done or W = Fs
However, if the object is moving at a
constant velocity we can make a simple
substitution to eliminate “s” from the
equation…
s = 0.5(v+u)t
s = vt (const velocity)
Pout  Fv
Power W Js-1
Force N
Velocity ms-1
useful
Eff 
total
Hence….
Work Done is Fs = Fvt
Power is then worked out as Fvt / t = Fv
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Index
Motive power Example
The power outputted by a powered object, such as
an engine or muscles, is sometimes called the motive
power.
If the powered object is moving at a constant speed
at a constant height:
power = force × speed
P = Fv
At constant speed and height, the force produced by
the powered object is equal but opposite to all
resistive forces acting on the object, such as friction
and air resistance.
Question...
What is the resistive force
on a cyclist who has leg
muscles of power 200 W
each and who reaches a top
speed of 10 ms-1 on a level
road?
P = Fv
F =P/v
= (200 × 2) / 10
= 40 N
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Index
Power: efficiency
Efficiency is the ratio of useful work done by
a device, to the total work done (or the ratio
of useful output energy to the total energy
input).
Question...
For example, what is the efficiency
of a 60 W filament lamp that gives
out 1 W of light?
efficiency = 1W / 60W
efficiency = useful work done / total
work done
efficiency = useful energy output /
total energy input
= 0.017
= 1.7%
Efficiency is often expressed as a
percentage.
Efficiency is always less than 100%, as no
device is perfect and some energy is always
lost.
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Index
10.4 Energy & Efficiency
1.
How can power be linked to velocity
2.
Into what form of energy is wasted energy almost
always converted?
3.
Can any device ever achieve 100% efficiency?
How Escalators Work
Outcomes …..
Be able to complete calculations to work out the output
Power of… (AO2b)
1.
2.
3.
4.
an electric motor (A-E)
A pulley system (A-E)
A winch (A-D) (involves efficiency calcs)
An Escalator (Higher Level) (A-C)
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Index
Key Points….
If we think of a machine at work that lifts an
object…..
Work Done or W = Fs
However, if the object is moving at a
constant velocity we can make a simple
substitution to eliminate “s” from the
equation…
s = 0.5(v+u)/t
s = vt (const velocity)
Pout  Fv
Power W
Force N
Velocity ms-1
useful
Eff 
total
Hence….
Work Done is Fs = Fvt
Power is then worked out as Fvt / t = Fv
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Index
Independent Study…
Work through the first two examples on page 155 in the text book
on the Electric Motor & Pulley System
For each one…
1.
2.
3.
4.
Draw a diagram
Write out the quantities
Write out the formulae you might require
Work through the calculations step-by-step
Higher Level
Now turn to page 156 and work through the calculations on the
winch in the same way. Move your work onto a higher level
with an efficiency calculation
Final Stage
If you are not so confident work through Summary Q1/2 then 3/4.
If you are confident do Q3/4 and then ask for a worksheet.
Mr Powell 2009
Index
Summary Q Extra Help…
Q1)
a)
Useful energy per second = Power
Power = Fv
= 30N x 15ms-1
=450Js-1
b) Useful energy is 450Js-1 which is
only 25% of the input energy.
Hence
= 450Js-1 x4 = 100% of the energy
=1800 Js-1
Q2)
a) E = Pt
E = 60Js-1 x 8s
E =480J
b) E = mgh
E = 20N x 2.5m
E = 50J
c) Efficiency =
= 50J / 480J
= 0.1
= 10%
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Index
Summary Q Extra Help…
3)
a) Efficiency of 35% gives us 200MW. Means that you need to produce
internally before the waste
Input Power = 1/0.35 * 200MW
= 571.4MW
= 570MW (2 s.f.)
If each kg of fuel produces 80MJkg-1
Your would need 571.4MW / 80MJkg per second = 7.1425kgs-1
24 hours is 60 x 60 x 24 = seconds in a day.
Total burn in a day = = 7.1425kgs-1 x 60 x 60 x 24 =617112kg = 6.2 x 105kg
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Index
Summary Q Extra Help…
4)
Pout = 6.2kW
a)
Fuel Burn 45MJkg-1
Vehicle travels 18km on one kg of fuel
What is the time taken to travel 18km at 30ms-1 when burning fuel at rate of
1kg per 18kg or 45MJ of energy per 1kg
t = s/v
= 18,000m / 30ms-1
= 600s
b) E = Pt
= 6.2kW x 600s
= 3720000J
= 3.72MJ
c)
Overall Efficiency = Input Power / Useful Power
= 3.72MJ / 45MJ
= 0.083 = 8%
Mr Powell 2009
Index
Does an escalator run at constant power?
Facts about a typical escalator



How Escalators Work
slopes at 30° to the horizontal
each step rises 15 m in 1.0 minute
efficiency of electric motor is 70%
When fully laden
 80 people get on and off in 1.0 min
 average mass of one person 70 kg
 a frictional force of 1.2 × 104 N, moving
30 m in 1.0 min, acts against the motion
1) Calculate the potential energy given each minute to the people on a
fully laden escalator. Take g = 9.8 N kg−1.
PE = mgh =
= 80 × 70kg × 9.8 N kg−1 × 15m
= 823 200
= 820 000 J (2 s.f.)
Mr Powell 2009
Index
Does an escalator run at constant power?
Facts about a typical escalator



How Escalators Work
slopes at 30° to the horizontal
each step rises 15 m in 1.0 minute
efficiency of electric motor is 70%
When fully laden
 80 people get on and off in 1.0 min
 average mass of one person 70 kg
 a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion
2) Show that the kinetic energy given to the people on the escalator is small
compared to the potential energy they receive. (Hint: find the speed up the
slope.)
15 m rise up a 30° slope is 30 m along the slope…
Distance travelled is … hyp = 15m/sin 30 = 30m
speed up slope = 30m/60s = 0.5 m s−1
KE of all people on escalator = 0.5mv2
= 80 × 0.5 x 70kg × (0.5ms-1)2
= 700 J
Mr Powell 2009
Index
Does an escalator run at constant power?
Facts about a typical escalator



How Escalators Work
slopes at 30° to the horizontal
each step rises 15 m in 1.0 minute
efficiency of electric motor is 70%
When fully laden
 80 people get on and off in 1.0 min
 average mass of one person 70 kg
 a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion
3) Calculate the work done per second by the frictional force.
P = Fv
= 1.2 × 104N ×30m/60ms-1
= 6000Js-1
= 6000 W
Mr Powell 2009
Index
Does an escalator run at constant power?
How Escalators Work
Facts about a typical escalator



slopes at 30° to the horizontal
each step rises 15 m in 1.0 minute
efficiency of electric motor is 70%
When fully laden
 80 people get on and off in 1.0 min
 average mass of one person 70 kg
 a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion
4) Using your answers to questions 1 and 3, calculate:
a) the total output power of the electric motor
b) the input power to the electric motor
c) the overall efficiency of the escalator.
a)
Total output power = increase in PE/sec + WD/sec against friction
= 823200J/60s + 6000Js-1
= 13 720Js-1 + 6000Js-1
= 19 720Js-1
= 20 000 W (2 s.f.)
Mr Powell 2009
Index
Does an escalator run at constant power?
How Escalators Work
Facts about a typical escalator



slopes at 30° to the horizontal
each step rises 15 m in 1.0 minute
efficiency of electric motor is 70%
When fully laden
 80 people get on and off in 1.0 min
 average mass of one person 70 kg
 a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion
4) Using your answers to questions 1 and 3, calculate:
a) the total output power of the electric motor
b) the input power to the electric motor
c) the overall efficiency of the escalator.
b)
Efficiency of electric motor is 70% or = 0.7
= useful power output/total input power
= 19720/ input power
Rearrange…
input power = 19720/0.7 = 28 171 ≈ 28.2 kW (3 s.f.)
Mr Powell 2009
Index
Does an escalator run at constant power?
How Escalators Work
Facts about a typical escalator



slopes at 30° to the horizontal
each step rises 15 m in 1.0 minute
efficiency of electric motor is 70%
When fully laden
 80 people get on and off in 1.0 min
 average mass of one person 70 kg
 a frictional force of 1.2 × 104 N, moving 30 m in 1.0 min, acts against the motion
4) Using your answers to questions 1 and 3, calculate:
a) the total output power of the electric motor
b) the input power to the electric motor
c) the overall efficiency of the escalator.
c)
Overall efficiency of escalator = useful power output/ total input power
= 13720Js-1/28171Js-1 (increase PE/sec Q4a / From Q4b
≈ 0.50 or 50% efficient
Mr Powell 2009
Index
10.5 Renewable Energy
Specification link-up 3.2.1: Conservation of energy
1.
How much power can we generate from different renewable
energy resources?
2.
Could we meet all our energy needs from renewable energy
resources?
3.
Can we meet our demand for energy and cut carbon emissions
at the same time?
AO1a/AO2b – Work out energy transfers for a Solar Panel (Basic)
1
2
v  V
2
AO1a/AO2b – Work out the energy transfers for a Tidal Power
(Medium)
AO1a/AO2b – Work out the Power of a Wind Turbine (Harder)
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Index
Crystalline Cells
There are two types of solar cell. These are
mono crystalline and poly crystalline.
These are wired in series to produce solar
panels
As each cell produces a voltage of between
0.5 and 0.6 Volts, 36 cells are needed to
produce an open-circuit voltage of about
20 V. This is sufficient to charge a 12 Volt
battery under most conditions.
A solar panel of area 1m2 in space
would absorb energy from the sun
at a rate of 1400Js-1 if all the
incident energy went into the cell
However, at the surface of the Earth
some of this energy is absorbed by
the atmosphere.
Water heating panels may absorb
500Js-1 and heat water to around
70C in a British Summer
The output from a solar cell is at its
greatest when the light hits the cell at right
angles.
They are made of a thin sandwich of Cover
glass, Transparent glue, Anti reflective
coating then Silicon and gallium arsenide
conductor layers to generate current flow
Mr Powell 2009
Index
Solar Cell efficiencies
http://en.wikipedia.org/wiki/Solar_cell
See how they efficiencies change by structure and year
of development!
Mr Powell 2009
Index
How does it work….
Imagine a Power Station has area of 10km x
10km (1 x108m2) and traps a tide of 6m
depth each day…
You would exchange a volume of water of..
1 x 108m2 x 6m = 6 x 108m3
With a density of 1000kgm-3
If the water drops a height on release of 1m
then energy released to turn a turbine
would be…
E = mgh
= 6 x 108m3 x 1000kgm-3 x 1m x 9.81Nkg-1
= 5.886 x 1012J
E = 5.9 x 1012J
1 m2 = 0.000001 (or 1 x 10−6) km2
Mr Powell 2009
Index
Other Methods...
Mr Powell 2009
Index
Wind
Issues/Facts
• The larger the blades or
wind the more energy
produced.
• Energy only when there is
wind.
• Low running costs
• Cause of noise pollution
• Cause of visual pollution
KE of wind
KE per unit volume
KE per unit of a
cylinder of air
swept out by
turbine
Volume of cylinder is
speed x area of
cone
KE per second or
Power as cylinder
of unit volume
area A passes
through turbine....
1 2
KE  mv
2
1 2
KE / V  v
2
1 2
KE  v V
2
V  vA
1 2
Power  v vA
2
1 3
Power  v A
2
NB: They will give you help on this in the exam!
Mr Powell 2009
Index
Summary Question Extra Help…
1)
2) Max Power  v3
Solar panel of areas 1m2 produces 200W.
1.2Mw when v = 10ms-1
Hence we can say that is 200Wm-2.
When speed is 15ms-1
To produce this much we need ….
v3 = (10ms-1)3 = 1000
v3 = (15ms-1)3 = 3375
200,000,0000Mw / 200Wm-2 = 1 x 107m2
This is 3.375x more.
Hence Power …
= 3.375 x 1.2
= 4.05MW
= 4MW
Mr Powell 2009
Index
Summary Question Extra Help…
3) Power Station Efficiency = 25% = 0.25
h = 650m
Volume water passing /s produces 200MW of power.
Density of water = 1000kgm-3
What volume of water would require this…
200MW is 25% of Input energy. Hence energy generated by the fall is 800MW.
Hence drop E = mgh
E / hg = m
m = 800,000,000W / (650m * 9.81) = 125.5 x 103kg
m/d = volume
= 125.5 x 103kg / 1000kgm-3
v = 125m3
Mr Powell 2009
Index
Summary Question Extra Help…
4)
Water trapped by tidal station is 200km2. Tide or h = 3m. Water released 6
hours.
a)
Mass of water is…. volume x density
m = area x depth x density
m = 200 x 106 m2 x 3m x 1050kgm-3
m = 6.3 x 1011 kg
b) This mass is then released over 6 hours
t = 60 x 60 x 6s = 21,600s
E = mgh / t
E = 6.3 x 1011 kg x 9.81 Nkg-1 x 3m / 21600s
E = 860MW
But Average PE is ½ this…
E = 430MW
Mr Powell 2009
Index