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MECHANICS MODULE 3:
Work and Energy
NOTES AND QUESTIONS
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Lesson 26 notes – Work and Energy
Objectives
(a) define work done by a force;
(b) define the joule;
(c) calculate the work done by a force using W = Fx and W = Fx cos θ;
Work done
Work is done by a force when the object concerned moves in the direction of the force,
and the force thereby transfers energy from one object to another.
Let us look at an example:
Imagine a cup of tea being lifted up to a very tall person’s mouth. The cup gains
gravitational potential energy as it is lifted. The force that must be used to lift the cup, if
it is lifted at a uniform velocity, is equal to the weight of the cup (we ignore air
resistance).
The force then equals mass x gravity. (Weight).
The gravitational potential that it has gained is mass x gravitational potential x height
to the tall person’s mouth.
So we can see that the energy gain is the force (needed to raise the cup) x distance
(to the tall person’s mouth).
If the cup is pushed along a bench towards a small person’s mouth, the cup will be
resisted because of friction. If the cup is pushed at a constant velocity, the force
needed to push the cup along the bench equals the resistive force due to friction. The
energy that is transferred will be heat energy and will equal the frictional force
multiplied by the distance to the small person’s mouth.
We call this energy the Work Done
Work Done (Joules) = Force (Newtons) x distance (m) (in direction of force)
W
=
Fx
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F

Direction of block
Imagine dragging an object along at an
angle so that it still moved along the
ground but not all of the force was in that
direction.
We would have to work out the
component of the force acting in the
direction of the block in order to work out
the Word done.
In other words we would have to work out what Fx was.
Using trig: cos = Fx / F
So Fx = Fcos 
F

So since W=Fx (where F is the force in the direction of
movement)
Fx
We get the fllowing equation that works for all situations:
W=Fxcos
Joule
We can then define 1 Joule as the energy transferred when a force of 1 Newton
moves through 1 metre.
1 J = 1 N m = 1 Kg ms-2 m = 1 Kg m2s-2
These two examples of forces which do no work are paradoxical. You can certainly
feel fatigued when doing either of these, so how can you say that no work is done?
In the case at left, no matter how hard or how long you have pushed, if the crate does
not move, then you have done no work on the crate. It is sitting still in the same place!
Why then do you feel fatigued (as well as frustrated, etc). The resolution to this
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dilemma comes in considering that when your muscles are used to exert a force on
something, the individual muscle fibers are in a continual process of contracting and
releasing to maintain the net collective result of a steady force on an external object.
That contracting and releasing involves force and motion, and constitutes internal work
in your body. The energy shows up as warming in your muscle tissue, but if the crate
doesn't move, there is still no net work on the crate.
The example at right is another standard paradox in the definition of work. If the box is
being carried at constant velocity, then no net force is necessary to keep it in motion.
The force exerted by the person is an upward force equal to the weight of the box, and
that force is perpendicular to the motion. If there is no motion in the direction of the
force, then no work in done by that force. Yet you certainly feel like you are doing work
if you carry a heavy box. The resolution of the paradox is similar to the first example your muscles must maintain an extra tension to stay upright under the load. This
requires a greater amount of internal contraction and release of our muscle fibers, and
hence internal work in our bodies. But the work done on the box is zero since by
moving in a straight line at constant speed, its energy is remaining the same.
There are many important examples of forces which do no work because they act
perpendicular to the motion. For circular motion, the centripetal force always acts at
right angles to the motion. It changes the direction of the motion, but it does no work
on the object. This can be applied to any circular orbit.
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Lesson 26 questions – Work Done
1.a) Travelling slowly, at about 5 miles an hour, the retarding forces acting on a cyclist are
about 5 N.
i)
What propulsive force must the cyclist provide to travel at constant speed?
Force = ……………N (1)
ii)
How much energy does the cyclist have to provide to cover 5 m?
Energy = …………… J (2)
b)
i)
At a higher speed, nearly 10 mph the retarding forces are much larger, nearly 8 N.
Write down the propulsive force required at this speed.
Force = ……………N (1)
ii)
Calculate the energy the cyclist must provide to cover 5 m at this constant speed.
Energy = …………… J (2)
c)
Climbing a hill at a steady speed of 5 mph the cyclist finds herself perspiring rather
more than on the flat. The mass of the cycle plus cyclist is 80 kg. The ergometer pedals fitted
report the energy supplied to the bicycle as 185 J to cover a 5 m stretch of road.
i)
How much energy is used to lift the cyclist uphill?
Energy = …………………J (2)
ii)
How much height does she gain in travelling along the 5 m stretch of road?
Height = …………………..m (3)
iii)
How much energy would she have to provide to cover this same stretch of road at 10
mph?
Energy = …………………. J (2)
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iv).
What is the retarding force acting on her, when travelling uphill at 10 mph?
Force = ……………….. N (2)
v)
Draw a vector diagram for the retarding forces (due to drag and gravity) acting on her at
10 mph.
(2)
d)i)
Calculate the retarding force due to the slope as a fraction of her weight.
……………% (2)
ii)
Comment on the advantages of using the slope of the hill rather than lifting bike and
rider vertically by the same distance
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (2)
Total [20]
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Lesson 27 notes – Energy changes
Objectives
(d) state the principle of conservation of energy;
(e) describe examples of energy in different forms, its conversion and conservation,
and apply the principle of energy conservation to simple examples;
(f) apply the idea that work done is equal to the transfer of energy to solve problems.
(a) select and apply the equation for kinetic energy Ek = ½ mv2;
(b) apply the definition of work done to derive the equation for the change in
gravitational potential energy;
Kinetic Energy
K.E. = ½ mv2
Units: J, joules
Kinetic energy stored in an object that is moving equals the amount of work done
accelerating it from stationary to that speed.
Work done = force x displacement
But F = ma, and displacement = s
So Work done = mas
Remember v2 = u2 + 2as
Reorganise v2 – u2 = 2as
But if the object is stationary at the start, u=0, so,
v2 = 2 as
This makes
= as
But W = m . as
So W = m .
Or Work done = Ek =
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Gravitational Potential Energy
Ep = P.E. = mgh
Units: J, Joules
Potential Energy is the energy stored in objects which have been lifted against a
gravitational field (or you can also have potential energy in magnetic or electric fields see later).
The work done lifting them = potential energy stored in them
Work done = Force x displacement
Here the force = weight = m x g
And the displacement = height = h
So Work done = m x g x h = P.E. gained
Note: this equation only applies if g remains constant. This is true for any movement
near the Earth’s surface.
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Lesson 27 questions – Energy calculations
Note: g = 9.8 N kg-1 = 9.8 m s-2
1
A jumper of mass 75 kg jumps a height of 1.9 m
(a)
Calculate the gravitational potential energy of the jumper at his highest point.
gravitational potential energy = ……………… (J) (2)
(b)
Calculate the vertical velocity of the jumper just before impact with the crash mat.
vertical velocity just before impact = …………………… (ms-1) (3)
(c)
Show that the kinetic energy associated with his vertical velocity is equal to the
gravitational potential energy at the top of his jump.
(3)
(d)
The jumper is stopped over a distance of 0.3 m when he lands on the crash mat.
Calculate the average force on the jumper as he comes to rest.
Average force = ……………… (N) (2)
Total [10]
2a)
Explain the quantities
i)
Gravitational potential energy
………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………… (2)
ii)
kinetic energy
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
………………………………………………………………………… (2)
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b)
Water leaves a reservoir and falls through a vertical height of 130m and causes a water
wheel to rotate. The rotating wheel is then used to produce 110 Joules per second of electrical
energy.
i)
Calculate the velocity of the water as it reaches the wheel, assuming that all the
gravitational potential energy is converted to kinetic energy.
velocity = ………………… ms-1 (3)
ii)
Calculate the mass of water flowing through the wheel per second, assuming the
production of electrical energy is 100% efficient.
mass of water per second = ……………….. unit……… (3)
iii)
State and explain two reasons why the mass of water flowing per second needs
to be greater than the value in (ii) in order to produce the amount of electrical energy stated.
………………………………………………………………………………………………
……………………………………………………………………………………………………
…………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………………………………
……………………………………………………………………………… (2)
Total [12]
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Lesson 28 notes – KE to GPE
Objectives
(c) select and apply the equation for the change in gravitational potential energy near
the Earth’s surface Ep= mgh;
(d) analyse problems where there is an exchange between gravitational potential
energy and kinetic energy;
(e) apply the principle of conservation of energy to determine the speed of an object
falling in the Earth’s gravitational field.
An Example: A slide
A child is playing with a nearly frictionless car down a smooth flexible track. She
arranges the identical length of track in two ways.
In both arrangements the track runs from the same place on the edge of a table to floor level.
1.
She thinks that the car runs off the track at very nearly the same speed in both arrangements
when started from rest at the top of the track (the point nearest the table). Is she right or wrong?
Give a reason for your answer.
Right, as both cars have the same amount of kinetic energy on leaving the track. A good
answer would relate this to the same drop in height, explicitly changing the change in
potential energy to the change in kinetic energy.
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2.
She thinks the car takes longer to run down the top arrangement than down the bottom
arrangement, when started from rest at the top of the track. Is she right or wrong? Again, justify
your answer.
Right, as the bottom arrangement gains kinetic energy more quickly, making the average
speed greater. A good answer would relate this more rapid change in kinetic energy to
the steepness of the slope–top arrangement gets faster sooner!
3.
Suppose she now lets a car of twice the mass of the previous one run from rest down the top
arrangement. Will this car, which has twice as great a mass, take a shorter, the same, or a
longer time to run down the track? Give a reason for your answer.
Same, assuming the frictional forces are really negligible as the change in kinetic energy
equated to the change in potential energy soon gives:
v  2gh .
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Lesson 29 notes - Power
Objectives
(d)state that the efficiency of a device is always less than 100% because of heat
losses;
(e) select and apply the relationship for efficiency Eff. = useful output energy/total input
energy ×100%
(f) interpret and construct Sankey diagrams.
Efficiency
You never get out what you put in. There are always losses due to heat and usually
sound, if nothing else. In other words nothing is 100% efficient.
Efficiency is a ratio of the useful energy or power output to the energy or power input.
Efficiency = useful output energy/total input energy ×100%
Sankey Diagrams
Sankey diagrams show all the energy losses in a system as it goes through the
different processes involved.
A more complicated diagram is shown for a car:
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Lesson 29 questions – Power
1
a)i)
Explain the concept of work and relate it to power.
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (2)
i)
Define the joule.
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (1)
b)
A cable car is used to carry people up a mountain. The mass of the car is 2000kg and it
carries 80 people, of average mass 60kg. The vertical height travelled is 900m and the time
taken is 5 minutes.
i)
Calculate the gain in gravitational potential energy of the 80 people in the car.
gain in gravitational potential energy = …………… J (2)
ii)
Calculate the minimum power required by a motor to lift the cable car and its
passengers to the top of the mountain.
power = ……………. unit …………. (3)
Total [8]
2.
A particularly macho mountain biker sets out to prove something. He attacks a 20% hill. (Goes
up 1m for every 5 across) The mass of the cyclist plus bicycle is 100 kg.
a)
i)
Draw the forces acting on the cyclist, whilst in motion.
(2)
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ii)
Show that the size of the retarding force due to gravity, acting along the slope is
equal to 196N. Assume g = 10 N kg-1
(3)
iii)
Calculate the energy he must supply to move 200 m up the slope.
Energy = ………………… J (2)
The cyclist covers this 200 m of road, whilst travelling up the hill, in 75 s. Previous tests show
that the retarding force at this speed is 15 N.
iv)
Calculate the energy he must also supply, just to cover any 200 m at this speed.
v)
Find his total power output.
Energy = ……………….. J (2)
Power = ……………….. W (3)
Total [12]
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Lesson 31 notes – Spring Constant
Objectives
(a) force in one dimension and can be tensile or compressive;
(b) describe the behaviour of springs and wires in terms of force, extension, elastic limit, Hooke’s
law and the force constant (ie force
per unit extension or compression);
(c) select and apply the equation F = kx, where k is the force constant of the spring or
the wire;
Hooke’s Law
Hooke’s Law describes how materials behave under force. As a force is applied,
materials that obey Hooke’s Law will extend in proportion to the load applied.
x  load
Tension (in the sample) F  extension
i.e. F = k x
This is a mathematical statement of Hooke’s law, where k is the stiffness, equal to
the tension per unit extension = F/x
Units N m-1. (The opposite of stiffness is the compliance.)
The stiffness k can be found from the slope of the linear part of the F-x graph –
taking the slope averages all the individual data points.
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x /m
F/N
A material obeying Hooke’s Law will have a graph as above. A straight line through
the origin shows that F is directly proportional to x. The gradient would give the
compliance of the material on this graph with 1/gradient giving the stiffness k.
If an unbalanced force is applied to a system then it will accelerate, so 2 forces must
be applied in order to an object for deformation to occur.
Elastic Plastic
If a material is deformed but can return to its original shape, it is called elastic
deformation. This will happen for low stresses below the elastic limit of that material. If
a material is deformed and a stress greater than the elastic limit and cannot go back to
its original shape than it is called plastic deformation.
Explaining stiffness and elasticity
The following diagrams explain why some materials are elastic or not.
Stretching any material stretches the bonds between molecules. These bonds can’t be
stretched that far before they break, so materials that can stretch a lot have folded
chains of molecules that unfold when stretched. If they are stretched too far, they stay
in this rotation and the material will not go back to its original shape (plastic
behaviour). Different ways of bonding give different material stiffness. For example:
polythene molecules can rotate and stretch out whilst polystyrene molecules cannot
rotate and so it is not as stretchy; Rubber chains can rotate and unfold but then can
refold and go back to its original shape, (elastic behaviour)
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Explaining stiffness and elasticity
Metals
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stretching has to pull bonds apart
a metal is an array of
positive ions bonded
by negative electron
'glue'
gaps open up a little
Young modulus
~1011 — 1012 Pa
Elastic extensibility ~ 0.1%
Stretching a metal stretches bonds — but not much.
Explaining stiffness and elasticity
Polythene
bond
rotates
chains are
folded
bond
rotates
polythene is a
long flexible
chain molecule
which folds up
stretching can rotate some bonds,
making the folded chain longer
Elastic extensibility ~ 1%
Young modulus
~108 — 109 Pa
Stretching polythene rotates bonds
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Rubber
sulphur cross-links
sulphur cross-links
In unstretched rubber, chains meander randomly
between sulphur cross-links.
In stretched rubber, the chain bonds rotate, and chains
follow straighter paths between cross-links. When let go,
the chains fold up again and the rubber contracts.
Elastic extensibility > 100%
Rubber stretches and contracts by chains uncoiling and coiling up again. Rubber is elastic, not plastic.
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Lesson 31 questions – Spring constant
1
Fig 1.1 shows a spring that is fixed at one end and is hanging vertically.
ai)
Calculate the spring constant of the spring.
b)
spring constant = …………….. unit ………. (3)
The mass M is pulled down a further 150mm by a force F additional to its weight.
i)
Determine the force F.
F = ……………… N (1)
ii)
State any assumptions made.
…………………………………………………………………………………………………….
……………………………………………………………………………………………… (1)
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Lesson 32 notes - Strain Energy
Objectives
(d) determine the area under a force against extension (or compression) graph to find
the work done by the force;
(e) select and use the equations for elastic potential energy
E = 1/2Fx and
E = ½ kx2
Work
Work = Force x distance
Work under a graph – move a distance of S with force F
Work = FxS
Strain energy
Strain energy is the energy stored in elastic objects that will return fully to their original
shape like a spring or elastic band
We can use the area under a graph for load verses extension but since load is
changing we must take the average load.
Take average force divided by extension this
gives us the stored strain energy in the
deformed object.
ie
Strain energy = ((F-0)/2) x extension
This is the same as the are under the graph:
W= ½Fx
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Extension: Curves
cut into small sections and work out
count squares under the line.
Hooke’s Law
Since Hooke’s Law says that F=kx, we can substitute this into the equation above and
we get:
E=1/2Fx
So E=1/2(kx)x
Simplifying: E=1/2kx2
(since F=kx)
So we can use either of the following equations to describe the energy in a material
that behaves elastically:
E = 1/2Fx and
E = ½ kx2
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Lesson 32 questions - Strain Energy
1)a) State Hooke’s Law:
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
b)
Fig 1.1 shows a graph of force F against extension e for a metal in the form of a wire.
The cross sectional area of the wire is 1.80x10-7 m2 and its length is 1.70m.
i)
fig 1.1
Calculate the Young modulus of the metal.
Young modulus = …………… Unit……….. (5)
ii)
Calculate the energy stored in the wire when it extended by 1.60mm.
energy stored = ………………….. J
Total [9]
2
a)
Hooke’s Law states that F=kx. What do F, k and x represent?
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………… (1)
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b)
A spring is compressed by applying a force. Fig 2.1 shows the variation of force F with
compression x.
Fig 2.1
i)
Calculate the spring constant.
spring constant = …………….. unit ……………. (2)
ii) Show that the work done in compressing the spring by 48mm is 2.9 J.
(2)
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Lesson 33 notes – Young Modulus
Objectives
(f) define and use the terms stress, strain, Young modulus and ultimate tensile
strength (breaking stress);
(g) describe an experiment to determine the Young modulus of a metal in the form of a
wire;
Stress
Stress is the Force per cross sectional area.
Stress = Tensile Force
cross sect. area
Or,  = F/A
Units N m-2 or Pascals (Pa)
Strain
Strain is the ratio of the extension of wire to its original length because of an applied
stress.
Strain = extension of wire
original length
Or, ε = ΔL/L
There are no units as it is a ratio.
The Young modulus
As greater stress is put onto a material it will increase the strain on it:
stress is proportional to strain: stress = constant x strain.
That constant is called the young modulus and is different for different materials. It is
given the symbol, E:
Young Modulus (E/Pa) = Stress (/Pa) / Strain (ε)
E = (F/A)/(ΔL/L)
E = (FL)/(AΔL)
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The Young modulus 2
large strain for little stress _
material is flexible, easy to
stretch
0
little strain for large stress
_ material is stiff, hard to
stretch
0
0
strain 
e.g. polymer
0
strain 
e.g. diamond, steel
The Young modulus is large for a stiff material (large stress, small strain). Graph is steep.
The Young modulus is a property of the material not the specimen. Units of the Young modulus
MN m–2 or MPa; for stiff materials GN m–2 or GPa. Same as units of stress, because strain is
a ratio of two lengths, e.g. extension is 1% of length
Ultimate Tensile Stress: (uts)
The highest stress a material can take before it breaks.
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Finding the Young Modulus
A load is attached to wire X to keep it taut. A load is attached
to wire Y sufficient to remove any kinks. (this is not counted
as part of the final load)
The original length of Y is measured from the support to the
vernier scale with a measuring tape or ruler. The diameter is
measured with a micrometer, (measured at different points
and averaged). From these, the mean radius is taken.
X
Y
The initial vernier reading is noted with no load on Y. A load
is then added and the new vernier reading taken. This
procedure is repeated adding loads in 10 Newton steps up to
a maximum of 60N. Loads are then removed and the vernier
reading recorded again.
The extension for each load is calculated by subtracting the
original vernier reading from the mean reading.
A graph is then plotted of load F(N) against extension ΔL(m).
The gradient = F/ΔL,
So E = gradient x L/πr2
The second wire makes sure that any temperature affects
are eliminated.
A long wire is used to increase the extension and reduce fractional uncertainty.
(reduce any errors made in measuring length and extension)
A similar procedure with slightly different equipment can also be used:
(resourcefulphysics.org)
Here, a pointer is placed on the material and read from a ruler. The length is
measured up to this marker and it is assumed that extension will be uniform
throughout the material. The extension is measured for different loads. The diameter
of the measured is again measured with a micrometer. The same graph is drawn and
again and so E = gradient x L/πr2.
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Lesson 33 questions - Young modulus
1
A strip of rubber originally 75 mm long is stretched until it is 100 mm long.
a)
What is the tensile strain?
Strain = ………….. (2)
b)
Why has the answer no units?
..........................................................................................................................................
.....................................................................................................................................(1)
2.
The greatest tensile stress which steel of a particular sort can withstand without breaking is about 10 9 N
m-2. A wire of cross-sectional area 0.01 mm2 is made of this steel. What is the greatest force that it can
withstand?
Force = …………. N (2)
4.
Find the minimum diameter of an alloy cable, tensile strength 75 MPa, needed to support a load of 15 kN.
diameter = ……………unit……. (3)
5.
Calculate the tensile stress in a suspension bridge supporting cable, of diameter of 50 mm, which pulls up
on the roadway with a force of 4 kN.
Stress = ………………. (3)
6.
Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force
of 20 N
Stress = ………………. (3)
7.
A large crane has a steel lifting cable of diameter 36 mm. The steel used has a Young modulus of 200
GPa. When the crane is used to lift 20 kN, the unstretched cable length is 25.0 m. Calculate the extension
of the cable.
extension = ……………. unit ………. (4)
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8
a) Define:
(i) stress
..........................................................................................................................................
.....................................................................................................................................(1)
(ii) strain
..........................................................................................................................................
.....................................................................................................................................(1)
b)
(i) Distinguish between elastic and plastic behavior when materials are stretched.
..........................................................................................................................................
..........................................................................................................................................
..........................................................................................................................................
....................................................................................................................... (2)
(ii) Define elastic limit
..........................................................................................................................................
.....................................................................................................................................(1)
c)
(i) State the SI unit of the Youngs modulus.
unit =......................................(1)
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(ii) Describe, with the aid of a diagram, an experiment to determine the Youngs modulus of
steel in the form of a wire. Explain how to use your readings to obtain the Youngs modulus.
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Lesson 34 notes - Material properties
Objectives
(h) define the terms elastic deformation and plastic deformation of a material;
(i) describe the shapes of the stress against strain graphs for typical ductile, brittle and
polymeric materials.
Material properties
Materials can be classified as follows:
Ductile materials stretch beyond their elastic limit,
Brittle materials snap when they reach their elastic limit,
Polymeric (Poly – lots of, mer – unit: So a material made from Lots of units) materials
don’t show linear behaviour.
Strong: A large stress is needed to break it.
Stiff: Not stretchy or bendy.
Tough: Materials undergo considerable plastic deformation before they break. They
therefore absorb a lot of energy before they break.
Hard: Resists indentation on impact.
Other important definitions:
Stress: The force per unit cross sectional area.
Strain: The change in length (extension) over original length of the material caused by
a stress.
Ultimate Tensile Stress: (uts)The highest stress a material can take before it breaks.
Elastic limit: (el)The stress limit at which a material will return to its original shape.
Stress – Strain Graphs
Ductile (copper)
Brittle (glass)
stress
Polymeric (rubber)
stress
stress
uts
uts
uts
el
strain
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el strain
el
RAB Plymstock School
strain
Explaining plasticity
Polythene
‘neck’
polythene strip 10 mm  100 mm
crystalline
amorphous
thin crystalline strip ‘pulled out
of’ wider region
new crystalline
region
When stretched plastically, the chains slip past
Polythene is semi-crystalline. Think of polythene
each other. More of the material has lined-up
as like cooked spaghetti. In amorphous regions
chains. More of it is crystalline.
the chains fold randomly. In crystalline regions
the chains line up.
Plastic extensibility > 100%
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Strengths of some materials
A logarithmic scale shows how the strength of materials ranges over several orders of magnitude.
Logarithmic scale of stress
100,000
105
diamond
10,000
104
glass
1,000
100
10
Each main division represents a factor of 10
103
Two main divisions represent a factor of
102 mild steel, 10  10 = 100 = 102
kevlar
101
wood
1
100
foamed polymers
0.1
10–1
0.01
10–2
You cannot have zero on a logarithmic scale.
At the lower end, it goes 0.1, 0.01, 0.001, and so on,
getting smaller and smaller but never reaching zero.
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Stress–strain graph for mild steel
The graph shows how the behaviour of steel changes as the stress increases.
elastic
region
plastic region,
extension uniform
along length
plastic region,
necking has
begun
400
300
200
+ fracture
100
0
0
0.1
0.2
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0.3
0.4
0.5
0.6
strain /%
0.7
39.0
39.1
RAB Plymstock School
39.2
Lesson 34 questions – material properties
1)
Here is the graphical output from tensile testing of five materials: Perspex, polypropylene, aluminium,
brass and mild steel. Load is on the Y axis and extension is on the X axis. The extension is not to same scale as
they vary a lot.
Perspex
Polypropylene
4000
4000
3000
3000
2000
2000
1000
1000
0
0
extension
extension
Aluminium
Brass
2000
8000
1500
6000
1000
4000
500
2000
0
0
extension
extension
Mild steel
5000
4000
2000
1000
0
a)
State which material has the highest elastic limit (yield load) and take a measurement of
this value
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b)
Which sample has the lowest maximum ultimate stress?
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Total [3]
2
Three materials P, Q and R have force-extension graphs as shown in Fig 2.1.
Fig 2.1
State and explain which of these three materials you would choose for the following
applications.
i)
making foundations of a building
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ii)
forming into a wire
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iii)
making an elastic rope (e.g. a bungee-jump rope)
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iv)
making a beam to support a roof
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