Download 2. Forces

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2. Forces
Study Design

Apply Newton’s three laws of motion in situations where two or more coplanar forces act
along a straight line and in two dimensions;
Types of forces
Forces can be divided into two major categories, field forces and contact forces
Forces that act at a distance are called
FIELD FORCES, (gravitational,
electrical or magnetic)
Forces created by travelling bodies are
called CONTACT FORCES.
The relationship between a force and the acceleration it causes was first understood by Isaac
Newton (1672 – 1727). Newton summarised all motion by three laws:
Newtons First Law
An important consequence of this law was the
realisation that an object can be in motion without a
force being constantly applied to it. When you throw a
ball, you exert a force to accelerate the ball, but once
it is moving, no force is necessary to keep it moving.
Prior to this realisation it was believed that a constant
force was necessary, and that this force was supplied by that the air pinching in behind the ball.
This model, first conceived by Aristotle, proved tenacious, and students still fall into the trap of
using it.
Newtons 1st law of motion
If an object has zero net force acting
on it, it will remain at rest, or continue
moving with an unchanged velocity.
Newton’s first law is commonly tested on the exam. This is achieved by the inclusion of statements
such as “An object is moving with a constant velocity” within questions. Whenever you see the
key words constant velocity in a question, you should highlight them. The realisation that the
object is travelling at a constant velocity, and hence that the net force on the object is zero, will be
essential for solving the problem.
Newton’s Second Law
Newtons 2nd law of motion
This law relates to the sum total of the
forces on the body ( ΣF ) the body's mass
(m) and the acceleration produced (a)
ΣF = ma.
a=
F
m
Note ΣF must have the same direction
as 'a'.
In words, Newton’s Second Law states that a force on
an object causes the object to accelerate (change its
velocity). The amount of acceleration that occurs
depends on the size of the force and the mass of the
object. Large forces cause large accelerations.
Objects with large mass accelerate less when they
experience the same force as a small mass. The
acceleration of the object is in the same direction as
the net force on the object.
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Newton’s Third Law
This law is the most commonly misunderstood.
You need to appreciate that these action/reaction
forces act on DIFFERENT OBJECTS and so you
do not add them to find a resultant force. For
example, consider a book resting on a table top as
shown in the diagram below. There are two forces
acting on the book: Gravity is pulling the book
downward and the tabletop is pushing the book upwards. These forces are the same size, and are
in opposite directions but THEY ARE NOT a Newton’s thirds law pair, because they both act on
the same object.
Newtons 3rd law of motion
For every action force acting on one object,
there is an equal but opposite reaction force
acting on the other object.
N
The best way of avoiding making a mistake using Newton’s third law
is to use the following statement.
FA on B = - FB on A
W
In the example of the book on the table the force Table on Book is a Newton third law pair with the
force Book on Table. Notice the first force is on the book and the second force is on the table. They do
not act on the same object. Similarly the weight force, which is the gravitational attraction of the
earth on the book, is a Newton third law pair with the gravitational force of the book on the earth.
The gravitational effect of the book on the earth is not apparent because the earth is so massive
that no acceleration is noticeable.
Drawing Force Diagrams
You will often be asked to draw diagrams illustrating forces. There are several considerations when
drawing force diagrams:
 The arrows that represent the forces should point in the direction of applied force. The
length of the arrow represents the strength of the force, so some effort should be made to
draw the arrows to scale.
 An arrow representing a field force should begin at the centre of the object.
 An arrow representing a contact force should begin at the point on contact where the force
is applied.
 All forces should be labelled.
Some sample force diagrams of common situations are drawn below.
Mass on a string
Mass in free flight
T
m
mg
mg
Velocity v = 0, so T = mg
Velocity v = constant upwards, so T = mg
Velocity v = constant downwards, so T = mg
Accelerating Upwards, T - mg = ma.
Acceleration Downwards, mg - T = ma.
F = mg = ma
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Mass pulled along a plane
Smooth (No Friction)
Rough (Friction)
N
a
N
a
m
m
T
Fr
m
m
mg
T = ma, N + mg = 0
T
mg
T - F = ma, N + mg = 0
Bodies with parallel forces acting
a
a
F1
m
m
F1
a
m
m
F2
F2
F1
F2 – F1 = ma
F1 + F2 = ma
m
F2
m
F1 + F2 = ma
Bodies with non-parallel forces acting
a
a
a
F1
F2
m
m
F1
m
m
F2
F1
F2
F1 + F2 = ma
m
m
F1 + F2 = ma
F1 + F2 = ma
The vectors need to be resolved in order to solve for the acceleration.
Inclined planes
Another example of forces acting at angles to each other is an object on an incline plane. There
are only three different types of examples of a body on an incline plane without a driving force.
A body accelerating
The component of the weight force acting down the plane is larger then the frictional forces. (This
is also true if there are no frictional forces). For these situations you would take down the plane to
be positive, the reason for this is that the acceleration is down the plane.
Forces perpendicular to the plane
Fnet = mgcos  - N = 0
N
mg
Forces parallel to the plane
Fnet = mgsin  - F = ma
Thus the acceleration is down the plane.
If there is not friction then the acceleration
is gsin 
F
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A body travelling at constant speed
This can be the when an object is not changing its speed whilst travelling down an incline or when
the object is at rest on the incline plane.
N
Forces perpendicular to the
plane
Fnet = mgcos - N = 0
Forces parallel to the plane
Fnet = mgsin - F = 0
mg
Thus the acceleration is zero
F
A body decelerating
For these situations you would choose up the plane to be positive, this is because this is the
direction of acceleration.
Forces perpendicular to the
plane
Fnet = mgcos - N = 0
N
mg
F
Forces parallel to the plane
Fnet = F - mgsin = ma
Thus the acceleration is up
the plane.
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Example 1
1973 Question 8
(1 mark)
A car has a maximum acceleration of 3.0 ms-2. What would its maximum acceleration be while
towing a car twice its own mass?
A train accelerates from rest at one station and travels to another station. The velocity-time for the
train is given below. The mass of the train is 5.0 x 105 kg. Assume that a constant frictional
resistance of 1.5 x 104 N acts on the train throughout its journey.
Example 2
1976 Question 1
(1 mark)
Calculate the distance between the two stations.
Example 3
1976 Question 2
(1 mark)
Calculate the net force acting on the train during the first 300 seconds.
Example 4
1976 Question 3
(1 mark)
Calculate the force exerted by the engine on the train during the first 300 seconds.
Example 5
1976 Question 4
(1 mark)
Calculate the power in kilowatt at which the engine works during the period of constant velocity.
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Example 6
(1 mark)
1976 Question 9
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A cyclist is accelerating downhill with his brakes off. Which arrow (A – F) best represents the
direction of the force of the road on the bicycle?
The cyclist now brakes so that his speed is constant.
Example 7
1976 Question 10
(1 mark)
The mass of the cyclist is m; the road makes an angle θ with the horizontal. The magnitude of the
vector sum of all forces acting on the cyclist is
A.
mg
B.
mg cos θ
C.
mg sin θ
D.
zero
Example 8
1976 Question 11
(1 mark)
Which arrow (A – F) now best represents the direction of the force of the road on the bicycle?
A car of mass 800 kg is towed along a straight road so that its velocity changes uniformly from
10 ms-1 to 20 ms-1 in a distance of 200 m. The frictional force is constant at 500N.
Example 9
1977 Question 1
(1 mark)
Calculate the acceleration of the car.
Example 10 1977 Question 2
(1 mark)
What is the magnitude of the net force on the car during this 200 m?
Example 11 1977 Question 3
(1 mark)
What is the magnitude of the force exerted on the car by the towing vehicle?
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Example 12 1977 Question 4
(1 mark)
-1
When the speed of 20 ms is reached, the towing force is adjusted so that the car now moves at
constant velocity. What is now the magnitude of the towing force?
Two masses A and B are accelerated together along a smooth surface by a force of 48 N, as
shown above. The acceleration of A and B is 4.0 ms-2.
The mass of A is 4.0 kg
Example 13 1979 Question 8
What is the mass of B?
(1 mark)
Example 14 1979 Question 9
(1 mark)
What is the magnitude of the force exerted by A on B?
Example 15 1979 Question 10
(1 mark)
What is the magnitude of the force exerted by B on A?
The bodies are now accelerated together along a smooth surface in the opposite direction by a
force of 48 N, as shown below.
Example 16 1979 Question 11
(1 mark)
What now is the magnitude of the force exerted by B on A?
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A 1.0 kg mass is suspended from a spring balance which is attached to the roof of a lift. The
balance is graduated in newton and reads 10 N when the lift is stationary.
Example 17 1980 Question 14
(1 mark)
What is the reading of the spring balance when the lift moves up with an acceleration of 2.0 ms -2?
Example 18 1980 Question 15
(1 mark)
What is the reading of the spring balance when the lift moves up with an upward constant velocity
of 2.0 ms-1?
Example 19 1981 Question 5
(1 mark)
A man of mass m is suspended from a parachute of mass M and descends at a constant speed.
What is the net force acting on the man?
A
mg
B
mg + Mg
C
mg - Mg
D
zero
Example 20 1981 Question 6
(1 mark)
A girl starts to run in a northerly direction across a floor. The frictional force of the floor on her
shoes is
A
in a northerly direction: it is the friction between the floor and her shoes which allows her to
travel in that direction.
B
in a southerly direction: friction always acts to oppose motion.
C
virtually zero, because the girl is running and not sliding across the floor.
D
zero, because action and reaction forces are equal and opposite.
For a block sliding down the frictionless inclined plane of height h metre and length L metre, it can
be shown that the acceleration, a ms-2 is given by
h
h
a
i.e. a = k
where k is a constant
L
L
Example 21 1981 Question 7
(1 mark)
What is the numerical value of the constant k?
A
1
B
5
C
10
D
20
E
100
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A rocket is drifting sideways from P to Q in outer space. It is not subject to any outside forces.
When the rocket reaches Q, its engine is fired to produce a constant thrust at right angles to PQ.
The engine is turned off again when it reaches R.
Example 22 1983 Question 4
(1 mark)
Which of the following (A, B, C, D, E, or F) best represents the path of the rocket?
Example 23 1984 Question 23a (1 mark)
A golf ball travels a distance of 250 m through the air as the result of a very powerful hit. If it is
travelling to the right, which of the diagrams below gives the best representation of the forces
acting on it when it is at the highest point of its flight. (Air resistance can be ignored.)
Example 24 1984 Question 23b (1 mark)
A golf ball travels a distance of 250 m through the air as the result of a very powerful hit. If it is
travelling to the right, which of the diagrams below gives the best representation of the forces
acting on it when it is at the highest point of its flight, if air resistance needs to be considered.
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A steel ball is dropped vertically on to a steel bench-top. W is the weight force acting on the ball
and FC is the contact force exerted on the ball by the bench at the instant of rebound when the
ball is at rest.
Example 25 1984 Question 24
(1 mark)
Which one or more of the following statements about W and FC are correct?
A
W and FC form an action-reaction pair and so are equal in magnitude.
B
The ball is instantaneously at rest, so W and FC cancel exactly to provide the necessary
zero resultant force.
C
the ball is accelerating upwards, so FC must be greater than W.
D
FC may be greater or less in magnitude than W depending on whether the collision is
elastic or inelastic.
Five identical blocks each of mass 1.0 kg are on a smooth, horizontal table.
A constant force of 1 N acts on the first block as shown in the figure below.
Example 26 1985 Question 14
(1 mark)
What force does block 4 exert on block 5?
Example 27 1985 Question 15
(1 mark)
What force does block 3 exert on block 4?
Newton’s third law of motion may be stated as follows: ‘To every action there is an equal and
opposite reaction, or the mutual reactions of two bodies upon each other are always equal and
directed in contrary directions’.
The figure above shows a block being pulled along a rough surface at constant velocity. Fa is the
applied force on the block, Fr the friction force between the block and the surface, W the weight
force on the block and Fc the normal contact force exerted by the surface on the block.
Example 28 1985 Question 17
(1 mark)
Which pair of forces (A – D) in this situation are action-reaction pairs in the sense of Newton’s
third law?
A
W and the gravitational force exerted by the block on the earth.
B
Fa and Fr .
C
D
W and Fc .
none of these.
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The graph above gives the velocity-time relationship for a block of mass 4.0 kg which slides across
a rough, horizontal floor, coming to rest after 1.0 s.
Example 29 1986 Question 1
(1 mark)
What is the magnitude of the frictional force of the floor on the block?
Example 30 1986 Question 2
(1 mark)
What is the magnitude of the frictional force of the block on the floor?
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The diagram above shows a velocity-time graph of the motion of a parachutist of mass 100 kg,
when he jumps from an aircraft.
After jumping, he waits for 2.0 seconds before pulling the ripcord, opening his parachute.
(Take g: 10 ms-2)
Example 31 1987Question 5
(1 mark)
What is his acceleration at time t =1.0 s?
Example 32
1987 Question 6
What is the value of the ratio:
(1 mark)
Force of gravity on the parachutist at t = 1.0 s
?
Force of gravity on the parachutist at t = 6.0 s
Example 33 1987 Question 7
(1 mark)
What is the resultant force on the parachutist at t =1.0s?
Example 34 1987 Question 8
(1 mark)
What is the resultant force on the parachutist at t = 20.0 s?
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A road train consists of a large truck towing a trailer, as shown below. The truck and the trailer
have a mass of 2.0 x 104 kg each. When moving along a level road, the truck and the trailer
experience a constant retarding force of 2.5 x 103 N each.
Example 35 1991 Question 4
(1 mark)
If the driving force on the road train when it is accelerating is 3.9 x 104 N, what is the magnitude of
the acceleration?
Example 36 1991 Question 5
(1 mark)
What is the tension force in the coupling between the truck and trailer when the road train is
moving at constant speed?
When the road train is travelling along a straight, level road at 20 m s-1 the truck is put into neutral
gear and the train allowed to roll to a stop.
Example 37 1991 Question 6
How far will it travel before coming to rest?
A car is tested on a straight level track on a day when there is no wind. When the car reaches
20.0 m s-1 the driver puts the car into neutral gear. The wheels are no longer driving the car, and it
gradually slows due to frictional forces. Measurements show that the car's speed decreases
uniformly from 20.0 m s-1 to 18.0 m s-1 in 4.0 s. The mass of the car is 1100 kg.
Example 38 1996 Question 5
(1 mark)
Calculate the magnitude of the net force on the car when it is slowing down.
Example 39 1996 Question 6
(1 mark)
Calculate the power required at the wheels of the car for it to be driven at a constant speed of
20.0 m s-1 on the straight level track.
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A car of mass 1000 kg is being towed on a level road by a van of mass 2000 kg. There is a
constant retarding force, due to air resistance and friction, of 500 N on the van, and 300 N on
the car. The vehicles are travelling at a constant speed.
Example 40 2004 Pilot Question 9
(2 marks)
What is the magnitude of the force driving the van?
Example 41 2004 Pilot Question 10
(2 marks)
What is the value of the tension, T, in the towbar?
When travelling at a speed of 15.0 ms-1 the van driver stops the engine, and the van and car slow
down at a constant rate due to the constant retarding forces acting on the vehicles.
Example 42 2004 Pilot Question 11
(3 marks)
How far will the van and car travel before coming to rest?
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The figure shows a car of mass 1600 kg towing a boat and trailer of mass 1200 kg.
The driver changes the engine power to maintain a constant speed of 72 km h-1 on a straight road.
The total retarding force on the car is 1400 N and on the boat and trailer 1200 N.
Example 43 2004 Question 1
(2 marks)
Calculate the driving force exerted by the car at this speed.
To overtake another car the driver accelerates at a constant rate of 1.20 m s-2 from 72 km h-1 until
reaching 108 km h-1
Example 44 2004 Question 2
(3 marks)
Calculate the distance covered during this acceleration.
Example 45 2004 Question 3
(3 marks)
Calculate the tension in the coupling between the car and trailer during the acceleration. (Assume
the same retarding forces of 1400 N and 1200 N respectively.)
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Two students are discussing the forces on the tyres of a car. Both agree that there must be a
friction force acting on the tyres of a car. The first student claims that the friction force acts to
oppose the motion of the car and slow it down, for example, when braking. The second student
claims that friction acts in the direction of motion as a driving force to speed the car up when
accelerating.
Example 46 2004 Question 8
(3 marks)
On the diagram of the front-wheel drive car in the figure below, clearly show all the forces acting
on the tyres of the car when it is accelerating forwards in a straight line. Use arrows for the force
vectors to show both the magnitude and point of action of the different forces.
Example 47 2004 Question 9
(2 marks)
On the diagram of the same car below clearly show all the forces acting on the tyres of the car
when it is braking in a straight line. Use arrows for the force vectors to show both the magnitude
and point of action of the different forces.
A recent Transport Accident Commission television advertisement explains the significant
difference between car stopping distances when travelling at 30 kmh-1 and 60 kmh-1.
Example 48 2000 Question 10
(2 marks)
The stopping distance, from when the brakes are applied, for a car travelling at 30 kmh-1 is 10 m.
Which one (A – D) is the best estimate of the stopping distance for the same car, under the same
braking, but travelling at 60 kmh-1?
A. 20 m
B. 30 m
C. 40 m
D. 90 m
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A car of mass 1300 kg has a caravan of mass 900 kg attached to it. The car and caravan move
off from rest. They have an initial acceleration of 1.25 m s-2.
Example 49 2000 Question 11
(2 marks)
What is the net force acting on the total system of car and caravan as it moves off from rest?
Example 50 2000 Question 12
(3 marks)
What is the tension in the coupling between the car and the caravan as they start to accelerate?
After some time the car reaches a speed of 100 kmh-1, and the driver adjusts the engine power to
maintain this constant speed. At this speed, the total retarding force on the car is 1300 N, and on
the caravan 1100 N.
Example 51 2000 Question 13
(2 marks)
What driving force is being exerted by the car at this speed?
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Anna is jumping on a trampoline. The figure below shows Anna at successive stages of her
downward motion.
a
b
c
d
Figure c shows Anna at a time when she is travelling downwards and slowing down.
Example 52 1999 Question 6
(2 marks)
What is the direction of Anna’s acceleration at the time shown in Figure 4c? Explain your
answer.
Example 53 1999 Question 7
(3 marks)
On Figure c draw arrows that show the two individual forces acting on Anna at this instant.
Label each arrow with the name of the force and indicate the relative magnitudes of the forces
by the lengths of the arrows you draw.
Connected bodies
Problems involving the motion of two bodies connected by strings are solved on the following
assumptions;
 the string is assumed light and inextensible so its weight can be neglected and
 there is no change in length as the tension varies.
In these cases, tension is the condition of a body subjected
to equal but opposite forces which attempt to increase its
length, and tension forces are pulls exerted by a string on
the bodies to which it is attached.
To solve these problems you need to consider the vertical
direction first.
m1g – T = m1a
The direction of this acceleration must be downwards.
This leads to: T = m1g – m1a
The tension in the string is the same in both directions, therefore T = m2a.
Since both bodies are connected by an inextensible string, both bodies must have the same
acceleration.
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The vertical forces acting on m2, (not shown) cancel each other out, and do not impact on its
motion.
Combing these two equations gives a =
T=
m1
m1 + m 2
m1m2
m1 + m 2
g
g
Two blocks, each of mass m, are connected by means of a string which passes over
a frictionless pulley. One is at rest on a frictionless table; the other is held at rest in the position
shown.
Example 54 1981 Question 9
(1 mark)
What is the force of the string on Block X?
A.
B.
C.
D.
zero
mg/2
mg
2mg
Example 55 1981 Question 10
(1 mark)
Block Y is now released and the blocks move. What will then be the net force on Y?
A.
B.
C.
D.
zero
mg/2
mg
2mg
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A block of mass M2 is held at rest on a horizontal frictionless table. A string is attached to a mass
M1 over a light frictionless pulley as shown below.
Example 56 1983 Question 6
(1 mark)
What is the tension in the string while the block is being held?
Example 57 1983 Question 7
(1 mark)
If the block is then released, which of the statements below best describes the subsequent state of
the block?
M + M2
A
The block starts to move with an acceleration 1
g
M2
M1
B
The block starts to move with an acceleration
g
M1 + M2
C
The block starts to move with an acceleration g
D
The direction of the motion depends on whether M2 is greater or less than M1.
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A mass M1 is accelerated from rest at X along a horizontal frictionless surface by a tight string
passing over a frictionless pulley and attached to mass M2.
M2 falls from its rest position a distance, d, where it strikes the floor. g is the acceleration due to
gravity.
Example 58 1986 Question 21
(1 mark)
What is the acceleration of M1 in the section XY?
Example 59 1986 Question 22
What is the speed of M1 at Y?
(1 mark)
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In a laboratory experiment two blocks, M1 (of mass 4.0 kg).and M2 (of mass 2.0 kg) are connected
by a light inextensible string as shown below. There is a constant frictional force of 2.0 N between
M1 and the table. All other friction forces should be ignored.
The masses are released from rest at the positions shown in the diagram.
Example 60 1988 Question 12
(1 mark)
What is the magnitude of the acceleration of the masses?
Example 61 1988 Question 13
(1 mark)
What is the magnitude of the tension in the string?
Example 62 1988 Question 14
(1 mark)
What is the speed of M1 at the moment when M2 hits the floor?
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Solutions
Example 1
1973 Question 8
F
Using a =
, the maximum acceleration is
m
caused by the maximum force that the car
can exert on the road, (hence the maximum
force that the road will exert on the car to
accelerate the car).
If Fmax is fixed, then tripling the mass will
result in 31 the original acceleration.
1 ms-2
(ANS)
Example 2
1976 Question 1
The distance travelled is the area under the
graph. Use the trapezium formula,
A = ½(a +b)h to get
d = ½ x (3000 + 3500) x 30
d = 97500 m
d = 9.8 x 104 m
(ANS)
(this is a bit difficult to express when you
take sig figs into consideration)
Example 3
1976 Question 2
Use F = ma to find the net force acting
The gradient of the graph over the first 300
seconds will give the acceleration.
a=
30
= 0.1
300
 F = 5.0 x 105 x 0.1
 F = 5.0 x 104 N
(ANS)
Example 4
1976 Question 3
The engine is required to provide a net force
of 5.0 x 104, so the engine must also provide
the force required to overcome the frictional
force of
1.5 x 104N.
 Fengine = 5.0 x 104 + 1.0 x 104
 Fengine = 6.5 x 104 N
(ANS)
Example 5
1976 Question 4
Power is the rate of doing work, it is given by
the formula P = Fv.
When the train is travelling at a constant
speed the engine just needs to overcome
the frictional forces.
 P = 1.5 x 104 x 30
 P = 4.5 x 105
 P = 450 kW
(ANS)
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Example 6
1976 Question 9
The force of the road on the bicycle is
always perpendicular to the surface of the
road.
C
(ANS)
The road cannot push sideways.
Example 7
1976 Question 10
Since the speed is constant the acceleration
must be zero.
D
(ANS)
Example 8
1976 Question 11
The sum of the forces must be zero,
therefore the force of the road on the bicycle
must be the sum of the normal reaction (C)
and the frictional force on the tyres (A).
B
(ANS)
Example 9
1977 Question 1
Use v2 = u2 + 2ax
202 = 102 + 2x a x 200
400 = 100 + 400a
300 = 400a
a = 0.75 ms-2
(ANS)
Example 10 1977 Question 2
The net force is found from F = ma
F = 800 x 0.75
F = 600 N (ANS)
Example 11 1977 Question 3
The towing car also needs to overcome the
frictional forces.
Ftowing car = 600 + 500
Ftowing car = 1100 N (ANS)
Example 12 1977 Question 4
Since the speed is constant, the net force
must equal zero. The magnitude of the net
force is the same as the magnitude of the
frictional force
F = 500 N (ANS)
Example 13 1979 Question 8
Using F = m x a
We get 48 = (4 + B) x 4
 12 = 4 + B
 B = 8 kg
(ANS)
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Example 14 1979 Question 9
The force that A exerts on B is the force that
accelerates B.
Using F = ma
F=8x4
 FA on B = 32 N
(ANS)
Example 15 1979 Question 10
This is the classic example of Newton’s third
law.
FB on A = - FA on B
We are asked for the magnitude,
 FB on A = 32 N
(ANS)
Example 16 1979 Question 11
Since the force is the same as before and
the total mass hasn’t changed, the
acceleration will also be 4.0 ms-2.
Therefore to accelerate 4.0 kg at 4.0 ms-2
requires a force of 16 N
 FB on A = 16 N
(ANS)
Example 17 1980 Question 14
The net force to accelerate a 1.0 kg mass at
2.0 ms-2 is 2N.
To overcome the weight, the spring balance
needs to supply a 10 N force.
Therefore the total force required to
accelerate the mass upwards at 2.0 ms-2 will
be 10 + 2 =12N
 12 N
(ANS)
Example 18 1980 Question 15
If everything is moving up at a constant
velocity, then the spring balance only needs
to overcome the weight (otherwise the mass
would fall to the floor of the lift).
 10 N
(ANS)
Example 19 1981 Question 5
The man and the parachute are travelling at
a constant speed, so the net force acting on
the man is zero
D
(ANS)
Example 20 1981 Question 6
The direction of the girl’s acceleration is
north. Therefore the net force acting on her
must be northerly.
A
(ANS)
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Example 21
1981 Question 7
N
g cosθ
g
g sinθ
In this case the unbalanced acceleration on
the block is given by g sinθ.
h
a= g
L
k=g
 C (ANS)
Example 22 1983 Question 4
With a constant thrust the rocket will
accelerate in the direction of the thrust. This
means that the answer needs to be D, E or
F. Once the thrust stops, there will not be
any forces acting on the rocket, so it will
continue to travel in the same direction.
 E (ANS)
Example 23 1984 Question 23a
Without air resistance the only force acting
on the golf ball is its weight. This acts
vertically downwards.
 A (ANS)
Example 24 1984 Question 23b
With air resistance, there are now two forces
acting on the golf ball its weight, and air
resistance in the opposite direction to the
motion. (to the left).
 D (ANS)
Example 25 1984 Question 24
The ball is momentarily stationary, but it is
about to rebound. This means that to get it
to rebound the net force acting on it must be
upwards.
 C (ANS)
Example 26 1985 Question 14
The force the block 4 exerts on block 5,
causes block 5 to accelerate. The system is
accelerating at 0.2 ms-2.
Therefore the net force on block 5 is
F = ma
F = 1 x 0.2
 0.2N
(ANS)
Bialik Physics Unit 3 2015
Forces notes
Example 27 1985 Question 15
The force the block 3 exerts on block 4,
causes blocks 4 and 5 to accelerate. The
system is accelerating at 0.2 ms-2.
Therefore the net force on block 4 is
F = ma
F = (1 + 1) x 0.2
 0.4N
(ANS)
Example 28 1985 Question 17
Newton’s third law states that
FA on B = -FB on A
The weight of an object can be written as
FEarth on Mass.
From an action reaction pair perspective, this
means that the opposite to the weight is:
FMass on Earth.
 A (ANS)
Example 29 1986 Question 1
Using F = ma, we need to find ‘a’.
The acceleration is the gradient of the
velocity-time graph.
The question asks for the magnitude. The
5
Δv
gradient is:
=
Δt
1
a = -5
Therefore F = 4 x 5
F = 20N
(ANS)
Example 30 1986 Question 2
This is equal to (in magnitude), but in the
opposite direction to the force of the floor on
the block.
F = 20N
(ANS)
Example 31 1987 Question 5 (72%)
The acceleration is the gradient of the
velocity time graph.
14.4
Δv
gradient is:
=
Δt
2
a = 7.2 ms-2
(ANS)
Example 32 1987 Question 6 (76%)
This ratio must always be 1.
Examiners comment
This question simply asked candidates to
realise that the force of gravity would be the
same irrespective of speed.
Example 33 1987 Question 7 (76%)
Using F = ma
F = 100 x 7.2
F = 720 N (ANS)
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Example 34 1987 Question 8 (77%)
At 20 secs, the velocity is constant, this
means that the net force must be zero.
Therefore the acceleration is zero.
a=0
(ANS)
Examiners comment
Questions 7 and 8 tested the relationship
between resultant force and acceleration.
These two questions, in conjunction with
others, seem to reflect many candidates' lack
of appreciation of Newton's laws. “lf the
velocity is constant there is no acceleration
and hence no resultant force.
Example 35 1991 Question 4
Driving force – friction forces = net force
FNet = ma
3.9 x 104 – (2 x 2.5 x 103) = 4.0 x 104 x a
3.9  104  0.5  10 4
a =
4.0 x 104
a = 0.85 ms-2
(ANS)
Example 36 1991 Question 5
The road train is moving at constant speed,
therefore, the net force acting on the trailer
must be zero.
The tension needs to overcome the frictional
force.
T = 2.5 x 103 N
(ANS)
Example 37 1991 Question 6
It is simplest to treat the truck and trailer as
the one system. The net retarding force is 5
x 103 N.
F
5  103
The deceleration is
=
m
4.0  104
= 0.125 ms-2
2
2
Now use v – u = 2ax
 -202 = 2 x 0.125 x ‘x’
 x = 400 ÷ 0.25
x = 1600 m
x = 1.6 x 103 m
(ANS)
Example 38
1996 Question 5
20  18
Δv
The acceleration is
=
Δt
4
= 0.5 ms-2
F = 1100 x 0.5
F = 5.5 x 102 N
(ANS)
Bialik Physics Unit 3 2015
Forces notes
Examiners comments
Students needed to calculate the acceleration
(change in velocity divided by time) and then
apply Newton's 2nd Law (F= ma) to
determine the net force, resulting in an
answer of 550 N.
Eighty percent of students were able to do
this correctly, a pleasing result indeed.
Example 39 1996 Question 6
Power is the rate at which work is being
Fd
done, P =
t
Power is found from P = Fv
P = 5.5 x 102 x 20
P = 1.1 x 104 W
(ANS)
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Example 42 2004 Pilot Question 11
With the driving force from the van = zero,
then the only forces acting are the frictional
forces that total 800N. This is acting on a
3000 kg mass. We treat both as one,
because the deceleration of both must be the
same as they are joined by the solid towbar.
Use v2 = u2 – 2ax
800
Let v = 0, u = 15 and a =
= 0.2666m/s2
3000
 0 = 225 – 0.53  x
 x = 421.9
= 422 m (ANS)
Make sure that you don’t round off to early in
this solution as it can lead to an answer of
417 m.
Examiners comments
In order to maintain a constant speed the
driving force needed to be equal and opposite
to the resistance force, as calculated in the
previous question. The power at the wheels is
then calculated using the formula, P = Fv.
where F is the driving force at the wheels of
the car. Assuming the correct answer to
the previous question, this calculation
resulted in a power of 11 000 W.
Consequentially correct answers, arising from
an incorrect answer to question 5, were also
scored as correct.
Forty per cent of students were able to
correctly answer this question. The most
common incorrect answer (220 000 W) arose
from students applying the formula, P = Fv,
but substituting in the weight force for F.
Example 43 2004 Question 1
Since the speed is constant, the net force is
zero. Hence the driving force exerted by the
car must equal the sum of the resistive
forces.
 1400 + 1200 = 2600 N
(ANS)
Example 40 2004 Pilot Question 9
Since they are travelling at a constant speed,
the net force must be zero. The only driving
force is due to the van’s wheels pushing the
van forward. This must be large enough to
overcome the two resistive forces.
 500 + 300 = 800N
(ANS)
Example 45 2004 Question 3
You need to convert both the 72 km h-1 and
108 km hr-1 to m s-1.
This is done by dividing by 3.6. (This should
be on your cheat sheet)
 72 km hr-1 = 20m s-1
and 108km hr-1 = 30m s-1.
Example 41 2004 Pilot Question 10
The tension in the towbar is the only force
acting forwards on the car being towed.
T = 300N, so that the net force acting on
the towed vehicle is zero, because it is
travelling at a constant speed.
T = 300N
(ANS)
Use v2 = u2 + 2ax
 302 = 202 + 2  1.20  x
 900 – 400 = 2.4  x
500

= 208 m. (ANS)
2.4
Example 44 2004 Question 2
The tension in the coupling (towbar) is the
only force acting forwards on the trailer.
The net force on the trailer must give rise to
its acceleration. Using F = ma gives F = 1200
 1.20 = 1440N.
This gives the net force acting on the trailer to
be 1440 (to provide the acceleration) and
1200 (to overcome friction)
 Tension = 1200 + 1440
= 2640N (ANS)
Bialik Physics Unit 3 2015
Example 46
Forces notes
2004 Question 8 (37%)
The net force acting on the car MUST be in
the direction of the acceleration. This net
force can only come from the frictional
contact between the tyres and the road. You
need to remember that the frictional forces
oppose the motion, but in this case, to get the
car to move forwards the tyre actually wants
to move backwards at ground level.
Examiners comment
The average score for Question 8 (1.1/3)
indicates that the concept of friction as a
driving force was poorly understood. While
many students were aware that there had to
be a net force acting forwards to accelerate
the car, the origin and point of application of
this force was rarely correctly sketched. In
fact, many students still sketched the road–
tyre friction force acting backwards so as to
oppose the motion of the car. It was also
disappointing to note the number of sketches
that did not show the correct point of
application of the weight, normal or frictional
forces. It is quite apparent that VCE Physics
students need more practice in drawing freebody force diagrams.
Example 47
2004 Question 9 (45%)
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Examiners comment
The average score for this question (0.9/2)
again indicates that students found this nearly
as difficult as the previous question and that
friction as a braking force was not thoroughly
understood.
Example 48 2000 Question 10 (57%)
C
This question was testing you understanding
of how speed and stopping distance is
related. Let’s assume the mass of the car,
and the force stopping it, are constant.
That is, if the car has a kinetic energy of
1
mv 2 and it requires a force by a distance to
2
stop it then:
1
mv 2 = F  d
2
if the speed of the car is doubled then the
stopping distance in 4 times larger. This is
known as a square relationship.
For this question the speed of the car is
30kmh-1 and the stopping distance is 10m.
The speed of the car is then doubled to
60kmh-1 then because the speed has
doubled the stopping distance is now 4
times the 10m, which is 40m so C is the
correct answer.
Examiners comment
Students needed to understand that work
done equals change in kinetic energy. If a car
is travelling at twice the speed then it will
have four times the kinetic energy and the
brakes will need to do four times as much
work in bringing it to rest. Hence, for the
same braking force the stopping distance will
be about four times as far.
Distance C (40 m) was the best estimate of
the stopping distance.
Example 49 2000 Question 11 (58%)
Just Fnet = ma
Using the mass of the system.
Fnet = (1300 + 900)  1.25
= 2.75  103N
(ANS)
The net force acting on the car MUST be in
the direction of the acceleration. This net
force can only come from the frictional
contact between the tyres and the road.
Since the car is decelerating the net force
must be in the opposite direction to the
motion.
Example 50 2000 Question 12 (58%)
For this question think of the caravan as an
object that is accelerating that is pulling it.
So F = ma
= 900  1.25
= 1.13  103N (ANS)
Bialik Physics Unit 3 2015
Forces notes
Examiners comment
Application of Newton’s Second Law for the
caravan alone resulted in an answer of 1.125
·x 103 N for the tension in the coupling.
Students needed to be aware that the driving
force for the caravan was provided solely by
the tension in the coupling.
Example 51 2000 Question 13 (63%)
If the car is travelling at a constant velocity
then there is no net force acting on the car.
Therefore the magnitude of the driving force
from the motor equals the retarding force of
the car and the caravan.
Fd = 1300 + 1100
= 2400N (ANS)
Examiners comment
Constant velocity implies a net force of zero.
Hence, the driving force must be equal and
opposite to the retarding forces, i.e. 2400 N.
Any errors were due to students not realising
that constant speed in a straight line implies a
net force of zero.
Example 52 1999 Question 6 (55%)
Up
At this time Anna is travelling downwards
and slowing down. This means that her
acceleration is up, because it is opposing
the motion (she is slowing down).
Examiners comment
Anna was travelling downwards and slowing,
hence the direction of her acceleration was
upwards. The main errors were to reason that
the net force was her weight and thus her
acceleration was
downwards, or that since she was moving
downwards then her acceleration must also
be downwards. It was certainly clear that
many students experience great difficulty in
distinguishing between velocity and
acceleration vectors, (a number of students
based their answer on the state of Anna's
hair at the instant).
28 of 30
Example 53
1999 Question 7 (50%)
The reaction force > weight, because she is
slowing down, ie. a net upward force.
Examiner’s comment
Students were expected to draw two force
arrows on Figure C. One arrow was the
weight force, acting through Anna’s centre of
mass and the other arrow being the normal
contact force, acting at her feet. The arrow
for the normal contact force should have
been longer than the weight force arrow.
The average mark for this question was
1.5/3, with the most common errors being in
either choosing the incorrect point of
application for each force or in not indicating
the relative magnitudes as specifically asked
in the question.
Example 54 1981 Question 9
Block X is initially stationary and remains
stationary until the hand supporting Block Y
is removed. Therefore the net force on Block
X is zero.
A
(ANS)
Bialik Physics Unit 3 2015
Forces notes
Example 55 1981 Question 10
When Block Y is released, its weight
accelerates it downward.
As the two blocks are connected by a string,
they are going to both have the same
acceleration when released.
The net force acting on Block Y is given by
mg – T = ma
The net force acting on Block X is given by T
= ma
Substituting for T gives mg – ma = ma
mg = 2ma
mg
 ma =
2
B
(ANS)
Example 56
1983 Question 6
29 of 30
Example 57
1983 Question 7
As the two blocks are connected by a string,
they are going to both have the same
acceleration when released.
The net force acting on Block M1 is given by
M1g – T = M1a
The net force acting on Block M2 is given by
T = M2a
Substituting for T gives M1g – M2a = M1a
 M1g = M1a + M2a
 M1g = (M1 + M2)a
M1g
a=
M1 +M2
B
(ANS)
Example 58 1986 Question 21
As the two blocks are connected by a string,
they are going to both have the same
acceleration when released.
The net force acting on Block M2 is given by
4g – T = 4a
The net force acting on Block M1 is given by
T = 1a
Substituting for T gives 4g – a = 4a
 4g = 4a + a
 4g = 5a
4 ×10
a=
5
8ms-2
(ANS)
Since the block is being held, M1 is
stationary, and going to remain stationary,
therefore the net force acting on it is zero.
 T = M1g
Example 59 1986 Question 22
Use v2 – u2 = 2ax
 v2 – 0 = 2 x 8 x 1
 v2 = 16
 v = 4 ms-1 (ANS)
Bialik Physics Unit 3 2015
Forces notes
Example 60 1988 Question 12
As the two blocks are connected by a string,
they are going to both have the same
acceleration when released.
The net force acting on Block M2 is given by
2g – T = 2a
The net force acting on Block M1 is given by
T – 2 = 4a
 T = 4a + 2
Substituting for T gives 2g – (4a + 2) = 2a
 2g = 2a + 4a + 2
 20 = 6a = 2
 18 = 6a
 a = 3 ms-2 (ANS)
Example 61 1988 Question 13
Using T = 4a + 2, give T = 4 x 3 + 2
 = 14 N
(ANS)
Example 62 1988 Question 14
Use v2 – u2 = 2ax
 v2 – 0 = 2 x 3 x 1.5
 v2 = 9
 v = 3 ms-1 (ANS)
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