Download friction newton`s third law

FRICTION
Frictional forces always act in the direction opposite to the motion. Since acceleration is
determined by the net force, and not just the applied force, frictional forces, if they exist,
must be taken into account.
NEWTON’S THIRD LAW
If Object A pushes against Object B with a certain force, then Object B will push back
against Object A with a force that is equal in magnitude but opposite in direction. The two
forces involved are called action/reaction forces.
Please note (this is very important): Action/reaction forces act on separate objects, never
do both forces act on only one object. (One force acts on Object A, the other acts on
Object B).
FAonB   FBonA
QUESTION 3
A book is sitting on a table.
(a)
What is the net force acting on the book?
(b)
Identify the individual forces acting on the book. Draw a diagram to show these forces.
(c)
Is the normal force of the table on the book, and the force of gravity, an action-reaction
pair in the sense of Newton’s Third Law? Give reasons to support your choice.
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The Essentials – Year 11 Physics – Book 1
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QUESTION 4
Two masses, 10 kg and 20 kg, are placed in contact on a rough surface as shown. A person
exerts a force of 45 N on the 10 kg mass. The magnitude of the frictional force acting on the
10 kg mass is 10 N and the magnitude of the frictional force acting on the 20 kg mass is
20 N.
(a)
What is the acceleration of the system of two masses?
(b)
What is the force that the 20 kg mass exerts on the 10 kg mass while they are
in motion?
(c)
What force does the 10 kg mass exert on the larger box while they are in motion?
(d)
Identify all the forces acting on the 10 kg mass?
Solution
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The Essentials – Year 11 Physics – Book 1
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FORCES ACTING ON AN INCLINED PLANE
An object of mass 10 kg is at rest on an plane that makes an angle of 30° to the horizontal.
We can fully specify all of the forces that act on the mass. (Note: Since the mass is at rest,
and remaining so, the net force acting on it must be zero).
Using highlighters or coloured pens can help greatly.
Reaction force = 86.6 N
m = 10 kg
Friction = 50 N
component of weight
along the plane = 50 N
30
30
0
component of weight
perpendicular to plane
= 86.6 N
0
Weight = mg
= 100 N
If the plane was frictionless then:
Acceleration along the plane = g sin 
Normal reaction force = m g cos 
where  is the angle of the plane above the horizontal.
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The Essentials – Year 11 Physics – Book 1
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ENERGY
WORK
The Simplest Case:
If a force, F, causes a displacement d, then work has been done.
F
F
d
If:
(a)
The applied force remains constant, and
(b)
The applied force is in the same direction as the resultant motion.
then:
W=F.d
Work done (joule, J) = Applied force (N), displacement (m)
Work is a scalar quantity. (It is the dot product of two vectors.)
Work is measured in joules. The joule is the standard unit of energy. Work and energy are
equivalent – whenever work is done, there is a conversion of energy.
Note: To calculate the work done, it is the applied force that concerns us, not the net force.
Another note:
The applied force must cause a displacement if work is to be done. If you exert a force but
do not cause any motion (by, say, pushing against a strong brick wall), you have done no
work.
If a graphical approach is taken, work is the area under a force-displacement graph.
When standard units are used for force and displacement, the work is in units of
Joules.
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The Essentials – Year 11 Physics – Book 1
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QUESTION 5
The graph represents the velocity-time relation for a block of mass 0.2 kg. The block was
given an initial push (at time t = 0), slid across a floor and came to rest.
(a)
(b)
(c)
(d)
What distance did the block travel in the 0.8 s time interval?
What was the frictional force of the floor on the block?
Draw a frictional force – time graph.
How much work was done in bringing the block to rest?
Solution
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The Essentials – Year 11 Physics – Book 1
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CONSTANT F, BUT F AND D IN
DIFFERENT DIRECTIONS
If the F and d vectors do not have the same direction, but subtend an angle  , thus:
F

d
then: W = F.d cos 
Again, this formula is only applicable if F is constant.
Note: If F has the same direction as d, then  = 0. In that case cos  =1, so W = F.dcos 
simply becomes W = F.d.
NON-CONSTANT F
If F is not constant (eg. as when compressing a spring), then the work done can be found by
calculating the area under the force-distance graph, no matter what the shape of the graph.
For Example:
Force (N)
3
2
1
1
2
3
4
5
6
Displacement (m)
W = Area of triangle = 9 J
(This result can be obtained by using the average force, 1.5 N, in the equation W=F.d)
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The Essentials – Year 11 Physics – Book 1
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KINETIC ENERGY
Any moving object has kinetic energy.
1
Ek = 2 mv2
1
Kinetic energy (J) = 2 x mass (kg) x (velocity (m/s))2
Note that EK  v 2 . This demonstrates that to double velocity requires four times the
energy.
Note that kinetic energy, just like any other type of energy, is a scalar quantity and is
measured in joules.
WORK AND KINETIC ENERGY
Consider an object of mass m with an initial velocity u. A force, F, acts on this object over a
distance d, causing it to accelerate from its initial velocity u to a final velocity v.
u
v
F
F
d
The work done is:
W = F.d
W = ma.d
W=m(
v2 - u2
)
2
(since F = ma)
(since ad =
v2 - u2
, from the equation v2 = u2 + 2 ad)
2
1
1
W = 2 mv2 - 2 mu2
W = (final Ek) - (initial Ek)
W =  Ek
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The Essentials – Year 11 Physics – Book 1
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QUESTION 6
A box of mass 20 kg, and initially at rest, is pulled 3.0 m along a rough floor with a force
of 100 N.
The friction force acting on the box is a constant 30 N.
What is the change in kinetic energy of the box?
Solution
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The Essentials – Year 11 Physics – Book 1
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GRAVITATIONAL POTENTIAL ENERGY
In order to lift an object of mass m, one must apply an average force equal to the object’s
weight, mg.
F
mass = m
weight = mg
Consider lifting the object is lifted to a height h.
F
h
F
The work done is:
W=F.d
W = mg . h
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Since work is done, energy is being converted – the work done in lifting an object of mass m
to a height h is the object’s gravitational potential energy, Ug, at that height.
W =  Ug
Ug = mgh
Gravitational potential energy(J) = Mass(kg) x Gravitational field strength
(N/kg) x Height (m).
Near the Earth’s surface, g = 9.8 N/kg. The value of g can be quite different elsewhere.
Note that Ug is zero if h is zero. Our choice for the zero level for h is entirely arbitrary,
i.e. We can choose to have our zero level wherever we find it most convenient – it could be
ground level, but usually the lowest point of the object's motion, wherever it may be, is the
best choice.
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ELASTIC POTENTIAL ENERGY
Energy is also stored when a spring is stretched or compressed. For the ideal spring the
relationship between the force and extension or compression is given by Hooke’s Law which
states:
F = -ks
Where:
s is the change in length of the material (either compression or extension).
k is the force constant or spring constant = Gradient.
The value of k is a measure of the stiffness of the spring.
As a vector quantity, the equation should read: F = -ks As the restoring force and extension
are always in the opposite direction.
We learnt earlier that the area under a force displacement graph represents the amount of
work done. The same holds true for a force – extension graph (which is basically the same).
Work done in compressing the spring = Ep
Ep =
area under graph = ½ x F x s = ½ x ks x s = ½ ks2
(units are Joules)
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TOTAL ENERGY
An object may have both kinetic energy and gravitational potential energy. If these are the
only types of energy it has, then the object's total energy is:
Etotal = Ek + Ug
Whenever work is done on an object, its energy changes. If the applied force changes the
object's speed, but not its height above the ground, then there is a change only in the
object's kinetic energy. In that case:
W =  Ek
If the applied force changes the object's height above the ground, but not its speed, then
there is a change only in the object's gravitational potential energy. In that case:
W =  Ug
Often, however, applied forces change both the object's speed and its height above the
ground. In that case:
W =  Etotal
W =  Ek +  Ug
CONSERVATION OF ENERGY
Energy can be transferred from one object to another, or it can be converted from one type
of energy into another type, but it cannot be created or destroyed. Therefore, the total
amount of energy in a closed system, Etotal, always remains constant.
What about a falling object? ... Say, from the Rialto……?
 The School For Excellence 2011
The Essentials – Year 11 Physics – Book 1
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QUESTION 7
In an amusement parlour a certain game uses a piston attached to a spring to fire a ball
around a tube, mounted in a vertical plane as shown in the figure below. The spring has a
stiffness of 300 Nm-1. The player draws the piston back a distance of 0.070 m. The ball,
which is resting against the piston, has a mass of 0.040 kg.
(a)
What is the magnitude of the force exerted by the player on the spring when the
spring is fully compressed?
(b)
How much work is done is compressing the spring?
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In order to play a shot, the player releases the piston and the ball moves around the tube.
Ignore friction in the tube.
(c)
Assuming half the energy in the spring is transferred to the ball, with what speed does
the ball leave the piston?
(d)
What is the speed of the ball as it leaves the tube at point P, 0.50 m higher than the
spring mechanism?
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