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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.1 Analysing Linear Motion
1.
Linear motion is motion in a straight line.
Distance And Displacement
1. Distance is the total length of the path traveled by an object.
2. Distance is a scalar quantity. It has magnitude but no direction.
3. Displacement is the distance of its final position from its initial position in specified direction.
4. Displacement is a vector quantity. It involves both magnitude and direction.
Example 1
To test your understanding of this distinction, consider the motion depicted in the diagram below. A physics teacher
walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North.
Even though the physics teacher has walked a total distance of 12 meters, her displacement is 0 meters. During the
course of her motion, she has "covered 12 meters of ground" (distance = 12 m). Yet, when she is finished walking, she
is not "out of place" – i.e., there is no displacement for her motion (displacement = 0 m). Displacement, being a vector
quantity, must give attention to direction. The 4 meters east is canceled by the 4 meters west; and the 2 meters south is
canceled by the 2 meters north.
Speed and Velocity
1. Speed is the ________________________________________________________. Speed during the course of a
motion is often computed using the following equation:
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2. Velocity is the ____________________________________________________________. Velocity is often
computed using the equation:
3. Speed is a ___________________ whereas velocity is __________________.
4. Both speed and velocity have the same SI unit. They are measured in meter per second or m s-1. Other unit may
be in cm s-1 or km h-1.
5. The average speed during the course of a motion is often computed using the following equation:
6. Meanwhile, the average velocity is often computed using the equation:
Example 1
Example 2
A man running in a race covers 60 m in 12 s.
(a) What is his speed in,
(i) m s-1
(ii) km h-1
(b) If he takes 40 s to complete the race, what is his
distance covered?
(c) Another man runs with a speed of 7.5 m s-1, how long
did he take to complete the race?
The physics teacher walks 4 meters East, 2 meters South,
4 meters West, and finally 2 meters North. The entire
motion lasted for 24 seconds. Determine the average
speed and the average velocity.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Acceleration And Deceleration
1. Acceleration is ________________________________________________________________________
2. It can be written as;
3. Acceleration is a _________________________________________. The SI unit for acceleration is m s-2.
4. The acceleration is positive if the velocity increases with time. The acceleration is negative if the velocity
decreases with time. It is also called deceleration.
5. Figure 2.1, shows that the car experiences acceleration, a constant velocity and then a deceleration.
Figure 2.1
Example 1
A vehicle accelerates uniformly from rest to a speed of 25 m s-1 in 100 s along a straight road. It then decelerates
uniformly at 0.2 m s-2 for 60 s. Find
(a) the initial acceleration
(b) the final speed
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Study Of Linear Motion
Ticker-timer
1.
In the laboratory, a ticker-timer as shown in figure 2.2, with a trolley is used to study the motion of an object for a
short time interval.
Figure 2.2
2.
A ticker-timer consists of a small electrical vibrator which vibrates at the frequency of 50 Hz.
3.
Hands-on Activity/page 13 in your practical book.
Analysing motion using a ticker timer
-
to identify the types of motion
-
to determine displacement, average velocity and acceleration.
4.
The time taken to make 50 dots on the ticker tape is 1 s. Hence, the time interval between 2 consecutive dots is
1
= 0.02 s
50
5.
To determine the time interval of motion of the object:
Time interval = Number of tick x 0.02 s
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
6.
The following shows the different types of motion recorded on the ticker tape and tape chart.
Ticker tape and chart
Characteristics
Inference
- The separation
between dots stays the
same.
- The length of the
strips of the tape chart
is equal.
- The distance between
the dots increases
uniformly.
- The length of the
strips of the tape
increase uniformly.
- The distance between
the dots and the length
of strips of the tape
decreases uniformly.
7.
Change in distance between dots indicates a changing velocity and thus, acceleration. A constant distance
between dots represents a ____________________________ and _________________________.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example 1
Figure 2.3 shows a ticker tape chart obtained in an experiment to study the motion of trolley on an inclined plane.
Calculate the acceleration of the trolley.
Time for each strip
= 5 x 0.02 s
= 0.1 s
Time interval, t
between u and v,
= (5-1) strips x 0.1 s
= 4 x 0.1 s
= 0.4 s
Initial velocity, u
=
2.0 cm
0.1 s
= 20 cm s-1
Acceleration,
a=
=
(100 − 20)
0.4
=
80
0.4
Final velocity, v
=
10.0 cm
0.1 cm
= 100 cm s-1
v−u
t
= 200 cm s-2 / 2 m s-2
Example 2
Figure 2.4 shows a strip of ticker tape pulled through a ticker-timer by a freely falling metal sphere. The ticker-timer
vibrates at a frequency of 50 Hz. Determine the acceleration of the sphere.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Equations of Liner Motion with Uniform Acceleration
For an object in linear motion with uniform acceleration, problems involving the displacement, velocity, acceleration
and time of motion can be solve by using the equation of motion.
1.
s = displacement
2.
u = initial velocity
v = final velocity
3.
a = constant acceleration
t = time interval
4.
Example 1
Ali is driving a car at velocity of 30 m s-1. On seeing a student crossing the road, Ali steps on his brakes to stop the
car. The speed of the car decreases uniformly and stops after traveling 150 m.
(a)
What is the deceleration of the car when the brakes are applied?
(b)
What is the time interval before the car stops?
Example 2
Example 1
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example 3
Example 4
Example 5
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example 6
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.2 Analysing Motion Graphs
Motion Graphs
1.
There are two main types of linear motion graphs:
(a)
displacement-time
(b)
velocity-time
Displacement-Time Graph
1.
We can analyze the velocity of an object by plotting a graph of displacement against time.
2.
A student walks at a constant velocity from position A to reach position B in 200 s. He rests for 100 s
at position B and then walks back to position A using the same straight path. He reaches position A
after 200 s.
Figure 2.1
3.
The graph below shows the change in the distance and direction with time of the student.
(a) On a displacement-time graph, the
_______________ of the graph is
equal to the ________________ of
the object.
(b) Gradient of the graph in section I,
=
∆y
∆x
=
=
Figure 2.2
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
(c) On a displacement-time graph, a horizontal line (gradient = 0) shows the object is
___________________
(d) In section II of the graph, the student remains at position B from 200 s to 300 s.
(e) Gradient of the graph in section III,
=-
∆y
∆x
=
=
(f)
The negative sign shows that the direction of motion is opposite to the original direction.
(velocity is a vector quantity)
At t = 500 s, the graphs intersects the t-axis. The displacement at this moment is zero that is the
student has returned to the original position.
4.
The various displacement-time graphs are shown as follows;
s
t
-
There is no change in the displacement over time.
-
The straight line graph is parallel to the time axis.
-
velocity = gradient of graph = 0 m s-1
-
The rate of change of displacement is constant.
-
The straight line graph has a constant gradient.
-
The rate of change of displacement is increasing.
-
The gradient of the curve is increasing showing that the velocity is
increasing.
-
The object is moving with constant acceleration.
0
(a) __________________________
s
t
0
(b) __________________________
s
0
t
(c) __________________________
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
s
t
0
-
The gradient is decreasing with time.
-
The velocity is decreasing with time.
-
The object is moving with constant deceleration.
(d) __________________________
Velocity-Time Graph
1.
A car starts from rest and accelerates for 20 s until it reaches a velocity of 60 m s-1. The driver
maintains this velocity for 20 seconds. The velocity of the car is then gradually reduced until
it stops at t = 60 s.
Figure 2.3
2.
The graph below shows how the velocity of the car changes over a certain period of time.
Figure 2.4
(a) On a velocity-time graph, the gradient of the graph is equal to the __________________ of the
object.
(b) In section I, the acceleration of the car;
a
=
∆y
∆x
=
=
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
(c) On a velocity-time graph, a horizontal line represents a _________________. The gradient of
the graph is equal to 0.
(d) In section II, the car travels at a constant velocity of 60 m s-1 from t = 20 s to t = 40 s.
(e) In section III, gradient of the graph;
=-
∆y
∆x
=
=
(f)
The negative sign indicates a deceleration.
(g) On the velocity-time graph, the area under the graph is equal to the ______________
traveled.
(h) In section I, the area under the graph (the shaded triangle),
=
=
=
(i)
In section II, area under the graph(the shaded rectangle),
=
=
(j)
In section III, area under the graph,
=
=
3.
The various velocity-time graphs are shown below.
v
0
-
The gradient of the graph represents acceleration.
-
The gradient of the graph is zero and hence the
acceleration of the object is always zero.
-
The area under the graph represents displacement
traveled.
-
The area under the graph is zero and hence the
displacement traveled is zero.
-
The velocity stays the same. The object is moving with
constant velocity.
-
Since the gradient is equal to zero, the acceleration is
zero.
t
(a) ___________________________
v
0
t
(b) ___________________________
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
v
0
-
The velocity is increasing uniformly with time.
-
The gradient of the graph is constant and hence the
object is moving with constant acceleration.
-
The gradient of the graph is increasing.
-
The acceleration of the object is increasing.
-
The gradient is decreasing with time. Therefore the
velocity is decreasing.
-
The object is moving with constant deceleration.
t
(c) ___________________________
v
0
t
(d) ___________________________
v
u
0
v
0
t
t
(e) ___________________________
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example 1
The velocity-time graph of a object starting from
rest and traveling towards the east is as shown in
figure 2.5.
(a)
how long does the object travel towards the
east?
(b)
how long does the object travel towards the
west?
(c)
find the average speed and the average
velocity.
Solution
Example 2
Figure 2.6 shows the motion of a motorcycle.
Describe the motion of the motorcycle.
Solution
In section OA, the motorcycle is moving at a
4
constant velocity, v =
= 2 m s-1
2
The motorcycle then travels in the opposite
direction at a velocity of – 4 m s-1 and reaches its
original position at time, t = 3 s, and continues to
travel in this direction at the same speed of -4 m s-1.
Figure 2.6
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Mastery Exercise
1.
The figure 2.7, shows the displacementtime graph of a moving object.
(a)
What is the velocity of the object
in the initial period of 3 seconds?
(b)
How long is the object
stationary?
(c)
At what point in time does the
particle return to its original
position?
(d)
Calculate
(i)
Figure 2.7
the average speed, and
(ii) the average velocity of the
moving object.
2.
The figure 2.8, shows the velocity-time
graph of a motorcycle starting from rest
and traveling towards the north.
(a)
What is the deceleration from t =
10 s to t = 13 s?
(b)
What is the displacement of the
motorcycle after 13 s?
(c)
For how long was the motorcycle
traveling towards the south?
(d)
What is the displacement of the
motorcycle at t = 20 s?
(e)
What is the average velocity of
the motorcycle for the whole
journey?
Figure 2.8
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.3
1.
Understanding Inertia
The diagrams below show the situations involving inertia.
Situation 1
Figure 1
When a driver inside a car applies brake suddenly (figure 1 a), the driver and the passengers are move
forwards. When the car suddenly accelerates (figure 1 b), the driver and passengers will move
backwards.
Situation 2
Figure 2
Passengers in a bus will fall backwards when a bus stats suddenly from rest.
Situation 3
Figure 3
Passengers in a moving bus fall forward when the bus stops suddenly.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Situation 4
Figure 4
When the table cloth is suddenly pulled horizontally, the dishes on the table top still remain on the
table.
2.
The situations above show that our body has an inbuilt resistance to any change in its state of rest or
motion. This reluctance to change is called inertia.
3.
If an object is at rest, it tends to stay in that position unless some force puts that object into motion. If
an object is moving, inertia makes the moving object continue to move at a constant speed in the same
direction unless some external force changes the object’s motion.
4.
The inertia of an object is the tendency of the object ___________________________ or, if moving,
_________________________________________________________________________________
5.
The concept of inertia was explained by Sir Isaac Newton in the _____________________________
6.
Newton’s first law of motion states that every object will _________________________________ or
___________________ unless it is acted by an external force.
Relationship between Mass and inertia
Figure 5
1.
A child and an adult each sit on similar swings as shown in figure 5. If they are given a push, which
one of them will be more difficult to be moved? When both swings are set in motion, which one of
them will be more difficult to stop?
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.
The adult who has more mass will show more reluctance to change her state of rest or motion. This
property of the mass of a body which resists change from its state of rest or motion is known as
inertia.
3.
The ______________ the ____________, the _______________ its ________________
Effect of Inertia
1.
The effect of inertia is often seen in our daily activities. We make use of the positive effects of inertia
to solve some of our daily problems. The negative effects of inertia that can endanger our lives need to
find ways to reduce them.
2.
Positive effect of inertia.
The head of a hammer can be tightened onto the
wooden handle by applying a knock on the
handle. The head of the hammer has big mass
and will remain in its state of motion, thus fitting
it tighter on the handle.
The inertia of the ice skater keeps her gliding on
the surface in a straight line.
In order to pour out the chili source, the bottle is
moved down fast with a sudden stop. The sauce
inside the bottle moves together with the bottle.
When the bottle stops suddenly, the sauces
continue in its state of motion due to the effect of
its inertia.
Droplets of water on a wet umbrella are spun off
when the umbrella is rotated and stopped
suddenly. The droplets of water initially move
with the rotating umbrella. The inertia of the
droplets of water causes them to continue
moving even when the umbrella has stopped
spinning.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
3.
Negative effect of inertia.
Furniture carried by a lorry normally is tied up
together by string. When the lorry starts to move
suddenly, the furniture is more difficult to fall off due
to their inertia because their combined mass has
increased.
Strong iron structure between the driver’s cabin and
the load to stop the inertial movement of the heavy
load towards the driver when the lorry is brought to
a halt suddenly.
If the car crashes while travelling, the inertia of the
passengers causes them to continue in motion. This is
a dangerous situation. Upon impact the passengers
will crash into parts of the car immediately in front of
them and suffer injuries.
4.
Ways to reduce the negative effects of inertia.
(a)
Wearing safety belts when driving.
(b)
An air begs is fitted inside the steering wheel. It provides a cushion to prevent the driver from
hitting the steering wheel or dashboard during a collision.
(c)
The oil tank of an oil tanker lorry is usually divided into a few smaller compartments so that the
effects of inertia can be reduced.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.4 Analysing Momentum
You can catch a fast moving ping-pong ball easily with your bare hand.
A softball keeper must wear a glove to catch a hard and fast moving softball.
Why is a slow moving softball much easier to catch?
1.
When an object is moving with a speed in a straight line, we say that it has linear momentum.
2.
The amount of linear momentum of the object depends on its mass and velocity.
3.
We define linear momentum as the _____________________________________________________
Momentum
p
=
Mass x velocity
=
mxv
4.
SI unit for momentum is given as ________________. It can also be written as N s (newton second)
5.
Momentum is a _______________ quantity. The direction of the momentum is the same as the
direction of the velocity.
Example:
A billiard ball A of mass 0.5 kg is moving from left to right with a velocity of 2 m s-1 while another
billiard ball B of equal mass is moving from right to left with the same speed.
Solution:
Momentum of ball A
Momentum of ball B
=
mAvA
=
mBvB
=
0.5 x 2
=
0.5 x (-2)
=
-1 kg m s-1
= 1 kg m s
-1
* Negative sign shows the object is moving in the opposite
direction.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Principle of Conservation of Momentum
1.
The of conservation of momentum states that the total momentum in a closed system is constant, if no
external force acts on the system, that is, the momentum of all objects before a collision ____________
the momentum of all objects after a collision.
2.
The principle of conservation of momentum is true for the following:
(a)
Collision of two objects
(b)
-
elastic collision
-
inelastic collision
Explosion
Elastic collision
-
When objects collide and bounce perfectly, the collision is said to be an elastic collision (figure 2.1)
Figure 2.1
-
Figure 2.2 shows the diagram of experiments to investigate the principle of conservation of
momentum in elastic collisions.
Figure 2.2
(i)
The runway is adjusted to compensate the friction.
(ii)
Trolley A with a spring-loaded piston is placed at the higher end of the runway and trolley B
is placed halfway down the runway and stayed at rest.
(iii)
Two ticker tapes are passed through the ticker timer, one attached to trolley A and another
attached to trolley B.
(iv)
The ticker-timer is switched on and trolley A is given a slight push so that it moves down
the runway at a uniform velocity and collides with trolley B which is stationery.
(v)
After collision, the two trolleys move separately.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Figure 2.3 shows the ticker tapes obtained from the experiment.
Figure 2.3
Inelastic collision
-
When objects collide and attached to one another or couple together after collision, the collision is
called an inelastic collision (figure 2.4).
Figure 2.4
-
Figure 2.5 shows the diagram of experiments to investigate the principle of conservation of
momentum in elastic collisions.
Figure 2.5
-
(i)
The runway is adjusted to compensate the friction.
(ii)
The spring loaded piston of trolley A is removed and some plasticine is pasted onto trolley
A and B.
(iii)
A ticker tape is attached to trolley A only.
(iv)
The ticker-timer is switched on and trolley A is given a slight push so that it moves down
the runway at a uniform velocity and collides with trolley B which is stationery.
(v)
After collision, the two trolleys are move together.
Figure 2.6 shows the ticker tapes obtained from the experiment.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Figure 2.6
Explosion
-
Means the separation of objects which are initially at rest.
Figure 2.7
Application of Conservation of Momentum
Launching of rockets.
Jet engine
25
FORCE AND MOTION
MOTION-CHAPTER 2]
PHYSICS FORM 4 [FORCE
Example 1
Block A of mass 5 kg is moving with velocity 2 m s-1 and collides with another stationery block B of
unknown mass. After the collision,
n, block A moves with velocity 0.5 m s-1. Given that the collision is elastic.
Find the momentum of block B after the collision.
Solution
Example 2
A truck travels at a velocity of 15 m s-11 collides head-on with a car that travels at 30 m s-1. The mas
mass of the
truck and the car are 6000 kg and 1500 kg respectively. What is the final velocity of the two vehicles after
the collision if they stick together?
Solution
Example 3
An instructor fires a pistol which has a mass of 1.5 kg. If the bullet weighs
weighs 10 g and it reaches a velocity of
300 m s-1 after shooting, what is the recoil velocity of the pistol?
Solution
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Mastery Exercises
1.
A bullet of mass 10 g is fired from a rifle of mass 2.0 kg. The velocity of the bullet is found to be 100 m s-1.
Find the recoil velocity of the rifle.
2.
Two cars have equal mass, m = 800 kg. One car moves at velocity of u = 4 m s-1 towards the other car which
is at rest. The two cars couple together after impact. What is their common velocity?
3.
A boy standing on a skateboard is moving at a constant speed of
10 m s-1 along a straight line, carrying a
bowling ball in his hand. The mass of the boy and the ball are 50 kg and 5 kg respectively. While moving, the
boy throws the ball horizontally and it is observed that his speed slows down to 9.5 m s-1. What is the speed of
the ball when the ball is thrown?
Figure 2.8
4.
The figure 2.8 shows a ticker-tape produced from a collision between trolleys A and B. The ticker tapes that
stick to trolley P is passed through a ticker timer which runs at a frequency of 50 Hz and collide with trolley B
which is stationery. After collision, trolley A sticks to trolley B and they move together. If the mass of trolley
A is 2.0 kg, what is the mass of trolley B?
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.5
Understanding the Effect of a Force
Figure 2.1 Effect of forces
1.
2.
3.
4.
We use force in our daily activities
Forces have magnitude and direction, that is, force is a vector quantity.
The SI unit of force is Newton, N.
Force can make things;
(a)
move a stationery object.
(b)
speed up (accelerate) or slow down (decelerate) a moving object.
(c)
change the direction of a moving object.
(d)
stop a moving object.
(e)
change the shape of an object.
Balanced Force
1.
2.
3.
An object may have several forces acting on it. But if the forces are in balance, they cancel each other
out. Then, the object behaves as if no force is applied to it.
When the forces are balanced, an object is either at rest, or moving at a constant velocity.
Example of balanced forces;
Books on the table remain at rest because normal
reaction force, R is cancel with weight of the books.
The net force acting on the books is zero.
An airplane can flaying horizontally with constant
speed because forward thrust, T provides by the
engine is cancel by a drag, F provides by the wind
and air resistance.
An airplane can flying at a constant height because a
lift force is balanced with its downward weight, W.
When these four forces are balanced, the net force
acting on the plane is zero.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Unbalanced Forces on an Object
1.
2.
When two or more forces acting on an object are not balanced, the object will accelerate. This net force
is known as the unbalanced force or the resultant force.
Example;
5N
(a)
10 N
When two opposite forces acting on a box, the box will move.
Relationship between Force, Mass and Acceleration
Figure 2.2
Figure 2.2 (a) shows a force, F
acting on a mass, m.
The mass moves with a
constant acceleration, a.
1.
Figure 2.3
Figure 2.4
If the same force, F is acting on
If the force acting on the same
mass is doubled, its acceleration a mass of 2m, the acceleration,
a, is reduced by half.
is also doubled.
The diagrams above show that the acceleration of an object is directly proportional to the unbalanced
force.
Acceleration,
2.
(F = ma)
a α Net force, F
The acceleration of an object is inversely proportional to the mass.
1
Acceleration, α
Mass, m
Combining the relationship above, we get;
a
3.
F = ma
4.
5.
The relationship between F, m and a, is known as Newton’s second law of motion.
Newton’s second law of motion states that the acceleration produced by a net force on an object is
directly proportional to the magnitude of the net force applied and is inversely proportional to the mass
of the object.
6.
In the formula, F = ma,
- F is the net force and a, is the acceleration in the same direction as the force.
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
7.
Net force, F is the force applied minus all other opposing forces, such as friction.
Examples;
(i)
Azhari applies a force of 50 N to move a 12 kg carton at a constant velocity. What is the
frictional force acting on the carton?
Figure 2.5
Solution
Frictional force is always acting in an opposite direction or opposing to the motion.
Constant velocity → acceleration, a = 0 m s-2
Net force in the direction of velocity, F
= F1 – Frictional force
= 50 – f
From F = ma
50 – f = 12 x 0
F = 50 N
2 kg
F = 0.8 N
f
Figure 2.6
(ii)
Figure 2.5 shows a trolley of mass 2 kg placed on a rough horizontal table and being pulled by a
force of 2.0 N. The trolley moves at constant velocity.
What is the frictional force between the trolley and the table?
(a)
(b)
The force, F is then increased to 12 N. what is the acceleration of the trolley.
Solution
(a)
Constant velocity → acceleration, a = 0 m s-2
Net force, F = ma
0.8 – f = 2 x 0
f = 0.8 N
(b)
Net force, F = ma
1.2 – 0.8 = 2 x a
0.4
a=
2
a = 0.2 m s-2
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PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Analysing Impulse And Impulsive Force
1.
2.
An impulsive force is a force that acts over a short period of time during collisions or explosions.
Consider an impulsive force, F and the following equations;
F = ma ………………………….. (1)
and
v −u
………………… (2)
a=
t
Substitute (2) into (1)
 v − u
F=m 

 t 
mv − mu
F=
………………. (3)
t
Hence,
Impulsive force also defined as the rate of change of momentum.
The SI unit for impulsive force is the Kgms-2 ot N.
3.
Impulse is the product between the impulsive force, F with the time of impact, t, that is,
mv − mu
Rearrange (3)……. F =
t
Ft = mv – mu
Hence,
Impulse, Ft also defined as the change of momentum.
The SI unit for impulse is Kgms-1 or Ns.
Example 1
Figure 1 shows a baseball approaching a bat with an initial velocity of -30 m s-1. A force is applied by
the bat to hit the ball and sends it in the opposite direction with a velocity of 50 m s-1.
Figure 1
If the mass of the baseball is 150 g and the time of contact between it and the bat is 1.6 x 10-3 s,
calculate,
(i)
the impulse applied to the ball.
(ii) the impulsive force exerted on the ball by the bat
31
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example 2
A 45 g golf ball is hit with a force of 5300 N. The time interval of interaction between the golf ball an d
the club is 0.6 s. Calculate the velocity of the golf ball immediately after it was hit.
Effect of Impulsive Forces in Daily Life
mv − mu
1. From the formula F =
, it can be seen the impulsive force , F is large if the time interval, t is
t
small,
1
Fα
t
2. This explains why in a car accident, the car experience a large impulsive force, which causes serious
damage to the car and also injuries to the passengers in the car.
Reducing Impulsive Forces
1.
2.
3.
4.
5.
From the formula of impulsive force, we know that impulsive force is inversely proportional to the time
interval.
An effective way to reduce the impulsive force in a collision is to extend the time interval of the
interaction.
A large impulsive force resulting from a car accident can cause serious injuries to its passengers. To
prevent this, cars are constructed with safety features to reduce the impact.
Motorcyclists should wear safety helmets and elbow and knee cushions/caps to prevent themselves from
falling down and knocking onto the hard surface.
In sports the effects of impulsive force are reduced to prevent injuries to the participants in the games.
(a)
Thick mattresses with a soft surface used in events such as the high jump, so that the time interval
of impact on landing is increased.
32
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
(b)
(c)
6.
Athletes who jump down from a height must bend their knees upon landing.
When a goalkeeper catches a fast moving ball, he ends his hands to increase the time of contact
when stooping the ball. With longer time of impact, a smaller impulsive force is acting on his
hand.
Polystyrene, cardboard and rubber foam are usually used in packing fragile goods. The soft surface of
these materials reduces the impulsive force during accidental dropping by lengthening the time of
impact.
Benefits Effects of Impulsive Forces
When a hammer is used to hit a nail into
wood, a large change in momentum occurs in
a short time interval. It produces a large
impulsive force which drives the nail into the
wood.
When foodstuff such as chillies are pounded
using a mortar and pestle (which are both
made from stone), the action produce large
impulsive force which crush the food. This
occurs when the pestle is brought down at a
high velocity onto the mortar with short
contact times.
33
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.7 SAFETY FEATURES IN VEHICLES
There are several safety features installed to a vehicle in order to reduce serious injury to the driver and
passengers.
(a) Dashboard – covered with soft materials to reduce injury due to passengers knocking on it.
(b) Air bags – will expand during collision so as to prevent the driver from smashing directly into the car
steering. The collision time will be lengthen and the impulsive force can be reduce and prevent
(c) Safety belts – to lengthen the collision time and prevent the passengers from being thrown due to their
inertial when the car is forced to slow down suddenly.
(d) Rubber bumper - absorbs impact in minor accidents, thus prevents damage to the car also lengthen the
collision time and to reduce the impulsive force.
(e) Front crumple zone – design to crumple upon impact. It increases the impact time and reduce the
impulsive force on the car.
(f) Anti-lock braking system (ABS) – will not immediately stop the car once the brakes are applied. The
car be momentarily brought to rest so that the force of impulsive is smaller.
34
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.8 GRAVITY
1.
2.
The concept of gravitational force was introduced by Sir Isaac Newton who, on seeing an apple falls on
his head.
According to Newton, objects fall because they are pulled towards the Earth by the force of gravity.
The pull of gravitational force;
(a) keeps things on the earth.
(b) brings things down to earth when they are thrown upwards
(c) holds the moon in its orbit round the Earth
(d) captures returning space capsules and pulls them into orbit.
3.
4.
The pull of gravity causes objects to fall with acceleration. This means that objects that fall are moving
with increasing velocity.
The magnitude of the acceleration due to gravity depends on the strength of the gravitational field.
35
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Free Fall
1.
An object is falling freely when it is falling under the force of gravity only.
Figure 1
2.
3.
4.
5.
6.
7.
8.
9.
Figure 1 show a free fall in a vacuum cylinder (where air resistance does not exist) of a coin and a
peace of paper. The both objects reach the bottom of the cylinder at the same time.
A piece of tissue paper (fall in atmosphere) does not fall freely because its fall is affected by air
resistance.
A heavier golf ball can be considered to be falling freely because the air resistance is small compared to
the pull of gravity and therefore is negligible.
Objects dropped under the influence of the pull of gravity accelerate at a constant rate.
This acceleration is known as the gravitational acceleration, g.
The value of g is 9.8 m s-2. In calculation, the value of g is often taken to be 10 m s-2 for simplicity.
The acceleration due to gravity does not depend on the mass and shape of the falling object.
All objects falling freely with the same acceleration.
Tips
-
when an object falls; g = 9.8 m s-2
when an object is thrown upwards; g = 9.8 m s-2
At the highest point, v = 0 m s-1
36
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
The experiment to determine the value of acceleration due to gravity, g.
1.
Apparatus/material:
- Ticker timer, ticker tape, 12 V a.c electrical power supply, retort stand, weights (50 g- 250 g), Gclamp, cellophane tape.
2.
Procedure:
- switch on the ticker timer and release the slotted weight from a height of about 1 m.
- calculate the acceleration from the dots on the ticker tape.
- repeat the steps for slotted weights of 100g, 150g, 200g and 250 g.
3.
Results:
Mass of slotted weight, m/g
Acceleration due to gravity,
g/ ms-2
50
100
150
200
250
-
gravitational acceleration, g is calculated by substituting u, v and t into the formula g =
v −u
t
Weight, W and Gravitational Field, g
1.
2.
Gravitational field is a region around the earth in which an
object experiences a force towards the centre of the earth.
This force is the gravitational attraction between the object
and the earth.
The gravitational field strength, g can be calculated as
follows:
g=
3.
F
m
Where F = Gravitational force
m = mass of body
The gravitational field strength at the surface of the earth is
9.8 N kg-1/kg ms-2.
37
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
4.
The weight of an object is defined as the gravitational force acting on the object and can be calculated
as follows;
Where, m = mass
W = mg
g = gravitational acceleration
5.
The unit of weight is Newton, N.
Example 1
A iron ball is dropped from the top of a building and
takes 2 s to reach the surface of the earth. What is
the height of the building? (g = 9.8 m s-2)
Solution
Example 3
Have you ever seen an astronaut walking on the
Moon? It is known that the acceleration due to
gravity near the surface of Moon is just 1/6 of that
on the surface of earth.
(a) Find the weight of a 50 kg man on the surface
of Moon.
(b) If the 50 kg man can jump to a height of 50 cm
on the Earth, find the maximum height reached
by him on the surface of moon. (Assume that
his initial speed is the same on the earth and on
the moon)
Solution
Example 2
A rock has a mass of 20.0 kg and weight of 90.0 N
on the surface of a planet.
(a) What is the gravitational field strength on the
surface of the planet?
(b) What are the mass and the weight of the rock on
the surface of the Earth where its gravitational field
strength is 9.8 N kg-1?
Solution
38
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
2.9 FORCES IN EQUILIBRIUM
Examples of forces in equilibrium
A book resting on a table
A skydiver falling at a constant
velocity
1.
When forces act upon an object and it remains stationery or moves at a constant velocity, the object is
said to be in a___________________________
2.
When equilibrium is reached, the _________________ acting on the object is __________ i.e there is
___________________________ acting upon it.
3.
Newton’s Third Law of Motion states that _______________________________________________
Figure 1
4.
Figure 1 shows a stationery block of wood resting on a table. The forces acting on the block of wood
are:
(a)
The weight, W, which is acting downwards. (Action)
(b)
The normal reaction, R, which is acting upwards. (Reaction)
5.
The Weight, W, is balanced by the Normal Reaction, R.
6.
Hence, the block of wood is in a __________________________.
39
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Figure 2
7.
Figure 2 shows a weight hanger attached to a string and which is in a stationery state. The forces acting
on the weight hanger are:
(a) The weight, W, which acts downwards. (Action)
(b) The tension, T, which acts upwards. (Reaction)
8.
The weight, W, is balanced by the tension T of the string.
9.
Hence, the weight hanger is in a state of equilibrium.
Figure 3
10. The cat resting on an inclined plane as shown in figure 3 is also in equilibrium. The three forces acting
on the cat cancel out each other so that the resultant force is zero.
40
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Addition of Forces
1.
A resultant force is a single force that represents the combined effect of two or more forces in
magnitude and direction. The directions of the forces have to be taken into considerations when forces
are added.
(a)
Two forces that act along the same direction.
Resultant Force, F
=8N+3N
= 11 N
The resultant force is found by adding the magnitude of the forces. Two forces acting in the
same direction with magnitudes 8 N and 3 N respectively will have a resultant force of 11 N, as
shown in figure 4. Resultant force, F = F1 + F2
(b)
Two forces that act in opposite directions
Resultant Force, F
=8N–3N
=5N
The two forces with magnitude 8 N and 3 N acting respectively in opposite directions will have
a resultant of 5 N, as shown in figure 5.
Resultant force, F = F1 – F2
(c)
When two forces act at a point at an angle to each other, the resultant force can be determined
by:
(i)
Adding forces using parallelogram of force.
(i)
Draw the forces F1 and F2 from a point with an angle of θ with each other.
(ii)
Draw another two lines to complete the parallelogram.
(iii)
Draw the diagonal of the parallelogram. The diagonal represents the resultant
force, F and its direction, α, can be determined by measuring the angle between
the diagonal with either one side of the parallelogram.
41
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
(ii)
(d)
Adding force using triangle of force (tip-to-tip method)
If the two forces are acting at right angle or perpendicular to each other, the resultant force can
be determined by using Pythagoras theorem and trigonometry.
Example
30 N
40 N
A carton is acted on by two forces of 30 N and 40 N which are at right angles to each other as
shown in figure 8. Determine the resultant force acting on the carton.
Solution
42
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Resolution Force
1.
A single force can be resolved into two perpendicular components.
2.
Example
A tourist is pulling a bag with a force 12.0 N at an angle of 300 to the horizontal floor. What is the
horizontal and vertical component of the force?
Solution
Fx and Fy are the vertical and horizontal components of the forces.
43
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example
The diagram above shows a block of wood (on a smooth surface) being pulled by a force of 12 N. What is
the horizontal component of the force?
Solution
Example
The figure shows a box is being pulled by a man.
What is the magnitude of the resultant force that causes the box to move horizontally?
44
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example
A workman pushes a carton of mass 50 kg up an inclined
plane into a lorry. The inclined plane makes an angle of
450 with the horizontal floor and the frictional force
between the inclined plane and the carton is 135 N. If the
workman pushes the cartoon with a force of 500 N
(a) Can the carton move up the inclined plane?
(b) What is the acceleration of the carton?
Solution
(a) Applied force on the carton towards the top
of the inclined plane, F = 500 N.
-
Opposing force along the entire length of
the inclined plane
= component of weight down the plane +
frictional force
= 500 sin 450 + 135 N
= 488.6 N
(b)
The resultant force acting on an inclined plane,
= 500 N – 488.6 N
= 11.4 N
From formula; F = ma
F
a=
m
11.4
=
50
= 0.23 m s-2
-
The carton is able to move upwards
because the applied force, F = 500 N is
larger than the opposing downward force.
Three forces in equilibrium.
Problems involving three forces in equilibrium can be solved either by,
(a)
Resolution of force, then follow by
Total force to the left = Total force to the right
Total force upwards = Total force downwards
(b)
Drawing a closed tail-to-head triangle of force.
Figure shows a wooden block supported by two strings. The tensions of the strings are T1 and T2
respectively. Since the wooden block is in equilibrium, the resultant force is zero.
Hence, by taking the horizontal components of forces, T1 sin α = T2 sin β
By taking the vertical components,
W = T1 cos α + T2 cos β
45
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
T1 cos α + T2 cos β
T2
T1
α β
T1 sin α
T2 sin β
W
Example
46
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Example
Figure 9 shows an aeroplane model with a mass of 450 g is hung from a ceiling with two strings.
Figure 9
Calculate the tension on each string. Assume that the system is in equilibrium. [g = 9.8 ms-2]
47
PHYSICS FORM 4 [FORCE AND MOTION-CHAPTER 2]
Solution
48
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
2.10 UNDERSTANDING WORK, ENERGY, POWER AND EFFICIENCY
Are you doing work when you are solving a physics problem?
Are you doing work when you pick up a pen that has fallen to the floor?
Photograph below shows that work is done.
A women is pushing the
stroller
A fisherman is pushing the
boat towards the beach.
The rocket engine produces an
upward thrust.
WORK
Figure 1
1.
Work, W is defined as the ______________________________________________________
in the direction of the force (figure 1).
Work, W = Force (F) x Distance (s) (in the direction of the force)
W=Fxs
The SI unit for work is Nm or Joule (J).
2.
Work is a _________________ quantity, that is, a quantity which has magnitude only but not
direction.
3.
If no distance is traversed or if there is no motion, then no work has been done.
Figure 2
49
[PHYSICS FORM 4]
4.
[CHAPTER 2 - FORCE AND MOTION]
When the force and the direction of motion of an object are perpendicular as shown in figure 2,
the work done is equal to zero.
Figure 3
5.
In such a case, the work can be determined by resolving the force along the direction of motion
as shown in figure 3.
Work, W = F cos θ x s
Example 1
A nurse is pushing a patient in a wheelchair with a force of 200 N over a distance of 1.5 m.
How much work is done by nurse?
Example 2
Diagram shows a workman pulling up a load onto a lorry using a smooth inclined plane. If the
tension in the rope is 600 N and the inclined plane is 3 m in length, how much work is being
done by the man?
50
[PHYSICS FORM 4]
6.
[CHAPTER 2 - FORCE AND MOTION]
If a graph of force versus (against) distance is drawn, then the work done is equal to the area
under the graph.
ENERGY
1.
When work is done, energy is consumed.
2.
Energy is the ability to do work.
3.
Mechanical energy is divided into potential energy and kinetic energy.
4.
Energy is the product of the force, F and the distance, d.
Energy
= Work done
= Force x Distance
E = Fxs
5.
The SI unit for energy is the Nm or Joule (J).
6.
Energy is a _________________ quantity
Kinetic Energy
1.
Kinetic energy is the energy possessed by a moving object. Only moving objects possess
kinetic energy.
2.
Assume a force, F is acting on a stationery trolley of mass, m kg moving on a smooth surface.
The force acting over a distance of s causes the trolley to achieve a velocity of v m s-1.
3.
The kinetic energy, Ek of an object of mass, m kg travelling at a velocity of v ms-1 is given as;
Kinetic energy, Ek
= ½ mv2
51
[PHYSICS FORM 4]
4.
[CHAPTER 2 - FORCE AND MOTION]
The factors that affect kinetic energy are:
(a)
mass of the object, m
(b)
velocity of the object, v
Example 1
A 0.6 kg trolley moves across the floor at a velocity of 0.5 m s-1. What is the kinetic energy of
the trolley?
Example 2
A car of mass 950 kg accelerates from a velocity of 20 m s-1 to a velocity of 35 m s-1. What is
the work done for the car to accelerate?
Potential Energy
1.
The potential energy of an object is the energy stored in the object because of its position or
state.
2.
There are two main types of potential energy.
(a) Gravitational potential energy
(b) Elastic potential energy
3.
The gravitational potential energy of any object is the energy stored in the object because of its
height above the earth’s surface.
4.
When an object of mass, m kg is raised to a height, h meters above the earth’s surface, the
object possesses gravitational potential.
52
[PHYSICS FORM 4]
5.
[CHAPTER 2 - FORCE AND MOTION]
The gravitational potential energy is equal to the work done to raise an object to a particular
height.
Work done, W = F x s
= mg x h
= mgh
Gravitional potential energy, Ep = W
Ep = mgh
6.
Elastic Potential Energy of an object is the energy stored in the object as result of stretching or
compressing it.
Example
The diagram above shows a mass of 6 kg being pulled by a force F up a smooth inclined plane.
After the body has been pulled 5 m up the ramp/platform, it is found that the vertical height of
the object is 2 m. Calculate the,
(a)
gravitational potential energy at 5 m.
(b) work done to lift the body/mass up the smooth inclined ramp.
(c)
value of F
53
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
Principle of Conservation of Energy
1.
The principle of conservation of energy states _____________________________________
__________________________________________________________________________
2.
Energy cannot be destroyed but can be converted from one from to another.
3.
When a coconut of mass, m kg falls from a height of h meters to the ground, it loses its
gravitational potential energy which is changed into kinetic energy of motion.
Example
Lim rides his bicycle down the slope of a hill 3 m high at an initial velocity of 2 m s-1, without
pedalling. At the foot of the hill, the velocity is 6 m s-1. Given that the mass of Lim with his
bicycle is 75 kg. Find
(a)
the initial kinetic energy of the bicycle.
(b) the initial potential energy of the bicycle
(c)
the work done against friction along the slope.
54
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
POWER
1.
2.
Power, P is the rate at which work is done or energy is changed or transferred.
Power is expressed as:
Power, P =
P=
3.
workd done
time taken
W
t
The SI unit of power is J s-1 or watt (W).
Example
A car is moving at a constant velocity of 30 m s-1 . If the car has to overcome a frictional force
of 700 N, what is the power of its engine?
EFFICIENCY
1.
Work can be done by a machine when input energy is supplied.
2.
If a machine does least work from the supplied energy, it is said to be non-efficient.
3.
Efficiency of a machine is expressed as:
Efficiency =
Useful energy output
x 100%
Energy input
4.
If a machine has the efficiency of 100% it is an excellent machine in which output work =
input work.
5.
In general, the efficiency for all machines is less than 100% because of the work done against
friction when operating a machine
Example
A 60 N load is lifted up to 0.30 m height with a machine which needs 25N of force and the
load moves 0.90 m. Calculate
(a) the work which is done by the force
(b) the work which is done by the machine
(c) the efficiency of the machine
55
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
Mastery Exercise
1.
In the diagram above, a fixed force F acts on a box of mass 800 g such that it moves up from
position A to position B. (Assume that the acceleration due to gravity, g = 10 ms-2)
(a)
What is the work done?
(b) Calculate the potential energy attained by the box when it reaches position B.
(c)
The box is released when it reaches position B. If the ramp is smooth and there is no
friction, what is the kinetic energy of the box when it slides back to position A?
(d) What is the velocity of the box when it reaches position A?
2.
The above diagram shows a particle A of mass 3 kg being released from position X on curved
ramp having a rough surface. (Assume that the acceleration due to gravity, g = 10 ms-2).
(a)
What is the energy possessed by A at the position X?
(b) Give another position where the particle A possesses the same energy as found in the
answer to (a) above.
(c)
Calculate the energy possessed by a particle A
(i)
at position X
(ii)
at position Z
(d) Explain why there is difference in energy as the answers in (c)(i) and (c)(ii).
(e)
Explain the changes occurring to particle A along its journey from position X to Z.
(f)
If the surface of the ramp is smooth and friction can be neglected
(i)
what is the velocity of particle A at position Y?
(ii)
does particle A stop before, after, or at position Z? Why?
56
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
2.12 ELASTICITY
Understanding Elasticity
1.
Some common devices like eraser and ruler will change its shape when an external force acting
on it. When the external force is removed, the objects return to its original shape and
dimensions.
A rubber band can be stretched due to
its elasticity.
A sponge can be compressed easily
due to its elasticity
2.
The property of an object that enables it to return to its original state once the applied force is removed.
is called elasticity
.
3.
The elasticity of solids is due to the strong intermolecular forces between the molecules of the
solid. Stretching a solid causes the molecules to be slightly displaced away from one another.
A strong attractive intermolecular force acts between the molecules to oppose the stretching as
shown in figure 2.64.
4.
When the external stretching force is removed, this strong attractive intermolecular force
brings the molecules back to their original positions. Therefore, the solid return to its original
shape and size.
5.
Compressing a solid causes the molecules in the solid to be closer to one another.
6.
When a solid is compressed, the strong force of repulsion will push the atoms or molecules of
the solid back to their original position as is shown in figure 2.65.
57
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
Relationship between force and extension of a spring
Title:
To investigate the relationship between force and the extension of a spring.
Hypothesis:
The tension of a spring is directly proportional to the applied force.
Variables:
Manipulated variable: Force
Responding variable: Extension of the spring
Materials and apparatus:
Steel spring, five 50 g slotted weight, half meter rule and retort stand with clamp.
Method
1. All the weights are removed from the spring. The original length of the spring is measured
using a half meter rule.
2. A 50 g weight is hung from the end of the spring. The length of the spring is measured again
and the extension is calculated.
3. Step 2 is repeated with 100 g, 150 g, 200 g, 250 g, and 300 g weights.
4. The graph of force against extension is plotted.
Result:
Mass of slotted
weight, m/g
Force on the spring,
(F=mg)/N
Length of spring,
L/cm
Extension, x = (LL0)/cm
50
100
150
200
250
300
58
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
HOOKE’S LAW
1.
Hooke’s law states that the extension of a spring is directly proportional to the applied force
provided that the elastic limit is not exceeded.
Force,F/N
Extension/cm
Figure 1
2.
The elastic limit of a spring is defined as the maximum force that can be applied to a spring
such that the spring will be able to be restored to its original length when the force is removed.
3.
If the elastic limit is exceeded, the length of the spring is longer than the original length even
though the force no longer acts on it / spring will not return to its original position.
Figure 2
4.
The results from experiment show that the graph is straight line passing through origin
(figure 1) meaning that the extension of a spring is directly proportional to the applied force if
the elastic limit is not exceeded.
5.
The mathematical expression for Hooke’s Law is,
Fαx
Therefore, F = kx ; k is constant of the spring/force constant
k=
F
with units N m-1.
x
6.
Spring constant is a measurement of the stiffness of the spring. A spring with a spring constant
of 12 N m-1 requires a force of 12 N to produce an extension of 1 cm.
7.
A spring with a large spring constant is harder to extend and is said to be stiffer.
59
[PHYSICS FORM 4]
8.
[CHAPTER 2 - FORCE AND MOTION]
A spring with a smaller force constant is easier to extend and is said to be less stiff or softer.
Figure 3
-
spring constant k = gradient of graph
a larger value of k indicates a stiffer spring
a steeper graph indicates a stiffer spring
Example
The length of a spring is increased from 23.0 cm to 28.0 cm when a mass of 4 kg was hung
from the end of a spring.
(a)
What the load on the spring in newtons?
(b)
What is the extension of the spring?
(c)
Calculate the force constant of the spring
(Assume g = 10 Nkg-1)
60
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
SYSTEM OF SPRING
1.
Two springs can be connected in series or in parallel.
2.
When two springs are connected in series, the applied force acts on each spring. The force due
to the 30 N load acts along the system such that each spring experience a force or tension of 30
N.
3.
When two springs are connected in parallel, the applied force is shared equally among the
springs. The force due to the 120 N load is shared by the springs. Therefore, the tension in each
spring is 60 N.
Example 2:
Figure above shows a spring extends by 1 cm when an 8 N force is applied on it. Similar
springs used to set up three systems.
Calculate the total extension in each system.
61
[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
Example 3:
Figure 3 shows identical springs. What is the value of Y?
ELASTIC POTENTIAL ENERGY
1.
Elastic potential energy is the energy stored in a spring when it is extended or compressed.
2.
When a force extends a spring, work is done. The work done on the spring is the energy
transferred to the spring and stored as elastic potential energy.
3.
Consider a force, F that produces an extension, x in a spring. The work done on the spring,
W = average force x extension
 F
=  x
2
=
1
Fx
2
From Hooke’s law, F = kx
Therefore, W =
1
(kx) x
2
= ½ kx2
Hence, the elastic potential energy stored in a stretched spring is given by,
Ep = ½ kx2
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[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
Area under the graph represents elastic potential energy
Example 4
A 2 kg load is hung form the end of a spring with a force constant of 160 Nm-1.
(a) What is the tension in the spring?
(b) What is the extension of the spring?
(c) Calculate the elastic potential energy stored in the spring.
[Assume g = 10 N kg-1]
FACTORS WHICH INFLUENCE THE ELASTICITY OF A SPRING
1.
Factors which influence the elasticity of a spring are as follows;
(a)
type of spring material
- a spring made from a hard material requires a larger force to stretch it. Hence, the
spring constant, k is greater. For example; steel spring is harder than the copper
spring.
(b)
diameter of the coil of spring
- a spring made of a larger diameter coil is ‘softer’.
(c)
diameter of the wire of the spring
- a spring coil of spring made from thicker wire is more difficult to stretch than a coil
of spring made from wire that is thinner.
(d)
arrangement of the spring
- a longer spring is easier to stretch compared to a shorter spring.
- springs arranged in series are easier to stretch when compared to springs arranged in
parallel.
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[CHAPTER 2 - FORCE AND MOTION]
Mastery Exercise
1.
A spring extends by 3 cm when it is hung with a 10g weight. Find the total extension in each of
the spring systems shown in the diagram.
2.
Diagram (a) shows a spring whose original length is 20 cm. When a weight of 20 g is placed
atop the spring as in diagram (b), the spring is stretched by 2 cm. If a weight of x g is placed
atop the same spring as in diagram (c), it is found that the spring is stretched by 8 cm. What is
the value of x?
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[CHAPTER 2 - FORCE AND MOTION]
3.
The above figure shows a spring which is compressed. Calculate the value of m if the original
length of the spring is 15 cm.
4.
The figure shows the pointer reading of a spring. When an additional load of 200 g is placed in
the pan, the pointer reading is 25 cm. What is the reading of the pointer when the total load is
removed?
5.
The above figure shows a system of two identical springs. Initially each spring is of length 12
cm and becomes 15 cm when it is loaded with a mass of 600 g. What is the value of y?
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[PHYSICS FORM 4]
[CHAPTER 2 - FORCE AND MOTION]
6.
M and N are two non-identical springs each measuring 12 cm. When subjected to loading, their
respective lengths are as shown in the above figure. What is the length y of the spring system if
M and N are arranged in series and the applied load is 600 g?
7.
The above figure shows a spring system comprised of identical springs. Each springs is of
length 18 cm and extends to 22 cm when subjected to a load of 200 g. What is the length of the
spring system in figures (a) and (b).
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[PHYSICS FORM 4]
8.
[CHAPTER 2 - FORCE AND MOTION]
The above figure is a graph of extension, x versus load, m for a spring.
(a)
what is the value of the spring constant, k?
(b)
what is the value of a?
(c)
what is the potential energy stored when the spring is extended by a cm?
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