Download Objective:

Form 4 Physics – Chapter 2 – Lesson 5
Objective:
1. Student will be able to describe the effects of balanced forces acting on
an object.
2. Student will be able to describe the effects of unbalanced forces acting
on an object.
3. Student will be able to determine the relationship between forces, mass
and acceleration, i.e. F  ma .
4. Student will be able to solve problems using F  ma .
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2.5
The Effects of Forces
2.5.1 Understanding the Effect of Forces
Force

A force is a push or a pull.

Forces have magnitude and direction – that is, force is a vector quantity.

The SI unit of force is newton (N).

Effects of Forces:
a) Moves a stationary object.
When a stationary toy car is pushed with the hand, the toy car start to
move.
b) Change the velocity of a moving object.
A toy car which is moving at a constant velocity accelerates when it is
pulled using an elastic string.
c) Slows down a moving object.
When a vehicle moves from a smooth surface to a rough surface, the
force of friction retards the motion of the vehicle.
d) Stops the motion of an object.
When a car collides with a wall, the car is stopped by the wall.
e) Change the direction of motion.
When a tennis ball is hit by a racket, the reactive force causes the tennis
ball to change its direction.
f) Change the shape of an object.
A straight ruler forms an arc when bending forces are applied to both ends.
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Form 4 Physics – Chapter 2 – Lesson 5
Balanced Forces

An object may have several forces acting on it.

If the forces are in balance, they cancel each other out. Then, the object
behaves as if no force is applied to it. F1  F2

With balanced forces on it, an object is either at rest, or moving at a constant
velocity.

Examples of balanced forces:
a) Balanced forces on a stationary gymnast.
b) Balanced forces on the aircraft result in it moving at constant velocity.
c) Balanced forces on a skater moving at a constant velocity.
Unbalanced Forces

The block either accelerates or decelerates since the forces are unbalanced or
the resultant forces is not zero. F1  F2
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Form 4 Physics – Chapter 2 – Lesson 5
2.5.2 Newton’s Second Law of Motion
1. The experiment shown that the acceleration, a , of an object depends on the
unbalanced force, F and the mass, m of the object.
(a) Acceleration, a  Net force, F
(b) Acceleration, a 
1
Mass, m
The symbol  stands for “is directly proportional to”.
2. Combining the relationship, a  F and a 
We get a 
 F  kma
F
m
or
1
,
m
F  ma
where k is a constant.
3. When a force of 1 N acts on a mass of 1 kg , the resultant acceleration is 1 m s 2 .
Substituting F  1N , m  1 kg , and a  1 m s 2 into formula F  kma , k  1 .
4. Then, the relationship between force, F , mass, m and acceleration, a can be
written as:
F  ma
5. The unit of force is Newton (N) or kg m s 1 in base unit.
Newton’s Second Law of Motion states that the acceleration
of a body, a , is directly proportional to the net force acting on
it, F , and inversely proportional to its mass, m .
F  ma
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Form 4 Physics – Chapter 2 – Lesson 5
Example 2.5.1:
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?
Solution:
Constant velocity  Acceleration, a  0
 Net force in the direction of velocity, F
 F1 - Frictional force
 50 - f
From
F  ma :
50 - f = 12 0
f = 50 N
This shows that the applied force of 50 N is just sufficient to overcome the
friction acting on the carton.
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Form 4 Physics – Chapter 2 – Lesson 5
Example 2.5.2:
A trolley of mass 2.0 kg placed on a rough horizontal table and being pulled by a
force of 2.0 N . The trolley moves at constant velocity.
(a) What is the frictional force between the trolley and the table?
(b) The force, F is then increased to 1.2 N . What is the acceleration of the trolley?
(c) The string is then broken when the trolley is moving. Find the subsequent
acceleration of the trolley?
Solution:
(a) Since the trolley moves at constant velocity, the acceleration, a  0 ,
Then, the unbalanced force, F  ma
0.8 N - G = 2.0 kg  0 
0.8 N - G = 0
Frictional force, G = 0.8 N
(b) Net force = 1.2 N - G = m  a
1.2 N - 0.8 = 2.0  a
a =
2 .4
= 0.2 m s 2
2 .0
The acceleration of the trolley = 0.2 m s 2
(c)
F  ma
0 N - 0.8 N = 2.0  a
a = 
0 .8
=  0.4 m s 2
2 .0
The acceleration is negative. This means that the trolley is decelerating or moving
with a retardation of 0.4 m s 2 .
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