PHYSICS KSSM FORM 4

1. a
F

2. OBJECTIVE
Students must be able to:
1. understand force as the rate of
change of momentum.
2. analyse situation to solve the
problem involving F=ma

3. The relationship between
force, F, mass, m and
acceleration, a for an
object in motion is

4. Carry out
the Activity
2.13

5. Carry out Activity 2.13 page 64
• Group A
Aim : To investigate the relationship
between a and F
(constant mass)
• Group B
Aim : To investigate the relationship
between a and m
(Constant Force)

6. Preparation before experiment :
1. Prepare a friction-compensated inclined
runway.
Adjust the runway so that the trolley
goes down the runway with a constant
speed when given a slight push.
2. Increasing the pulling force on the trolley
Always stretch the elastic cord until the
end of the trolley.
Practice by pulling one cord, two cords
and three cords.

7. Preparation before experiment :
1. Prepare a friction-compensated inclined
runway.
Adjust the runway so that the trolley goes down
the runway with a constant speed when given a
slight push.
2. Pull the trolley with a constant force.
Always stretch the elastic cord until the end of
the trolley.
Practice by pulling one trolley, two trolleys and
three trolleys.

8. TYPES OF GRAPH
m constant
F constant

11. Newton’s Second Law
states that the rate of change
of momentum is directly
proportional to the force and
acts in the direction of the
applied force.
F∝ma,
F = kma, k is constant.

12. In S.I. units, 1 N is the force
that produces an acceleration
of 1 m s-2 when applied on a
mass of 1 kg.
As such,
1 N = k × 1 kg × 1 m s-2
k = 1
Therefore, F = ma

13. What force is required to move a 2 kg
object with an acceleration of 3 m s-2, if
(a) the object is on a smooth surface ?
(b) the object is on a surface where the
average force of friction acting on
the object is 2 N ?
Example 1
solution
(a)
Net force, F = ma
= 2 (3)
= 6 N
(b)
Net force, F = ma
F – 2 = 2(3)
= 6 N
F = 8 N
F F2 N

14. A car of mass 1200 kg travelling at 15 m s-1
is brought to rest over a distance of 30 m.
Find
(a) the average deceleration, and
(b) the average braking force
Example 2
solution
(a)u=15, v = 0, s=30
Using v2=u2+2as
0 = 152+2a(30)
60a = -152
a = -3.75 m s-2
Deceleration=3.75 m s-2
(b) F = ma
F = 1200(-3.75)
= - 4500 N
Braking force= 4500 N

15. THINK-PAIR-SHARE
 By using Newton Second Law and
previous knowledge, answer 2 question
in pair in 10 minutes
 Randomly, teacher will call out your
name and solve on the white board and
explain to your friends

16. QUESTION 1
A 10 kg object accelerates from rest to 5
ms-1 in 2 s. what is the net force acting on
the object?
QUESTION 2
A net force of 10 N acts on a 2 kg
stationary object. Calculate the distance
travelled by the object in 5 s.

17. SUMMARY
1. Force is a vector quantity.
2. SI unit of force is kg m s-2 or newton,
N
3. If an object is at rest or moving with
constant velocity resultant force F = 0.
4. If an object is moving with an
acceleration, resultant force F = ma

18. CONCLUSION