Forces &
Movement
Nishan Wanasinghe
Force, mass & acceleration
Newton’s 1st law
If there is no unbalanced force, the stationary objects will
remain stationary and moving objects will move with a steady
speed on a straight path.
Newton’s 2nd law
The acceleration of an object is directly proportional to the
unbalanced force and inversely proportional to the mass of the
object.
𝒂∝𝑭
𝑭 = 𝒎𝒂
𝟏
𝒂∝
𝒎
Newton’s 3rd law
Every action has an equal and an opposite reaction.
Unbalanced force,
mass and acceleration
The acceleration of an object directly
proportional to the unbalanced force.
𝒂∝𝑭
𝒂
𝒂
F
m
𝒂
m
F
If mass remain constant;
Large acceleration → large force
Small acceleration→ small force
(m=Constant mass)
F
The acceleration of an object inversely
proportional to the mass of the object.
𝟏
𝒂∝
𝒎
(F=Constant Force)
𝒂
𝒂
F
m
m
If force remain constant;
𝒂
M
F
Large acceleration → small mass
Small acceleration→ large mass
𝟏
𝒂∝
𝒎
𝒂
1
𝑥=
𝑚
𝒂∝𝒙
𝟏
𝒎
Concept learning questions
The direction of the acceleration and
the unbalanced force are the same.
Find the acceleration.
𝑎
F= ma
200 = 5a
a= 40 𝑚/𝑠 2
200 N
5 kg
𝑎
200 N
500 N
10 kg
F= ma
500-200 = 10a
300 = 10a
a= 30 𝑚/𝑠 2
Weight(W)
• The gravitational pull acting on the object towards the earth center is
called the weight.
Weight = mass × gravitational field strength.
W=m×g
• The gravitational force acting on 1 kg of object is called the gravitational field
strength (g = 10 N/kg)
Concept learning questions
• Find the weight.
5 kg
W = mg
= 5 × 10 = 50 N
(g= 10 N/kg)
• Find the mass.
W = mg
m
12 = m×10
m = 1.2 kg
W=12 N
Concept learning questions
Find the acceleration.
𝑎
5 kg
Drag is the friction force caused by air(air resistance).
The drag force always acting in the opposite direction
to the motion.
Find the drag force
2 𝑚/𝑠 2
𝐷𝑟𝑎𝑔(𝐷)
5 kg
𝑊 = 𝑚𝑔
F = ma
50 = 5 × a
a= 10 𝑚/𝑠 2 = g
g= Gravitational acceleration.
𝑊
F = ma
50-D = 5 × 2
D = 50- 10 = 40 N
Concept learning questions
i)
F=ma
W - R = ma
800 – R = 80× 5
R = 800 – 400
R = 400 N
ii) F = ma
800 – R = 80 × 10
R=0
iii) F = ma
R – W = ma
R – 800 = 80 × 5
R = 400 + 800
R = 1200 N
Find the Normal reaction (R) force acting
on the person who is in the moving lift, if
the acceleration of the lift is,
(g=10 N/kg)
i) 5 m/𝑠 2 down
ii) 10 m/𝑠 2 down
iii) 5 m/𝑠 2 up
R
𝑎
W
A lift
𝑎
80 kg
Lab Practical:
Investigating Newton’s second law
Experiment 1: Investigating the relationship between acceleration and
unbalanced force.
Step 1:
Measure the mass of
the trolley.
Step 2:
Arrange the setup.
use light gates and
data logger to
measure time taken to
pass known length at
two points.
Step 3:
Calculate acceleration
for each applied force.
Take at least 6
readings.
Alternative: A video camera can be used instead of light gates and data loggers.
Step 4:
Plot a graph between
acceleration and force
Why is an inclined plane used for the practical?
Then the weight can minimize the effect of the friction on the trolley.
(A component of weight balances the friction force)
Hence the only unbalanced force acting on the trolley is the applied force by the weights.
𝑹
R
𝑭
𝑭
𝑻
W
𝑾
𝑻
When investigating the relationship between acceleration and unbalanced
force, the mass should remain constant. Explain how to keep the mass
constant?
Ex: Mass of the trolley = 300 g
Total mass(M) =mass of the
trolley + mass on the trolley +
mass of the weights in hanger
1N
M = (300 + 400 + 100) g = 800 g
M = (300 + 300 + 200) g = 800 g
2N
Total mass of the system remain
constant.
Ans: Force is increased by transferring one of the masses from trolley to the mass hanger.
Measurements and Calculations-Experiment -1
Mass of the trolley=…………
Length of the cardboard=………
Force(N)
t1(s)
Total mass of the system= ……………
t2(s)
𝒖 (m/s)
𝒗 (m/s)
𝒂(m/𝒔𝟐 )
Experiment 2: Investigating the relationship between acceleration and mass.
Change the mass on
the trolley and
calculate acceleration.
Plot a graph between
1
acceleration and
𝑚𝑎𝑠𝑠
How to find
acceleration?
Use light gates and
data logger or video
camera
Experiment 2: Investigating the relationship between acceleration and mass.
Ex: Mass of the trolley = 300 g
Total mass(M) =mass of the
trolley + mass on the trolley +
mass of the weights in hanger
2N
M = (300 + 200 + 200) g = 700 g
M = (300 + 300 + 200) g = 800 g
2N
Total mass of the system
increases.
Force remains constant.
Measurements and Calculations-Experiment-2
Mass of the trolley=…………
Constant force= ……………..
1/mass
(𝒌𝒈−𝟏 )
t1(s)
Length of the cardboard=………
t2(s)
𝒖 (m/s)
𝒗 (m/s)
𝒂(m/𝒔𝟐 )
Safety
Precautions
Weights may fall on your foot. –
Wearing safety shoes.
Trolley may move with higher
acceleration and hit the end of
the track with a large force- a
box filled with bubble wraps can
be placed at the end of the track.