Newton’s Second Law
Edexcel A-Level Physics
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Outline
1
Revision of Newton’s First Law
2
Deriving the Second Law
3
The Law: F = ma
4
Worked Example 1: Thrust of a Car
5
Worked Example 2: Effect of Towing an Identical Car
6
Worked Example 3: Resultant Force on a Barge
7
Summary
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Newton’s First Law
Newton’s First Law states:
An object remains at rest or in uniform motion in a straight line unless
acted on by an external resultant force.
Implication: No force no change in velocity.
This law introduces the concept of inertia — resistance to changes in
motion.
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Deriving Newton’s Second Law
Apparatus: Trolley, pulley, string, hanging mass, light gate.
Steps:
Connect trolley to a hanging mass over a pulley.
Measure acceleration using a light gate.
Change the hanging mass (force), keeping total mass constant.
Record acceleration each time.
Result: Acceleration increases with force; decreases with more mass.
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Experimental Setup
Pulley
Tension
Light Gate
Trolley
Mass
Weight
Trolley and pulley system showing direction of forces and motion.
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Conclusion
Acceleration is directly proportional to force:
Increasing the hanging mass increases the net force.
This results in a greater acceleration.
a∝F
Acceleration is inversely proportional to mass:
Adding mass to the trolley decreases acceleration.
This shows the greater the mass, the less the acceleration for the same
force.
1
a∝
m
This experiment confirms Newton’s Second Law:
F = ma
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Newton’s Second Law
The acceleration of an object depends on:
The net force acting on it
Its mass
Expressed mathematically:
F = ma
Where:
F : Net force (N)
m: Mass (kg)
a: Acceleration (m/s2 )
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F = ma
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Example 1: Thrust of a Car
Problem: A 1.5 tonne car accelerates from 60 km/h to 100 km/h in 5.0
seconds.
(a) Calculate the thrust required:
Convert velocities: 60 km/h = 16.7 m/s, 100 km/h = 27.8 m/s
Calculate acceleration:
v − v0
27.8 − 16.7
a=
=
= 2.22 m/s2
t
5.0
Apply F = ma:
F = (1.5 × 103 ) × 2.22 = 3300 N
(b) Drag Forces at Constant Velocity:
The car reaches constant velocity, meaning net force is zero.
The drag forces must balance the thrust, so the magnitude of the
collective drag forces is 3300 N.
These forces act in the opposite direction of motion.
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Example 2: Towing an Identical Car
Problem: A car can accelerate at 3.00 m/s2 . Given the same engine
thrust, determine the acceleration if it is towing another identical car.
Solution:
The total mass is now doubled, so the new mass is 2m.
From F = ma, the force F remains constant.
Acceleration is inversely proportional to mass:
a=
F
1
= × 3.00 = 1.50 m/s2
2m
2
Conclusion: The acceleration is halved when the mass doubles.
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Example 3: Resultant Force on a Barge
Problem: A barge is being towed along a canal. The forces acting on the
barge are shown in the diagram below.
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(a)Resultant Force: The forces are at right angles, so we use
Pythagoras’ theorem to calculate the resultant force:
F =
p
6002 + 6002 =
√
720000 = 850 N
(b) Calculate the acceleration of the barge:
The mass of the barge is 350 kg.
Using F = ma, the acceleration is:
a=
Edexcel A-Level Physics
F
850
2
=
= 2.43 m/s
m
350
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Summary
Newton’s First Law explains that motion remains unchanged unless a
force acts.
Newton’s Second Law: F = ma, force causes acceleration.
Thrust, drag, and mass all affect the acceleration of objects.
Worked examples show how to apply these concepts in real situations.
Homework: Solve problems involving varying forces, masses, and
acceleration.
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