CBSE Class 9 Physics Force and Laws of Motion Notes Set 07

Official Class 9 Physics Revision Material: CBSE Class 9 Physics Force and Laws of Motion Notes Set 07

Access comprehensive revision notes for Force and Laws of Motion using the CBSE Class 9 Physics Force and Laws of Motion Notes Set 07. Designed to align with the 2026-27 academic syllabus for Class 9 Physics, these concept summaries help students streamline their exam preparation and review complex topics efficiently.

Chapter-wise Concept Summaries: Force and Laws of Motion

View or download the dedicated Force and Laws of Motion revision resource below. Engaging with these structured concept notes daily ensures continuous academic progress and mastery of the 2026-27 curriculum for Physics.

Force and the Four Fundamental Interactions

 

A force is a push or a pull that changes, or tries to change, the state of rest or of uniform motion of a body. It arises from contact (normal reaction, tension, friction, spring force) or from a distance (gravity, electric and magnetic forces).
Every force seen in nature is a form of four basic interactions: gravitational, electromagnetic, strong nuclear and weak nuclear. Gravity is the weakest; the strong nuclear force is the strongest but acts only up to about 10⁻¹⁵ m.
Contact forces such as tension, friction, normal reaction and spring force are all electromagnetic in origin.

 

Contact Forces

 

1. Normal reaction N: the component of the contact force perpendicular to the surfaces; it adjusts itself to whatever value stops the surfaces from penetrating, and is not always equal to mg.
2. Tension T: the pulling force along a string or rope. In a light string passing over a smooth pulley, the tension is the same throughout.
3. Spring force: F = −kx, proportional to the extension or compression x. A spring balance reads the force at one end only.
4. Friction: static friction adjusts up to a limit fs ≤ μs N; kinetic friction is fk = μk N and opposes motion. Friction on a rolling or walking body acts forward on the driving parts, because it opposes the slipping tendency of the contact point.

 

Free Body Diagram and Problem Solving

 

1. Choose the body (or system) and draw only the forces acting on it: weight, normal reaction, tension, friction and applied forces. Do not draw forces the body exerts on others.
2. Choose axes, often along and perpendicular to the motion or the incline, and resolve each force.
3. Write F = ma along each axis. Constraints from strings (equal accelerations) and from pulleys give the remaining equations.
4. Solve the equations and check units and limiting cases.

 

Newton's Laws of Motion

 

1. First law: a body stays at rest or in uniform motion in a straight line unless a net external force acts on it. This defines inertia and inertial frames.
2. Second law: the net force equals the rate of change of momentum, \( \vec F=\frac{d\vec p}{dt} \), which is \( m\vec a \) for constant mass. Force in newton: 1 N = 1 kg m/s².
3. Third law: to every action there is an equal and opposite reaction. The two forces act on different bodies, so they never cancel each other, and they are equal in magnitude in every frame of reference.

 

Useful Results

 

1. Pushing blocks in contact: the force between two blocks equals the mass ahead of the junction times the common acceleration.
2. Atwood machine: \( a=\frac{(m_1-m_2)g}{m_1+m_2} \), \( T=\frac{2m_1m_2g}{m_1+m_2} \).
3. Lift: apparent weight N = m(g + a) for upward acceleration a, m(g − a) for downward acceleration, and mg for uniform motion. In free fall N = 0.
4. Force from a stream of bullets or water that stops: F = (mass per second) × v; if it rebounds with the same speed, F = (mass per second) × 2v.
5. Block on a rough incline just slipping: tan θ = μs.
6. Impulse = F × t = change of momentum, equal to the area under the force–time graph.

 

Worked Example 1

 

Two blocks of 4 kg and 2 kg lie in contact on a smooth table and a force of 12 N pushes the 4 kg block. Find the contact force.
Acceleration of both: a = 12/6 = 2 m/s². Contact force = mass ahead × a = 2 × 2 = 4 N.

 

Worked Example 2

 

A man of 60 kg stands on a weighing machine in a lift that accelerates upward at 2 m/s². Find the reading (g = 10 m/s²).
N − mg = ma, so N = 60 × 12 = 720 N, which the machine shows as 72 kg.

 

Worked Example 3

 

A 5 kg block is pressed against a wall by a horizontal force of 100 N, with μ = 0.5. Does it slide? (g = 10 m/s²)
Limiting friction = 0.5 × 100 = 50 N, equal to the weight 50 N, so the block is just in equilibrium. Friction acts upward and equals the weight.

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Yes, our CBSE Class 9 Physics Force and Laws of Motion Notes Set 07 include 50% competency-based questions with focus on core logic, keyword definitions, and the practical application of Physics principles which is important for getting more marks in 2026 CBSE exams.

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Yes, our CBSE Class 9 Physics Force and Laws of Motion Notes Set 07 provide a detailed, topic wise breakdown of the chapter. Fundamental definitions, complex numerical formulas and all topics of CBSE syllabus in Class 9 is covered.

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