NEB Class 11 • Physics • Mechanics

Dynamics: NEB Class 11 Physics Guide

Connect changes in motion to interactions through Newton’s laws, free-body diagrams, friction and a systematic net-force method for NEB-ready problems.

Concepts before equations and isolated free-body diagrams
Worked horizontal, inclined and connected-body examples
Checks, common mistakes, practice tasks and FAQs

Curriculum and source boundary

From describing motion to explaining its change

Kinematics describes position, velocity and acceleration; dynamics asks what interactions produce the acceleration. Review the Kinematics guide if sign conventions or acceleration meaning are weak and the Vectors guide if resolving forces is difficult.

The current CDC Physics Grade 11 page is the Nepal scope checkpoint. The OpenStax Newton’s-laws introduction, Newton’s second-law section and friction reference support the underlying mechanics. Follow current NEB and college instructions for prescribed breadth and notation.

Interaction changes motion

Force, mass, inertia and net force

A force is an interaction capable of changing an object’s velocity. Mass measures inertia: resistance to acceleration for a given net force. A body may have several forces while its acceleration is zero because the vector sum is zero. Conversely, a constant non-zero net force produces acceleration even if the body was initially at rest.

Net force is not a new extra force. It is the vector sum of all external forces on the chosen system. Changing the system boundary changes which interactions are external. For a book on a table, Earth pulls the book downward and the table pushes it upward. If those forces balance, the book has zero acceleration.

Force	Typical direction	Modelling note
Weight W = mg	Toward Earth	Acts on mass through gravity
Normal reaction N	Perpendicular to contact surface	Not automatically equal to weight
Tension T	Along a taut string away from the body	Ideal-string assumptions must be stated
Friction f	Opposes relative or impending sliding	Static friction adjusts up to a limit
Applied force	As specified	Resolve if angled to chosen axes

Three connected laws

Newton’s laws with correct interpretation

First law

In an inertial frame, zero net external force means constant velocity. Rest is one constant-velocity case, not the only one.

Second law

Net external force equals rate of momentum change; for constant mass, ΣF = ma. Apply components along declared axes.

Third law

Interaction forces between two bodies are equal in magnitude and opposite in direction. They act on different bodies and do not cancel on one free-body diagram.

Newton’s first law corrects the idea that continuous force is needed for continuous motion. Force is needed for acceleration—a change in velocity. On a nearly frictionless surface, an object can continue with constant velocity after the push stops. In everyday life friction often hides this.

The second law is a vector equation. If horizontal forces cancel but vertical forces do not, acceleration is vertical. Choose axes that make components convenient, then write ΣF_x = ma_x and ΣF_y = ma_y. The unit newton is kg m s⁻².

For the third law, name both bodies: “table on book” and “book on table.” Weight and normal on the same book are not a third-law pair, even when equal. Their partners act on Earth and the table respectively.

The most important diagram

How to draw a free-body diagram

Choose the system. Name the one body or combined group you will analyse.
Replace it with a point or simple box. Remove distracting scenery.
Draw external forces only. Label each by type and direction.
Choose axes. On an incline, parallel and perpendicular axes often reduce components.
Resolve angled forces. Do not split a force and then also keep the original arrow in equations.
Write component equations. Include signs from the diagram.

A free-body diagram is not a picture of motion. An object moving right may have a net force left if it is slowing. Arrow lengths may indicate relative force size, but direction and identity are more important than artistic scale.

Audit question: For every force arrow, ask “which external body exerts this force on my chosen system?” If you cannot name the interaction, the arrow may be invented or mislabelled.

Contact with a surface

Static friction, kinetic friction and normal reaction

Static friction acts when contacting surfaces do not slide relative to one another. Its magnitude adjusts from zero up to a limiting value; it is not automatically μ_sN. Kinetic friction applies during sliding and is often modelled as f_k = μ_kN. The coefficients are dimensionless idealizations for specified surfaces and conditions.

The normal reaction is determined by perpendicular force balance or acceleration. On a horizontal surface with no other vertical components and zero vertical acceleration, N = mg. If a rope pulls upward at an angle, the normal force can be smaller; if an applied force pushes downward, it can be larger. On an incline, a common result is N = mg cos θ only under the corresponding conditions.

Friction opposes relative motion or the tendency to slide, not necessarily the object’s overall direction of travel. A driven wheel, walking foot or block on a moving surface can require more careful interaction reasoning. Draw the impending relative motion before choosing the friction direction.

Zero acceleration

Translational equilibrium and balanced forces

Equilibrium means vector net force is zero, so acceleration is zero. The body may be at rest or moving with constant velocity. In two dimensions, both ΣF_x = 0 and ΣF_y = 0 must hold.

For a sign suspended by two cables, resolve each tension into components. Horizontal components can cancel while vertical components support weight. Do not assume tensions are equal unless geometry and loading justify it. A symmetric arrangement often produces equal tensions; an asymmetric arrangement generally does not.

Equilibrium example

A 100 N lamp hangs from two identical cables, each making 30° above the horizontal. Horizontal components cancel. Vertical balance gives 2T sin 30° = 100 N, so T = 100 N in each cable.

Check: Each vertical component is 50 N. If the cables became more horizontal, a larger tension would be required to supply the same vertical support.

Choose useful axes

Forces on an inclined plane

For a block on an incline at angle θ, choose x parallel to the plane and y perpendicular. Weight resolves into mg sin θ down the plane and mg cos θ into the plane. The normal force balances the perpendicular component only if there is no perpendicular acceleration and no additional perpendicular force.

Along the plane, include applied force, tension and friction with signs. For sliding down, kinetic friction points up the plane. For a block at rest that would otherwise slide down, static friction also points up the plane and takes the value needed until its maximum is reached.

Frictionless incline

A 5.0 kg block slides on a frictionless 30° incline. Down-plane net force is mg sin 30°. Therefore a = g sin 30° = 4.9 m s⁻² using g = 9.8 m s⁻². Mass cancels.

Check: At θ = 0°, acceleration is zero; at θ = 90°, it approaches g. The result lies between those limits.

Worked examples

Apply free-body diagrams and component equations

Example 1: horizontal net force

A 12 kg trolley is pushed east with 50 N while friction is 14 N west. Net force = 50−14 = 36 N east. Acceleration = 36/12 = 3.0 m s⁻² east.

Check: The acceleration direction matches the larger applied force, not necessarily the trolley’s current velocity.

Example 2: angled pull and friction

A 10 kg crate is pulled by 40 N at 30° above horizontal. Take g = 9.8 m s⁻² and μ_k = 0.20. Vertical equilibrium gives N = mg − 40 sin30° = 98−20 = 78 N. Kinetic friction = 0.20×78 = 15.6 N. Horizontal pull = 40 cos30° ≈ 34.6 N, so net horizontal force ≈ 19.0 N and a ≈ 1.90 m s⁻².

Check: The upward pull reduces normal force and therefore reduces friction compared with μmg.

Example 3: two connected blocks

Blocks of 2 kg and 3 kg are connected on a frictionless horizontal surface. A 20 N force pulls the 3 kg block. Treat both as one system: a = 20/(2+3) = 4 m s⁻². For the 2 kg block alone, tension provides the horizontal force, so T = 2×4 = 8 N.

Check: For the 3 kg block, 20−8 = 12 N = 3×4 N.

Example 4: third-law identification

A student pushes a wall. The student exerts a force on the wall; the wall exerts an equal and opposite force on the student. These forces act on different bodies. On the student’s free-body diagram only the wall-on-student force appears, along with weight, floor forces and any others.

Reliable problem solving

The system–diagram–equation–check method

Name the system and whether you will analyse one body or a combined group.
Draw separate free-body diagrams when internal forces must be found.
Choose axes and resolve forces once.
Write ΣF = ma by component before inserting values.
Solve the combined system first when it removes internal tension.
Check direction, dimensions, limiting cases and consistency across bodies.

Use the PhET Forces and Motion: Basics simulation for prediction–observation–explanation practice. Predict acceleration after changing one force, mass or friction setting, observe, then explain the result using net force.

Protect conceptual marks

Common Dynamics mistakes

Force follows velocity

Net force follows acceleration. An object can move right while net force points left during slowing.

Third-law forces cancel

They act on different bodies. Forces cancel in one equation only when they act on the same chosen system.

N always equals mg

Normal force depends on perpendicular forces and acceleration; angled pulls and inclines change it.

Static friction equals μN

Static friction adjusts up to its maximum. Use the required equilibrium value and compare it with the limiting value.

Try without copying

Practice and self-check

Draw a free-body diagram for a book at rest on a table and identify third-law partners outside the diagram.
A 6 kg body experiences 20 N east and 8 N west. Find acceleration.
Explain how a moving object can have zero net force.
A 4 kg block rests on a 25° incline. Find the down-plane weight component and normal force under simple equilibrium assumptions.
A 30 N force pulls a crate 20° above horizontal. Explain qualitatively how it changes normal force and friction.
Two blocks are connected. Describe when analysing them as one system is useful and when separate diagrams are required.

Readiness: You should identify every external force, justify its direction, write component equations and check the result across a second body or limiting case. For online or physical tuition, call 9846662070.

Frequently asked questions

Questions students ask about Dynamics

What is Dynamics in Class 11 Physics?

Dynamics studies how interactions and net forces relate to changes in motion, commonly through Newton’s laws and momentum ideas.

Does an object need a force to keep moving?

No. In an inertial frame, zero net force gives constant velocity. A net force is required to change velocity.

Why do action and reaction forces not cancel?

They act on different bodies. Only forces acting on the same chosen system are combined in one free-body equation.

Is normal force always equal to weight?

No. Normal force depends on the perpendicular component balance. Inclines, angled applied forces and vertical acceleration can change it.

Is static friction always μsN?

No. Static friction adjusts to the value needed up to a maximum μsN. Equality applies only at impending slip under the model.

Where can I get help with NEB Class 11 Dynamics?

For current online or physical tuition options, call 9846662070 and confirm schedule, class mode, teacher availability and fees.

References and next steps

Sources and related study guides

Use the Kinematics practice set to review acceleration and the Physics revision roadmap to plan the remaining mechanics topics. Sources and scope were checked on 2 August 2026; follow current CDC, NEB and college instructions if assessment requirements change.

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