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Laws of Motion and Friction Questions for AGNIVEER VAYU

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Why this topic matters · 8 min read
Laws of Motion (Newton's 3 laws) and friction are core mechanics topics tested heavily in Agniveer Vayu. Expect 2-3 questions combining inclined planes, connected bodies, and friction coefficients. Questions test conceptual clarity on action-reaction, limiting friction vs kinetic friction, and problem-solving with free-body diagrams. Weightage: ~8-10% of physics paper.

Newton's Three Laws of Motion

Newton's laws form the foundation of classical mechanics. The First Law states that an object at rest stays at rest, and an object in motion stays in motion unless acted upon by an external force — this is inertia. The Second Law quantifies force as the rate of change of momentum (F = ma). The Third Law states that for every action, there is an equal and opposite reaction. In Agniveer Vayu exams, the third law often confuses students because action-reaction pairs act on different bodies and never cancel out. For example, when you push a wall, the wall pushes back on you with equal force — but these forces act on different objects, so they don't neutralize.

  • First Law: Inertia — object maintains state unless external force acts
  • Second Law: F = ma — force causes acceleration proportional to mass
  • Third Law: Action-reaction pairs are equal, opposite, and act on different bodies
  • Action-reaction pairs NEVER cancel because they act on different objects
  • Net force on a system determines acceleration of center of mass
  • In equilibrium, net force = 0 and acceleration = 0

Static Friction and Limiting Friction

Static friction is the friction force that prevents an object from moving when an external force is applied. It can vary from zero up to a maximum value called limiting friction. The limiting friction depends on the normal force and the coefficient of static friction (μs). A key exam trap: static friction is NOT always equal to μs × N; it only reaches this maximum when the object is on the verge of sliding. Before that, static friction equals the applied force (up to the limit). Once motion begins, kinetic friction takes over, which is typically less than limiting friction and is constant at μk × N.

  • Static friction adjusts to match applied force (up to maximum)
  • Limiting friction = μs × N (maximum static friction)
  • Kinetic friction = μk × N (constant during motion)
  • Usually μk < μs — kinetic friction is weaker than limiting friction
  • Static friction acts when object is stationary; kinetic when moving
  • Direction of friction opposes the direction of motion or impending motion
Key formulas
Limiting Friction
f_limit = μs × N
When: Maximum static friction before object starts sliding
Kinetic Friction
f_k = μk × N
When: Friction force during motion (constant)
Normal Force on Horizontal Surface
N = mg
When: When object rests on horizontal surface with no vertical component of applied force
Worked examples

A 5 kg block on a horizontal surface (μs = 0.4, μk = 0.3). Applied force = 15 N. Is it moving? Limiting friction = 0.4 × 50 = 20 N. Since 15 N < 20 N, block does NOT move. Static friction = 15 N (balances applied force).

Same block, applied force = 25 N. Since 25 N > 20 N, block slides. Kinetic friction = 0.3 × 50 = 15 N. Net force = 25 - 15 = 10 N. Acceleration = 10/5 = 2 m/s².

Friction on Inclined Planes

Inclined plane problems combine gravity, normal force, and friction. The weight mg acts vertically downward and must be resolved into components parallel and perpendicular to the plane. The component perpendicular to the plane (mg cos θ) determines the normal force. The component parallel to the plane (mg sin θ) tries to pull the object down. Friction acts up the plane (opposing motion). For the object to remain stationary, the static friction must balance the parallel component of weight. This is a high-frequency Agniveer Vayu question type.

  • Weight component perpendicular to plane: mg cos θ (determines normal force)
  • Weight component parallel to plane: mg sin θ (tries to slide object down)
  • Normal force N = mg cos θ (on frictionless incline)
  • For equilibrium on incline: friction force = mg sin θ
  • Critical angle: tan θ_c = μs (angle at which object just starts to slide)
  • If θ > θ_c, object slides; if θ < θ_c, object remains stationary
Key formulas
Normal Force on Incline
N = mg cos θ
When: Object on inclined plane at angle θ
Parallel Component of Weight
F_parallel = mg sin θ
When: Component of weight along the incline
Critical Angle
tan θ_c = μs
When: Angle at which limiting friction equals parallel weight component
Acceleration Down Incline (with friction)
a = g(sin θ - μk cos θ)
When: Object sliding down incline with kinetic friction
Worked examples

Block on 30° incline, μs = 0.5. Is it stationary? tan 30° = 0.577 > 0.5, so YES, it remains stationary. Static friction = mg sin 30° = 0.5 mg.

Block on 45° incline, μk = 0.3. Acceleration down? a = g(sin 45° - 0.3 cos 45°) = g(0.707 - 0.212) = 0.495g ≈ 4.85 m/s².

Connected Bodies and Pulley Systems

Agniveer Vayu frequently tests connected bodies — two or more objects linked by strings or resting on each other. The key is to treat the system as a whole first to find acceleration, then analyze individual forces on each body using free-body diagrams. For pulley systems, assume the string is inextensible (doesn't stretch) and massless, and the pulley is frictionless. The tension is the same throughout the string. If two bodies are connected by a string over a pulley, they have the same magnitude of acceleration but in opposite directions.

  • For connected bodies, find total acceleration of system first
  • Then apply Newton's second law to each body individually
  • Tension in string is same throughout (ideal string, ideal pulley)
  • If masses are different, heavier mass accelerates downward
  • Constraint: acceleration magnitudes are equal for connected bodies
  • Free-body diagram is essential — draw forces on each body separately
Key formulas
Atwood Machine (two masses over pulley)
a = g(m1 - m2)/(m1 + m2)
When: Two masses m1 > m2 connected by string over frictionless pulley
Tension in Atwood Machine
T = 2m1 m2 g/(m1 + m2)
When: Tension in string connecting two masses
Worked examples

Two masses: 5 kg and 3 kg over a pulley. Acceleration = g(5-3)/(5+3) = 2g/8 = 0.25g ≈ 2.5 m/s². Tension = 2 × 5 × 3 × 10 / 8 = 37.5 N.

Block of 4 kg on horizontal surface (μk = 0.2) connected to 2 kg hanging mass. System acceleration = g(2 - 0.2×4)/(4+2) = g(2 - 0.8)/6 = 1.2g/6 = 0.2g ≈ 2 m/s².

⚠ Common mistakes to avoid
  • Confusing action-reaction pairs with balanced forces. Action-reaction act on different bodies and never cancel. Balanced forces act on the SAME body and cancel.
  • Assuming static friction always equals μs × N. Static friction varies from 0 to μs × N depending on applied force. It only reaches maximum when object is about to slide.
  • Forgetting to resolve weight into components on an incline. Many students use mg directly instead of mg sin θ and mg cos θ.
  • Treating tension as different in different parts of a string. In ideal problems (massless, inextensible string), tension is uniform throughout.
  • Neglecting friction in connected body problems. Always check if friction is mentioned and include it in the net force calculation.
  • Misidentifying the direction of friction. Friction opposes motion (or impending motion), not the applied force direction.
🧠 Memory aids
  • Newton's 3 Laws: INF — Inertia (1st), Newton's force (2nd), Newton's pairs (3rd). Action-Reaction are Newton's Pairs, not balanced forces.
  • Friction types: SLICK — Static Limiting Is Coefficient-based; Kinetic is lower. Remember μs > μk always.
  • Incline trick: Perpendicular = cos, Parallel = sin. Think: cos is 'close to vertical', sin is 'slanted'.
  • Connected bodies: TENSION SAME — Tension is uniform in ideal string. Acceleration is same magnitude but opposite direction for two connected masses.
  • Critical angle: tan θ_c = μ — At critical angle, friction force exactly balances weight component. Above this angle, object slides.
🎯 AGNIVEER VAYU exam tips
  • Agniveer Vayu physics paper typically has 1-2 direct questions on friction and 1-2 on inclined planes. Expect mixed questions combining both concepts.
  • Free-body diagrams are your best friend. Examiners test whether you can identify all forces correctly. Draw them even if not asked.
  • Inclined plane + friction questions often ask for critical angle or acceleration. Know both formulas cold.
  • Connected body questions appear in 'medium' difficulty section. They test conceptual understanding of tension and constraint. Practice Atwood machine variations.
  • Recent Agniveer Vayu papers show preference for numerical problems over theory. Expect 2-3 calculation-based questions requiring substitution into formulas.
  • Watch out for 'trick' questions where friction is zero or where the object is already moving (use kinetic friction, not static).
  • Time management: friction + incline problems take 3-4 minutes each. Allocate accordingly in your 2-hour paper.

Sample questions

Q1 · medium · AI-verified
Two blocks of masses 3 kg and 5 kg are connected by a string and placed on a frictionless surface. A force of 16 N is applied to the 5 kg block. What is the tension in the string?
  1. 4 N
  2. 10 N
  3. 6 N
  4. 8 N
Q2 · hard · AI-verified
A car of mass 1200 kg is moving at 72 km/h. The driver applies brakes and the car stops in 50 m. What is the coefficient of kinetic friction between the tyres and the road? (g = 10 m/s²)
  1. 0.6
  2. 0.5
  3. 0.3
  4. 0.4
Q3 · easy · AI-verified
Which of Newton's laws states that 'every object continues in its state of rest or uniform motion in a straight line unless acted upon by an external force'?
  1. Law of Conservation of Momentum
  2. Newton's First Law
  3. Newton's Third Law
  4. Newton's Second Law
Q4 · hard · AI-verified
A block of mass 4 kg slides down a frictionless inclined plane of angle 45° from rest and reaches the bottom in time t₁. The same block slides down a rough inclined plane of the same angle from rest and reaches in time t₂. If μk = 0.5, the ratio t₂/t₁ is:
  1. 1.5
  2. 2
  3. √3
  4. √2
Q5 · medium · AI-verified
Which of the following is the correct statement of Newton's Third Law of Motion?
  1. The rate of change of momentum is equal to the net applied force.
  2. For every action, there is an equal and opposite reaction acting on a different body.
  3. A body remains at rest until an external force acts on it.
  4. Force equals mass times acceleration.
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