CBSE Class 9 Science Question 9 of 17

Force and Laws of Motion — Question 10

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Question 10

Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer

As the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. Hence, the magnitude of opposing frictional force is equal to the force exerted on the cabinet i.e., 200 N. So, the total friction force is 200 N.

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CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
Science | Chapter 8: Force and Laws of MotionWeb Content — Quick Reference

Chapter 8: Force and Laws of Motion — Quick Reference

Newton laws inertia momentum F=ma conservation of momentum action reaction

Quick Revision Points

  • Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
  • First Law (Inertia): object stays at rest/uniform motion unless external force acts
  • Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
  • 1 N = force giving 1 kg an acceleration of 1 m/s²
  • Third Law: every action has equal and opposite reaction (on different bodies)
  • Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)
Exam Tips for Chapter 8
  • Read the detailed chapter notes for complete coverage of all NCERT topics.
  • Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
  • Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
  • For numericals (if applicable), practice at least 20 problems of varying difficulty.
  • Refer to the practice question bank (200+ questions) for thorough preparation.
Related Resources
  • Detailed Notes: ch08-force-and-laws-of-motion.html
  • Practice Questions: 100+ questions with answers in 05-practice-questions/
  • Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
  • Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html
Web Content: Motion | Bright Tutorials
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BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
Science | Chapter 7: MotionWeb Content — Quick Reference

Chapter 7: Motion — Quick Reference

distance displacement speed velocity acceleration equations of motion graphs circular motion

Quick Revision Points

  • Distance (scalar, total path) vs Displacement (vector, shortest path)
  • Speed = distance/time (scalar); Velocity = displacement/time (vector)
  • Acceleration a = (v−u)/t; unit: m/s²
  • Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
  • d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
  • 1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
  • Uniform circular motion: constant speed, changing velocity (direction changes)
Exam Tips for Chapter 7
  • Read the detailed chapter notes for complete coverage of all NCERT topics.
  • Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
  • Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
  • For numericals (if applicable), practice at least 20 problems of varying difficulty.
  • Refer to the practice question bank (200+ questions) for thorough preparation.
Related Resources
  • Detailed Notes: ch07-motion.html
  • Practice Questions: 100+ questions with answers in 05-practice-questions/
  • Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
  • Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html