4
Question Question 4
Certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1. Calculate the work done by the force.
Given,
Initial velocity u = 5 ms-1
Mass of the body = 20 kg
Final velocity v = 2 ms-1
Initial kinetic energy
Ei = mu2
Substituting we get,
Ei = x 20 x 52
= 10 × 25
= 250 J
Final kinetic energy
Ef = mv2
= x 20 x 22
= 10 × 4
= 40 J
As, Work done = Change in kinetic energy = Ef – Ei
Work done = 40 J - 250 J
Work done = -210 J
Hence, work done by the force = -210 J.
BRIGHT TUTORIALS
BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
Science | Chapter 10: Work and EnergyWeb Content — Quick Reference
Chapter 10: Work and Energy — Quick Reference
Quick Revision Points
- Work W = Fs cosθ (joule); W = 0 when F ⊥ s or displacement = 0
- KE = ½mv²; PE = mgh; Work-energy theorem: W = ΔKE
- Conservation: energy transforms but total remains constant; PE + KE = constant in free fall
- Power P = W/t (watt = J/s); 1 HP = 746 W
- 1 kWh = 3.6 × 10⁶ J = 1 unit of electricity
Exam Tips for Chapter 10
- 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: ch10-work-and-energy.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