ICSE Class 10 Physics: Work, Energy and Power — Important Questions with Answers 2026

Short Answer Questions (2-3 Marks)

• Q: Define work. State the conditions necessary for work to be done.
  Ans: Work is done when a force acts on a body and the body moves in the direction of the applied force. W = F × d × cos θ. Conditions: (1) A force must act on the body, (2) The body must be displaced, (3) There must be a component of force along the displacement.
• Q: What is the SI unit of work? Define one joule.
  Ans: SI unit of work is Joule (J). One joule is the work done when a force of 1 Newton displaces a body through 1 metre in the direction of the force. 1 J = 1 N × 1 m.
• Q: State the work-energy theorem.
  Ans: The work-energy theorem states that the net work done on a body equals the change in its kinetic energy. W_net = ΔKE = ½mv² - ½mu². If net work is positive, KE increases; if negative, KE decreases.
• Q: Define kinetic energy. Derive the expression KE = ½mv².
  Ans: Kinetic energy is the energy possessed by a body due to its motion. Derivation: Work done W = F × s. From v² = u² + 2as, s = (v² - u²)/2a. And F = ma. So W = ma × (v² - u²)/2a = ½m(v² - u²). For u = 0: KE = ½mv².
• Q: Define potential energy. A body of mass 2 kg is raised to a height of 10 m. Find its PE. (g = 10 m/s²)
  Ans: Potential energy is the energy possessed by a body due to its position or configuration. PE = mgh. Given: m = 2 kg, h = 10 m, g = 10 m/s². PE = 2 × 10 × 10 = 200 J.

Long Answer / Application Questions (4-6 Marks)

• Q: State the law of conservation of energy. Explain with a freely falling body.
  Ans: The law of conservation of energy states: Energy can neither be created nor destroyed, only transformed from one form to another. Total energy remains constant. For a freely falling body: At height h: PE = mgh, KE = 0, Total = mgh. At height x: PE = mgx, KE = mg(h-x), Total = mgh. At ground: PE = 0, KE = mgh, Total = mgh. Total energy remains constant.
• Q: Define power. What is 1 watt?
  Ans: Power is the rate of doing work. P = W/t. SI unit: Watt (W). 1 Watt is the power when 1 Joule of work is done in 1 second. 1 W = 1 J/s. Also, P = F × v (force × velocity).
• Q: A machine does 5000 J of work in 20 seconds. Find its power.
  Ans: Given: W = 5000 J, t = 20 s. Power P = W/t = 5000/20 = 250 W.
• Q: Differentiate between kinetic energy and potential energy.
  Ans: Kinetic Energy: energy due to motion, KE = ½mv², depends on mass and velocity, is always positive. Potential Energy: energy due to position, PE = mgh (gravitational), depends on mass, height and g, can be positive or negative depending on reference.
• Q: A ball of mass 200 g is thrown vertically upward with velocity 20 m/s. Find (i) KE at the point of projection, (ii) PE at maximum height.
  Ans: (i) KE = ½mv² = ½ × 0.2 × (20)² = ½ × 0.2 × 400 = 40 J. (ii) At maximum height, all KE converts to PE (conservation of energy). PE = 40 J.

Exam Tips for This Chapter

• Revise all definitions and laws from Work, Energy and Power — they are commonly asked as 1-2 mark questions
• Practice diagrams related to Work, Energy and Power — neat labelled diagrams carry 2-3 marks
• For numericals, always show formula → substitution → answer with correct units
• Previous year analysis shows Work, Energy and Power carries 8-12 marks in the board exam