7.4.1 Kinetic Energy

Last updated at September 16, 2026 by

What is Kinetic Energy?
  • The energy possessed by an object due to its motion is called kinetic energy.
  • All moving objects possess kinetic energy.
    Example:
    • A moving bicycle
    • A rolling ball
  • GIF Watch: what is Kinetic Energy Animation for "What is Kinetic Energy". A short looping GIF, three or four beats, drawn in the flat classroom style of the chapter. Beat 1 — The energy possessed by an object due to its motion is called kinetic energy. Beat 2 — All moving objects possess kinetic energy. Beat 3 — An object that does not move has zero kinetic energy. Label every arrow and quantity, name the direction each force or motion acts in, and hold the last frame for a moment before the loop starts again.
  • An object that does not move has zero kinetic energy.
What is the formula of Kinetic Energy?
  • Kinetic energy of an object of mass m moving with velocity v is
  • K = ½ mv²
  • The SI unit of kinetic energy is the joule (J) — same as work.
  • Kinetic energy has no direction — it is just a number.
How do we derive the formula K = ½mv²? (Derivation)
  • We start with the Work-Energy Theorem —
    • Work done on an object = change in its energy
  • For a moving object, this energy is its kinetic energy.
  • So if we find the work done on the object, we will get the change in its kinetic energy.
  • Lets calculate this work.
Suppose a constant force F acts on an object of mass m. Its velocity changes from initial velocity u to final velocity v over a displacement s.
  • Step 1 — Write the equation for Work Done
    • From the definition of work
    • Work = Force × Displacement
    • W = F × s … (Eq. 1)
  • Step 2 — Substitute Force using Newton's Second Law
    • Newton's second law says
    • Force = mass × acceleration
    • F = ma
    • Putting this in Eq. 1
    • W = ma × s … (Eq. 2)
  • Step 3 — Rearrange the Kinematic Equation
    • From Chapter 4, the third equation of motion is
    • v² = u² + 2as
    • Rearranging it for s
    • s = (v² − u²) / 2a … (Eq. 3)
  • Step 4 — Combine the equations
    • Put Eq. 3 into Eq. 2
    • W = ma × (v² − u²) / 2a
    • a cancels out
    • W = ½ m(v² − u²)
    • Or, W = ½mv² − ½mu²
  • Now compare with the Work-Energy Theorem —
    • Work done = change in energy
    • W = ½mv² − ½mu²
    • So, change in energy = ½mv² (final) − ½mu² (initial)
  • This means the energy of the object due to its motion is ½ × mass × velocity²
  • This proves — Work done = Final kinetic energy − Initial kinetic energy
How do we get K = ½mv² from this?
  • Suppose the object starts from rest — initial velocity u = 0.
  • Then, Work done = ½mv² − ½m(0)² = ½mv²
  • All this work becomes the kinetic energy of the object.
  • So kinetic energy of an object of mass m moving with velocity v is
  • K = ½ mv²
How does work change kinetic energy?
Work done on object Velocity Kinetic energy
Positive work Increases Increases
Negative work Decreases Decreases
No work (W = 0), velocity unchanged Same Remains constant
📘 Example 7.4 (NCERT)
If the velocity of a vehicle doubles in magnitude, what will its kinetic energy be compared to its original value?
  • Let mass = m, initial velocity = v
  • Initial kinetic energy = ½mv²
  • New velocity = 2v
  • New kinetic energy = ½m(2v)² = ½m × 4v² = 4 × ½mv²
  • New kinetic energy is 4 times the previous value.
  • Note — velocity doubles, but kinetic energy becomes 4 times (because of the square).
📘 Example 7.5 (NCERT)
In one of their fastest deliveries, an Indian cricketer bowled a cricket ball with an approximate mass of 0.2 kg at a velocity of about 154.8 km h⁻¹. Calculate the kinetic energy of the ball at the time of its delivery.
  • Mass of the cricket ball = 0.2 kg
  • Velocity of the ball = 154.8 km h⁻¹
  • Converting to m s⁻¹
    • 1 km h⁻¹ = 1000 m / 3600 s = 5/18 m s⁻¹
    • v = 154.8 × 5/18
    • v = 43 m s⁻¹
  • K = ½mv²
  • K = ½ × 0.2 kg × (43 m s⁻¹)²
  • K = ½ × 0.2 × 1849
  • K = 184.9 J
📘 Example 7.6 (NCERT)
A jet aircraft of mass 15000 kg lands on the deck of an aircraft carrier. To stop the aircraft within the short length of the deck, a hook on the aircraft's tail is caught in a wire stretched across the deck. The wire exerts an approximately constant backward force of 367500 N and stops the jet within 100 m. What was the velocity of the aircraft just before the wire caught the hook?
  • (We saw this jet qualitatively in the Work-Energy Theorem section. Now lets calculate.)
  • Let the aircraft approach with velocity v
  • Initial kinetic energy = ½ × 15000 kg × v²
  • Final kinetic energy = 0 J (the jet stops)
  • Change in kinetic energy = 0 − ½ × 15000 × v²
  • Work done by the wire = F × s = 367500 N × (−100 m) (negative, force opposite to displacement)
  • By the work-energy theorem
    • Change in kinetic energy = Work done by the wire
    • − ½ × 15000 × v² = − 367500 × 100
    • v² = (367500 × 100 × 2) / 15000
    • v² = 4900 m² s⁻²
    • v = 70 m s⁻¹
  • v = 70 m s⁻¹ = 252 km h⁻¹ towards the aircraft carrier.
  • 🤔 Pause and Ponder 4 (from the book)
    Two objects A and B of mass m and 4m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?
    View answer Hide answer
    Answer
    • Kinetic energy of A = Kinetic energy of B
    • ½ × m × v_A² = ½ × 4m × v_B²
    • v_A² = 4 v_B²
    • v_A = 2 v_B
    • Ratio of velocities of A and B = 2 : 1
  • 🤔 Pause and Ponder 5 (from the book)
    Does the kinetic energy of an object which moves with constant velocity change with its position?
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    Answer
    • No.
    • Kinetic energy depends only on mass and velocity — K = ½mv².
    • It does not depend on position.
    • If velocity is constant, kinetic energy stays the same at every position.
  • 📋 Revise, Reflect, Refine, Q8 (Page 137)
    A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? (Assume energy transfer is entirely due to fuel, no losses.)
    View answer Hide answer
    Answer
    • Fuel used = energy given to the scooter = kinetic energy gained (starting from rest)
    • Day 1 — mass = 60 + 100 = 160 kg
      • Energy = ½ × 160 × v²
    • Day 2 — mass = 60 + 100 + 40 = 200 kg
      • Energy = ½ × 200 × v²
    • Ratio = 160 : 200
    • Ratio of fuel used = 4 : 5
  • 📋 Revise, Reflect, Refine, Q11 (Page 138)
    A 10.0 kg block is moving on horizontal floor with negligible friction. A variable force is applied on the block in its direction of motion from 0 m till 4 m (graph: force rises from 0 to 50 N over 0–1 m, stays 50 N from 1–3 m, falls back to 0 at 4 m). If the block had a kinetic energy of 180 J at 0 m, find the block's speed (i) at 0 m, and (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
    View answer Hide answer
    Answer
    • (i) Speed at 0 m
      • K = ½mv²
      • 180 = ½ × 10 × v²
      • v² = 36
      • v = 6 m s⁻¹
    • (ii) Speed at 4 m
      • Work done = area under force-displacement graph
      • Area = triangle (0–1 m) + rectangle (1–3 m) + triangle (3–4 m)
      • Area = (½ × 1 × 50) + (2 × 50) + (½ × 1 × 50)
      • Area = 25 + 100 + 25 = 150 J
      • New kinetic energy = 180 + 150 = 330 J
      • 330 = ½ × 10 × v²
      • v² = 66
      • v ≈ 8.1 m s⁻¹
    • Negative acceleration?
      • No. The force always acts in the direction of motion (it is never negative).
      • So the block never has negative acceleration — it only speeds up or moves at constant speed.
  • 📋 Revise, Reflect, Refine, Q13 (Page 138)
    A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. (Graph: speed constant at 35 m s⁻¹ from A at t = 0 to B at t = 1 s, then decreasing to 0 at C at t = 3 s.)
    View answer Hide answer
    Answer
    • (i) How does the car move between A and B?
      • At constant speed of 35 m s⁻¹.
      • This is the reaction time — the driver has seen the obstruction but the brakes have not been applied yet.
    • (ii) Kinetic energy of the car at A
      • K = ½mv²
      • K = ½ × 1000 × (35)²
      • K = ½ × 1000 × 1225
      • K = 612500 J
    • (iii) Work done by the brakes between B and C
      • The car goes from 612500 J of kinetic energy to 0.
      • Work done by brakes = change in kinetic energy = 0 − 612500
      • Work done by brakes = − 612500 J (negative — brake force is opposite to motion)
    • (iv) What does the kinetic energy transform into?
      • Mainly thermal energy (heat) — due to friction between the brakes and wheels, and tyres and road.
      • Some of it becomes sound energy.
📝 Important points — 7.4.1 Kinetic Energy
Point Detail
Kinetic energy Energy due to motion
Formula K = ½mv²
SI unit joule (J)
Direction None — kinetic energy is just a number
Object at rest K = 0
Velocity doubles K becomes 4 times
Work-KE link Work done = change in kinetic energy
💡 Worth remembering
  • The square matters — 2× velocity means 4× kinetic energy, 3× velocity means 9× kinetic energy. This is why fast cars are so dangerous.
  • Between two objects with the same kinetic energy, the lighter one moves faster.
  • Work done can be read directly from the area under a force-displacement graph, then converted to speed via K = ½mv².
✅ Quick self-check
  1. A 2 kg ball moves at 3 m s⁻¹. Find its kinetic energy.
    View Answer Hide Answer
    • K = ½mv²
    • K = ½ × 2 × (3)²
    • K = ½ × 2 × 9
    • K = 9 J
  2. A car's speed becomes 3 times. How many times does its kinetic energy become?
    View Answer Hide Answer
    • Kinetic energy depends on the square of velocity.
    • New K = ½m(3v)² = 9 × ½mv²
    • Kinetic energy becomes 9 times.
  3. What is the kinetic energy of a parked truck?
    View Answer Hide Answer
    • A parked truck is not moving, so v = 0.
    • K = ½m(0)² = 0
    • Kinetic energy = 0 J

Key terms and units

Term Meaning Unit
Kinetic energy (K) Energy due to motion, K = ½mv² joule (J)
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