The Work-Energy Theorem & Kinetic Energy (NCERT 5.1, 5.2, 5.4)
1. What the Theorem Claims
Chapter 4 taught you that forces change motion; this chapter prices that change. The claim, in one line: the net work done by all forces on a particle equals the change in its kinetic energy — Wnet = Kf − Ki. NCERT builds this slowly (sections 5.1–5.2): first the everyday word "work" is given a precise meaning — force's scalar delivery along a displacement — then kinetic energy is defined as the balance that the delivery fills or drains. The payoff is enormous: a single scalar equation that needs no directions, no components, no free-body geometry — just the speeds at the start and the end. Every braking-distance question, every "with what speed does it reach the bottom" question, and half of all spring problems are this theorem wearing a costume.
Keep one caution from the start: "net" is doing real work in that sentence. The theorem consumes the total work of all forces — applied pushes, gravity, friction, normal — and a student who feeds it only the applied force's work gets answers that are confidently wrong. The bookkeeping discipline is section 4 of this page.
2. Complete Theory: Kinetic Energy and the Derivation
Kinetic energy (NCERT 5.4). A body of mass m moving with speed v has K = ½ m v². The units are joules (1 J = 1 kg m² s⁻²), the dimensions are M L² T⁻² — the same as work, which is exactly the point — and K is a scalar that is never negative. Speed, not velocity, enters the formula: a car rounding a bend at constant 20 m/s has constant kinetic energy even though its velocity changes every instant. Two identities earn their keep all chapter: K = p²/2m (from p = mv, so K = ½m(v²) = p²/2m), and the comparison consequences — equal momenta give the lighter body more kinetic energy; equal speeds give the heavier body more.
The derivation (learn it — it is a theory item). For a constant net force F along the line of motion, the second law gives a = F/m. Take the kinematic identity v² − u² = 2as and multiply both sides by m/2:
½ m v² − ½ m u² = m a s = (F) s = Wnet
The left side is Kf − Ki; the right side is the net work. Hence Wnet = ΔK. The derivation openly borrows Chapter 2's equation of motion — which is why the theorem and the kinematics equations must (and do) agree on every braking and launch numerical, a cross-check NTA occasionally asks you to perform explicitly.
Generality. Although derived for a constant force, the theorem is true far beyond its derivation: for a variable force it reads Kf − Ki = ∫F dx, and for non-conservative forces it simply includes their work in the net total. Its one hard condition is the frame: work and speed must be measured in the same inertial frame, because kinetic energy is frame-dependent — a suitcase at rest on a station platform has enormous kinetic energy in a passing train's frame.
3. Visualising the Energy Ledger
Figure 5.1 — The work-energy theorem as a ledger
Left bar: initial kinetic energy ½ × 2 × 4² = 16 J. Red segment: friction's work, negative. Right bar: final kinetic energy. The bar heights are the audit.
Exam read-out: the theorem is an audit, not a slogan — every joule is accounted for. Note the arithmetic: ½ × 2 × 3² = 9 J, and 16 − 9 = 7 J of friction work over the slide. When NTA asks "how much energy became heat?", it is asking for exactly the red segment.
4. Solved Examples
Example 1 — Launch speed from applied work
A 4 kg crate at rest on a frictionless floor is pushed by a steady 8 N horizontal force through 9 m. Find its final speed by the work-energy theorem.
Solution (step by step): Only the push does work (floor frictionless; normal and weight ⊥ displacement). W = F s = 8 × 9 = 72 J. Theorem: ΔK = 72 J with Ki = 0, so ½ × 4 × v² = 72 → v² = 36 → v = 6 m/s. Cross-check by forces: a = 8/4 = 2 m s⁻²; v² = 2as = 2 × 2 × 9 = 36 ✓. Both routes agree — as they must, since the derivation is the kinematic equation.
Example 2 — Stopping distance from friction's work
A 2 kg block slides along a rough floor at 4 m/s. Kinetic friction is 4 N. How far does it slide before stopping?
Solution: Only friction does work along the motion: W = −f s = −4s. Theorem: −4s = 0 − ½ × 2 × 4² = −16 → s = 4 m. Kinematic cross-check: a = −f/m = −2 m s⁻²; 0 = 16 + 2(−2)s → s = 4 m ✓. The kinetic energy that vanished (16 J) became heat — the same ledger as Figure 5.1. Notice the mass cancels in this particular setup: any block at 4 m/s with deceleration 2 m s⁻² stops in 4 m.
5. Practice Questions
Q1. Find the kinetic energy of a 500 kg car moving at 72 km h⁻¹. Also find its KE at 36 km h⁻¹ and state the ratio.
ANSWER: 72 km h⁻¹ = 20 m/s → K = ½ × 500 × 400 = 100,000 J = 100 kJ. At 36 km h⁻¹ = 10 m/s → K = ½ × 500 × 100 = 25 kJ. Halving the speed quarters the KE: ratio 4 : 1 — the v² in ½mv² at work.
Q2 (MCQ). If the speed of a body is tripled, its kinetic energy becomes: (a) 3× (b) 6× (c) 9× (d) 27×
ANSWER: (c) — K ∝ v² at fixed mass: 3² = 9 times. This scaling is why braking distance (which must absorb the KE) grows roughly as speed squared, the fact behind every highway-safety question.
Q3. An 800 kg car speeds up from 54 km h⁻¹ to 72 km h⁻¹. Find the net work done on it.
ANSWER: 54 km h⁻¹ = 15 m/s, 72 km h⁻¹ = 20 m/s. Wnet = ΔK = ½ × 800 × (400 − 225) = 400 × 175 = 70,000 J = 70 kJ. No force diagram, no engine detail needed — endpoints only; that is the theorem's whole economy.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| K = ½ m v² | Speed in a chosen inertial frame; K ≥ 0; unit joule |
| K = p²/2m | Equal p → lighter body has more K; equal K → heavier body has more p |
| Wnet = ΔK | All forces count; same inertial frame for work and speeds |
| Wnet = 0 ⇒ v constant | Speed constant, direction may change (uniform circular motion) |
| Kf − Ki = ∫F dx | Variable-force form (next page develops the area rule) |
| [K] = M L² T⁻² | Same dimensions as work — the two belong on one ledger |
The theorem converts force problems into energy problems, but it does not conserve mechanical energy — that needs the potential-energy pages ahead. Kinetic energy alone is just the balance that net work fills or drains.
7. Frequently Asked Questions
State the work-energy theorem.
The work-energy theorem says that the net work done by all forces acting on a particle equals the change in its kinetic energy: W_net = K_f − K_i = ΔK. 'Net' means the sum of the work of every force — applied, gravity, friction, normal. It is a scalar equation, needs no force directions, holds for constant and variable forces alike, and is valid in any inertial frame provided the same frame is used for both the work and the speeds.
Derive the work-energy theorem for a constant force.
For a constant net force F along the displacement, Newton's second law gives a = F/m. Multiplying the kinematic identity v² − u² = 2as by m/2 gives ½mv² − ½mu² = mas. But mas = (ma)s = F s, which is the work done by the constant net force. Hence W = ½mv² − ½mu² = K_f − K_i, which is the theorem. The derivation openly recycles the Chapter 2 kinematic equations — that is why the theorem and kinematics cross-check each other on braking-distance and launch-speed problems.
Does the work-energy theorem hold when friction acts or the force varies?
Yes to both — the theorem is completely general. With friction, simply include friction's (usually negative) work in the net total: W_applied + W_gravity + W_friction = ΔK. With a variable force, the same statement survives as K_f − K_i = ∫F dx, the area under the force-displacement curve; NCERT develops this in section 5.6. What the theorem never promises is conservation of mechanical energy — that is a separate claim that fails whenever friction is skimming energy out of the system.
How is kinetic energy related to momentum?
Since p = mv, kinetic energy can be written as K = ½mv² = p²/2m. The exam-useful consequences: at equal momentum, the lighter body carries the larger kinetic energy (K ∝ 1/m); at equal kinetic energy, the heavier body carries the larger momentum (p = √(2mK)); and a ninefold increase in kinetic energy corresponds to a threefold increase in speed. This single line resolves a whole family of NTA comparison MCQs without any computation.
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