The Conservation of Mechanical Energy (NCERT 5.8)
1. The Claim, and Its One Condition
Watch a ball fall: height drains, speed grows. Watch it rise on a smooth track: speed drains, height grows. The numbers refuse to wander — their sum stays put. The law: if only conservative forces do work on a system, its total mechanical energy E = K + U is constant, or between any two states,
Ki + Ui = Kf + Uf
The italicised condition is the entire law, and it is where NTA plants its traps: "only conservative forces do work" means no friction, no drag, no driving motors between the two snapshots. When the condition holds, this one equation replaces whole chains of force diagrams — it is the single most productive line in school mechanics, and the exam's favourite one-equation numerical generator.
2. Complete Theory: Derivation, the Friction Audit, and the Recipe
Derivation (three lines — a guaranteed theory item). Begin with the work-energy theorem: Wtotal = ΔK. Split the total work by force type: Wcons + Wnc = ΔK. By the definition of potential energy (previous page), Wcons = −ΔU. Substitute: −ΔU + Wnc = ΔK, so
ΔK + ΔU = Wnc
Set Wnc = 0 — no non-conservative work — and ΔK + ΔU = 0: the gain of one is exactly the loss of the other, K + U = constant. The general line ΔK + ΔU = Wnc is worth memorising in its own right: it is the conservation law and its correction term in one sentence.
The friction audit. With kinetic friction sliding for a distance s, Wnc = −f s and the ledger closes as Ef = Ei − f s. The mechanical energy that vanished is not destroyed — it left as heat, sound and deformation (touch the block after a rough slide). NTA's standard rough-incline numerical is exactly this audit: compute mgh on the left, ½mv² on the right, and the difference is f·s.
The recipe (why this chapter converts so fast). One: declare the zero of U. Two: write K + U at the start snapshot. Three: write K + U at the end snapshot. Four: equate them (conservation) or correct by Wnc (friction audit). No accelerations, no time, no force directions — endpoints only. When a question supplies or demands forces instead, that is the cue to switch back to free-body methods; energy buys speeds, heights and compressions, not contact forces.
3. Visualising: The Falling Ball's Constant Ledger
Figure 5.5 — Three snapshots, one total
Blue = potential energy U = mgh; green = kinetic energy K = ½mv². Each bar's total height is the same 50 J at every height — only the split moves.
Exam read-out: check the arithmetic: at h = 1.8 m, U = 1 × 10 × 1.8 = 18 J and K = 50 − 18 = 32 J → v = √(2 × 32/1) = 8 m/s, matching free fall's v² = 2g(5 − 1.8) = 64. The kinematic free-fall results and the energy ledger are the same physics in two languages.
4. Solved Examples
Example 1 — Speed at a given height during a fall
A 1 kg stone is dropped from 20 m. Using energy conservation, find its speed at a height of 5 m (g = 10 m s⁻²).
Solution (step by step): Choose U = 0 at the ground. Top snapshot: K = 0, U = 1 × 10 × 20 = 200 J → E = 200 J. At h = 5 m: U = 50 J, so K = 200 − 50 = 150 J → ½ × 1 × v² = 150 → v² = 300 → v = √300 = 10√3 ≈ 17.3 m/s. Cross-check by kinematics: the stone has fallen 15 m, v² = 2 × 10 × 15 = 300 ✓. Same number, one line versus three.
Example 2 — The friction audit on a rough slide
A 2 kg block slides from rest down a rough incline of height 5 m and reaches the bottom at 8 m s⁻¹. How much mechanical energy became heat, and what was friction's work?
Solution: Initial (all U): Ei = mgh = 2 × 10 × 5 = 100 J. Final (all K): Ef = ½ × 2 × 8² = 64 J. Audit: Ei = Ef + heat → heat = 100 − 64 = 36 J, and Wfriction = −36 J. Momentum-style check: had the incline been frictionless the arrival speed would be √(2 × 10 × 5) = 10 m/s; the rough floor delivered only 8 — the deficit's energy is the 36 J of heat. One equation, no incline angle needed.
5. Practice Questions
Q1. A 1 kg pendulum bob on a 1 m string is released from rest with the string horizontal. Find its speed at the lowest point (g = 10 m s⁻²).
ANSWER: The drop is h = 1 m; string tension is radial (⊥ motion) and does no work, so only gravity books anything: mgh = ½mv² → v² = 2 × 10 × 1 = 20 → v = √20 ≈ 4.5 m/s. Note the mass and the string length's role: length sets h, mass cancels — a bob of any mass released from horizontal arrives at √(2gL).
Q2 (MCQ). A ball is thrown vertically upward. Ignoring air resistance, which statement is correct at every point of the flight? (a) K + U decreases (b) K + U increases (c) K + U stays constant (d) K and U each stay constant
ANSWER: (c) — only gravity (conservative) does work, so K + U is constant: on the way up, K drains into U at the rate of mg per metre climbed, and the trade reverses on the way down. This is the stock NEET one-liner; the drag-included version of the same question has answer (a), which is how NTA separates the careful from the pattern-matched.
Q3. A 0.5 kg ball falls from rest through 8 m while air resistance does 6 J of negative work. Find its kinetic energy and speed just before impact (g = 10 m s⁻²).
ANSWER: Gravity's deposit: mgh = 0.5 × 10 × 8 = 40 J; air resistance bills −6 J → K = 40 − 6 = 34 J → v = √(2 × 34/0.5) = √136 ≈ 11.7 m/s. Compare the clean-drop value √160 ≈ 12.6 m/s: the audit line Ef = Ei − 6 handles the leak in one step.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| Ki + Ui = Kf + Uf | Only conservative forces do work between the snapshots |
| ΔK + ΔU = Wnc | The general line: conservation plus its correction term |
| Ef = Ei − f s | Friction audit; the deficit is heat, sound, deformation |
| v = √(2 g h) | Frictionless drop of h from rest; mass- and path-independent |
| ½mv²top = mghswing | Pendulum/track chains: tension and normal do no work (radial) |
| Endpoints only | Declare U = 0, write two snapshots, equate — no force algebra |
The conservation law is a conditional, not a creed: check "who did work?" before writing it. Gravity and springs alone → conserve. Friction present → audit. Motor or push present → include its work explicitly.
7. Frequently Asked Questions
State the law of conservation of mechanical energy.
If only conservative forces do work on a system, its total mechanical energy E = K + U stays constant: energy sloshes between kinetic and potential forms but the total never changes. Written between any two states, K_i + U_i = K_f + U_f. The condition is the whole law: the moment a non-conservative force like kinetic friction does work, mechanical energy leaks out as heat and the conservation statement must be replaced by the audit E_f = E_i + W_nc, with W_nc negative for friction.
Derive the conservation of mechanical energy from the work-energy theorem.
Start from W_total = ΔK, the work-energy theorem. Split the total work into conservative and non-conservative parts: W_cons + W_nc = ΔK. By the definition of potential energy, W_cons = −ΔU. Substituting: −ΔU + W_nc = ΔK, or ΔK + ΔU = W_nc. If no non-conservative force does work (W_nc = 0), then ΔK + ΔU = 0, which says K + U is constant. One derivation, three exam items: the theorem, the ΔU = −W definition, and the condition.
What happens to mechanical energy when friction acts?
Friction audits the account rather than closing it: E_f − E_i = W_friction, which is negative and equals −f times the sliding distance. The missing mechanical energy reappears as heat, sound and deformation — total energy is still conserved in the grand ledger, but mechanical energy is not. Exam procedure: write E_i = E_f + f·s for a rough slide, placing the f·s heat term visibly, and never claim K + U is constant while a block grinds along a rough floor.
Why does v = √(2gh) not depend on the path or the mass?
For a frictionless descent through height h starting from rest, conservation gives mgh = ½mv², and the mass cancels from both sides: v = √(2gh) — a mass-independent result. Path independence follows because gravity is conservative: a straight drop, a slide down any shaped ramp or a curved chute all descend the same height, deposit the same mgh, and therefore deliver the same speed. Only the travel time differs. This is why the same formula answers the dropped stone and the roller-coaster dip alike.
Saved on this device only — no account, no sign-in.