Work, Energy and Power

2.1 Core concepts

• The scalar product of two vectors is defined as A·B = AB cos θ, where θ is the angle between the vectors; it is a scalar quantity even though A and B have directions (NCERT §5.1.1, p. 72).
• Geometrically, A·B is the product of the magnitude of A with the projection of B on A (B cos θ), or equivalently the magnitude of B with the projection of A on B (A cos θ) (NCERT §5.1.1, p. 72).
• The scalar product obeys the commutative law (A·B = B·A) and the distributive law [A·(B + C) = A·B + A·C], and for unit vectors î·î = ĵ·ĵ = k̂·k̂ = 1 and î·ĵ = ĵ·k̂ = k̂·î = 0 (NCERT §5.1.1, p. 72).
• For two vectors A = Ax î + Ay ĵ + Az k̂ and B = Bx î + By ĵ + Bz k̂, A·B = Ax Bx + Ay By + Az Bz; also A·A = A² (NCERT §5.1.1, p. 72).
• Starting from v² − u² = 2 a·d and multiplying by m/2, one obtains (½)mv² − (½)mu² = F·d, which motivates definitions of kinetic energy K and work W (NCERT §5.2, p. 73).
• Work-energy (WE) theorem: the change in kinetic energy of a particle equals the work done on it by the net force, Kf − Ki = W (NCERT §5.2, Eq. 5.3, p. 73).
• For a constant force, W = (F cos θ)d = F·d; no work is done if (i) displacement is zero, (ii) force is zero, or (iii) force is perpendicular to displacement (NCERT §5.3, Eq. 5.4, p. 74).
• Work can be positive (0 ≤ θ < 90°), zero (θ = 90°), or negative (90° < θ ≤ 180°); friction opposing motion does negative work since cos 180° = −1 (NCERT §5.3, p. 74).
• Dimensions of work and energy are [ML²T⁻²]; SI unit is the joule (J), named after James Prescott Joule (NCERT §5.3, p. 74).
• Kinetic energy of a body of mass m moving with velocity v is K = (½) m v·v = (½) m v²; it is a scalar and measures the work the body can do by virtue of its motion (NCERT §5.4, Eq. 5.5, p. 74).
• For a variable force F(x) in 1-D, the work done over a small displacement Δx is ΔW = F(x) Δx; summing infinitesimal rectangles gives W = ∫F(x) dx from xi to xf, i.e. the area under the F–x curve (NCERT §5.5, Eqs. 5.6–5.7, p. 75).
• WE theorem for variable force: starting from dK/dt = Fv = F(dx/dt), integration gives Kf − Ki = ∫F dx, so the theorem also holds for variable forces (NCERT §5.6, Eq. 5.8, p. 76).
• Potential energy is "stored energy" by virtue of position or configuration; the body releases it as kinetic energy when constraints are removed (NCERT §5.7, p. 77).
• Gravitational PE near the earth: V(h) = mgh, and the gravitational force F = −dV/dh = −mg (downward); when released, V(h) converts to KE so that (½)mv² = mgh just before striking the ground (NCERT §5.7, p. 77).
• A force F(x) is conservative if (i) it can be derived from a scalar V(x) as F(x) = −dV/dx, (ii) work done depends only on end points, or (iii) work done over a closed path is zero (NCERT §5.7 and §5.8, p. 78).
• Conservation of mechanical energy: if only conservative forces act, then ΔK + ΔV = 0 ⇒ K + V = constant; i.e. Ki + V(xi) = Kf + V(xf) (NCERT §5.8, Eqs. 5.10–5.11, p. 78).
• For a ball dropped from height H: total energy EH = mgH at the top, Eh = mgh + (½)mvh² at intermediate height h, and E0 = (½)m vf² at the ground; conservation gives vf = √(2gH) and vh = √[2g(H − h)] (NCERT §5.8, Eq. 5.11, p. 79).
• Spring force obeys Hooke's law Fs = −kx, where k is the spring constant; the force is variable but conservative (NCERT §5.9, p. 80).
• Work done by the spring force when stretched/compressed from 0 to xm is Ws = −k xm²/2 (area of the F–x triangle); over a cyclic path Ws = 0 (NCERT §5.9, Eqs. 5.15–5.18, p. 80–81).
• Spring PE: V(x) = kx²/2 (zero taken at the equilibrium position); the speed and KE are maximum at x = 0, with vm = √(k/m) · xm (NCERT §5.9, Eq. 5.19, p. 81).
• The zero of potential energy is arbitrary (a matter of convenience), but must be consistently used throughout a problem (NCERT §5.9 remarks, p. 82).
• In presence of a non-conservative force Fnc, mechanical energy is no longer conserved: Ef − Ei = Wnc, where Wnc depends on the path (NCERT §5.9, p. 82–83).
• Power is the time rate of doing work: average power Pav = W/t and instantaneous power P = dW/dt = F·v (NCERT §5.10, Eqs. 5.20–5.21, p. 83–84).
• Power is a scalar; SI unit watt (W) = 1 J s⁻¹; dimensions [ML²T⁻³]. 1 hp = 746 W. 1 kWh = 3.6 × 10⁶ J — a unit of energy, not power (NCERT §5.10, p. 84).
• Collisions: in every collision total linear momentum is conserved (follows from Newton's third law: F12 = −F21 so Δp1 + Δp2 = 0); kinetic energy is not in general conserved (NCERT §5.11.1, p. 85).
• Elastic collision: both momentum and kinetic energy are conserved. Inelastic collision: only momentum is conserved; some KE is lost as heat, sound, deformation. Completely inelastic: the two bodies stick and move together (NCERT §5.11.1, p. 85).
• 1-D completely inelastic collision (m2 at rest): m1 v1i = (m1 + m2) vf, giving vf = m1 v1i/(m1 + m2); loss in KE ΔK = (½) m1 m2 v1i²/(m1 + m2) (NCERT §5.11.2, Eq. 5.22, p. 85).
• 1-D elastic collision (m2 at rest): final velocities are v1f = (m1 − m2)/(m1 + m2) · v1i and v2f = 2 m1/(m1 + m2) · v1i (NCERT §5.11.2, Eqs. 5.26–5.27, p. 85).
• Special cases: if m1 = m2, then v1f = 0 and v2f = v1i (first body stops, second moves with v1i); if m2 ≫ m1, then v1f ≈ −v1i and v2f ≈ 0 (heavier body undisturbed, lighter rebounds) (NCERT §5.11.2, p. 85).
• 2-D collisions: linear momentum conservation gives m1 v1i = m1 v1f cos θ1 + m2 v2f cos θ2 and 0 = m1 v1f sin θ1 − m2 v2f sin θ2; for elastic case the KE equation provides a third equation, but one unknown (e.g. θ1) must still be supplied (NCERT §5.11.3, Eqs. 5.28–5.30, p. 86).
• When two equal masses undergo a glancing elastic collision with one initially at rest, after collision they move at right angles to each other (NCERT §5.11.3, Example 5.12, p. 86).
• Different forms of energy and their interconvertibility: the law of conservation of energy, in its broadest sense, says the total energy of an isolated system is constant, although energy may change form (mechanical, heat, electrical, chemical, nuclear). Mechanical-energy conservation is a special case valid when only conservative forces act (NCERT §5.8, p. 79).
• Mass–energy equivalence (Einstein): every form of mass m has an associated rest energy E = mc². In ordinary mechanics this rest energy is gigantic (1 g ≡ 9 × 10¹³ J) but unchanging, so we treat KE and PE alone. In nuclear reactions, however, small fractions of mass convert into kinetic energy of products — the basis of fission and fusion (NCERT §5.8 remark; broader development in Class XII Nuclei).
• Coefficient of restitution e = (relative velocity of separation)/(relative velocity of approach). e = 1 for an elastic, 0 < e < 1 for an inelastic, and e = 0 for a perfectly inelastic collision — these cases are the three referred to throughout §5.11 (NCERT §5.11.1, p. 85).
• Vertical-circle motion of a bob on a string: combining centripetal-force at the top (T_top + mg = mv_top²/L) with energy conservation gives v_bottom = √(5gL) and v_top = √(gL); these critical-speed results recur in NCERT exercises (NCERT Fig. 5.6, p. 79).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig. 5.1 (a), (b), (c) (p. 72): Geometric meaning of the scalar product — A·B as A times the projection of B onto A, and vice versa.
• Fig. 5.2 (p. 73): A constant force F acting through displacement d, used to define W = F·d.
• Fig. 5.3 (a), (b) (p. 75): Work done by a variable force F(x) — sum of rectangles tending to the area under the curve as Δx → 0.
• Fig. 5.4 (p. 76): Force applied by a woman pushing a trunk versus the opposing friction f = 50 N; areas give WF = 1750 J and Wf = −1000 J.
• Fig. 5.5 (p. 78): Ball dropped from a cliff of height H — illustrates conversion of PE to KE with E remaining constant.
• Fig. 5.6 (p. 79): Bob on a light string of length L describing a vertical circle — used to derive v0 = √(5gL) and vC = √(gL).
• Fig. 5.7 (a)–(d) (p. 80): Spring on a smooth surface — equilibrium, stretched (Fs < 0), compressed (Fs > 0), and the Fs vs x plot whose triangular area equals the work done.
• Fig. 5.8 (p. 81): Parabolic plots of V = kx²/2 and K for a spring–mass system, complementary, with E = K + V constant.
• Fig. 5.9 (p. 82): Forces on a car colliding with a spring in presence of friction — used to obtain the quadratic for xm.
• Fig. 5.10 (p. 84): Two-body collision of m1 (moving with v1i) with m2 at rest — basis for 1-D and 2-D collision analysis.

2.4 Common confusions / NTA trap points

• Students confuse the magnitude of force with work: pushing hard against a rigid wall does NO work since displacement is zero (NCERT §5.3, p. 74).
• For a body moving in a circle (e.g. moon around earth), the gravitational force is perpendicular to velocity, so the work done over one orbit is zero — easy to misjudge.
• Newton's third law gives equal and opposite forces, but the works done by them need not be equal and opposite (e.g. cycle stops on road — road does −2000 J on cycle, cycle does 0 J on road since road doesn't move) (NCERT Ex. 5.3, p. 74).
• Distinguish carefully: in an inelastic collision only momentum is conserved (KE is lost), while in an elastic collision both momentum and KE are conserved. Many students wrongly assume KE is always conserved.
• For the elastic 1-D collision formulae v1f = (m1 − m2)v1i/(m1 + m2) and v2f = 2 m1 v1i/(m1 + m2), watch for the sign — if m2 > m1, v1f is negative (incident body rebounds).
• 1 kWh is a unit of energy, not power; students often mark it as a power unit (NCERT §5.10, p. 84).
• Spring potential energy V = kx²/2 is the same whether the spring is stretched or compressed by the same magnitude (sign of x doesn't matter).
• The zero of potential energy is arbitrary; only the change in PE has physical meaning. Standard choices: ground for gravity, equilibrium position for spring.
• Centripetal force does no work. For uniform circular motion the force is perpendicular to v, so W = 0 over any closed orbit; mistaking F × distance-around for work is a routine NTA trap.
• In a perfectly inelastic collision, KE is lost but not all of it. A common error is to write ΔK = ½m₁v₁i²; the actual loss is ½ (m₁m₂)/(m₁+m₂) v₁i² (the rest survives as KE of the combined mass).
• Power output of a vehicle at constant speed equals F_drag × v. Doubling v on a flat road requires (approximately) double the engine power if drag is constant — useful for highway-mileage questions.
• Spring PE depends on x², so doubling compression quadruples PE. Students sometimes write V ∝ x or V ∝ k², getting the dependence wrong.

2.5 Key formulas table