Formula Bank
Every formula for Work, Energy & Power — with dimensional analysis, conditions of use, and common mistakes. Searchable.
W = F · d · cos θ [W] = [ML²T⁻²] = Joule W = F⃗ · d⃗ = Fₓdₓ + F_yd_y + F_zd_z W = ∫F(x) dx from x₁ to x₂ W_spring = ½k(x₁² − x₂²) W_ext = +½kx² W_f = −μmg · d (horizontal) W_g = mgh (downward displacement = +h) W_net = ΔKE = ½mv² − ½mu² Special cases: W = 0 when (1) F = 0, (2) d = 0, (3) θ = 90°. Normal force, centripetal force, and magnetic force on a charge always do zero work.
KE = ½mv² [KE] = [ML²T⁻²] KE = p²/(2m) p = √(2mKE) KE₂/KE₁ = (v₂/v₁)² KE₂/KE₁ = (p₂/p₁)² (same mass) KE_total = ½mv² + ½Iω² When two objects have the same KE: lighter one has smaller momentum (p = √2mKE, p ∝ √m). When two objects have the same momentum: lighter one has larger KE (KE = p²/2m, KE ∝ 1/m). Flip your intuition — this is counter-intuitive.
PE = mgh [PE] = [ML²T⁻²] PE = ½kx² U = −GMm/r F = −dU/dx KE₁ + PE₁ = KE₂ + PE₂ KE₁ + PE₁ = KE₂ + PE₂ + W_friction Stable: dU/dx=0, d²U/dx²>0 | Unstable: d²U/dx²<0 1/k_eff = 1/k₁ + 1/k₂ k_eff = k₁ + k₂ P_avg = W/t [P] = [ML²T⁻³] P = dW/dt = F⃗ · v⃗ = Fv cos θ P = Fv (F || v, θ = 0°) η = (P_output/P_input) × 100% E = P × t 1 HP = 746 W ≈ 750 W Power problems: (1) If constant speed → F_engine = F_friction → P = f×v. (2) If accelerating → net force is non-zero → P = (F_engine − F_friction)×v + KE gained per second. The second case is JEE Advanced territory.
Dimensional Analysis — Work, Energy & Power
Dimensional analysis is used to: (1) verify formulas, (2) derive unknown relations, (3) convert units. Boards ask: "Write dimensional formula of Work." JEE asks you to use dimensional analysis to find missing variables.
| Quantity | Formula | Dimensional Formula | SI Unit |
|---|---|---|---|
| Work | W = Fd cos θ | [ML²T⁻²] | Joule (J) |
| Kinetic Energy | KE = ½mv² | [ML²T⁻²] | Joule (J) |
| Potential Energy | PE = mgh | [ML²T⁻²] | Joule (J) |
| Power | P = W/t | [ML²T⁻³] | Watt (W) |
| Spring Constant | F = kx | [MT⁻²] | N/m |
| Force | F = ma | [MLT⁻²] | Newton (N) |
| Momentum | p = mv | [MLT⁻¹] | kg·m/s |
| Impulse | J = Ft | [MLT⁻¹] | N·s = kg·m/s |
| Torque | τ = r × F | [ML²T⁻²] | N·m |
| Pressure | P = F/A | [ML⁻¹T⁻²] | Pascal (Pa) |
Same Dimensions as Work/Energy
All of these have dimensions [ML²T⁻²]:
- • Work (Fd cos θ)
- • Kinetic Energy (½mv²)
- • Potential Energy (mgh)
- • Torque (r × F)
- • Moment of couple
Torque and Work have same dimensions but different physical meanings. You cannot use dimensional analysis to distinguish them — context matters.
Key Dimensional Relations
[F] = [MLT⁻²], [d] = [L], cos θ = dimensionless
[W] = [MLT⁻²][L] = [ML²T⁻²] ✅
[m] = [M], [v²] = [L²T⁻²]
[KE] = [M][L²T⁻²] = [ML²T⁻²] ✅
[W] = [ML²T⁻²], [t] = [T]
[P] = [ML²T⁻²]/[T] = [ML²T⁻³] ✅
[k] = [MT⁻²] (from F = kx), [x²] = [L²]
[PE] = [MT⁻²][L²] = [ML²T⁻²] ✅
Use calculators to verify your manual calculations during practice. If your answer doesn't match, recheck your angle convention and unit consistency.