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Physics formulas

Mechanics, fluids, heat, electricity, magnetism, optics and modern physics: formulas, units and worked examples.

Acceleration

a = (v−u)/t

u and v are initial and final velocities in m/s. This expression gives average acceleration over the time interval.

Worked example

From 5 m/s to 17 m/s in 4 s gives a=3 m/s².

Constant acceleration

v=u+at; s=ut+½at²; v²=u²+2as

These one-dimensional relationships assume constant acceleration. Choose a positive direction and keep signs consistent.

Worked example

Starting from rest at 2 m/s² for 3 s gives v=6 m/s and displacement s=9 m.

Force

F = ma

F is net force in newtons, m mass in kg and a acceleration in m/s². Use the resultant of all forces.

Worked example

A net force of 6 N on 2 kg produces 3 m/s² acceleration.

Work and energy

W = Fs cos θ; K = ½mv²

For constant force, θ is the angle between force and displacement. Both work and kinetic energy are measured in joules.

Worked example

A 10 N force acting parallel to a 3 m displacement does 30 J of work. At 90°, it does no work.

Potential energy

ΔU = mgΔh

Near Earth’s surface with approximately constant g, raising an object increases its gravitational potential energy.

Worked example

For 2 kg raised 3 m using g=9.81 m/s², ΔU=58.86 J.

Power

Average power = work / time

Power measures the rate of energy transfer. One watt equals one joule per second.

Worked example

600 J transferred in 20 s corresponds to 30 W average power.

Electricity

V = IR; P = VI

Ohm’s law applies to an ohmic resistor under unchanged physical conditions. V is volts, I amperes and R ohms.

Worked example

Across a 6 Ω resistor, 12 V produces 2 A and dissipates 24 W.

Waves

v = fλ

Wave speed v is in m/s, frequency f in Hz and wavelength λ in metres. Use all values for the same wave and medium.

Worked example

A 50 Hz wave with wavelength 2 m travels at 100 m/s.

Heat

Q = mcΔT

This relates heat to temperature change without a phase transition; c is specific heat capacity in J/(kg·K).

Worked example

For m=0.5 kg, c=4200 J/(kg·K) and ΔT=10 K, Q=21,000 J.

Momentum

p = mv; impulse = Δp

Momentum is a vector measured in kg·m/s. Total momentum is conserved for an isolated system.

Worked example

A 2 kg cart moving at 3 m/s has momentum 6 kg·m/s in its direction of travel.

Density and pressure

ρ = m/V; pressure = normal force/area

Density uses kg/m³ and pressure uses pascals (N/m²). Convert area to square metres before substitution.

Worked example

A normal force of 100 N spread over 0.5 m² creates 200 Pa pressure.

Units, dimensions and constants

Quantity	SI unit	Relationship
Length	metre (m)	Base quantity
Mass	kilogram (kg)	Base quantity
Time	second (s)	Base quantity
Current	ampere (A)	Charge/time
Temperature	kelvin (K)	K=°C+273.15
Force	newton (N)	kg·m·s⁻²
Energy	joule (J)	N·m
Power	watt (W)	J/s
Pressure	pascal (Pa)	N/m²
Charge	coulomb (C)	A·s

Constants: c=299,792,458 m/s exactly; h=6.62607015×10⁻³⁴ J·s exactly; elementary charge e=1.602176634×10⁻¹⁹ C exactly. Use G≈6.67430×10⁻¹¹ N·m²/kg². Standard gravity is 9.80665 m/s²; local gravitational acceleration varies. Dimensional consistency is necessary but does not prove a formula is correct.

Vectors and projectile motion

x=u cosθ·t; y=u sinθ·t−½gt²

Assume uniform gravity, no drag and launch at the origin. For landing at launch height: T=2u sinθ/g, range R=u²sin2θ/g, maximum height H=u²sin²θ/(2g). Horizontal velocity is constant; vertical acceleration is −g.

Worked example

For u=20 m/s, θ=30° and g=10 m/s²: T=2 s, H=5 m and R=20√3 m≈34.64 m.

Common mistake: Using the equal-height range formula when landing above or below launch.

Circular motion and gravitation

aᶜ=v²/r=ω²r; F=Gm₁m₂/r²

Centripetal acceleration points inward. Centripetal force is the inward net force, not an extra force. For spherical bodies outside their surfaces, r is centre-to-centre distance. Circular orbital speed is √(GM/r) when the central mass dominates.

Worked example

A 2 kg body at 3 m/s on radius 1.5 m needs inward net force 12 N.

Common mistake: Using height above Earth instead of centre-to-centre orbital radius.

Friction and equilibrium

Static friction f≤μₛN; sliding friction f≈μₖN

Normal reaction N is not always mg. Static friction adjusts up to its limiting value. Equilibrium requires zero net force and zero net torque. Torque magnitude τ=rFsinθ.

Worked example

On a horizontal surface with m=5 kg, g=10 and μₛ=0.4, limiting friction is 20 N. A 12 N horizontal pull is balanced by 12 N static friction.

Common mistake: Always setting static friction equal to μₛN.

Rotational motion

τ=Iα; L=Iω; rotational K=½Iω²

These scalar forms apply about a suitable fixed axis. Point mass I=mr²; solid disc about symmetry axis I=½MR²; thin hoop I=MR²; uniform rod about centre perpendicular to length I=ML²/12.

Worked example

I=0.5 kg·m² and α=4 rad/s² gives torque 2 N·m. At ω=6 rad/s the rotational energy is 9 J.

Common mistake: Using a moment of inertia for the wrong axis.

Fluids, buoyancy and flow

p=p₀+ρgh; buoyant force=ρfluid g Vdisplaced

Hydrostatic pressure uses depth h and constant density. Steady incompressible flow satisfies A₁v₁=A₂v₂. Bernoulli p+½ρv²+ρgh=constant applies along a streamline for ideal steady incompressible flow without dissipation or added work.

Worked example

At 2 m water depth with ρ=1000 kg/m³ and g=10, gauge pressure is 20,000 Pa. Displacing 0.003 m³ of water gives buoyancy 30 N.

Common mistake: Confusing gauge and absolute pressure or using object density in the buoyancy formula.

Elasticity and simple harmonic motion

Young modulus Y=(F/A)/(ΔL/L); spring period T=2π√(m/k)

The elastic formula assumes small strain in the linear region. An ideal spring oscillator has a=−ω²x, ω=√(k/m). For a small-angle simple pendulum T=2π√(L/g), neglecting damping and bob size.

Worked example

m=1 kg and k=100 N/m gives T≈0.628 s. Doubling mass multiplies period by √2.

Common mistake: Using the small-angle pendulum formula for large oscillations.

Thermal expansion and phase changes

ΔL=αL₀ΔT; Q=mL

α is the linear expansion coefficient in K⁻¹. For isotropic solids and small expansions the volume coefficient is about 3α. Latent heat L in J/kg describes a phase change at fixed temperature under suitable equilibrium conditions.

Worked example

Melting 0.2 kg ice at 0°C using L=334,000 J/kg needs 66,800 J before further warming.

Common mistake: Using Q=mcΔT for melting at unchanged temperature.

Gas laws and thermodynamics

PV=nRT; ΔU=Q−W

Use absolute pressure and kelvin; R≈8.314 J/(mol·K). Here Q is heat added and W is work done by the system. At fixed gas amount, PV is constant at fixed T and V/T is constant at fixed P for an ideal gas.

Worked example

A gas receiving 500 J and doing 180 J work gains 320 J internal energy. Heating from 300 K to 600 K at constant pressure doubles ideal-gas volume.

Common mistake: Mixing work-by-system and work-on-system sign conventions.

Electrostatics and capacitance

F=k|q₁q₂|/r²; E=F/q; C=Q/V

Point-charge force in vacuum uses k≈8.99×10⁹ N·m²/C². Like charges repel. An ideal parallel-plate capacitor has C=εA/d, ignoring edges; energy U=½CV².

Worked example

A 2 μF capacitor at 10 V stores charge 20 μC and energy 0.0001 J.

Common mistake: Confusing electric field (N/C) and electric potential (J/C).

Circuits and electrical energy

Case	Relationship	Conditions
Series resistors	R=R₁+R₂+…	Same current
Parallel resistors	1/R=1/R₁+1/R₂+…	Same voltage
Uniform wire	R=ρL/A	Fixed material conditions
Joule heating	E=I²Rt	Constant I and R
Electrical energy	E=Pt	Constant power
Kirchhoff junction rule	Current in=current out	Charge conservation
Kirchhoff loop rule	Sum of potential changes=0	Include any induced emf where applicable

For 6 Ω and 3 Ω in parallel, 1/R=1/6+1/3=1/2 and R=2 Ω. A 1000 W appliance used for 2 h consumes 2 kWh=7.2 MJ. Kilowatt-hour measures energy, not power.

Magnetism and electromagnetic induction

F=qvBsinθ; F=BILsinθ; induced emf=−dΦ/dt

The moving-charge force is perpendicular to velocity and field. For a straight wire L is length in uniform field. Flux Φ=BAcosθ uses angle to the surface normal. The negative induction sign expresses Lenz’s law.

Worked example

A 0.5 m wire carrying 2 A perpendicular to 0.3 T has force 0.3 N. Losing 0.02 Wb flux in 0.1 s induces average emf magnitude 0.2 V in one loop.

Common mistake: Using angle to the surface rather than surface normal for flux.

Alternating current and transformers

Vᵣₘₛ=V₀/√2; Iᵣₘₛ=I₀/√2; Vₛ/Vₚ=Nₛ/Nₚ

RMS formulas here assume sinusoidal signals. Average real power P=VᵣₘₛIᵣₘₛcosφ. The turns ratio assumes an ideal transformer with alternating flux; ideal input and output powers are equal.

Worked example

An ideal transformer with 1000 primary and 100 secondary turns reduces 230 V AC to 23 V AC. A resistive load has power factor 1.

Common mistake: Using the √2 relation for non-sinusoidal waveforms.

Reflection, refraction, lenses and mirrors

n₁sin i=n₂sin r; thin lens: 1/f=1/v−1/u

Cartesian sign convention: incident light travels left to right, distances rightward positive. A real object left of a lens has u<0. Lens magnification m=v/u. Spherical mirror: 1/f=1/v+1/u and m=−v/u. Lens power P=1/f with f in metres gives dioptres.

Worked example

Convex lens f=+10 cm, object u=−30 cm: 1/v=1/10−1/30=1/15; v=15 cm and m=−0.5, a real inverted image.

Common mistake: Mixing mirror and lens equations or using centimetres for lens power.

Interference and photons

Young fringe width β=λD/d; E=hf=hc/λ

Young’s expression assumes coherent light, small angles and D much greater than slit separation d. Photoelectric maximum kinetic energy Kmax=hf−φ when photon energy exceeds work function φ.

Worked example

λ=600 nm, D=2 m, d=0.5 mm gives β=2.4 mm. Higher-frequency photons carry more energy.

Common mistake: Assuming greater intensity increases each photon’s energy at fixed frequency.

Radioactivity and semiconductors

N=N₀e^(−λt); half-life=ln2/λ; E=mc²

Decay is statistical; this predicts expected undecayed nuclei. Activity A=λN is in becquerels. A p–n junction generally conducts much more readily under forward bias than reverse bias; real diodes have thresholds and breakdown limits.

Worked example

After three half-lives, an initial 80 g of an isotope has 10 g of that isotope remaining, excluding daughter products from this figure.

Common mistake: Confusing isotope remaining with the entire sample including daughter products.

Practice with answers

Find net force for 4 kg at 2.5 m/s².

F=ma=10 N.

12 V across 4 Ω: find I and P.

I=3 A; P=36 W.

Lens focal length +25 cm: find power.

f=+0.25 m; P=+4 dioptres.

Wave f=200 Hz, λ=1.5 m: find speed.

v=300 m/s.

Half-life 5 days: fraction after 20 days?

Four half-lives leave 1/16.

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