Complete List of All Physics Formulas with Examples for Students

Physics can look scary with all its math, but it’s really just about how things move and work. To make it easier, here’s a list of the most useful physics formulas—with simple examples you see every day.

Physics Formulas Cheat Sheet – Equations & Examples

Physics is the study of matter—what it’s made of, how it behaves, and how forces and energy interact with it. But here’s the thing: physics isn’t just about memorizing formulas. It’s about seeing how these equations explain real-world phenomena, from a bouncing ball to the glow of a lightbulb.

To truly grasp physics, you need two things:

1. A solid understanding of the concepts – Without knowing the “why” behind the formulas, they’re just random letters and numbers.
2. Familiarity with units – Keeping track of S.I. units (like meters, seconds, and kilograms) helps you apply formulas correctly.

And guess what? Physics and math go hand in hand. Many physics problems rely on basic algebra and geometry, so if you’re comfortable with math, you’re already one step ahead!

In this guide, we’ll break down key physics formulas with simple examples—so you can see how they work in everyday life. No jargon, no confusion—just clear, practical learning.

List of All Physics Formulas

Given below is the list of all Physics formulas:

Other Important Physics Formulas

Below are the most important Physics formulae list:

• Planck constant h = 6.63 × 10−34 J.s = 4.136 × 10-15 eV.s
• Gravitation constant G = 6.67×10−11 m3 kg−1 s−2
• Boltzmann constant k = 1.38 × 10−23 J/K
• Molar gas constant R = 8.314 J/(mol K)
• Avogadro’s number NA = 6.023 × 1023 mol−1
• Charge of electron e = 1.602 × 10−19 C
• Permittivity of vacuum 0 = 8.85 × 10−12 F/m
• Coulomb constant 1/4πε0 = 8.9875517923(14) × 109 N m2/C2
• Faraday constant F = 96485 C/mol
• Mass of electron me = 9.1 × 10−31 kg
• Mass of proton mp = 1.6726 × 10−27 kg
• Mass of neutron mn = 1.6749 × 10−27 kg
• Stefan-Boltzmann constant σ = 5.67 × 10−8 W/(m2 K4)
• Rydberg constant R∞ = 1.097 × 107 m−1
• Bohr magneton µB = 9.27 × 10−24 J/T 
• Bohr radius a0 = 0.529 × 10−10 m 
• Standard atmosphere atm = 1.01325 × 105 Pa 
• Wien displacement constant b = 2.9 × 10−3 m K .
• Wave = ∆x ∆t wave = average velocity ∆x = displacement ∆t = elapsed time.
• Vavg = (vi + vf*)2
  Vavg = The average velocity 
  vi = initial velocity 
  vf = final velocity
• a = ∆v ∆t,
  a = acceleration 
  ∆v = change in velocity 
  ∆t = elapsed time.
• ∆x = vi∆t + 1/2 a(∆t)2
  ∆x = the displacement 
  vi = the initial velocity 
  ∆t = the elapsed time 
  a = the acceleration 
• ∆x = vf∆t − 1/2 a(∆t)2
  ∆x = displacement 
  vf = is the final velocity 
  ∆t = elapsed time
  a = acceleration 
• F = ma 
  F = force 
  m = mass 
  a = acceleration
• W = mg 
  W = weight 
  m = mass 
  g = acceleration which is due to gravity.
• f = µN 
  f = friction force 
  µ = coefficient of friction 
  N = normal force 
• p = mv
• W = F d cos θ or W = F!d 
  W = work t
  F = force 
  d = distance 
  θ = angle between F and the direction of motion
• KE  = 1/2 mv2 K
  KE = kinetic energy
  m = mass
  v = velocity
• PE = mgh 
  PE = potential energy 
  m = mass 
  g = acceleration due to gravity 
  h = height
• W = ∆(KE) 
  W = work done 
  KE = kinetic energy. 
• P = W ∆t 
  P = power
  W = work
  ∆t = elapsed time

Solved Physics Examples

Calculate the DC voltage drop if the circuit length is 240 cm and in it 6 A of current flows for 12 seconds.

• Length (L) = 240 cm = 2.4 m
• Current (I) = 6 A
• Time (T) = 12 s

Formula:
V=L×ITV = \frac{L \times I}{T}V=TL×I​

V=2.4×612=14.412=1.2 VV = \frac{2.4 \times 6}{12} = \frac{14.4}{12} = 1.2 \, \text{V}V=122.4×6​=1214.4​=1.2V

A spring has a spring constant of 80 N/m and is stretched by 10 cm. Calculate the potential energy stored in the spring.

• k=80N/m
• x=10 cm=0.1 mx = 10 \, \text{cm} = 0.1 \, \text{m}x=10cm=0.1m

Formula:
PE=12kx2PE = \frac{1}{2} k x^2PE=21​kx2

PE=12×80×(0.1)2=40×0.01=0.4 JPE = \frac{1}{2} \times 80 \times (0.1)^2 = 40 \times 0.01 = 0.4 \, \text{J}PE=21​×80×(0.1)2=40×0.01=0.4J

A body moves along the x-axis following the relation x=5+4t+2t2x = 5 + 4t + 2t^2x=5+4t+2t2. Find its acceleration when t=2 st = 2 \, \text{s}t=2s.

x(t)=5+4t+2t2x(t) = 5 + 4t + 2t^2x(t)=5+4t+2t2

First derivative (velocity):
v=dxdt=4+4tv = \frac{dx}{dt} = 4 + 4tv=dtdx​=4+4t

Second derivative (acceleration):
a=dvdt=4 m/s2a = \frac{dv}{dt} = 4 \, \text{m/s}^2a=dtdv​=4m/s2
Acceleration is constant.

What is the weight of a 70 kg object on the Moon?

• Mass (m) = 70 kg
• Moon gravity (g) = 1.6 m/s²

Formula:
W=m⋅g=70×1.6=112 NW = m \cdot g = 70 \times 1.6 = 112 \, \text{N}W=m⋅g=70×1.6=112N

Find the displacement of an object that accelerates from rest to 40 m/s in 4 seconds

• u=0u = 0u=0, v=40 m/sv = 40 \, \text{m/s}v=40m/s, t=4 st = 4 \, \text{s}t=4s

Acceleration:
a=v−ut=404=10 m/s2a = \frac{v – u}{t} = \frac{40}{4} = 10 \, \text{m/s}^2a=tv−u​=440​=10m/s2

Displacement:
S=ut+12at2=0+12×10×42=5×16=80 mS = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 10 \times 4^2 = 5 \times 16 = 80 \, \text{m}S=ut+21​at2=0+21​×10×42=5×16=80m

A person walks from Point A to Point B in 15 seconds and returns in 13 seconds. The distance between A and B is 40 m. Find average speed.

Total distance: 2×40=80 m2 \times 40 = 80 \, \text{m}2×40=80m
Total time: 15+13=28 s15 + 13 = 28 \, \text{s}15+13=28s

Average Speed:
8028≈2.86 m/s\frac{80}{28} \approx 2.86 \, \text{m/s}2880​≈2.86m/s

An object is moving at 4 m/s and has a kinetic energy of 64 J. Find its mass.

KE = 64 J, v=4 m/sv = 4 \, \text{m/s}v=4m/s

Formula:
KE=12mv2⇒64=12m⋅16⇒64=8mKE = \frac{1}{2} m v^2 \Rightarrow 64 = \frac{1}{2} m \cdot 16 \Rightarrow 64 = 8mKE=21​mv2⇒64=21​m⋅16⇒64=8m

m=648=8 kgm = \frac{64}{8} = 8 \, \text{kg}m=864​=8kg

A rod of length 80 cm carries a charge of 4 mC uniformly. Find its linear charge density.

q=4×10−3 Cq = 4 \times 10^{-3} \, \text{C}q=4×10−3C, l=0.8 ml = 0.8 \, \text{m}l=0.8m

Formula:
λ=ql=4×10−30.8=5×10−3 C/m\lambda = \frac{q}{l} = \frac{4 \times 10^{-3}}{0.8} = 5 \times 10^{-3} \, \text{C/m}λ=lq​=0.84×10−3​=5×10−3C/m

Calculate the kinetic energy of a truck with mass 500 kg moving at 15 m/s.

m=500 kgm = 500 \, \text{kg}m=500kg, v=15 m/sv = 15 \, \text{m/s}v=15m/s

Formula:
KE=12mv2=12⋅500⋅225=250⋅225=56,250 JKE = \frac{1}{2} mv^2 = \frac{1}{2} \cdot 500 \cdot 225 = 250 \cdot 225 = 56,250 \, \text{J}KE=21​mv2=21​⋅500⋅225=250⋅225=56,250J

A substance requires 600 kcal of heat to undergo a phase change. If its mass is 3 kg, calculate its latent heat.

Q=600 kcalQ = 600 \, \text{kcal}Q=600kcal, m=3 kgm = 3 \, \text{kg}m=3kg

Formula:
L=Qm=6003=200 kcal/kgL = \frac{Q}{m} = \frac{600}{3} = 200 \, \text{kcal/kg}L=mQ​=3600​=200kcal/kg

Physics Help Desk: Answers to Students’ Most Asked Questions

Can I memorise Physics formulas faster?

Yes everyone can easily memories physics formulas please checkout School Finder guide for physics formulas.

Where can I find the important formulas of Physics?

The list of all important physics formulas can be found here on the official website of schoolsfinders.com.

Is Physics a hard subject?

Physics can be interesting and easy to understand with the right approach. Regular practice and application help make tricky formulas and concepts easier to grasp.