Physics

Physics Problem Solver with Steps

Kinematics, forces, energy, and circuits solved with the governing equation named first and units carried through every line.

Solve a physics problem now

First solution free

A physics answer is a number with a unit attached, and the unit is not decoration. It is the cheapest error check available: if the algebra returns metres per second where the question wanted joules, something upstream is wrong and you can find it without rechecking a single digit. Every line of working here keeps its units for that reason.

The other habit worth borrowing is naming the principle before touching the numbers. Deciding between kinematics, Newton's second law, and conservation of energy takes about ten seconds and determines whether the problem takes two lines or twenty. Describe the situation in plain language, give the values with their units, and say what you are solving for.

What this solver handles

One-dimensional kinematics: the constant-acceleration equations, free fall, and reading motion graphs.
Projectile motion, where the two axes are handled independently and the time of flight solves a quadratic.
Newton's laws: free-body analysis, friction, inclines, tension, and connected bodies over a pulley.
Work, kinetic and potential energy, power, and conservation with friction accounted for.
Momentum and impulse, elastic and inelastic collisions in one and two dimensions.
Circular motion, centripetal force, banked curves, and Newtonian gravitation.
DC circuits: Ohm's law, series and parallel combination, Kirchhoff's rules as a system of equations, and electrical power.
Waves, simple harmonic motion, and sound, including the Doppler effect.
Thermodynamics: specific heat capacity, latent heat, and the gas laws, often set as a paragraph to translate before any physics starts.

State the idealisations your course is making. Frictionless surfaces, massless strings, ideal ammeters, and negligible air resistance are assumptions, not facts, and a solution that carefully includes a resistance you were told to ignore is not the one your marker wants. Say which value of $g$ you use, and if the geometry is ambiguous, say which direction counts as positive — otherwise the answer may be right with the opposite sign to the one on your mark scheme.

Three problems, worked

Worked example

A ball is thrown vertically upward at 20 m/s . How high does it rise?

1
List the knowns with signs. Taking upward as positive makes gravity negative, and at the top of the flight the velocity is momentarily zero.
$v_{0} = + 20 m/s, v = 0, a = - 9.8 m/s^{2}$
2
Time is neither given nor wanted, so use the kinematic equation that omits it.
$v^{2} = v_{0}^{2} + 2 a Δ y$
3
Substitute with units attached and rearrange for the displacement.
$0 = (20 m/s)^{2} + 2 (- 9.8 m/s^{2}) Δ y$
4
Divide. The units work out as metres squared per second squared over metres per second squared, which leaves metres — the confirmation that the right equation was used.
$Δ y = \frac{400 m ^{2} / s ^{2}}{19.6 m/s ^{2}} = 20.4 m$
5
The time to reach that point follows from v = v₀ + at, giving 20/9.8 = 2.04 s. The ball returns to the thrower's hand after twice that, since the motion is symmetric.
$t_{up} = 2.04 s$

Answer

Δ y = 20.4 m

Worked example

A 5.0 kg block is pushed with 12 N against 3.0 N of friction.

1
Identify the horizontal forces. The applied force drives the motion; friction opposes it, so the two subtract.
$F_{net} = 12 N - 3.0 N = 9.0 N$
2
Apply Newton's second law along the direction of motion. The vertical forces balance, so they contribute nothing here.
$a = \frac{F _{net}}{m} = \frac{9.0 N}{5.0 kg}$
3
Expand the newton into its base units to see the answer's units appear: one newton is one kilogram metre per second squared, so the kilograms cancel.
$a = 1.8 \frac{kg \cdot m/s ^{2}}{kg} = 1.8 m/s^{2}$
4
With the acceleration known, kinematics takes over. Starting from rest, after 4.0 s the block is moving at:
$v = a t = (1.8 m/s^{2}) (4.0 s) = 7.2 m/s$
5
And it has travelled a distance of one half a t squared.
$Δ x = \frac{1}{2} (1.8) (4.0)^{2} = 14.4 m$

Answer

a = 1.8 m/s^{2}

Worked example

A 2.0 kg block slides 3.0 m down a 3 0^{\circ} incline, μ_{k} = 0.25.

1
Find the vertical drop, since gravitational potential energy depends on height and not on distance along the slope.
$h = (3.0 m) sin 3 0^{\circ} = 1.5 m$
2
Compute the energy released by that drop.
$E_{p} = m g h = (2.0) (9.8) (1.5) = 29.4 J$
3
The normal force balances only the perpendicular component of the weight, which is where the cosine enters.
$N = m g cos 3 0^{\circ} = (19.6 N) (0.866) = 17.0 N$
4
Kinetic friction is the coefficient times the normal force, and the work it does is that force over the distance travelled along the slope.
$W_{f} = μ_{k} N d = (0.25) (17.0 N) (3.0 m) = 12.7 J$
5
Mechanical energy is not conserved here, so subtract the energy lost to friction from the energy released.
$E_{k} = 29.4 J - 12.7 J = 16.7 J$
6
Finally invert the kinetic energy formula. Joules per kilogram give metres squared per second squared, whose square root is a speed.
$v = \frac{2 E _{k}}{m} = \frac{2 ( 16.7 )}{2.0} = 4.1 m/s$

Answer

v \approx 4.1 m/s

Test an answer at a limiting case before trusting it. Set the coefficient of friction to zero in that third problem and the working collapses to $v = 2 g h = 2 (9.8) (1.5) = 5.4 m/s$ . The frictionless case has to be faster than the 4.1 m/s with friction, and it is. When a result behaves absurdly as a parameter goes to zero or grows without bound, it is wrong, and the limiting case finds that out faster than rechecking every line.

Dimensions do the same job for a symbolic answer. Anything standing for a speed must reduce to metres per second once the units are substituted, which is why the division in the first problem was written with its units attached. An expression that resolves to kilograms per second is a rearrangement error announcing itself before a single number has been plugged in.

Choosing the principle before the equation

Most physics problems can be forced through kinematics eventually, and most should not be. The clue is usually in what the question omits: no mention of time points at energy, a collision points at momentum, and a request for tension points at a free-body diagram.

Which principle does this problem want?

What is the question asking for?

Harder questions need two principles in sequence rather than one. Energy gives the speed at the bottom of a slope; kinematics then takes that speed and finds how far the object slides across the flat ground beyond it. Solve in stages, write the intermediate result on its own line with its units, and treat it as a given for the next stage. Trying to carry a single algebraic expression through both stages is where most long physics solutions collapse, and it also makes partial credit impossible to award.

Where students go wrong

Leaving units unconverted

A mass given as 500 g must enter $F = ma$ as $0.5 kg$ , and a speed in km/h must become m/s before it meets an acceleration in m/s squared. Convert everything to base SI units in a single line at the top, before any physics happens, and the rest of the problem stops producing answers that are wrong by exactly a thousand.

Losing the sign convention

Pick a positive direction, write it down, and keep it for the whole problem. If up is positive then $g = - 9.8 m/s^{2}$ , the initial velocity of a thrown ball is positive, and a displacement below the start is negative. Half of all kinematics errors are a sign chosen twice, differently.

Assuming the normal force equals the weight

It does on flat ground with no vertical push. On an incline $N = m g cos θ$ , and with a force applied at an angle the vertical component adds to or subtracts from it. Since friction is proportional to $N$ , this error propagates into every subsequent line.

Conserving energy when friction is present

Mechanical energy is only conserved when no non-conservative force does work. Friction, air resistance, and anything that heats a surface all remove energy from the mechanical account. Subtract that work explicitly rather than ignoring it, or use Newton's second law instead.

An incline with one wrong force

One of these lines is wrong. Click it.

Every one of those four is invisible in the final number. A net force of 11.8 newtons looks exactly as respectable as 12.8, and a calculation done in grams instead of kilograms is wrong by a factor of a thousand without appearing strange on the page. Carrying units through each line and declaring the positive direction before the first equation are the two habits that make these errors visible while you are still making them, and they generalise into the wider set of verification tricks worth running on any quantitative answer.

Formulas worth knowing cold

Physics reference

Tap any formula with a derivation to see where it comes from.

Constant-acceleration kinematics

Forces and energy

DC circuits

Practice

Four to try

Answers are checked here — nothing is sent anywhere.

1
$A car goes 0 \to 24 m/s in 6.0 s. a = ? (m/s^{2})$
2
$Weight of a 12 kg mass, g = 9.8 m/s^{2} (N)$
3
$4 Ω and 6 Ω in parallel (Ω)$
4
$E_{k} of a 3.0 kg object at 4.0 m/s (J)$

Frequently asked questions

Is the physics solver free?

The first solution is free with no account required. A free account then allows three solutions a day, and Gauth Plus is $11.99 a month with a three-day free trial for unlimited working.

Does it show the steps or just the answer?

Every solution starts by naming the governing equation and why it applies, then carries units through each line. A bare number with no units and no equation earns almost nothing on a marked physics script.

Which value of g does it use?

9.8 metres per second squared unless your question specifies otherwise. Write 9.81 or 10 in the question and that value is used instead, since courses differ and the marks follow the convention you were taught.

Can it draw or interpret a free-body diagram?

It lists every force acting, with direction and the component along each axis, which is the information a free-body diagram encodes. Describe the setup in words or photograph the diagram from your sheet.

Does it handle significant figures?

Answers are rounded to match the least precise value you supplied, with the unrounded figure shown alongside so intermediate steps stay accurate.

Which topics does it cover?

Kinematics, dynamics, energy, momentum, circular motion, gravitation, simple harmonic motion, waves, DC circuits, and introductory thermodynamics — roughly a first-year college or A-level syllabus.

Keep going

Turning words into equationsPhysics problems are word problems with units attached.Exponent and radical rulesScientific notation and unit powers, handled without slips.Chemistry solverThe same unit discipline, applied to moles and concentrations.Quadratic formula calculatorFor projectile times of flight and stopping-distance problems.System of equations solverKirchhoff loops and connected-body problems, solved as systems.