Physics practice guide

Physics Formula Worksheets

Practice electricity, energy and motion with clear unit conversions, worked examples and eight original questions. Print a student sheet, then check the separate worked answer key.

Four worked examples · Three quantitative illustrations · Prefilled calculators

Open practice worksheet

A repeatable method for a physics worksheet

A useful solution shows why a formula applies, how the units fit and what the result means. Start by naming the unknown. A question about energy is different from a question about power, and a question about displacement is different from a question about distance. Read the physical conditions before searching for an equation.

Write the givens. Keep each quantity, value and unit together. For motion, choose a positive direction and keep that choice throughout the solution.
Choose the model. State whether the resistor is ohmic, the power is steady, or the acceleration is constant. These conditions justify the relationship you use.
Convert, then substitute. Put a unit-conversion line before the calculation. Rearrange the formula for the requested unknown before entering numerical values.
Check and report. Keep guard digits, round the final answer, include units and explain its sign or scale. Substitute back when a second relationship is available.

The examples below are hypothetical classroom calculations. They cover introductory electricity and classical translational motion, with assumptions visible beside the formulas. They do not form a complete reference for thermodynamics, waves, fields, rotational dynamics or every examination syllabus.

Four-step physics worksheet workflow: choose a physical model, convert units, substitute values and check units, signs and precision. — Every worked solution keeps the model, units and interpretation visible. The printable student sheet has a separate worked answer key.

Physics formula reference with conditions

In electricity, $V$ is potential difference, $I$ current, $R$ resistance, $P$ power and $E$ transferred energy. In mechanics, $m$ is mass, $K$ translational kinetic energy, $u$ initial velocity, $v$ final velocity or speed as labeled, $a$ acceleration, $t$ elapsed time and $\Delta x$ displacement. The same letter can describe different quantities in different topics; its definition and units resolve that ambiguity.

Choose a relationship whose assumptions match the question
Quantity or model	Relationship	Use it when
Ohmic resistor	$V=IR$	Resistance stays constant under the stated conditions. The circuit examples use positive magnitudes.
Steady DC power	$P=VI=I^2R=\frac{V^2}{R}$	The resistor alternatives require an ohmic resistor; dividing by resistance requires it to be positive.
Constant-power energy	$E=Pt$	Power is constant over the stated elapsed interval, or is the correctly determined average power.
Classical kinetic energy	$K=\frac12mv^2$	Translational energy of a positive mass, at speed far below light speed, in a chosen inertial frame.
Constant acceleration	$v=u+at$	Signed velocity and acceleration along one axis; time is elapsed from the initial state.
Signed displacement	$\Delta x=ut+\frac12at^2$	The same constant-acceleration interval. Displacement need not equal total distance.

The electrical relationships are supported by OpenStax on ohmic materials and electrical power and energy. The kinetic and motion models are introduced in their respective worked sections.

Keep a separate SI conversion line

Prefixes and time units are a common source of large errors. Use coherent units for a substitution, or show an explicitly compatible alternative. In the energy example, watts with seconds produce joules, while kilowatts with hours produce kilowatt-hours. Mixing those numerical values without conversion changes the answer.

Defined conversion factors used in this worksheet
Conversion	Exact relationship
Milliamperes to amperes	$1\,\mathrm{mA}=10^{-3}\,\mathrm A$
Kilohms to ohms	$1\,\mathrm{k\Omega}=10^3\,\Omega$
Kilometers per hour to SI speed	$1\,\mathrm{km/h}=\frac{5}{18}\,\mathrm{m/s}$
Hours to seconds	$1\,\mathrm h=3600\,\mathrm s$
Power and energy units	$1\,\mathrm W=1\,\mathrm{J/s}$
Kilowatt-hours to joules	$1\,\mathrm{kWh}=3.6\times10^6\,\mathrm J$

NIST lists SI units and prefixes; its conversion tables include practical speed and energy units. Defined conversions do not introduce a new measurement precision limit. Keep their exact factors until you round the final result.

Worked example 1: Ohm's law and resistor power

An ohmic resistor carries a steady current of $20.0\,\mathrm{mA}$ and has resistance $0.470\,\mathrm{k\Omega}$ under fixed conditions. Find voltage and dissipated power. Here current and voltage are positive magnitudes for the passive load, rather than signed terminal conventions.

A constant resistance gives a straight line

This model holds $R=470\,\Omega$ fixed. It is not measured data. The axes use different units, so visual slope alone is not a resistance until those units are included.

1. Normalize the prefixes

I=20.0\,\mathrm{mA}=0.0200\,\mathrm A

R=0.470\,\mathrm{k\Omega}=4.70\times10^2\,\Omega

2. Calculate voltage and power

V=IR=(0.0200)(470)\,\mathrm V=9.40\,\mathrm V

P=VI=(9.40)(0.0200)\,\mathrm W=0.188\,\mathrm W

3. Cross-check using a second relationship

I^2R=(0.0200)^2(470)\,\mathrm W=0.188\,\mathrm W

Both power methods agree in the nominal model. If they do not, check the current prefix and whether the entire current was squared. The illustrated straight line assumes resistance stays constant; a nonohmic component or a changing temperature may require a different relationship. Check this voltage and power in the Ohm's Law Calculator.

Worked example 2: DC power and electrical energy

A steady DC load draws $2.00\,\mathrm A$ at $12.0\,\mathrm V$ for $1.00\,\mathrm h$ . Find power, then energy in joules and kilowatt-hours. Assume those voltage and current values remain constant for the full interval; a nameplate rating alone would not establish actual energy use.

1. Calculate the rate of energy transfer

P=VI=(12.0)(2.00)\,\mathrm W=24.0\,\mathrm W

2. Convert elapsed time and calculate energy

t=1.00\,\mathrm h=3.60\times10^3\,\mathrm s

E=Pt=(24.0)(3600)\,\mathrm J=8.64\times10^4\,\mathrm J

3. Check the same energy in practical units

E=(0.0240\,\mathrm{kW})(1.00\,\mathrm h)=0.0240\,\mathrm{kWh}

The two energy values agree after conversion. Power describes how fast energy transfers; energy also depends on duration. For a load that switches or varies, use the appropriate time history or average power rather than silently treating the maximum rating as constant. Open the prefilled Electrical Power Calculator.

Worked example 3: kinetic energy after a speed conversion

A $2.00\,\mathrm{kg}$ object moves at $36.0\,\mathrm{km/h}$ in a chosen inertial reference frame. Find its classical translational kinetic energy. This is a low-speed model; it excludes rotational energy and relativistic corrections. OpenStax defines this kinetic-energy model.

The speed is squared

For the same $2.00\,\mathrm{kg}$ mass, the bars start at zero and show $25$ , $100$ and $400\,\mathrm J$ . The ratio follows squared speeds, not equal energy increments.

1. Convert speed before squaring

v=36.0\frac{\mathrm{km}}{\mathrm h}\frac{1000\,\mathrm m}{1\,\mathrm{km}}\frac{1\,\mathrm h}{3600\,\mathrm s}

v=10.0\,\mathrm{m/s}

2. Substitute into the energy formula

K=\frac12(2.00)(10.0)^2\,\mathrm J

K=1.00\times10^2\,\mathrm J

3. Check units and the speed dependence

\mathrm{kg}\left(\frac{\mathrm m}{\mathrm s}\right)^2=\mathrm J

Doubling the speed to $20.0\,\mathrm{m/s}$ gives $4.00\times10^2\,\mathrm J$ for the same mass. The speed doubles, but energy quadruples. Solving in reverse gives $v=\sqrt{2K/m}$ with the nonnegative root; energy alone cannot tell you which direction the object moves. Check the energy and doubled-speed comparison.

Worked example 4: constant acceleration and displacement

Take right as positive. An object begins at rest and accelerates at a constant $3.00\,\mathrm{m/s^2}$ for $5.00\,\mathrm s$ . The initial time is zero and $u=0\,\mathrm{m/s}$ . Find final velocity and displacement using the constant-acceleration model.

1. Find final velocity

v=u+at=0+(3.00)(5.00)\,\mathrm{m/s}

v=15.0\,\mathrm{m/s}

2. Find displacement over the same interval

\Delta x=ut+\frac12at^2

\Delta x=\frac12(3.00)(5.00)^2\,\mathrm m=37.5\,\mathrm m

The object keeps moving right after it starts, so total distance is also $37.5\,\mathrm m$ . As a check, the average velocity for this constant acceleration is $(u+v)/2=7.50\,\mathrm{m/s}$ ; multiplying by time gives the same displacement. Inspect the velocity and position timeline.

3. Test a case that reverses direction

Now start with $u=12.0\,\mathrm{m/s}$ and constant $a=-3.00\,\mathrm{m/s^2}$ for $8.00\,\mathrm s$ . The same positive axis still points right. The object stops at $t_*=-u/a=4.00\,\mathrm s$ , then moves left. Negative acceleration slows it before that point and increases its speed afterward.

Signed areas cancel; path lengths add

Above-zero area is $+24.0\,\mathrm m$ ; the hatched below-zero area is $-24.0\,\mathrm m$ . Their sum is zero displacement. Their absolute magnitudes add to $48.0\,\mathrm m$ of distance.

v=12.0+(-3.00)(8.00)=-12.0\,\mathrm{m/s}

\Delta x=(12.0)(8.00)+\frac12(-3.00)(8.00)^2

\Delta x=0\,\mathrm m,\qquad L=48.0\,\mathrm m

Each triangular part of the velocity-time graph has area magnitude $24.0\,\mathrm m$ . Add the signed areas for displacement; add their magnitudes for distance. Average velocity is $0\,\mathrm{m/s}$ , while average speed is $48.0/8.00=6.00\,\mathrm{m/s}$ .

The definitions of distance and displacement explain why the two answers differ. Open the direction-reversal timeline. Its rest point separates the two path segments; simply taking the absolute value of the net displacement would lose both traveled lengths.

Printable physics formula worksheet

Eight questions · Electricity, energy and one-dimensional motion

Name: __________________________ Date: __________________

Choose A4 or Letter in your print preview. You can also save the worksheet as PDF.

Use the worked examples first, then attempt the questions before opening the key. In print preview choose portrait orientation and check paper size, scaling and margins. Save as PDF if you want a local copy; no account is required. The question-only version excludes answers, calculator links and the rest of this guide.

Show a conversion or substitution line and label every answer. Report nonzero final results to three significant figures. Use an ohmic resistor or steady DC model for circuit questions, classical translational energy for energy questions, and constant acceleration with right positive for motion questions.

1. Current and resistor power
An ohmic resistor has $R=3.00\,\mathrm{k\Omega}$ and a DC potential difference of $V=9.00\,\mathrm V$ . Find current and dissipated power.
2. Voltage from current
A constant $I=0.250\,\mathrm A$ flows through an ohmic $R=48.0\,\Omega$ resistor. Find its voltage drop and power.
3. Minutes to electrical energy
A steady DC load draws $I=0.500\,\mathrm A$ at $V=6.00\,\mathrm V$ for $t=30.0\,\mathrm{min}$ . Find power and energy in joules and kilowatt-hours.
4. Hours and kilowatt-hours
A steady DC load draws $I=2.00\,\mathrm A$ at $V=20.0\,\mathrm V$ for $t=2.50\,\mathrm h$ . Find power and energy in kilowatt-hours and joules.
5. Classical kinetic energy
A $0.500\,\mathrm{kg}$ object has speed $4.00\,\mathrm{m/s}$ in the chosen reference frame. Find its translational kinetic energy.
6. Solve for speed
A $10.0\,\mathrm{kg}$ object has classical translational kinetic energy $45.0\,\mathrm J$ . Find its speed, not its direction.
7. Accelerate along a line
Choose right as positive. An object starts at $u=2.00\,\mathrm{m/s}$ with constant $a=2.00\,\mathrm{m/s^2}$ for $t=3.00\,\mathrm s$ . Find final velocity and displacement.
8. Reverse direction
Choose right as positive. An object starts at $u=6.00\,\mathrm{m/s}$ with constant $a=-2.00\,\mathrm{m/s^2}$ for $t=6.00\,\mathrm s$ . Find final velocity, displacement and total distance.

Toolarithm · Physics Formula Worksheets · toolarithm.com/blog/physics-formula-worksheets

Worked answer key

Open an answer after attempting its question. Each calculator link restores the intended inputs and units. Calculators retain guard digits and display six significant figures; the answers below report three significant figures for the nonzero measured inputs.

1. Current and resistor power

I=\frac{9.00}{3.00\times10^3}\,\mathrm A=3.00\,\mathrm{mA}

P=VI=27.0\,\mathrm{mW}

Convert kilohms to ohms before dividing. Current and power are magnitudes for this passive resistor.

Check question 1 in the calculator

2. Voltage from current

V=IR=12.0\,\mathrm V

P=I^2R=3.00\,\mathrm W

The squared current is part of the power calculation. Squaring only a unit or only the number would be inconsistent.

Check question 2 in the calculator

3. Minutes to electrical energy

P=VI=3.00\,\mathrm W

E=P(1800\,\mathrm s)=5.40\times10^3\,\mathrm J

E=0.00150\,\mathrm{kWh}

Use seconds with watts to obtain joules, or kilowatts with hours to obtain kilowatt-hours. Both methods describe the same energy.

Check question 3 in the calculator

4. Hours and kilowatt-hours

P=VI=40.0\,\mathrm W

E=(0.0400\,\mathrm{kW})(2.50\,\mathrm h)=0.100\,\mathrm{kWh}

E=3.60\times10^5\,\mathrm J

The watt is a unit of power; the kilowatt-hour is a unit of energy. Do not divide by elapsed hours.

Check question 4 in the calculator

5. Classical kinetic energy

K=\frac12(0.500)(4.00)^2\,\mathrm J=4.00\,\mathrm J

Square the speed before multiplying by half the mass. The numerical one-half is exact.

Check question 5 in the calculator

6. Solve for speed

v=\sqrt{2K/m}=\sqrt{2(45.0)/10.0}\,\mathrm{m/s}

v=3.00\,\mathrm{m/s}

Choose the nonnegative root because speed is a magnitude. Kinetic energy alone does not determine a velocity direction.

Check question 6 in the calculator

7. Accelerate along a line

v=u+at=8.00\,\mathrm{m/s}

\Delta x=ut+\frac12at^2=15.0\,\mathrm m

Motion stays in the positive direction throughout, so total distance also equals the positive displacement.

Check question 7 in the calculator

8. Reverse direction

v=-6.00\,\mathrm{m/s},\qquad \Delta x=0\,\mathrm m

t_*=-u/a=3.00\,\mathrm s

L=9.00\,\mathrm m+9.00\,\mathrm m=18.0\,\mathrm m

Split the path at the rest point. The two signed displacements cancel, but their positive lengths add.

Check question 8 in the calculator

Check the answer before you move on

Read the unit. A power answer cannot be reported in an energy unit. Converting a speed does not convert its square; convert the speed first and square the complete quantity.
Check the sign. A negative final velocity can be valid. A negative speed or classical kinetic energy for a positive mass is inconsistent with these definitions. Name your positive axis.
Test the conditions. Acceleration must stay constant for the motion substitutions. A voltage-current line for an ohmic resistor assumes fixed resistance, rather than applying to every electrical device.
Keep the final precision honest. The worksheet givens carry three significant figures. Defined conversion factors and the numerical one-half are exact. Extra calculator digits help verify arithmetic; they do not remove measurement uncertainty.
Respect zero denominators. Average velocity and average speed over an interval require positive elapsed time. Solving speed from energy requires positive mass. Do not replace an undefined ratio with zero.

The reversal examples have exact cancellation within the stated nominal model. In a real experiment, uncertain inputs would need an uncertainty analysis; the symbolic zero here does not prove a measured object returned to an exactly identical position.

For related practice, use the physics calculator collection, Speed Converter, Energy Converter and Lab Report Calculations guide. The Significant Figures Calculator can help check final reporting.

FAQ

Frequently Asked Questions

What is included in these physics formula worksheets?

The worksheet contains eight original practice problems on ohmic resistance, steady DC power and energy, classical translational kinetic energy and one-dimensional constant acceleration. Four worked examples and a direction-reversal example prepare you for the questions. A separate answer key shows substitutions, units and reasoning. This is a focused introductory set, rather than a complete formula sheet for every physics course or examination.

How do I print the questions without the answers?

Use Print questions in the practice section to open your browser's print dialog with only the student worksheet. Print with answers includes the student sheet followed by the worked key. Choose A4 or Letter and check the preview before printing or saving as PDF; browser margins can change pagination. Without JavaScript, the article, questions and native answer disclosures still work; browser printing uses the full guide.

Why should I convert units before substituting?

A formula combines physical quantities, not bare numbers. In coherent SI units, amperes multiplied by ohms give volts, and watts multiplied by seconds give joules. Prefixes change numerical scale:

20.0\,\mathrm{mA}

0.0200\,\mathrm A

, not twenty amperes. Keep one visible conversion line so that a prefix error can be separated from a formula or algebra error when checking your work.

Can I use these DC power examples for AC circuits?

The worked circuit examples assume steady DC conditions and use

P=VI

. For single-phase sinusoidal AC, average real power is

P_{\mathrm{avg}}=V_{\mathrm{rms}}I_{\mathrm{rms}}\cos\phi

, so RMS values and the voltage-current phase angle matter. A purely resistive load has unity power factor. Peak values cannot simply replace RMS values in that expression. Select the appropriate AC model in the electrical power calculator and consult the cited AC reference.

Why does doubling speed quadruple kinetic energy?

With the same mass in the same reference frame, classical translational kinetic energy is proportional to the square of speed. Therefore

K(2v)=4K(v)

for a nonzero starting speed. Our example changes speed from

10.0\,\mathrm{m/s}

20.0\,\mathrm{m/s}

and energy from

100\,\mathrm J

400\,\mathrm J

. This model applies at speeds far below the speed of light and does not add rotational kinetic energy.

How can displacement be zero when the object has moved?

Displacement compares final position with initial position, while distance adds every part of the path. In the reversal example the object travels

24.0\,\mathrm m

right, then

24.0\,\mathrm m

left. Its displacement is zero and its distance is

48.0\,\mathrm m

. On a velocity-time graph, signed areas cancel for displacement; adding their absolute magnitudes gives distance. Average speed and average velocity consequently differ.

How many significant figures should I report?

The nonzero measured givens in this worksheet carry three significant figures. Keep guard digits through intermediate calculations and report the final nonzero results to three significant figures. Defined unit conversions and the numerical one-half do not limit that precision. Zero results arising from exact cancellation in the nominal model are shown explicitly. A calculator's six-digit display helps check arithmetic but does not establish measurement uncertainty or justify extra experimental precision.

When do the motion equations stop being valid?

The worked motion equations require constant acceleration along one chosen axis throughout the stated interval. Negative acceleration indicates direction relative to that axis; it does not always mean slowing down. If acceleration changes, divide the motion into valid intervals or use a suitable variable-acceleration method. An average defined by division needs positive elapsed time. Setting

t=0

does not define average speed or average velocity over an interval.

Sources and verification method

The references support the definitions, physical assumptions and conversion factors. All worked and practice inputs are original hypothetical examples. An independent exact-arithmetic specification checks the conversions, substitutions, inverse speed, rest times, signed areas and total distances. The three illustrations use coordinates derived from their declared models, rather than schematic values presented as measurements.

OpenStax: Ohm's law
The proportional relationship for ohmic materials and its limits.
OpenStax: electrical energy and power
Electrical power, resistor relations and constant-power energy.
OpenStax: kinetic energy
Classical translational kinetic energy, speed and reference frames.
OpenStax: constant acceleration
The one-dimensional constant-acceleration equations and initial conditions.
OpenStax: displacement and average velocity
The distinction between signed displacement and total distance traveled.
NIST: SI units and prefixes
Ampere, volt, ohm, joule, watt and SI prefix factors.
NIST: conversion factors, Appendix B.8
Kilowatt-hour to joule conversion and speed-unit factors.
OpenStax: power in an AC circuit
RMS quantities, average real power and power factor for sinusoidal AC.

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Ownership and review

Written and maintained by the Toolarithm editorial team. The method was last reviewed on October 30, 2026. No independent professional review is claimed.

Methodology

The editorial team checks definitions, examples, linked tools, and cited references. Dates shown here come from the page's maintained publication record.

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Dates and sources

Published June 10, 2026. Updated October 30, 2026.