Physics · Mechanics

Mechanical power

Q: How do you calculate mechanical power?

Divide work performed during the selected interval by its positive duration: $\overline P=\frac{W}{\Delta t}$. This is average power, not automatically its instantaneous value. Performing 3000 J of lifting work in 60 s gives an average of 50 W; delivering the same work in 20 s gives 150 W. Instantaneous power of a force is $P=\vec F\cdot\vec v$, using its application-point velocity. The shorter form $P=Fv$ requires force and velocity to point in the same direction. Perpendicular force contributes no power; opposing force contributes negative power. For vehicles, identify the application point and the energy transfer being counted. Convert minutes to seconds before calculating power in watts.

Q: What does the formula $P = F\cdot{}v$ mean and when do you use it?

$P = Fv$ gives instantaneous power of a force when force and application-point velocity are aligned. In general, $P = F\cdot{}v = Fv \cos{} \alpha{}$. Divide $dW = F\cdot{}dr$ by dt to obtain this relation; constant force is not required. Use values from the same instant. A 400 N pull at $\frac{15 m}{s}$ supplies 6 kW instantaneously, while an equal opposing force performs −6 kW. Other forces may transfer energy at the same time. Substituting average velocity does not generally give average power when force also varies with time. Instead, integrate P(t) over the interval and divide by its duration.

Q: What is the difference between work and power?

Work is energy transfer in joules; power describes its rate in watts. In an ideal lifting model with equal initial and final speeds, raising 100 kg by 10 m at $g=9.81\,\mathrm{m\,s^{-2}}$ requires 9810 J of useful energy. Over 5 s this is 1962 W of average useful power; over 120 s it is about 81.8 W. These values concern energy delivered to the load, not automatically a crane’s electrical consumption or a person’s metabolic energy rate. Efficiency and other transfers are needed for those quantities. In general, $\overline P=\frac{W}{\Delta t}$. For varying power, area under the power curve gives total work.

Q: How do you convert horsepower to watts and where does the unit come from?

Metric horsepower, denoted PS, equals 735.49875 W, commonly rounded to 735.5 W. Thus 100 PS is about 73.55 kW. Divide a value in kW by 0.73549875 to obtain PS. This unit is defined through 75 kgf·m/s using standard gravity $g_{0} =\frac{ 9.80665 m}{s^{2}}$, not local gravitational acceleration. English-language mechanical horsepower is different: approximately 745.6999 W. The word horsepower alone therefore does not specify a conversion. Use the SI unit watt for physics calculations and identify the convention for non-SI inputs. Do not silently translate PS into mechanical hp while retaining the metric conversion factor.

Q: How do you calculate power when climbing stairs or walking uphill?

Average rate of potential-energy increase is mgΔh/Δt. Δh is vertical rise, not distance along a slope. For 70 kg rising 5 m in 30 s with $g =\frac{ 9.81 m}{s^{2}}$, this gives about 114.45 W. It is average lifting power within the chosen body–Earth model, not the person’s full metabolic energy rate. Muscle efficiency, internal motion and other transfers require separate treatment. Equal initial and final speeds avoid an additional net kinetic-energy increase. This calculation does not determine instantaneous peak output or sustainable athletic performance. To compare people or machines, specify whether the quantity is useful mechanical output or total energy input.

Mechanical power is the rate of work transfer; $\frac{W}{t}$ gives average power over duration t.

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Formula

Calculate

P = \frac{W}{t}

P in watts [W] = [J/s] · W in joules [J] · t in seconds [s]

Variables & units

Variables & units
Symbol	Meaning	Unit
$P$	Average power over the selected duration	W
$W$	Work performed during the interval	J
$t$	Positive duration of the interval	s

LaTeX

P = \frac{W}{t}

QUANTA STUDYquanta-study.de

Work–time curve W=40t² with joules and seconds. A secant connects (1 s,40 J) and (3 s,360 J); its slope is 160 W. The steeper tangent at 3 s has slope 240 W. — Original example W(t) = 40 J·(t/s)²: average power from 1 s to 3 s is 160 W. The tangent at 3 s gives instantaneous power of 240 W.

Mechanical power calculator

Find average power, work or time when work is done over a given time interval.

Find

Work W (signed)

Work done in the interval, e.g. about 1800 J to climb 3 m with 60 kg; negative when energy is removed.

Time interval Δt

Duration of the process, greater than zero, e.g. 6 s for one floor.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for P, substitute, evaluate

1. Rearrange for P
$\overline P=\frac W{\Delta t}$
2. Substitute values (in SI units)
$\overline P=\frac{1800}{6}=300$
3. Result
P = 300 W

Result: Average power P

P = 300 W

With W = 1800 J and Δt = 6 s, P = 300 W.

Change Time interval Δt

If Δt doubles, P halves: they are inversely proportional.

Quick tips

Formula triangle: W on top, P and t below. Cover the quantity you need, e.g. P = W ÷ t or W = P · t.
kW is power, kWh is energy: 1 kWh = 1 kW · 1 h = 3.6 MJ. So enter a value in kWh as MJ in the work field, e.g. 2 kWh as 7.2 MJ.

Assumptions and limits

Work and elapsed time Δt > 0 belong to the same process and energy boundary. P is interval-average power, not arbitrary instantaneous power. W and P may be negative for energy removal.

Derivation & background

Average power is $\bar P=\frac{W}{\Delta t}$ ; instantaneous power is $P =\frac{ dW}{dt} = F\cdot{}v$ . Aligned force gives $P = Fv$ without requiring force to be constant in time. A watt is a joule per second. A kilowatt-hour is energy: $1 kWh = 3.6\cdot{}10^{6} J$ . Distinguish input and useful output; steady operation without changing stored energy gives $P_{\mathrm{in}}=P_{\mathrm{out}}+P_{\mathrm{loss}}$ .

Exam blueprint

Validity range

$\frac{W}{\Delta{}t}$ is average power. Instantaneous power is $F\cdot{}v = Fv \cos{} \alpha{}$ , even for time-dependent force. Force and velocity must refer to the same instant and application point.

Derivation steps

Average power is work per elapsed interval; the derivative describes instantaneous transfer rate.

1Specify work W and positive duration Δt.
2Define average power: $\bar P=\frac{W}{\Delta t}$ .
3For an infinitesimal interval, $dW = F\cdot{}dr$ .
4Divide by dt and substitute $\frac{dr}{dt} = v$ .
5Instantaneously, $P =\frac{ dW}{dt} = F\cdot{}v = Fv \cos{} \alpha{}$ .
6Aligned force gives $P = Fv$ ; W = PΔt requires constant P, otherwise integrate.

Rearrangements

Work from power and time

W = P \cdot t

The basis of energy billing: $kWh = kW$ times hours.

Instantaneous power from force and speed

P = F \cdot v

Explains why more engine power is needed at high speed.

Task variant

A motor does $W = 6\,000 J$ at $P = 200 W$ . How long does it take?

$t=\frac WP=\frac{6000}{200}\,\mathrm s=30\,\mathrm s$ .

A car sustains $v =\frac{ 15 m}{s}$ with a driving force of $F = 400 N$ . What is its power?

$P=Fv=400\cdot15\,\mathrm W=6000\,\mathrm W=6\,\mathrm{kW}$ .

Common mistakes

Confusing power with work or energy.

Power is the rate in watts; energy is power times time.

Substituting time in minutes.

For watts, t must be in seconds: $5 \min{} = 300 s$ .

Treating kWh as a unit of power.

The kilowatt hour is a unit of energy $(1 kWh = 3.6\times{}10^{6} J)$ .

Exam context

Typical in efficiency and engine problems: lifting work per time, driving resistance times speed, or comparing human and machine.

Formula cluster

Power

Mechanical and electric power follow the same definition $P =\frac{ E}{t}$ .

W = F·s·cos(α)

Mechanical Work

P = U·I

Electric Power

Ekin = ½mv²

Kinetic Energy

Worked example

Climbing stairs, a person performs $W = 3\,000 J$ of lifting work in $t = 60 s$ : $P =\frac{ 3\,000}{60} = 50 W$ .

Applications

Engine power, sports diagnostics (watt measurement in cycling), lift design, pump power

How do you calculate with Mechanical Power?

Here is how to work through a typical Mechanical Power (P = W/t) task step by step:

1
Task
A motor does $W = 6\,000 J$ at $P = 200 W$ . How long does it take?
Solution path
$t=\frac WP=\frac{6000}{200}\,\mathrm s=30\,\mathrm s$ .
2
Task
A car sustains $v =\frac{ 15 m}{s}$ with a driving force of $F = 400 N$ . What is its power?
Solution path
$P=Fv=400\cdot15\,\mathrm W=6000\,\mathrm W=6\,\mathrm{kW}$ .

Understand and apply the equation

Two drives raise the same load through the same height. The faster one transfers the required energy in less time and has greater average useful power. Power therefore describes the rate of energy transfer. How much energy is required overall is a separate question.

How do you calculate average mechanical power?

Choose a process with a clear beginning and end. If mechanical work $W$ is performed over duration $\Delta t>0$ , average mechanical power is $\overline P=\frac{W}{\Delta t}$ . Performing the same work in half the time doubles average power. The average describes the entire interval, not fluctuations during starting or at its end.[1]

\overline P=\frac{W}{\Delta t}

Here the quantity W means work in joules; the unit symbol W means watts.

Original duration check: 600 J in 3 s gives 200 W; the same 600 J in 12 s gives 50 W. Total energy is identical. A more powerful motor need not consume more total energy when it finishes the same task sooner at equal efficiency.

When does power equal force times velocity?

To find power at one instant, consider a very small displacement. Work over that displacement is the force–displacement dot product. Divide by elapsed time: displacement per time becomes velocity. Force and velocity must refer to the same instant and the same application point. For pure rotation about a fixed axis, axial torque takes this role: $P=\tau\omega$ . Use torque and angular velocity for the same axis.[1][3]

\begin{aligned}\mathrm dW&=\vec F\cdot\mathrm d\vec r\\P&=\frac{\mathrm dW}{\mathrm dt}\\&=\vec F\cdot\frac{\mathrm d\vec r}{\mathrm dt}=\vec F\cdot\vec v\end{aligned}

The force need not be constant for this step.[1]

P(t)=\frac{\mathrm dW}{\mathrm dt}=\vec F(t)\cdot\vec v(t)=Fv\cos\alpha

Here α is the angle between force and velocity.

Aligned force and velocity give positive $P=Fv$ ; opposite directions give negative power; perpendicular directions give zero. A radial force on a circular path can therefore change direction without transferring energy. Sum individual powers; particle net power equals dK/dt.[2]

What distinguishes watts, kilowatts and kilowatt-hours?

\begin{aligned}&1\,\mathrm W=1\,\mathrm{J\,s^{-1}}\\&1\,\mathrm{kW}=1000\,\mathrm W\\&1\,\mathrm{kWh}=3.6\cdot10^6\,\mathrm J\end{aligned}

Watts and kilowatts measure power; a kilowatt-hour measures energy.

An idealised drive drawing constant 0.6 kW for 20 minutes consumes $(0.6\,\mathrm{kW})\frac{20}{60}\,\mathrm h=0.2\,\mathrm{kWh}$ , or 720 kJ. Do not drop the h: 0.2 kW is a different quantity. $W=P\Delta t$ assumes constant power; otherwise use area under P(t).

\begin{aligned}&W=\int_{t_1}^{t_2}P(t)\,\mathrm dt\\&\overline P=\frac{1}{t_2-t_1}\int_{t_1}^{t_2}P(t)\,\mathrm dt\end{aligned}

Integrate to obtain work, then divide by duration to obtain average power.

Specify the convention for horsepower: metric horsepower PS is about 735.5 W, whereas mechanical horsepower is about 745.7 W. Do not relabel an unchanged numerical value between these units.[4]

How do you calculate lifting power and power from an angled force?

Lifting a load with efficiency

Given: Load mass 200 kg; rise 3 m in 15 s; g = 9.81 m/s². Equal initial and final speeds; overall efficiency over the operation 0.80.
Find: Find average useful output and input power.

Approach: With equal initial and final speeds, useful energy is $\Delta E_{\mathrm{pot}}=mg\Delta h$ ; use $\overline P_{\mathrm{out}}=\frac{mg\Delta h}{\Delta t}$ and $\overline P_{\mathrm{in}}=\frac{\overline P_{\mathrm{out}}}{\eta}$ . Calculate required potential-energy increase.
$W_{\mathrm{useful}}=200\cdot9.81\cdot3=5886\,\mathrm J$
Divide by lifting duration.
$\overline P_{\mathrm{out}}=\frac{5886}{15}=392.4\,\mathrm W$
Apply energy efficiency to the same interval.
$\overline P_{\mathrm{in}}=\frac{392.4}{0.80}=490.5\,\mathrm W$

Result: Average useful power is 392.4 W; average input is 490.5 W.

Check: Input energy is 7357.5 J; the difference from useful energy is 1471.5 J. Peak starting power is not determined.

Power of an angled force

Given: A 60 N force acts at 60° to an instantaneous velocity of magnitude 2 m/s.
Find: Find its instantaneous power and, if these values remain constant, its work over 5 s.

Approach: Instantaneous force power is $P=\vec F\cdot\vec v=Fv\cos\theta$ ; for constant power use $W=P\Delta t$ . Find the force component along motion.
$F_\parallel=60\cos60^\circ=30\,\mathrm N$
Multiply by speed.
$P=30\cdot2=60\,\mathrm W$
Integrate over the assumed constant interval.
$W=60\cdot5=300\,\mathrm J$

Result: Instantaneous power is +60 W; a constant five-second interval transfers 300 J.

Check: $W=Fs\cos\theta=60\cdot10\cdot\cos60^\circ\,\mathrm J=300\,\mathrm J$ . Distance is then 10 m. Other forces may offset this energy gain.

Why does constant speed still require driving power?

Pull a sled at constant speed

Given: A sled is pulled over level ground by a horizontal rope at constant 3.0 m/s. Total opposing sliding resistance is 120 N. Rope tension has the same magnitude; no other horizontal forces act. The sled does not rotate.
Find: Find mechanical power of the pulling force, resistance power and net power.

Approach: take each force’s dot product with the same velocity separately.
$P_{\mathrm{drive}}=F_{\mathrm{drive}}v,\qquad P_{\mathrm{res}}=-F_{\mathrm{res}}v$
Substitute aligned driving force and opposing resistance.
$\begin{aligned}P_{\mathrm{drive}}&=(120\,\mathrm N)(3.0\,\mathrm{m\,s^{-1}})=360\,\mathrm W\\P_{\mathrm{res}}&=-360\,\mathrm W\\P_{\mathrm{net}}&=P_{\mathrm{drive}}+P_{\mathrm{res}}=0\end{aligned}$

Result: The pulling force supplies 360 W of mechanical power; net power changing translational motion is zero.

Check: In 5.0 s the displacement is 15 m. Driving work is 120 N times 15 m, or 1800 J; the same energy follows from 360 W times 5.0 s. Resistance converts this amount into other energy forms while kinetic energy stays constant.

How do useful power, input power and efficiency relate?

\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}=\frac{\overline P_{\mathrm{out}}}{\overline P_{\mathrm{in}}}

The power ratio uses the same interval as the energy ratio.

Identify electrical input, mechanical shaft output or useful power at the load. Compare matching quantities. If an energy store charges or discharges, instantaneous output may exceed instantaneous input; that does not imply efficiency above 100% over a complete storage cycle. For the corresponding steady conversion with $0<\eta\le1$ , conversion efficiency gives input power as $P_{\mathrm{in}}=\frac{P_{\mathrm{out}}}{\eta}$ ; this rearrangement is invalid at $\eta=0$ .

For a steady drive without changing stored energy, $P_{\mathrm{in}}=P_{\mathrm{out}}+P_{\mathrm{loss}}$ . Heat and sound represent other energy transfers or storage, not destroyed energy. Efficiency refers to an operating point or an entire process; it need not remain identical at every load.

What do time-dependent and negative power mean?

Original profile: transferred power rises linearly from zero to 100 W over 4 s. Triangular area gives work $\frac12(4\,\mathrm s)(100\,\mathrm W)=200\,\mathrm J$ and average power 50 W. Applying the final 100 W for the whole interval would incorrectly give 400 J.

The original illustration instead plots accumulated work $W(t)=(40\,\mathrm J)\left(\frac t{1\,\mathrm s}\right)^2$ . From 1 s to 3 s, work rises from 40 to 360 J; secant slope is $\frac{(360-40)\,\mathrm J}{2\,\mathrm s}=160\,\mathrm W$ . Tangent slope at 3 s is $\frac{\mathrm dW}{\mathrm dt}=240\,\mathrm W$ . On a W(t) graph, slope gives power; on a P(t) graph, area gives work.

A braking force has negative mechanical power on a body moving opposite the force. Regenerative braking may recover some of it as electrical energy. Recovered electrical power need not equal the magnitude of mechanical braking power because other energy paths and losses exist.

Slope or area? Read the axes

If the vertical axis already shows work in joules, slope gives the rate at which work increases: power. If the vertical axis shows power in watts, contributions must be accumulated over time: area gives work. A steeper work curve and a larger area under a power curve are therefore different graphical properties.[1]

\begin{aligned}\overline P_{1\to3}&=\frac{W(3\,\mathrm s)-W(1\,\mathrm s)}{3\,\mathrm s-1\,\mathrm s}\\&=\frac{(360-40)\,\mathrm J}{2\,\mathrm s}=160\,\mathrm W\\P(3\,\mathrm s)&=\left.\frac{\mathrm dW}{\mathrm dt}\right|_{3\,\mathrm s}=240\,\mathrm W\end{aligned}

For the quadratic work history above: the secant refers to an interval, the tangent to an instant.[1]

Average power between two instants is not generally the arithmetic mean of their instantaneous powers. That would require a suitable intervening history, for example a linear one. For unequal operating intervals, each power level must be weighted by its own duration; the exercises below examine this distinction.[1]

Work as area under power

QUANTA STUDYquanta-study.de

In this example P(t) = 80t watts for t in seconds. From 1 to 3 s, 320 J is transferred: average power of 160 W times 2 s. Instantaneous power at 3 s is instead 240 W. This area equals the increase in the work graph W(t) = 40t².[1]

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What a power rating does not decide

$P=Fv$ alone does not determine starting force. At v = 0 a finite force may act while instantaneous mechanical power is zero. $F=\frac Pv$ is invalid there. Torque, gearing and operating limits require separate information.

For pure rotation about a fixed axis, the corresponding relation is $P=\tau\omega$ using axial torque τ. Translation uses the relevant linear velocity instead. Mechanical output during exercise also differs from a person’s metabolic energy rate.[3]

Five problems with solutions

Calculate · Foundation

A drive performs 1.8 kJ of work in 6 s. Find average power.

Show hint

Convert kilojoules to joules.

Learning goal: Find mean power from work and duration.

Show solution and calculation

$\overline P=\frac W{\Delta t}=\frac{1800}{6}\,\mathrm W=300\,\mathrm W$ .

\overline P=300\,\mathrm W

Calculate · Application

An 80 N force opposes a body moving at 3 m/s. Find instantaneous power.

Show hint

Force and velocity make an angle of 180°.

Learning goal: Calculate negative power of an opposing force.

Show solution and calculation

$P=Fv\cos\theta=80\cdot3\cdot\cos180^\circ\,\mathrm W=-240\,\mathrm W$ .

P=-240\,\mathrm W

Transfer · Application

A steady lifting system takes 1.2 kW input and delivers 900 W useful output. Find efficiency and loss power.

Show hint

Use watts for both powers.

Learning goal: Balance useful power, input power, efficiency and losses.

Show solution and calculation

$\eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}=\frac{900}{1200}=0.75=75\,\%$ .
Loss power: $P_{\mathrm{loss}}=1200\,\mathrm W-900\,\mathrm W=300\,\mathrm W$ .

\begin{aligned}\eta&=0.75\\{} P_{\mathrm{loss}}&=300\,\mathrm W\end{aligned}

Interpret · Application

A body moves uniformly on a horizontal circle. An inward force of 25 N acts while speed is 4.0 m/s. A diagram shows force perpendicular to velocity. Find this force’s instantaneous power and explain why the body is nevertheless accelerating. Other forces need not be balanced for this question.

Show hint

Power uses the dot product of force and velocity. Acceleration can change velocity direction rather than its magnitude.

Learning goal: Understand how a large force can deliver zero mechanical power because of direction.

Show solution and calculation

Approach: insert the angle between force and instantaneous velocity into the power formula.
Cosine of 90° is zero, giving zero watts rather than 100 W.
The force redirects motion and produces radial acceleration.

Check: constant speed leaves kinetic energy constant, consistent with zero power transfer from this force.

\begin{aligned}P&=\vec F\cdot\vec v=Fv\cos\theta\\&=(25\,\mathrm N)(4.0\,\mathrm{m/s})\cos90^\circ=0\,\mathrm W\\\frac{\mathrm dE_{\mathrm{kin}}}{\mathrm dt}&=0\end{aligned}

[1][2]

Error analysis · Application

A drive supplies constant mechanical power 100 W for 2 s and then 300 W for 6 s. A solution calls the simple average 200 W the mean power over all 8 s. Correct this using a work balance. The stated values already represent useful mechanical power.

Show hint

Add work from each interval first. Mean power is total work divided by total time, not an unweighted average of power values.

Learning goal: Weight power intervals correctly when their durations differ.

Show solution and calculation

Approach: calculate power times duration for each constant-power interval.
Work contributions are 200 J and 1800 J, totaling 2000 J.
Distributed over eight seconds this gives 250 W.

Check: the longer 300 W interval must pull the average closer to 300 W than to 100 W. A simple average would work only for equal interval durations.

\begin{aligned}W&=P_1\Delta t_1+P_2\Delta t_2\\&=(100)(2)\,\mathrm J+(300)(6)\,\mathrm J=2000\,\mathrm J\\\overline P&=\frac W{\Delta t_1+\Delta t_2}=\frac{2000\,\mathrm J}{8\,\mathrm s}=250\,\mathrm W\end{aligned}

[1]

What to take away

Work describes energy transfer; power describes its rate. Distinguish averages over a duration from instantaneous power at an application point. Label input, output and useful power. High driving power can balance resistance without accelerating an object. Power is slope on a work graph; work is area on a power graph.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Exposed average, instantaneous and driving power through question headings; clarified the steady efficiency rearrangement and linked work, torque and efficiency in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the introduction and learning progression; added explanatory intermediate steps, an application example and a summary. Retained existing examples and all five exercises.

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Quanta. Mechanical Power. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/leistung-mechanisch

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Frequently asked questions about Mechanical Power

How do you calculate mechanical power?+

Divide work performed during the selected interval by its positive duration: $\overline P=\frac{W}{\Delta t}$ . This is average power, not automatically its instantaneous value. Performing 3000 J of lifting work in 60 s gives an average of 50 W; delivering the same work in 20 s gives 150 W. Instantaneous power of a force is $P=\vec F\cdot\vec v$ , using its application-point velocity. The shorter form $P=Fv$ requires force and velocity to point in the same direction. Perpendicular force contributes no power; opposing force contributes negative power. For vehicles, identify the application point and the energy transfer being counted. Convert minutes to seconds before calculating power in watts.

What does the formula

P = F\cdot{}v

mean and when do you use it?+

$P = Fv$ gives instantaneous power of a force when force and application-point velocity are aligned. In general, $P = F\cdot{}v = Fv \cos{} \alpha{}$ . Divide $dW = F\cdot{}dr$ by dt to obtain this relation; constant force is not required. Use values from the same instant. A 400 N pull at $\frac{15 m}{s}$ supplies 6 kW instantaneously, while an equal opposing force performs −6 kW. Other forces may transfer energy at the same time. Substituting average velocity does not generally give average power when force also varies with time. Instead, integrate P(t) over the interval and divide by its duration.

What is the difference between work and power?+

Work is energy transfer in joules; power describes its rate in watts. In an ideal lifting model with equal initial and final speeds, raising 100 kg by 10 m at $g=9.81\,\mathrm{m\,s^{-2}}$ requires 9810 J of useful energy. Over 5 s this is 1962 W of average useful power; over 120 s it is about 81.8 W. These values concern energy delivered to the load, not automatically a crane’s electrical consumption or a person’s metabolic energy rate. Efficiency and other transfers are needed for those quantities. In general, $\overline P=\frac{W}{\Delta t}$ . For varying power, area under the power curve gives total work.

How do you convert horsepower to watts and where does the unit come from?+

Metric horsepower, denoted PS, equals 735.49875 W, commonly rounded to 735.5 W. Thus 100 PS is about 73.55 kW. Divide a value in kW by 0.73549875 to obtain PS. This unit is defined through 75 kgf·m/s using standard gravity $g_{0} =\frac{ 9.80665 m}{s^{2}}$ , not local gravitational acceleration. English-language mechanical horsepower is different: approximately 745.6999 W. The word horsepower alone therefore does not specify a conversion. Use the SI unit watt for physics calculations and identify the convention for non-SI inputs. Do not silently translate PS into mechanical hp while retaining the metric conversion factor.

How do you calculate power when climbing stairs or walking uphill?+

Average rate of potential-energy increase is mgΔh/Δt. Δh is vertical rise, not distance along a slope. For 70 kg rising 5 m in 30 s with $g =\frac{ 9.81 m}{s^{2}}$ , this gives about 114.45 W. It is average lifting power within the chosen body–Earth model, not the person’s full metabolic energy rate. Muscle efficiency, internal motion and other transfers require separate treatment. Equal initial and final speeds avoid an additional net kinetic-energy increase. This calculation does not determine instantaneous peak output or sustainable athletic performance. To compare people or machines, specify whether the quantity is useful mechanical output or total energy input.

Quanta exam set

Curated exam set for "Mechanical Power":

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Which formula describes Mechanical Power?

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How do you rearrange Mechanical Power for Work from power and time?

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Which common mistake happens with Mechanical Power?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §7.4 Power

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

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Symbol

Meaning

Unit

P

Average power over the selected duration

W

Work performed during the interval

t

Positive duration of the interval

Derivation & background

Exam blueprint

Validity range

Derivation steps

Average power is work per elapsed interval; the derivative describes instantaneous transfer rate.

1Specify work W and positive duration Δt.
2Define average power: $\bar P=\frac{W}{\Delta t}$ .
3For an infinitesimal interval, $dW = F\cdot{}dr$ .
4Divide by dt and substitute $\frac{dr}{dt} = v$ .
5Instantaneously, $P =\frac{ dW}{dt} = F\cdot{}v = Fv \cos{} \alpha{}$ .
6Aligned force gives $P = Fv$ ; W = PΔt requires constant P, otherwise integrate.

How do you calculate with Mechanical Power?

Here is how to work through a typical Mechanical Power (P = W/t) task step by step:

Task

A motor does $W = 6\,000 J$ at $P = 200 W$ . How long does it take?

Solution path

$t=\frac WP=\frac{6000}{200}\,\mathrm s=30\,\mathrm s$ .

Task

A car sustains $v =\frac{ 15 m}{s}$ with a driving force of $F = 400 N$ . What is its power?

Solution path

$P=Fv=400\cdot15\,\mathrm W=6000\,\mathrm W=6\,\mathrm{kW}$ .

How do you calculate average mechanical power?

\overline P=\frac{W}{\Delta t}

Here the quantity W means work in joules; the unit symbol W means watts.

When does power equal force times velocity?

\begin{aligned}\mathrm dW&=\vec F\cdot\mathrm d\vec r\\P&=\frac{\mathrm dW}{\mathrm dt}\\&=\vec F\cdot\frac{\mathrm d\vec r}{\mathrm dt}=\vec F\cdot\vec v\end{aligned}

The force need not be constant for this step.[1]

P(t)=\frac{\mathrm dW}{\mathrm dt}=\vec F(t)\cdot\vec v(t)=Fv\cos\alpha

Here α is the angle between force and velocity.

What distinguishes watts, kilowatts and kilowatt-hours?

\begin{aligned}&1\,\mathrm W=1\,\mathrm{J\,s^{-1}}\\&1\,\mathrm{kW}=1000\,\mathrm W\\&1\,\mathrm{kWh}=3.6\cdot10^6\,\mathrm J\end{aligned}

Watts and kilowatts measure power; a kilowatt-hour measures energy.

\begin{aligned}&W=\int_{t_1}^{t_2}P(t)\,\mathrm dt\\&\overline P=\frac{1}{t_2-t_1}\int_{t_1}^{t_2}P(t)\,\mathrm dt\end{aligned}

Integrate to obtain work, then divide by duration to obtain average power.

Specify the convention for horsepower: metric horsepower PS is about 735.5 W, whereas mechanical horsepower is about 745.7 W. Do not relabel an unchanged numerical value between these units.[4]

How do you calculate lifting power and power from an angled force?

Lifting a load with efficiency

Given: Load mass 200 kg; rise 3 m in 15 s; g = 9.81 m/s². Equal initial and final speeds; overall efficiency over the operation 0.80.
Find: Find average useful output and input power.

Approach: With equal initial and final speeds, useful energy is $\Delta E_{\mathrm{pot}}=mg\Delta h$ ; use $\overline P_{\mathrm{out}}=\frac{mg\Delta h}{\Delta t}$ and $\overline P_{\mathrm{in}}=\frac{\overline P_{\mathrm{out}}}{\eta}$ . Calculate required potential-energy increase.
$W_{\mathrm{useful}}=200\cdot9.81\cdot3=5886\,\mathrm J$
Divide by lifting duration.
$\overline P_{\mathrm{out}}=\frac{5886}{15}=392.4\,\mathrm W$
Apply energy efficiency to the same interval.
$\overline P_{\mathrm{in}}=\frac{392.4}{0.80}=490.5\,\mathrm W$

Result: Average useful power is 392.4 W; average input is 490.5 W.

Check: Input energy is 7357.5 J; the difference from useful energy is 1471.5 J. Peak starting power is not determined.

Power of an angled force

Given: A 60 N force acts at 60° to an instantaneous velocity of magnitude 2 m/s.
Find: Find its instantaneous power and, if these values remain constant, its work over 5 s.

Approach: Instantaneous force power is $P=\vec F\cdot\vec v=Fv\cos\theta$ ; for constant power use $W=P\Delta t$ . Find the force component along motion.
$F_\parallel=60\cos60^\circ=30\,\mathrm N$
Multiply by speed.
$P=30\cdot2=60\,\mathrm W$
Integrate over the assumed constant interval.
$W=60\cdot5=300\,\mathrm J$

Result: Instantaneous power is +60 W; a constant five-second interval transfers 300 J.

Check: $W=Fs\cos\theta=60\cdot10\cdot\cos60^\circ\,\mathrm J=300\,\mathrm J$ . Distance is then 10 m. Other forces may offset this energy gain.

Why does constant speed still require driving power?

Pull a sled at constant speed

Given: A sled is pulled over level ground by a horizontal rope at constant 3.0 m/s. Total opposing sliding resistance is 120 N. Rope tension has the same magnitude; no other horizontal forces act. The sled does not rotate.
Find: Find mechanical power of the pulling force, resistance power and net power.

Approach: take each force’s dot product with the same velocity separately.
$P_{\mathrm{drive}}=F_{\mathrm{drive}}v,\qquad P_{\mathrm{res}}=-F_{\mathrm{res}}v$
Substitute aligned driving force and opposing resistance.
$\begin{aligned}P_{\mathrm{drive}}&=(120\,\mathrm N)(3.0\,\mathrm{m\,s^{-1}})=360\,\mathrm W\\P_{\mathrm{res}}&=-360\,\mathrm W\\P_{\mathrm{net}}&=P_{\mathrm{drive}}+P_{\mathrm{res}}=0\end{aligned}$

Result: The pulling force supplies 360 W of mechanical power; net power changing translational motion is zero.

How do useful power, input power and efficiency relate?

\eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}=\frac{\overline P_{\mathrm{out}}}{\overline P_{\mathrm{in}}}

The power ratio uses the same interval as the energy ratio.

What do time-dependent and negative power mean?

Slope or area? Read the axes

\begin{aligned}\overline P_{1\to3}&=\frac{W(3\,\mathrm s)-W(1\,\mathrm s)}{3\,\mathrm s-1\,\mathrm s}\\&=\frac{(360-40)\,\mathrm J}{2\,\mathrm s}=160\,\mathrm W\\P(3\,\mathrm s)&=\left.\frac{\mathrm dW}{\mathrm dt}\right|_{3\,\mathrm s}=240\,\mathrm W\end{aligned}

For the quadratic work history above: the secant refers to an interval, the tangent to an instant.[1]

Work as area under power

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What a power rating does not decide

Five problems with solutions

Calculate · Foundation

A drive performs 1.8 kJ of work in 6 s. Find average power.

Show hint

Convert kilojoules to joules.

Learning goal: Find mean power from work and duration.

Show solution and calculation

$\overline P=\frac W{\Delta t}=\frac{1800}{6}\,\mathrm W=300\,\mathrm W$ .

\overline P=300\,\mathrm W

Calculate · Application

An 80 N force opposes a body moving at 3 m/s. Find instantaneous power.

Show hint

Force and velocity make an angle of 180°.

Learning goal: Calculate negative power of an opposing force.

Show solution and calculation

$P=Fv\cos\theta=80\cdot3\cdot\cos180^\circ\,\mathrm W=-240\,\mathrm W$ .

P=-240\,\mathrm W

Transfer · Application

A steady lifting system takes 1.2 kW input and delivers 900 W useful output. Find efficiency and loss power.

Show hint

Use watts for both powers.

Learning goal: Balance useful power, input power, efficiency and losses.

Show solution and calculation

$\eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}=\frac{900}{1200}=0.75=75\,\%$ .
Loss power: $P_{\mathrm{loss}}=1200\,\mathrm W-900\,\mathrm W=300\,\mathrm W$ .

\begin{aligned}\eta&=0.75\\{} P_{\mathrm{loss}}&=300\,\mathrm W\end{aligned}

Interpret · Application

A body moves uniformly on a horizontal circle. An inward force of 25 N acts while speed is 4.0 m/s. A diagram shows force perpendicular to velocity. Find this force’s instantaneous power and explain why the body is nevertheless accelerating. Other forces need not be balanced for this question.

Show hint

Power uses the dot product of force and velocity. Acceleration can change velocity direction rather than its magnitude.

Learning goal: Understand how a large force can deliver zero mechanical power because of direction.

Show solution and calculation

Approach: insert the angle between force and instantaneous velocity into the power formula.
Cosine of 90° is zero, giving zero watts rather than 100 W.
The force redirects motion and produces radial acceleration.

Check: constant speed leaves kinetic energy constant, consistent with zero power transfer from this force.

\begin{aligned}P&=\vec F\cdot\vec v=Fv\cos\theta\\&=(25\,\mathrm N)(4.0\,\mathrm{m/s})\cos90^\circ=0\,\mathrm W\\\frac{\mathrm dE_{\mathrm{kin}}}{\mathrm dt}&=0\end{aligned}

[1][2]

Error analysis · Application

A drive supplies constant mechanical power 100 W for 2 s and then 300 W for 6 s. A solution calls the simple average 200 W the mean power over all 8 s. Correct this using a work balance. The stated values already represent useful mechanical power.

Show hint

Add work from each interval first. Mean power is total work divided by total time, not an unweighted average of power values.

Learning goal: Weight power intervals correctly when their durations differ.

Show solution and calculation

Approach: calculate power times duration for each constant-power interval.
Work contributions are 200 J and 1800 J, totaling 2000 J.
Distributed over eight seconds this gives 250 W.

Check: the longer 300 W interval must pull the average closer to 300 W than to 100 W. A simple average would work only for equal interval durations.

\begin{aligned}W&=P_1\Delta t_1+P_2\Delta t_2\\&=(100)(2)\,\mathrm J+(300)(6)\,\mathrm J=2000\,\mathrm J\\\overline P&=\frac W{\Delta t_1+\Delta t_2}=\frac{2000\,\mathrm J}{8\,\mathrm s}=250\,\mathrm W\end{aligned}

[1]

What to take away

Sources for this explanation

Quanta · Updated October 1, 2026

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Change history

October 1, 2026Exposed average, instantaneous and driving power through question headings; clarified the steady efficiency rearrangement and linked work, torque and efficiency in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Revised the introduction and learning progression; added explanatory intermediate steps, an application example and a summary. Retained existing examples and all five exercises.

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Quanta. Mechanical Power. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/leistung-mechanisch

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Frequently asked questions about Mechanical Power

How do you calculate mechanical power?+

What does the formula

P = F\cdot{}v

mean and when do you use it?+

What is the difference between work and power?+

How do you convert horsepower to watts and where does the unit come from?+

How do you calculate power when climbing stairs or walking uphill?+