Example 1
A kettle delivers 300,000 J to the water in 150 s. The quotient is 2,000 W, or 2 kW. This is the reference example on the page.
P = 2,000 W
Power from Work and Time Calculator
Enter the work in joules and the time in seconds. The calculator finds average power from P = W/t and also reports watts, watt-hours, and the power at half the time and at twice the time. For 300,000 J and 150 s the result is 2,000 W, or 2 kW. The model does not include electric power U·I or device efficiency.
Result
Enter the work and a time greater than zero. Power, watt-hours, and the P(t) chart will appear here.
Work from a constant force is on the W = F·s calculator, circuit power is on the P = U·I calculator, and time from the same definition is on the t = W/P calculator.
How it works
W300,000 J
t150 s
P2,000 W
Average power for a steady transfer of energy is the quotient P = W/t. The unit is the watt: 1 W = 1 J/s. For 300,000 J and 150 s: 300,000 ÷ 150 = 2,000 W, or 2 kW. The same 300,000 J in 300 s gives 1,000 W, and in 75 s it gives 4,000 W. At zero time the quotient is not defined.
The calculator uses work W in joules and time t in seconds. A comma and a period are the same separator. A time of zero or a negative time is rejected. Zero work gives a power of 0 W. Negative work means energy leaving the system, so the power comes out negative.
What the result panel shows. Besides the power in kilowatts or watts, the calculator reports the same value in watts, the entered work, the entered time, the same energy in watt-hours, and the power at half the time and at twice the time. For 300,000 J and 150 s that is 2,000 W, or 2 kW, and the energy is 83.33 Wh. Half the time, 75 s, gives 4 kW. Twice the time, 300 s, gives 1 kW. The chart draws five bars of the same work at 0.25 t, 0.5 t, t, 2 t, and 4 t. For these numbers those bars are 8,000 W, 4,000 W, 2,000 W, 1,000 W, and 500 W.
Limits of the model. The formula is the average over the whole interval t. The calculator does not evaluate the derivative dW/dt or an efficiency. There is no voltage and no current: circuit power P = U·I is on the P = U·I page. There is no force and no distance: work W = F·s is on the mechanical-work page. Time from the same three numbers, t = W/P, is on the time-from-work-and-power page. The product W = P·t is on the engine-work page. Power from force and velocity, P = F·v, is on the P = F·v page.
The unit switch in the header converts the work field from joules to ft·lb and converts the number already entered. Time stays in seconds. In US mode, at 2,000 W, the panel adds 2.682 hp, because it divides watts by 745.699872. That is US mechanical horsepower. The calculator does not use the metric horsepower of 735.5 W. Values in the text stay in SI units.
The examples on the page use the same quotient. 300,000 J in 150 s gives 2,000 W. The Fill this example button enters those numbers into the fields and refreshes the panel and the chart. The fields start empty.
Formula
P = W / t
[P] = J / s = W
Model: average power, the rate of doing work. Textbooks write P = W/Δt (OpenStax College Physics 2e, section 7.7; Young and Freedman). The watt is named for James Watt (1736–1819), who compared engines with horsepower. The quotient itself is not a theorem with a separate surname: it is the definition. Order of calculation: the quotient W/t in watts, then the kW form, the same energy in Wh, and the power at t/2 and at 2t.
Theory, derivation, and domain
Definition and the watt in SI
Average power over a time t is P = W/t. It is the energy transferred per unit time, not a measure of effort. The SI unit is the watt: 1 W = 1 J/s = 1 kg·m²/s³. The definition is in the SI Brochure (BIPM, 9th edition). In the United States, NIST SP 330 describes the same SI units.
For 300,000 J and 150 s the order is short. Step 1: work W = 300,000 J. Step 2: time t = 150 s. Step 3: the quotient P = 300,000 / 150 = 2,000 W = 2 kW. Those 300,000 J are 83.33 Wh, because 1 Wh = 3,600 J, and 0.08333 kWh, because 1 kWh = 3,600,000 J. The same work in 75 s gives 4,000 W, and in 300 s it gives 1,000 W, because time sits in the denominator.
Inverse proportion
At fixed work, doubling the time halves the power, and halving the time doubles the power. That is the hyperbola P(t) = W/t, not a straight line. The five bars on the chart stand at 0.25 t, 0.5 t, t, 2 t, and 4 t, so for 150 s the times are 37.5 s, 75 s, 150 s, 300 s, and 600 s, and the powers are 8,000 W, 4,000 W, 2,000 W, 1,000 W, and 500 W.
The watt, horsepower, and James Watt
James Watt introduced horsepower in order to compare engines with the work of a horse. SI later adopted the watt as the name of the unit of power. In US mode this calculator divides watts by 745.699872 and adds hp. At 2,000 W that is 2.682 hp. The calculator does not use the metric horsepower of 735.5 W and does not convert the result with that constant.
Force, velocity, and other forms of the same power
When the force is constant and parallel to the velocity, instantaneous power is often written P = F·v. If 2,000 J of work came from 200 N · 10 m and lasted 10 s, the quotient is 200 W, and the product F·v at v = 1 m/s is also 200 W. This form does not take force or velocity, so it does not check that equality. The work and the time are entered directly. The P = F·v page computes that product from its own fields.
Outside the formula
The formula does not cover power that changes with time, except that it reports the average over the whole interval. It does not cover efficiency: if only part of a kettle’s energy reaches the water, that part has to be entered as the work, not a default fraction. It also does not cover mains voltage or current. The calculator does not apply a relativistic correction.
How to use
- Enter the work in joules, for example 300000. After the header is switched to US, the field takes ft·lb. Kilowatts of the result are not an input of this field.
- Enter the time in seconds, for example 150. Zero and a negative number will not pass, because the calculator divides by t.
- Click Calculate or wait for the live result. For 300,000 J and 150 s the calculator shows 2 kW, which is 2,000 W.
- Check the panel: watts, work, time, watt-hours, and the power at half the time and at twice the time. The chart is five bars of the same work.
- Clearing the form removes the entered numbers and hides the panel. The examples enter ready pairs, including 300,000 J and 150 s, 15,000 J and 60 s, and 36,000 J and 3,600 s.
Symbol glossary
P is the average power, W is the work, and t is the time. For 300,000 J and 150 s the result is 2,000 W.
- P
- Average power [W], P = W/t. The panel also shows kW when the value is at least 1,000 W.
- W
- Work [J], the first field. It is not the watt of the result. 2 kW appears after the division, not as a number in this field.
- t
- Time [s]. Zero and a negative value are rejected. The field stays in seconds after the header is switched to US.
- wat
- The SI unit of power. 1 W = 1 J/s. 2,000 W = 2 kW.
- Wh
- Watt-hour, 1 Wh = 3,600 J. For 300,000 J the panel shows 83.33 Wh.
- hp
- US mechanical horsepower, 745.699872 W. The panel adds it only after the header is switched to US. The constant 735.5 W is not used.
Worked examples
Example 2
A travel kettle at a campsite: 120,000 J in 180 s. The quotient is 666.67 W. The same tea, a lower average power.
P = 666.67 W
Example 3
An elevator lifts the car one floor: 50,000 J in 20 s. The quotient is 2,500 W, or 2.5 kW.
P = 2,500 W
Example 4
A cyclist on a climb: 15,000 J in 60 s. The quotient is 250 W.
P = 250 W
Example 5
A vacuum cleaner for ten minutes: 900,000 J in 600 s. The quotient is 1,500 W, or 1.5 kW.
P = 1,500 W
Example 6
An LED lamp for one hour: 36,000 J in 3,600 s. The quotient is 10 W.
P = 10 W
Example 7
A hair dryer for five minutes: 600,000 J in 300 s. The quotient is 2,000 W, the same 2 kW as the kettle, from a different pair of numbers.
P = 2,000 W
Example 8
A tire pump: 800 J in 40 s. The quotient is 20 W.
P = 20 W
Example 9
A winch lifts a load: 12,000 J in 15 s. The quotient is 800 W.
P = 800 W
Example 10
A run up a short hill: 20,000 J in 50 s. The quotient is 400 W.
P = 400 W
Example 11
A mixer for three minutes: 54,000 J in 180 s. The quotient is 300 W.
P = 300 W
Example 12
A crate lifted onto a shelf: 2,000 J in 5 s. The quotient is 400 W. A short time raises the power when the work is small.
P = 400 W
Common questions
What does this calculator compute, and from which formula?
The calculator finds average power from P = W/t. It divides the entered work by the entered time. For 300,000 J and 150 s the result is 2,000 W, or 2 kW.
Which units should I enter for work and time?
Enter work in joules and time in seconds. The result is in watts. 1,000 W is 1 kW, and 1 W is 1 J/s. After the header is switched to US units, work is in ft·lb, time stays in seconds, and the panel adds hp using the constant 745.699872 W. A comma and a period are the same decimal separator. Convert kilowatt-hours to joules first: 1 kWh = 3,600,000 J.
Why does the same work in less time give a higher power?
Because time is in the denominator. At constant work, power is inversely proportional to time. 300,000 J in 150 s is 2 kW, in 75 s it is 4 kW, and in 300 s it is 1 kW. The chart shows that relation as five bars.
How is this result different from P = U·I?
Here power comes from work and time. Electric power is the product of voltage and current. Both results are in watts, and the fields are different. 2 kW on the circuit page needs volts and amperes, not 300,000 in the work field.
How do I reverse the calculation when I know the power or I need the time?
This calculator computes only P. Time t = W/P is on the time-from-work-and-power page. The product W = P·t is on the engine-work page. All three statements are the same definition, solved for a different quantity. This form has no power field on the input.
What do zero and a minus sign mean?
A time of zero or a negative time is rejected, because division by zero is not defined. Zero work gives 0 W. Negative work gives negative power: energy leaves the system. Both fields need a number before the panel appears. An empty field is not a zero.
Does the result include the efficiency of a device?
No. The quotient uses the work and the time you enter. If only part of the energy is useful, enter that part in the work field. The calculator has no efficiency field and does not multiply the result by η.
Where do I get the work if I know a force, a mass, or heat?
From the problem, or from a neighboring calculator. The product F·s is on the mechanical-work page. Energy from speed is on the kinetic-energy page. On a lift the work is often close to m g h, but this form does not take mass or height. Heat is also measured in joules, so it can go in the work field when that energy is what you divide by time.
What do the panel and the chart show?
The panel reports the power, the same value in watts, the work, the time, the watt-hours, and the power at half the time and at twice the time. For 300,000 J and 150 s that is 2 kW, or 2,000 W, and 83.33 Wh. Half the time gives 4 kW, and twice the time gives 1 kW. The chart draws bars at 0.25 t, 0.5 t, t, 2 t, and 4 t. The example button enters the numbers and refreshes the panel and the chart. The fields start empty.
Does the formula describe instantaneous power?
No. P = W/t is the average over the whole interval. A peak in the middle of the interval can be higher, and the result is still the total work divided by the total time. Power that changes continuously is the derivative dW/dt, not this quotient.
How do I calculate power with P = W/t?
Divide work in joules by time in seconds. For 300,000 J and 150 s: 300,000/150 = 2,000 W.
What is P for 300,000 J done in 150 s?
2,000 W, or 2 kW. With the same W in 75 s the result would be 4,000 W, because time is in the denominator.
How do I convert minutes to seconds before the formula?
Multiply minutes by 60. 2.5 min is 150 s. The time field expects seconds, not minutes.
How is P = W/t different from P = U·I?
Here power comes from work and time. P = U·I is electric power from voltage and current. This form has no U or I fields.
Does the calculator solve for work or time from a given power?
Not with this field layout. You enter W and t, and the result is P. The rearranged forms W = P·t and t = W/P are on companion pages.
Sources
The definition of average power and the watt come from physics and metrology sources. The page does not invent publication identifiers.
- OpenStax College Physics 2e — Power (section 7.7)
- Wikipedia — Power (physics)
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 330 — The International System of Units
- Young, H. D.; Freedman, R. A., University Physics, chapter on work and kinetic energy (average power P = W/Δt).
Page updated November 3, 2026.