Physics · Mechanics

Mechanical Power

Power states how fast work is done: the work performed divided by the time needed.

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Formula

P = \frac{W}{t}

LaTeX: P = \frac{W}{t}

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

Variables & units – Mechanical Power

Symbol	Meaning	Unit
P	Power	W (Watt)
W	Work done	J
t	Time needed	s

Derivation & background – Mechanical Power

The unit watt commemorates James Watt, who compared steam engines with the pulling power of horses, which is where the old unit horsepower comes from (1 hp(M) = 735.5 W). For a constant force along the direction of motion, P = F·v also holds, useful for driving resistances. The same formula P = E/t links mechanical and electric power.

Exam blueprint

Validity range

P = W/t gives the average power over the interval t. The instantaneous power is P = F·v and fluctuates when force or speed change.

Derivation steps

Power measures how fast work is done, work per time interval.

1Definition: P = W/t with W in joules and t in seconds.
2With W = F·s this becomes P = F·(s/t) = F·v.

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 = W/P = 6,000/200 = 30 s.

A car sustains v = 15 m/s with a driving force of F = 400 N. What is its power?

P = F·v = 400 × 15 = 6,000 W = 6 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 = 3.6×10⁶ J).

Exam context

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

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Formula cluster

Power

Mechanical and electric power follow the same definition P = E/t.

W = F \cdot s \cdot cos (α)

Mechanical Work

P = U \cdot I

Electric Power

E_{kin} = \frac{1}{2} m v^{2}

Kinetic Energy

Worked example

Climbing stairs, a person performs W = 3,000 J of lifting work in t = 60 s: P = 3,000/60 = 50 W.

Applications

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

Quanta exam set

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Which formula describes Mechanical Power?

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Question (front)

How do you rearrange P = W/t for Work from power and time?

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Question (front)

Which common mistake happens with Mechanical Power?

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Scientific sources

[1]Tipler, P.A., & Mosca, G. (2015). Physik für Wissenschaftler und Ingenieure. Springer Spektrum.

Common notations & search queries

P=W/tP = W durch tP=F*vLeistung Formel Physikmechanische Leistung berechnenWatt berechnenpower formulaPS in Watt

Related formulas

W = F \cdot s \cdot cos (α)

Mechanical Work

P = U \cdot I

Electric Power

E_{kin} = \frac{1}{2} m v^{2}

Kinetic Energy

More Physics formulas

F = m \cdot a

Newton's Second Law

F_{G} = G \cdot \frac{m _{1} \cdot m _{2}}{r ^{2}}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^{2}

Kinetic Energy

E_{p o t} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Frequently asked questions about Mechanical Power

How do you calculate mechanical power?+

Divide the work done by the time needed: P = W/t. Work is in joules, time in seconds, and the result in watts. Example: climbing stairs, a person does 3,000 J of lifting work in 60 s, so P = 50 W. If she runs up the same stairs in 20 s, the work is the same but the power is 150 W. That is exactly the core of the power concept: not how much work is done, but how fast. For moving vehicles the variant P = F·v is practical when driving force and speed are known. Always convert minutes to seconds.

What does the formula P = F·v mean and when do you use it?+

It is the instantaneous form of power: from P = W/t and W = F·s follows P = F·(s/t) = F·v, valid when the force acts along the direction of motion. It answers questions like: what power does a car need to overcome the driving resistance at constant speed? Example: F = 400 N of total resistance at v = 15 m/s gives P = 6,000 W = 6 kW. The formula also explains why consumption rises at high speed: air drag grows quadratically with v, so the required power grows cubically; double the speed demands eight times the power against air drag. On inclines the climbing power m·g·v_vertical is added.

What is the difference between work and power?+

Work measures the amount of energy transferred, in joules; power measures how fast this transfer happens, in watts = joules per second. The same work can be done at very different power levels: lifting 100 kg by 10 m always costs about 9,810 J; a crane manages it in 5 s (1,962 W), a pulley with muscle power in 2 minutes (82 W). Everyday language often mixes the two, for instance meaning "lots of power" while counting energy. Rule of thumb: if a time appears in the problem or "how fast" is asked, it is about power; if it is "how much in total", it is about work or energy.

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

One (metric) horsepower equals 735.5 W, so: hp × 735.5 = power in watts. A 100 hp car therefore delivers about 73.5 kW; conversely divide kilowatts by 0.7355 to get hp. The unit goes back to James Watt, who wanted to make the output of his steam engines tangible for customers and compared it with the sustained output of brewery horses, defined as lifting 75 kg by 1 m in 1 s, i.e. 75 × 9.81 ≈ 736 W. In the SI system only the watt is valid, but hp still appears in car brochures. In exams the conversion is a popular bonus step.

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

What counts is the lifting work against gravity: W = m·g·h with the vertical height difference h, not the slanted path length. The power follows from P = m·g·h/t. Example: a 70 kg person climbs stairs with a 5 m height difference in 30 s: P = 70 × 9.81 × 5 / 30 ≈ 114 W. Athletic people manage several hundred watts briefly; professional cyclists sustain about 400 W uphill for an hour. The horizontal part of the path ideally costs no lifting work; in reality friction and muscle efficiency add to it. This problem is an exam classic because it tests work, power and energy conversion in a realistic context.

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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 = W/P = 6,000/200 = 30 s.
2
Task
A car sustains v = 15 m/s with a driving force of F = 400 N. What is its power?
Solution path
P = F·v = 400 × 15 = 6,000 W = 6 kW.