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.

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.

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.

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.

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.