Example 1
F = 500 N, v = 20 m/s -> P = 10000 W.
Type force F [N] along the motion and speed v [m/s]. The calculator computes instantaneous power P = F·v. 15,000 N at 25 m/s is 375,000 W, or 375 kW. A bike at 200 N and 2 m/s is 400 W.
Force from this pair: F = P/v. Average from work: P = W/t. Force work: W = F·s.
Force F (N) and Speed v (m/s). The result shows up here.
Instantaneous mechanical power is force times speed when F runs along v. P = F·v. A newton times a meter per second is a watt. If the problem puts force at an angle, cos φ appears. Here φ = 0, the full product. Electrical P = U·I is also in watts, but the inputs differ: volts and amperes there.
Car: F = 15,000 N, v = 25 m/s (90 km/h). P = 375,000 W = 375 kW. That scale is a strong car on a straight, when the whole force goes into motion. Bike: 200 N and 2 m/s give 400 W, legs on a short climb. Lifting 1 kg at a steady 1 m/s: F ≈ 9.81 N, P ≈ 9.81 W.
Twice the speed at the same F is twice the watts. 500 N and 20 m/s is 10,000 W = 10 kW. 1000 N and 10 m/s is also 10,000 W: same product, different split. 800 N at 5 m/s is 4000 W. 3000 N at 15 m/s is 45 kW.
The field wants m/s, not km/h. 90 km/h = 25 m/s. If you type 90 as v with F = 15,000 N, you get 1.35 MW, a factor 3.6 too high. F and v must be positive: the model is “something pulls and it moves.” The result is in watts, with kilowatts beside it when the number is large.
This is power at this F and this v, not an average over a whole stretch. The average from energy and time is P = W/t. When you know kW and speed from a nameplate and want force, go to F = P/v. That page keeps the engine-power-kw address.
The chart of P(v) at fixed F is a straight line. On a car, air drag grows with v², so the real F grows too. Here you hold one F, as in a school problem. A comma and a period in 9.81 both parse.
P = F·v
F > 0, v > 0. Angle 0°. Unit of P: watt.
Instantaneous power at angle 0°: P = F·v. A car at 15000 N and 25 m/s is 375000 W, or 375 kW; a bike at 200 N and 2 m/s is 400 W.
F = 500 N, v = 20 m/s -> P = 10000 W.
F = 1000 N, v = 10 m/s -> P = 10000 W.
F = 200 N, v = 2 m/s -> P = 400 W.
F = 15000 N, v = 25 m/s -> P = 375000 W.
F = 50 N, v = 1 m/s -> P = 50 W.
F = 800 N, v = 5 m/s -> P = 4000 W.
F = 9.81 N, v = 1 m/s -> P ≈ 9.81 W.
F = 3000 N, v = 15 m/s -> P = 45000 W.
Power is 375,000 W, or 375 kW. The product of force and speed, car-on-a-highway order, not a bicycle.
You get 400 W. That is cyclist order: two hundred newtons of drag at two metres per second.
Force F [N], speed v [m/s]. Result P [W]; 1000 W is 1 kW.
Yes. P = F·v takes the force component in the direction of v. A force perpendicular to the velocity does no work of this kind.
Here you have instantaneous power from F and v. There you have an average from work and time. 400 W for 10 s is 4000 J on the W = P·t card.
On the force-from-engine-power page. Here the unknown is P, there it is F.
Yes. You are standing still, so the product F·v vanishes. Both fields still need a number before the calculator computes.
Yes. 2.5 and 2,5 are the same speed. A comma and a period mean the same value.
Power is 10,000 W, or 10 kW. A thousand newtons at ten metres per second.
On the work-of-a-force page. There you have distance instead of speed. Here you stay with P = F·v.
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.