Average power measures energy transfer over a finite time interval. When the transfer rate changes continuously, the instantaneous power is

$$\boxed{P=\frac{dW}{dt}}.$$

For a force $\mathbf F$ acting at a point whose velocity is $\mathbf v$, the small work done during displacement $d\mathbf r$ is

$$dW=\mathbf F\cdot d\mathbf r.$$

Dividing by $dt$ gives

$$P=\mathbf F\cdot\frac{d\mathbf r}{dt},$$

so

$$\boxed{P=\mathbf F\cdot\mathbf v}.$$

Only the component of force along the instantaneous velocity transfers mechanical energy at that instant.

Sign of power

If force has a component in the direction of motion,

$$P>0,$$

and that force is transferring energy into the object's mechanical motion.

If the force opposes the motion,

$$P<0,$$

and it is removing mechanical energy from that motion.

If force and velocity are perpendicular,

$$P=0.$$

A force can therefore be large while doing no instantaneous work, as with an ideal force that only bends a trajectory without changing speed.

Worked example

A vehicle moves forward at

$$v=20,\mathrm{m/s}$$

while its drive force is

$$F=1500,\mathrm N$$

in the same direction. The mechanical power delivered by that force is

$$P=Fv=(1500)(20)=30{,}000,\mathrm W=30,\mathrm{kW}.$$

If an aerodynamic drag force of $500,\mathrm N$ acts backward at the same instant, its power is

$$P_{\rm drag}=-(500)(20)=-10,\mathrm{kW}.$$

The net mechanical power associated with these two forces is therefore

$$P_{\rm net}=20,\mathrm{kW}.$$

Connection to kinetic energy

The work-energy theorem gives

$$dW_{\rm net}=dK.$$

Taking the time derivative,

$$\boxed{P_{\rm net}=\frac{dK}{dt}}.$$

Thus net power tells how quickly kinetic energy is changing.

Instantaneous power depends on both force and velocity. The same force can deliver little power at low speed and much more power at high speed, which is why force and power impose different limits on real machines and vehicles.