For moving objects, the quantity of work/time (power) is calculated. Thus, at any instant, the rate of the work done by a force (measured in joules/second, or watts) is the scalar product of the force (a vector), and the velocity vector of the point of application. This scalar product of force and velocity is classified as instantaneous power. Just as velocities may be integrated over time to obtain a total distance, by the fundamental theorem of calculus, the total work along a path is similarly the time-integral of instantaneous power applied along the trajectory of the point of application.

Work is the result of a force on a point that moves through a distance. As the point moves, it follows a curve X, with a velocity v, at each instant. The small amount of work δW that occurs over an instant of time dt is calculated as

δ W = F ⋅ d s = F ⋅ v d t {\displaystyle \delta W=\mathbf {F} \cdot d\mathbf {s} =\mathbf {F} \cdot \mathbf {v} dt}

where the F ⋅ v is the power over the instant dt. The sum of these small amounts of work over the trajectory of the point yields the work,

W = ∫ t 1 t 2 F ⋅ v d t = ∫ t 1 t 2 F ⋅ d s d t d t = ∫ C F ⋅ d s , {\displaystyle W=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot \mathbf {v} dt=\int _{t_{1}}^{t_{2}}\mathbf {F} \cdot {\tfrac {d\mathbf {s} }{dt}}dt=\int _{C}\mathbf {F} \cdot d\mathbf {s} ,}

where C is the trajectory from x(t1) to x(t2). This integral is computed along the trajectory of the particle, and is therefore said to be path dependent.

If the force is always directed along this line, and the magnitude of the force is F, then this integral simplifies to

W = ∫ C F d s {\displaystyle W=\int _{C}F\,ds}

where s is distance along the line. If F is constant, in addition to being directed along the line, then the integral simplifies further to

W = ∫ C F d s = F ∫ C d s = F s {\displaystyle W=\int _{C}F\,ds=F\int _{C}ds=Fs}

where s is the distance travelled by the point along the line.

This calculation can be generalized for a constant force that is not directed along the line, followed by the particle. In this case the dot product F ⋅ ds = F cos θ ds, where θ is the angle between the force vector and the direction of movement, that is

W = ∫ C F ⋅ d s = F s cos θ . {\displaystyle W=\int _{C}\mathbf {F} \cdot d\mathbf {s} =Fs\cos \theta .}

In the notable case of a force applied to a body always at an angle of 90° from the velocity vector (as when a body moves in a circle under a central force), no work is done at all, since the cosine of 90 degrees is zero. Thus, no work can be performed by gravity on a planet with a circular orbit (this is ideal, as all orbits are slightly elliptical). Also, no work is done on a body moving circularly at a constant speed while constrained by mechanical force, such as moving at constant speed in a frictionless ideal centrifuge.