Physical Applications

Work

For moving objects, the quantity of work/time (power) is integrated along the trajectory of the point of application of the force. 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 known 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 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

\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=\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=\int _{C}F\,ds

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

W=\int _{C}F\,ds=F\int _{C}ds=Fs

where s is the displacement of 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=\int _{C}\mathbf {F} \cdot d\mathbf {s} =Fs\cos \theta .

When a force component is perpendicular to the displacement of the object (such as when a body moves in a circular path under a central force), no work is done, since the cosine of 90° 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.

Work done by a variable force

Calculating the work as "force times straight path segment" would only apply in the most simple of circumstances, as noted above. If force is changing, or if the body is moving along a curved path, possibly rotating and not necessarily rigid, then only the path of the application point of the force is relevant for the work done, and only the component of the force parallel to the application point velocity is doing work (positive work when in the same direction, and negative when in the opposite direction of the velocity). This component of force can be described by the scalar quantity called scalar tangential component ( $F cos(θ)$ , where $θ$ is the angle between the force and the velocity). And then the most general definition of work can be formulated as follows:

Work of a force is the line integral of its scalar tangential component along the path of its application point.

If the force varies (e.g. compressing a spring) we need to use calculus to find the work done. If the force is given by