Vector Calculus, Kinetic Energy

Kinetic Energy

Let a particle with mass m move through a force field f along a path p, without thrust or friction. The speed of the particle is affected only by the field. Think of a marble that gathers speed as it rolls down hill; or think of a marble that loses speed as it rolls up hill, having been given an initial push. A roller coaster starts at the top of a large hill and gains and loses and gains and loses speed as it moves with and against the force of gravity. Whatever the model, the particle traces the path and is pushed or pulled by the force field. We would like to compute the change in kinetic energy.

Start with Newton's law, which says force = mass times accelleration. Write this as f = ma. We're only interested in the force tangent to the path. The perpendicular component tries to push the particle off the track, or out of the wire, or whatever. There's no friction, so I'm not going to worry about that. The tangential component of force is the dot product of f and the direction vector, or f.p′/|p′|. By Newton's law, this is the same as mass times tangential accelleration, or mass times the change in speed. Let s(t) indicate speed as a function of time, hence f.p′/s = m×s′. Multiply through by speed and obtain f.p′ = m×s×s′. Integrate the left side and get the line integral. Integrate the right side and get ½ms². This is the formula for kinetic energy. Therefore the kinetic energy gained or lost by a particle as it moves (without friction or thrust) through a force field is given by the line integral.