Editing Work (physics) (section)

===Derivation for a particle moving along a straight line===
In the case the [[resultant force]] {{math|'''F'''}} is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration ''a'' along a straight line.<ref>{{cite web| url=http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html|title=Work–energy principle| website=www.wwu.edu| access-date=2012-08-06|archive-url=https://web.archive.org/web/20120530075449/http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Work/WorkEngergyTheorem.html|archive-date=2012-05-30|url-status=dead}}</ref> The relation between the net force and the acceleration is given by the equation {{math|1=''F'' = ''ma''}} ([[Newton's second law]]), and the particle [[displacement (vector)|displacement]] {{mvar|s}} can be expressed by the equation
<math display="block">s = \frac{v_2^2 - v_1^2}{2a}</math>
which follows from <math>v_2^2 = v_1^2 + 2as</math> (see [[Equations of motion]]).

The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains:
<math display="block"> W = Fs = mas = ma\frac{v_2^2-v_1^2}{2a} = \frac{1}{2} mv_2^2- \frac{1}{2} mv_1^2 = \Delta E_\text{k}</math>

Other derivation:
<math display="block"> W = Fs = mas = m\frac{v_2^2 - v_1^2}{2s}s = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 = \Delta E_\text{k}</math>

In the general case of rectilinear motion, when the net force {{math|'''F'''}} is not constant in magnitude, but is constant in direction, and parallel to the velocity of the particle, the work must be integrated along the path of the particle:
<math display="block"> W = \int_{t_1}^{t_2} \mathbf{F}\cdot \mathbf{v}dt = \int_{t_1}^{t_2} F \,v \, dt = \int_{t_1}^{t_2} ma \,v \, dt = m \int_{t_1}^{t_2} v \,\frac{dv}{dt}\,dt = m \int_{v_1}^{v_2} v\,dv = \tfrac12 m \left(v_2^2 - v_1^2\right) .</math>