relation between Integration and work?

Question

Can someone tell me why it's possible to use integration when dealing with work?

Answer 1

Work done is usually the area under a graph of force vs distance or something like that (cant remember what the labels for the axis should be i havent touched physics in a long time). Using integration you can calculate the area under a graph within certain boundaries and hence calculate the work done

Answer 2

It's not merely "possible," it is in general ABSOLUTELY ESSENTIAL !

The infinitesimal work done when a force f(x) moves something (a particle or an object) through an infinitesimal distance dx is f(x) dx.

Expressed in this way, the force when the particle or object is at the position ' x ' could be ANYTHING, ranging from possibly zero though a non-zero constant to a quite general function of position ' x '. No matter what it is, in this general formulation, the work done in moving the particle or object from x = x1 to x = x2 (through a FINITE CHANGE in x) is:

Integral (from x1 to x2) of f(x) dx.

If that is POSITIVE, work is done ON the particle or object. If it's NEGATIVE, work is done BY the particle or object.

Note : changing the x-coordinate may not be the only thing that happens in the motion. In general the work done could involve more complicated integrals such as the integral of f dx + g dy + h dz between appropriate limits for x, y and z, where (f, g, h) [all functions of (x, y, z)] are the components of a total force vector, with all of them being functions of position variables (x, y, z).

Nor are we liimited to such perpendicular components or variables to denote both the "position" and/or "force" terms. In many problems, work may be done by ANGULARLY accelerating something. In that case the work done involves a TORQUE moving something through infinitesimal and then finite ANGLES, so that the integral then looks like Integral (from theta1 to theta2) of F_theta d(theta).

Finally, in Lagrangian or Hamiltonian methods, yet more varied or even abstract GENERALISED COORDINATES can be employed.

But the general idea is always the same. To move an object or a complicated system through a small change in some coordinate q involves some infinitesimal work (F_q) dq, and thus to move it through a finite change requires evaluation of the corresponding integral between the relevant limits (from the beginning to the end of the motion).

Live long and propser.

Answer 3

Renesis is pretty much right. Work is, by definititon, force applied through a distnce. So if you label the x-axis as distance and the y-axis as force (the 'usual' choice), the area under the curve is proportional to the total work done.

Hope that helps.


Doug