Work Done by Forces that Vary In general, forces may vary in magnitude and direction at points in space, and paths between two points may be curved. The infinitesimal work done by a variable force can be expressed in terms of the components of the force and the displacement along the path, dW = F2dx + Fydy + F2dz. Here, the components of the force are functions of position along the path, and the displacements depend on the equations of the path. The total work is defined as a line integral, or the limit of a sum of infinitesimal amounts of work. The physical concept of work is straightforward: you calculate the work for tiny displacements and add them up. Sometimes the mathematics can seem complicated, but the following example demonstrates how cleanly they can operate. Work Done by a Variable Force over a Curved Path An object moves along a parabolic path y = (0.5 m-1) x2 from the origin A = (0, 0) to the point B= (2 m, 2 m) under the action of a force F = (5 N/m) yi + (10N/m) xj . Calculate the work done. y(m). (2. 2) y(x) F(x, y) (0, 0) ×(m) The parabolic path of a particle acted on by a given force. Strategy The components of the force are given functions of x and y. We can use the equation of the path to express y and dy in terms of x and dx; namely, y= (0.5m-1) z2 and dy = 2(0.5m-1)zda. Then, the integral for the work is just a definite integral of a function of x.
Solution The infinitesimal element of work is dW =F,dx+ F,dy=(5N/m) ydx +(10 N/m)xdy = (5 N/m) (0.5m 1) a2 dx + (10 N/m) 2 (0.5 m 1) x3 dx = (12.5 N/m2) x2 da. The integral of x2 is 2:3 /3, so O 2 m (12.5 N/m2) 2 de = (12.5 N/m2) 2.3 |2 ml 3 = (12.5 N/m2) 8 3 = 33.3 J. Significance This integral was not hard to do. You can follow the same steps, as in this example, to calculate line integrals representing work for more complicated forces and paths. In this example, everything was given in terms of x- and y-components, which are easiest to use in evaluating the work in this case. In other situations, magnitudes and angles might be easier. You saw in the example above that to evaluate a line integral, you could reduce it to an integral over a single variable or parameter. Usually, there are several ways to do this, which may be more or less convenient, depending on the particular case. In the example above, we reduced the line integral to an integral over x, but we could equally well have chosen to reduce everything to a function of y. We didn't do that because the functions in y involve the square root and fractional exponents, which may be less familiar, but for illustrative purposes, we do this now. Solving