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Check Your Understanding Can Earth’s gravity ever be a constant force for all paths?

No, it’s only approximately constant near Earth’s surface.

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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,

d W = F x d x + F y d y + F z d z .

Here, the components of the force are functions of position along the path, and the displacements depend on the equations of the path. (Although we chose to illustrate dW in Cartesian coordinates, other coordinates are better suited to some situations.) [link] defines the total work 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 ) x 2 from the origin A = ( 0 , 0 ) to the point B = ( 2 m, 2 m ) under the action of a force F = ( 5 N/m ) y i ^ + ( 10 N/m ) x j ^ ( [link] ). Calculate the work done.

A graph of y in meters versus x in meters is shown. A parabolic path labeled as y of x starts at 0, 0 and curves up and to the right. The point (2, 2) is on the parabola. Vector F of x, y is shown at a point between the origin and coordinate 2, 2. Vector F points to the right and up, at some angle to the curve y of x.
The parabolic path of a particle acted on by a given force.


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 = ( 0.5 m −1 ) x 2 and d y = 2 ( 0.5 m −1 ) x d x .

Then, the integral for the work is just a definite integral of a function of x .


The infinitesimal element of work is

d W = F x d x + F y d y = ( 5 N/m ) y d x + ( 10 N/m ) x d y = ( 5 N/m ) ( 0.5 m 1 ) x 2 d x + ( 10 N/m ) 2 ( 0.5 m 1 ) x 2 d x = ( 12.5 N/m 2 ) x 2 d x .

The integral of x 2 is x 3 / 3 , so

W = 0 2 m ( 12.5 N/m 2 ) x 2 d x = ( 12.5 N/m 2 ) x 3 3 | 0 2 m = ( 12.5 N/m 2 ) ( 8 3 ) = 33.3 J .


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.

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Check Your Understanding Find the work done by the same force in [link] over a cubic path, y = ( 0.25 m −2 ) x 3 , between the same points A = ( 0 , 0 ) and B = ( 2 m, 2 m ) .

W = 35 J

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You saw in [link] 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 [link] , 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 for x and dx , in terms of y , along the parabolic path, we get

x = y / ( 0.5 m −1 ) = ( 2 m ) y and d x = ( 2 m ) × 1 2 d y / y = d y / ( 2 m −1 ) y .

The components of the force, in terms of y , are

F x = ( 5 N/m ) y and F y = ( 10 N/m ) x = ( 10 N/m ) ( 2 m ) y ,

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