## Using Integration to Calculate the Work Done by Variable Forces

A force is said to do work when it acts on a body so that there is a displacement of the point of application in the direction of the force. Thus, a force does work when it results in movement.

The work done by a constant force of magnitude F on a point that moves a displacement \(\mathrm{Δx}\) in the direction of the force is simply the product

\[\mathrm{W=F⋅Δx}\]

In the case of a variable force, integration is necessary to calculate the work done. For example, let’s consider work done by a spring. According to the Hooke’s law the restoring force (or spring force) of a perfectly elastic spring is proportional to its extension (or compression), but opposite to the direction of extension (or compression). So the spring force acting upon an object attached to a horizontal spring is given by:

\[\mathrm{Fs=−kx}\]

that is proportional to its displacement (extension or compression) in the x direction from the spring’s equilibrium position, but its direction is opposite to the x direction. For a variable force, one must add all the infinitesimally small contributions to the work done during infinitesimally small time intervals dt (or equivalently, in infinitely small length intervals dx=v_{x}dt). In other words, an integral must be evaluated:

\[\mathrm{W_s=\int_0^tFs⋅vdt=\int_0^t−kxv_xdt=\int_{x_o}^x−kxdx=−\dfrac{1}{2}kΔx^2}\]

This is the work done by a spring exerting a variable force on a mass moving from position x_{o} to x (from time 0 to time t). The work done is positive if the applied force is in the same direction as the direction of motion; so the work done by the object on spring from time 0 to time t, is:

\[\mathrm{W_a=\int_0^tFa⋅vdt=\int_0^t−F_s⋅vdt=\dfrac{1}{2}kΔx^2}\]

in this relation \(\mathrm{F_a}\) is the force acted upon spring by the object. \(\mathrm{F_a}\) and \(\mathrm{F_s}\) are in fact action- reaction pairs; and \(\mathrm{W_a}\) is equal to the elastic potential energy stored in spring.

### Using Integration to Calculate the Work Done by Constant Forces

The same integration approach can be also applied to the work done by a constant force. This suggests that *integrating *the product of force and distance is the general way of determining the work done by a force on a moving body.

Consider the situation of a gas sealed in a piston, the study of which is important in Thermodynamics. In this case, the Pressure (Pressure =Force/Area) is constant and can be taken out of the integral:

\[\mathrm{W=\int_a^bPdV=P \int_a^bdV=PΔV}\]

Another example is the work done by gravity (a constant force) on a free-falling object (we assign the y-axis to vertical motion, in this case):

\[\mathrm{W=\int_{t_1}^{t_2} F⋅vdt=\int_{t_1}^{t_2} mgv_ydt=mg \int_{y_1}^{y_2} dy=mgΔy}\]

Notice that the result is *the same* as we would have obtained by simply evaluating the product of force and distance.

### Units Used for Work

The SI unit of work is the joule (J), which is defined as the work done by a force of one newton moving an object through a distance of one meter.

Non-SI units of work include the erg, the foot-pound, the foot-pound, the kilowatt hour, the liter-atmosphere, and the horsepower-hour.