## Work Done by a Variable Force

Integration is used to calculate the work done by a variable force.

### Learning Objectives

Describe approaches used to calculate work done by a variable force

### Key Takeaways

#### Key Points

- The work done by a constant force of magnitude F on a point that moves a displacement d in the direction of the force is the product: W = Fd.
- Integration approach can be used both to calculate work done by a variable force and work done by a constant force.
- The SI unit of work is the joule; non- SI units of work include the erg, the foot-pound, the foot-poundal, the kilowatt hour, the litre-atmosphere, and the horsepower-hour.

#### Key Terms

**work**: A measure of energy expended in moving an object; most commonly, force times displacement. No work is done if the object does not move.**force**: A physical quantity that denotes ability to push, pull, twist or accelerate a body, which is measured in a unit dimensioned in mass × distance/time² (ML/T²): SI: newton (N); CGS: dyne (dyn)

### 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 [latex]\Delta \text{x}[/latex] in the direction of the force is simply the product

[latex]\text{W}=\text{F}\cdot \Delta \text{x}[/latex]

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:

[latex]\mathbf{\text{F}_{\text{s}}}=-\text{k}\mathbf{\text{x}}[/latex]

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:

[latex]\text{W}_{\text{s}}=\int_0^\text{t}\mathbf{\text{F}_{\text{s}}}\cdot\mathbf{\text{v}}\text{dt} =\int_0^\text{t} -\text{kx} \hspace{3 pt} \text{v}_\text{x} \text{dt} =\int_{\text{x}_\text{o}}^\text{x} -\text{kx} \hspace{3 pt} \text{dx}= -\frac{1}{2}\text{k}\Delta \text{x}^2[/latex]

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:

[latex]\text{W}_{\text{a}}=\int_0^\text{t}\mathbf{\text{F}_{\text{a}}}\cdot\mathbf{\text{v}}\text{dt} =\int_0^\text{t}-\mathbf{\text{F}_{\text{s}}}\cdot\mathbf{\text{v}}\text{dt} = \frac{1}{2}\text{k}\Delta \text{x}^2[/latex]

in this relation [latex]\mathbf{\text{F}_{\text{a}}}[/latex] is the force acted upon spring by the object. [latex]\mathbf{\text{F}_{\text{a}}}[/latex] and [latex]\mathbf{\text{F}_{\text{s}}}[/latex] are in fact action- reaction pairs; and [latex]\mathbf{\text{W}_{\text{a}}}[/latex] 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:

[latex]\text{W}=\int_\text{a}^\text{b}{\text{P}}\text{dV}=\text{P}\int_\text{a}^\text{b} \text{dV}=\text{P} \Delta \text{V}[/latex]

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):

[latex]\text{W}=\int_{\text{t}_1}^{\text{t}_2}\mathbf{\text{F}}\cdot\mathbf{\text{v}}\text{dt} = \int_{\text{t}_1}^{\text{t}_2}\text{mg} \hspace{3 pt} \text{v}_\text{y} \text{dt} = \text{mg} \int_{\text{y}_1}^{\text{y}_2} \text{dy}=\text{mg}\Delta \text{y}[/latex]

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.