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Work and Power

Definition of Work

Terms and Formulae

Problems

Work, though easily defined mathematically, takes some explanation to grasp conceptually. In order to build an understanding of the concept, we begin with the most simple situation, then generalize to come up with the common formula.

The Simple Case

Consider a particle moving in a straight line that is acted on by a constant force in the same direction as the motion of the particle. In this very simple case, the work is defined as the product of the force and the displacement of the particle. Unlike a situation in which you hold something in place, exerting a normal force, the crucial aspect to the concept of work is that it defines a constant force applied over a distance. If a force F acts on a particle over a distance x , then the work done is simply:

W = Fx    

Since w increases as x increases, given a constant force, the greater the distance during which that force acts on the particle, the more work is done. We can also see from this equation that work is a scalar quantity, rather than a vector one. Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account.

What are the units of work? The work done by moving a 1 kg body a distance of 1 m is defined as a Joule. A joule, in terms of fundamental units, is easily calculated:

W = Fx = (m) =

The joule is a multipurpose unit. It serves not only as a unit of work, but also of energy. Also, the joule is used beyond the realm of physics, in chemistry, or any other subject dealing with energy.

In dynamics we were able to define a force conceptually as a push or a pull. Such a concise definition is difficult to generate when dealing with work. To give a vague idea, we can describe work as a force applied over a distance. If a force is to do work, it must act on a particle while it moves; it cannot just cause it to move. For instance, when you kick a soccer ball, you do no work on the ball. Though you produce a great deal of motion, you have only instantaneous contact with the ball, and can do no work. On the other hand, if I pick the ball up and run with it, I do work on the ball: I am exerting a force over a certain distance. In technical jargon, the "point of application" of the force must move in order to do work. Now, with a conceptual understanding of work, we can move on to define it generally.

The General Case

In the last section we came up with a definition of work given that the force acted in the same direction as the displacement of the particle. How do we calculate work if this is not the case? We simply resolve the force into components parallel and perpendicular to the direction of displacement of the particle (see Vectors, Component Method). Only the force parallel to the displacement does work on the particle. Thus, if a force is applied at an angle θ to the displacement of the particle, the resulting work is defined by:

W = (F cosθ)x    

This new equation has similar form to the old equation, but provides a more complete description. If θ = 0 , then cosθ = 1 and we have our first equation. Also, this equation ensures that it does not take into account any forces acting on a moving particle that do not do any work. Consider the normal force acting on a ball rolling across a horizontal floor. The normal force is perpendicular to the motion, implying that θ = 90 and cosθ = 0 . Thus there is no work done on the ball by the normal force. In this sense, work can be seen as produced by any force that aids or hinders the motion of the particle. Stationary forces and forces perpendicular to the motion do not cause work.

To see an example of work in a simple system, let us consider the work done by a gravitational force on a falling object. The gravitational force is simply mg , and let us denote the distance of the fall by h . Clearly if the object simply falls straight down, the work done is given by W = mgh . But what if the object falls at an angle θ from vertical, as seen below?

Figure %: An object falling at an angle
If the object falls the same height, then the distance traveled is given by x = . The work, then, is given by:

W = Fx cosθ = (mg)()(cosθ) = mgh

As long as the object falls h distance, the work done on the object falling at an angle is the same as if the object were falling straight down. This fact, special to gravity and other forces, is significant in the study of energy, but for now suffices to demonstrate how to calculate work.

Work is commonly misunderstood because of its common definition. Most people think that it takes a lot of work to hold a 100 pound weight in the air. The weight is not moving, though, so in the sense of physics no work is done. It is important to realize how our definition differs from a common one, and stick to the physical understanding of work. From this definition of work, we will be able to bring in a concept of energy, and greatly simplify many aspects of classical mechanics.

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