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The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to energy—whenever work is done, energy is transferred.
For work, in the scientific sense, to be done, a force must be exerted and there must be displacement in the direction of the force.
Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the direction of motion times the distance through which the force acts . For one-way motion in one dimension, this is expressed in equation form as
where $W$ is work, $\mathbf{d}$ is the displacement of the system, and $\theta $ is the angle between the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$ , as in [link] . We can also write this as
To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the motion into one-way one-dimensional segments and add up the work done over each segment.
The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts . For one-way motion in one dimension, this is expressed in equation form as
where $W$ is work, $F$ is the magnitude of the force on the system, $d$ is the magnitude of the displacement of the system, and $\theta $ is the angle between the force vector $\mathbf{F}$ and the displacement vector $\mathbf{d}$ .
To examine what the definition of work means, let us consider the other situations shown in [link] . The person holding the briefcase in [link] (b) does no work, for example. Here $d=0$ , so $W=0$ . Why is it you get tired just holding a load? The answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the “briefcase-Earth system”—see Gravitational Potential Energy for more details). There must be displacement for work to be done, and there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase on level ground in [link] (c) does no work on it, because the force is perpendicular to the motion. That is, $\text{cos}\phantom{\rule{0.25em}{0ex}}\text{90}\text{\xba =}\phantom{\rule{0.25em}{0ex}}0$ , and so $W=0$ .
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