4.2 Kinetic energy and the work-energy theorem  (Page 8/6)

Work transfers energy

What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if a lawn mower is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on a briefcase by a person carrying it up stairs is stored in the briefcase-Earth system and can be recovered at any time. In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.

In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.

Net work and the work-energy theorem

We know from the study of Newton’s laws that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.

Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, net work    is the work done by the net external force ${\mathbf{F}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$ . In equation form, this is $W_{\mathrm{n}\mathrm{e}\mathrm{t}}=F_{\mathrm{n}\mathrm{e}\mathrm{t}}d$ .

Consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in [link] .

The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force ${\mathbf{F}}_{\text{app}}$ and the horizontal friction force $\mathbf{f}$ . Thus, as expected, the net force is parallel to the displacement and the net work is given by

The effect of the net force ${\mathbf{F}}_{\text{net}}$ is to accelerate the package from $v_0$ to $v$ . By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting $F_{\text{net}}=\text{ma}$ from Newton’s second law gives

To get a relationship between net work and the speed given to a system by the net force acting on it, we take $d=x-x_0$ and use the equation studied in "Motion Equations for Constant Acceleration in One Dimension" for the change in speed over a distance $d$ if the acceleration has the constant value $a$ ; namely, $v^2={v_0}^2+2\text{ad}$ (note that $a$ appears in the expression for the net work). Solving for acceleration gives $a=\frac{v^2-{v_0}^2}{2d}$ . When $a$ is substituted into the preceding expression for $W_{\text{net}}$ , we obtain