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 the lawn mower in Figure 7.2(a) 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 the briefcase by the person carrying it up stairs in Figure 7.2(d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Figure 7.2(e). 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 in Dynamics: Force and Newton's Laws of Motion 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 ${\mathbf{F}}_{\text{net}}$. In equation form, this is $W_{\text{net}}=F_{\text{net}}d\text{cos}\theta$ where $\theta$ is the angle between the force vector and the displacement vector.

Figure 7.3(a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an $F\text{cos}\theta$ vs. $d$ graph. In this case, $F\text{cos}\theta$ is constant. You can see that the area under the graph is $Fd\text{cos}\theta$, or the work done. Figure 7.3(b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force $(F\text{cos}\theta {)}_{i(\text{ave})}$. The work done is $(F\text{cos}\theta {)}_{i(\text{ave})}{d}_i$ for each strip, and the total work done is the sum of $W_i$. Thus the total work done is the total area under the curve, a useful property to which we will refer later.

Figure 7.3 (a) A graph of $F\text{cos}\theta$ vs. $d$, when $F\text{cos}\theta$ is constant. The area under the curve represents the work done by the force. (b) A graph of $F\text{cos}\theta$ vs. $d$ in which the force varies. The work done for each interval is the area of each strip; thus, the total area under the curve equals the total work done.

Net work will be simpler to examine if we 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 Figure 7.4.

Figure 7.4 A package on a roller belt is pushed horizontally through a distance $\mathbf{d}$.

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 when 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, so that $\theta =\mathrm{0º}$ and $\text{cos}\theta =1$, 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 $v$. The kinetic energy of the package increases, indicating that the net work done on the system is positive (see Example 7.2). 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

The $d$ cancels, and we rearrange this to obtain

This expression is called the work-energy theorem, and it actually applies in general—even for forces that vary in direction and magnitude—although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity $\frac{1}{2}{\text{mv}}^2$. This quantity is our first example of a form of energy.

The quantity $\frac{1}{2}{\text{mv}}^2$ in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass $m$ moving at a speed $v$. Translational kinetic energy is distinct from rotational kinetic energy, which is considered later. In equation form, the translational kinetic energy,

is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects that move together.

We are aware that it takes energy to get an object, like a car or the package in Figure 7.4, up to speed; however, it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.