9.2.1 Work

Now let’s turn to the integral on the right-hand side of Equation 9.6. This integral is telling us by how much the kinetic energy changes due to the force. That is, it is the energy transferred to or from the system by the force. Earlier we said that a process that transfers energy to or from a system by mechanical means—by forces—is called work. So the integral on the right-hand side of Equation 9.6 must be the work W done by force $\vec{F}$ .

Having identified the left side of Equation 9.6 with the changing kinetic energy of the system and the right side with the work done on the system, we can rewrite Equation 9.6 as

(9.9)

The equation for the energy principle of a one-particle system.

Full Alternative Text

This is our first version of the energy principle. Notice that it’s a cause-and-effect statement: The work done on a one-particle system causes the system’s kinetic energy to change.

We’ll study work thoroughly in the next section, but for now we’re considering only the simplest case of a constant force parallel to the direction of motion (the s-axis). A constant force can be factored out of the integral, giving

\begin{array}{l} W & = \int_{s_{i}}^{s_{f}} F_{s} d s = F_{s} {\int_{s_{i}}^{s_{f}} d s = F_{s} s |}_{s_{i}}^{s_{f}} = F_{s} (s_{f} - s_{i}) \\ = F_{s} Δ s \end{array}

(9.10)

The unit of work, that of force multiplied by distance, is the $N m$ . Recall that $1 N = 1 kg m / s^{2}$ . Thus

1 N m = 1 (kg m / s^{2}) m = 1 {kg m}^{2} {/s}^{2} = 1 J

Thus the unit of work is really the unit of energy. This is consistent with the idea that work is a transfer of energy. Rather than use N m we will measure work in joules.

EXAMPLE 9.1 Firing a cannonball

A 5.0 kg cannonball is fired straight up at $35 m/s$ . What is its speed after rising 45 m?

MODEL Let the system consist of only the cannonball, which we model as a particle. Assume that air resistance is negligible.

VISUALIZE FIGURE 9.3 is a before-and-after pictorial representation. Notice two things: First, for problem solving we use numerical subscripts instead of the generic i and f. Second, the v symbols are speeds, not velocities, so there’s no x or y in the subscripts. Before-and-after representations are usually simpler than the pictorial representation used in dynamics problems, so you can include known information right on the diagram instead of making a Known table.

The figure shows a before-and-after representation of a cannonball fired straight up.

FIGURE 9.3 Full Alternative Text

SOLVE Using work and energy is not the only way to solve this problem. You could solve it as a free-fall problem. Or you might have previously learned to solve problems like this using potential energy, a topic we’ll take up in the next chapter. But using work and energy emphasizes how these two key ideas are related, and it gives us a simple example of the problem-solving process before we get to more complex problems. The energy principle is $Δ K = W$ , where work is done by the force of gravity. The cannonball is rising, so its displacement $Δ y = y_{1} - y_{0}$ is positive. But the force vector points down, with component $F_{y} = - m g$ . Thus gravity does work

W = F_{y} Δ y = - m g Δ y = - (5.0 kg)(9.80 {m/s}^{2}) (45 m) = - 2210 J

as the cannonball rises 45 m. A negative work means that the system is losing energy, which is what we expect as the cannonball slows.

The cannonball’s change of kinetic energy is $Δ K = K_{1} - K_{0}$ . The initial kinetic energy is

K_{0} = \frac{1}{2} m v_{0}^{2} = \frac{1}{2} (5.0 {kg)(35 m/s)}^{2} = 3060 J

Using the energy principle, we find the final kinetic energy to be $K_{1} = K_{0} + W = 3060 J - 2210 J = 850 J$ . Then

v_{1} = \sqrt{\frac{2 K_{1}}{m}} = \sqrt{\frac{2 (850 J)}{5.0 kg}} = 18 m / s

REVIEW $35 m/s \approx 70 mph$ . A cannonball fired upward at that speed is going to go fairly high. To have lost half its speed at a height of $45 m \approx 150 ft$ seems reasonable.