第7章仕事と運動エネルギー

The simplest work to evaluate is that done by a force that is constant in magnitude and direction. In this case, we can factor out the force; the remaining integral is just the total displacement, which only depends on the end points A and B, but not on the path between them:

Figure 7.3(a) shows a person exerting a constant force $\overrightarrow{F}$ along the handle of a lawn mower, which makes an angle $\theta$ with the horizontal. The horizontal displacement of the lawn mower, over which the force acts, is $\overrightarrow{d}.$ The work done on the lawn mower is$W=\overrightarrow{F}·\overrightarrow{d}=Fd\text{cos}\theta$, which the figure also illustrates as the horizontal component of the force times the magnitude of the displacement.

Figure 7.3(b) shows a person holding a briefcase. The person must exert an upward force, equal in magnitude to the weight of the briefcase, but this force does no work, because the displacement over which it acts is zero.

In Figure 7.3(c), where the person in (b) is walking horizontally with constant speed, the work done by the person on the briefcase is still zero, but now because the angle between the force exerted and the displacement is $90\text{°}$ ($\overrightarrow{F}$ perpendicular to $\overrightarrow{d}$) and $\text{cos}90\text{°}=0$.

When you mow the grass, other forces act on the lawn mower besides the force you exert—namely, the contact force of the ground and the gravitational force of Earth. Let’s consider the work done by these forces in general. For an object moving on a surface, the displacement $d\overrightarrow{r}$ is tangent to the surface. The part of the contact force on the object that is perpendicular to the surface is the normal force $\overrightarrow{N}.$ Since the cosine of the angle between the normal and the tangent to a surface is zero, we have

The normal force never does work under these circumstances. (Note that if the displacement $d\overrightarrow{r}$ did have a relative component perpendicular to the surface, the object would either leave the surface or break through it, and there would no longer be any normal contact force. However, if the object is more than a particle, and has an internal structure, the normal contact force can do work on it, for example, by displacing it or deforming its shape. This will be mentioned in the next chapter.)

The part of the contact force on the object that is parallel to the surface is friction, $\overrightarrow{f}.$ For this object sliding along the surface, kinetic friction ${\overrightarrow{f}}_{\text{k}}$ is opposite to $d\overrightarrow{r},$ relative to the surface, so the work done by kinetic friction is negative. If the magnitude of ${\overrightarrow{f}}_{\text{k}}$ is constant (as it would be if all the other forces on the object were constant), then the work done by friction is

where $\left|l_{AB}\right|$ is the path length on the surface. The force of static friction does no work in the reference frame between two surfaces because there is never displacement between the surfaces. As an external force, static friction can do work. Static friction can keep someone from sliding off a sled when the sled is moving and perform positive work on the person. If you’re driving your car at the speed limit on a straight, level stretch of highway, the negative work done by air resistance is balanced by the positive work done by the static friction of the road on the drive wheels. You can pull the rug out from under an object in such a way that it slides backward relative to the rug, but forward relative to the floor. In this case, kinetic friction exerted by the rug on the object could be in the same direction as the displacement of the object, relative to the floor, and do positive work. The bottom line is that you need to analyze each particular case to determine the work done by the forces, whether positive, negative or zero.

The other force on the lawn mower mentioned above was Earth’s gravitational force, or the weight of the mower. Near the surface of Earth, the gravitational force on an object of mass m has a constant magnitude, mg, and constant direction, vertically down. Therefore, the work done by gravity on an object is the dot product of its weight and its displacement. In many cases, it is convenient to express the dot product for gravitational work in terms of the x-, y-, and z-components of the vectors. A typical coordinate system has the x-axis horizontal and the y-axis vertically up. Then the gravitational force is $\text{−}mg\widehat{j},$ so the work done by gravity, over any path from A to B, is

The work done by a constant force of gravity on an object depends only on the object’s weight and the difference in height through which the object is displaced. Gravity does negative work on an object that moves upward ($y_B>y_A$), or, in other words, you must do positive work against gravity to lift an object upward. Alternately, gravity does positive work on an object that moves downward ($y_B<y_A$), or you do negative work against gravity to “lift” an object downward, controlling its descent so it doesn’t drop to the ground. (“Lift” is used as opposed to “drop”.)