第6章積分の応用

We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.

In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate $1$ kilogram of mass at the rate of $1$ m/sec². Thus, the most common unit of work is the newton-meter. This same unit is also called the joule. Both are defined as kilograms times meters squared over seconds squared $\left(\text{kg}·{\text{m}}^2\text{/}{\text{s}}^2\right).$

When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.

Suppose we have a variable force $F(x)$ that moves an object in a positive direction along the x-axis from point $a$ to point $b.$ To calculate the work done, we partition the interval $\left[a,b\right]$ and estimate the work done over each subinterval. So, for $i=0,1,2\text{,…},n,$ let $P=\left\{x_i\right\}$ be a regular partition of the interval $\left[a,b\right],$ and for $i=1,2\text{,…},n,$ choose an arbitrary point $x_i^{*}\in \left[x_{i-1},x_i\right].$ To calculate the work done to move an object from point $x_{i-1}$ to point $x_i,$ we assume the force is roughly constant over the interval, and use $F(x_i^{*})$ to approximate the force. The work done over the interval $\left[x_{i-1},x_i\right],$ then, is given by

Therefore, the work done over the interval $\left[a,b\right]$ is approximately

Taking the limit of this expression as $n\to \infty$ gives us the exact value for work:

Thus, we can define work as follows.

Note that if F is constant, the integral evaluates to $F·(b-a)=F·d,$ which is the formula we stated at the beginning of this section.

Now let’s look at the specific example of the work done to compress or elongate a spring. Consider a block attached to a horizontal spring. The block moves back and forth as the spring stretches and compresses. Although in the real world we would have to account for the force of friction between the block and the surface on which it is resting, we ignore friction here and assume the block is resting on a frictionless surface. When the spring is at its natural length (at rest), the system is said to be at equilibrium. In this state, the spring is neither elongated nor compressed, and in this equilibrium position the block does not move until some force is introduced. We orient the system such that $x=0$ corresponds to the equilibrium position (see the following figure).

According to Hooke’s law, the force required to compress or stretch a spring from an equilibrium position is given by $F(x)=kx,$ for some constant $k.$ The value of $k$ depends on the physical characteristics of the spring. The constant $k$ is called the spring constant and is always positive. We can use this information to calculate the work done to compress or elongate a spring, as shown in the following example.