6.9

Work Done by a Variable Force

The work W done by a constant force (constant in both magnitude and direction) is given by Equation 6.1 as . Quite often, situations arise in which the force is not constant but changes with the displacement of the object. For instance, Figure 6-22a shows an archer using a high-tech compound bow. This type of bow consists of a series of pulleys and strings that produce a force-versus-displacement graph like that in Figure 6-22b. One of the key features of the compound bow is that the force rises to a maximum as the string is drawn back, and then falls to 60% of this maximum value when the string is fully drawn. The reduced force at makes it much easier for the archer to hold the fully drawn bow while aiming the arrow.

Figure 6-22 (a) A compound bow. (b) A plot of versus s as the bowstring is drawn back. (a, Ray Laskowitz/Brand X-X Image/Image State)

When the force varies with the displacement, as in Figure 6-22b, we cannot use the relation to find the work, because this equation is valid only when the force is constant. However, we can use a graphical method. In this method we divide the total displacement into very small segments, , , and so on (see Figure 6-23a). For each segment, the average value of the force component is indicated by a short horizontal line. For example, the short horizontal line for segment is labeled in Figure 6-23a. We can then use this average value as the constant-force component in Equation 6.1 and determine an approximate value for the work done during the first segment: . But this work is just the area of the colored rectangle in the drawing. The word “area” here refers to the area of a rectangle that has a width of and a height of ; it does not mean an area in square meters, such as the area of a parcel of land. In a like manner, we can calculate an approximate value for the work for each segment. Then we add the results for the segments to get, approximately, the work W done by the variable force:

The symbol “≈” means “approximately equal to.” The right side of this equation is the sum of all the rectangular areas in Figure 6-23a and is an approximate value for the area shaded in color under the graph in Figure 6-23b. If the rectangles are made narrower and narrower by decreasing each , the right side of this equation eventually becomes equal to the area under the graph. Thus, we define the work done by a variable force as follows: The work done by a variable force in moving an object is equal to the area under the graph of F cos versus s. Example 14 illustrates how to use this graphical method to determine the approximate work done when a high-tech compound bow is drawn.

Figure 6-23 (a) The work done by the average-force component during the small displacement is , which is the area of the colored rectangle. (b) The work done by a variable force is equal to the colored area under the -versus-s curve.

Problem solving insight

| Example 14 | | | Work and the Compound Bow

Find the work that the archer must do in drawing back the string of the compound bow in Figure 6-22 from 0 to 0.500 m. Reasoning The work is equal to the colored area under the curved line in Figure 6-22b. For convenience, this area is divided into a number of small squares, each having an area of . The area can be found by counting the number of squares under the curve and multiplying by the area per square.
Solution We estimate that there are 242 colored squares in the drawing. Since each square represents 0.250 J of work, the total work done is

When the arrow is fired, part of this work is imparted to it as kinetic energy.

Test your understanding of the material in Sections 6.6, 6.7, 6.8 and 6.9:

· | Nonconservative Forces and the Work–Energy Theorem
· | Power
· | Work Done by a Variable Force

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.