6.1

Work Done by a Constant Force

Work is a familiar concept. For example, it takes work to push a stalled car. In fact, more work is done when the pushing force is greater or when the displacement of the car is greater. Force and displacement are, in fact, the two essential elements of work, as Figure 6-1 illustrates. The drawing shows a constant pushing force that points in the same direction as the resulting displacement .* In such a case, the work W is defined as the magnitude F of the force times the magnitude s of the displacement: . The work done to push a car is the same whether the car is moved north to south or east to west, provided that the amount of force used and the distance moved are the same. Since work does not convey directional information, it is a scalar quantity.

Figure 6-1 Work is done when a force pushes a car through a displacement .

The equation indicates that the unit of work is the unit of force times the unit of distance, or the newton·meter in SI units. One newton·meter is referred to as a joule (J) (rhymes with “cool”), in honor of James Joule (1818–1889) and his research into the nature of work, energy, and heat. Table 6-1 summarizes the units for work in several systems of measurement.

Table 6-1 Units of Measurement for Work

System | Force | × | Distance | = | Work
SI | newton (N) |  | meter (m) |  | joule (J)
CGS | dyne (dyn) |  | centimeter (cm) |  | erg
BE | pound (lb) |  | foot (ft) |  | foot·pound (ft·lb)

The definition of work as does have one surprising feature: If the distance s is zero, the work is zero, even if a force is applied. Pushing on an immovable object, such as a brick wall, may tire your muscles, but there is no work done of the type we are discussing. In physics, the idea of work is intimately tied up with the idea of motion. If the object does not move, the force acting on the object does no work.
Often, the force and displacement do not point in the same direction. For instance, Figure 6-2a shows a suitcase-on-wheels being pulled to the right by a force that is applied along the handle. The force is directed at an angle relative to the displacement. In such a case, only the component of the force along the displacement is used in defining work. As Figure 6-2b shows, this component is , and it appears in the general definition below:

| Definition of Work Done by a Constant* Force
| The work done on an object by a constant force is

| (6.1)

where F is the magnitude of the force, s is the magnitude of the displacement, and is the angle between the force and the displacement.
SI Unit of Work:

When the force points in the same direction as the displacement, then , and Equation 6.1 reduces to . Example 1 shows how Equation 6.1 is used to calculate work.

Figure 6-2 (a) Work can be done by a force that points at an angle relative to the displacement . (b) The force component that points along the displacement is .

| Example 1 | | | Pulling a Suitcase-on-Wheels

Find the work done by a 45.0-N force in pulling the suitcase in Figure 6-2a at an angle for a distance . Reasoning The pulling force causes the suitcase to move a distance of 75.0 m and does work. However, the force makes an angle of 50.0° with the displacement, and we must take this angle into account by using the definition of work given by Equation 6.1.
Solution The work done by the 45.0-N force is

The answer is expressed in newton·meters or joules (J).

The definition of work in Equation 6.1 takes into account only the component of the force in the direction of the displacement. The force component perpendicular to the displacement does no work. To do work, there must be a force and a displacement, and since there is no displacement in the perpendicular direction, there is no work done by the perpendicular component of the force. If the entire force is perpendicular to the displacement, the angle in Equation 6.1 is 90°, and the force does no work at all.
Work can be either positive or negative, depending on whether a component of the force points in the same direction as the displacement or in the opposite direction. Example 2 illustrates how positive and negative work arise.

| Example 2 | | | Bench-Pressing

The weight lifter in Figure 6-3a is bench-pressing a barbell whose weight is 710 N. In part b of the figure, he raises the barbell a distance of 0.65 m above his chest, and in part c he lowers it the same distance. The weight is raised and lowered at a constant velocity. Determine the work done on the barbell by the weight lifter during (a) the lifting phase and (b) the lowering phase.

Figure 6-3 (a) In the bench press, work is done during both the lifting and lowering phases. (b) During the lifting phase, the force does positive work. (c) During the lowering phase, the force does negative work. (a, Rachel Epstein/The Image Works)

Reasoning To calculate the work, it is necessary to know the force exerted by the weight lifter. The barbell is raised and lowered at a constant velocity and, therefore, is in equilibrium. Consequently, the force exerted by the weight lifter must balance the weight of the barbell, so . During the lifting phase, the force and displacement are in the same direction, as Figure 6-3b shows. The angle between them is . When the barbell is lowered, however, the force and displacement are in opposite directions, as in Figure 6-3c. The angle between the force and the displacement is now . With these observations, we can find the work.
Solution

a. | During the lifting phase, the work done by the force is given by Equation 6.1 as

b. | The work done during the lowering phase is

since . The work is negative, because the force is opposite to the displacement. The physics of positive and negative “reps” in weight lifting. Weight lifters call each complete up-and-down movement of the barbell a repetition, or “rep.” The lifting of the weight is referred to as the positive part of the rep, and the lowering is known as the negative part.

Example 3 deals with the work done by a static frictional force when it acts on a crate that is resting on the bed of an accelerating truck.

| Example 3 | | | Accelerating a Crate

Figure 6-4a shows a 120-kg crate on the flatbed of a truck that is moving with an acceleration of along the positive x axis. The crate does not slip with respect to the truck as the truck undergoes a displacement whose magnitude is . What is the total work done on the crate by all of the forces acting on it?

Figure 6-4 (a) The truck and crate are accelerating to the right for a distance of . (b) The free-body diagram for the crate.

Reasoning The free-body diagram in Figure 6-4b shows the forces that act on the crate: (1) the weight of the crate, (2) the normal force exerted by the flatbed, and (3) the static frictional force , which is exerted by the flatbed in the forward direction and keeps the crate from slipping backward. The weight and the normal force are perpendicular to the displacement, so they do no work. Only the static frictional force does work, since it acts in the x direction. To determine the frictional force, we note that the crate does not slip and, therefore, must have the same acceleration of as does the truck. The force creating this acceleration is the static frictional force, and knowing the mass of the crate and its acceleration, we can use Newton’s second law to obtain its magnitude. Then, knowing the frictional force and the displacement, we can determine the total work done on the crate.
Solution From Newton’s second law, we find that the magnitude fs of the static frictional force is

The total work is that done by the static frictional force and is

| (6.1)

The work is positive, because the frictional force is in the same direction as the displacement .

Need more practice?

Interactive LearningWare6.1
A cyclist coasts down a hill, losing 20.0 m in height, and then coasts up an identical hill to the same initial vertical height. The bicycle and rider have a mass of 73.5 kg. Two forces act on the bicycle/rider system: the weight of the bicycle and rider and the normal force exerted on the tires by the ground. Determine the work done by each of these forces during (a) the downward part of the motion and (b) the upward part of the motion.
Related Homework: Problem 6

Check Your Understanding 1

A suitcase is hanging straight down from your hand as you ride an escalator. Your hand exerts a force on the suitcase, and this force does work. Which one of the following statements is correct? (a) The work is negative when you ride up the escalator and positive when you ride down the escalator. (b) The work is positive when you ride up the escalator and negative when you ride down the escalator. (c) The work is positive irrespective of whether you ride up or down the escalator. (d) The work is negative irrespective of whether you ride up or down the escalator. (The answer is given at the end of the book.)
Background: This question deals with work and its relationship to force and displacement. Pay close attention to the directions of the force and displacement vectors.
For similar questions (including calculational counterparts), consult Self-Assessment Test 6.1. This test is described at the end of Section 6.2.

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