6.2

The Work–Energy Theorem and Kinetic Energy

CONCEPTS AT A GLANCE Most people expect that if you do work, you get something as a result. In physics, when a net force performs work on an object, there is always a result from the effort. The result is a change in the kinetic energy of the object. As we will now see, the relationship that relates work to the change in kinetic energy is known as the work–energy theorem. This theorem is obtained by bringing together three basic concepts that we’ve already learned about, as the Concepts-at-a-Glance chart in Figure 6-5 illustrates. First we’ll apply Newton’s second law of motion , which relates the net force ∑F, to the acceleration a of an object. Then, we’ll determine the work done by the net force when the object moves through a certain distance. Finally, we’ll use Equation 2.9, one of the equations of kinematics, to relate the distance and acceleration to the initial and final speeds of the object. The result of this approach will be the work–energy theorem.

Figure 6-5 CONCEPTS AT A GLANCE In this section, Newton’s second law, the definition of work, and the equations of kinematics are brought together to formulate the work–energy theorem. In hitting a softball, a player does work on it. This work causes the ball’s kinetic energy to change in accordance with the work–energy theorem. (Adam Pretty/Allsport/Getty Images Sport Services)

To gain some insight into the idea of kinetic energy and the work–energy theorem, look at Figure 6-6, where a constant net external force acts on an airplane of mass m. This net force is the vector sum of all the external forces acting on the plane, and, for simplicity, it is assumed to have the same direction as the displacement . According to Newton’s second law, the net force produces an acceleration a, given by . Consequently, the speed of the plane changes from an initial value of v0 to a final value of vf.* Multiplying both sides of by the distance s gives

The left side of this equation is the work done by the net external force. The term as on the right side can be related to v0 and vf by using (Equation 2.9) from the equations of kinematics. Solving this equation to give and substituting into shows that

This expression is the work–energy theorem. Its left side is the work W done by the net external force, while its right side involves the difference between two terms, each of which has the form . The quantity is called kinetic energy (KE) and plays a significant role in physics, as we will soon see.

Figure 6-6 A constant net external force acts over a displacement and does work on the plane. As a result of the work done, the plane’s kinetic energy changes.

| Definition of Kinetic Energy
| The kinetic energy KE of an object with mass m and speed v is given by

| (6.2)

SI Unit of Kinetic Energy: joule(J)

The SI unit of kinetic energy is the same as the unit for work, the joule. Kinetic energy, like work, is a scalar quantity. These are not surprising observations, for work and kinetic energy are closely related, as is clear from the following statement of the work–energy theorem.

| The Work–Energy Theorem
| When a net external force does work W on an object, the kinetic energy of the object changes from its initial value of KE0 to a final value of KEf, the difference between the two values being equal to the work:

| (6.3)

The work–energy theorem may be derived for any direction of the force relative to the displacement, not just the situation in Figure 6-6. In fact, the force may even vary from point to point along a path that is curved rather than straight, and the theorem remains valid. According to the work–energy theorem, a moving object has kinetic energy, because work was done to accelerate the object from rest to a speed vf.* Conversely, an object with kinetic energy can perform work, if it is allowed to push or pull on another object. Example 4 illustrates the work–energy theorem and considers a single force that does work to change the kinetic energy of a space probe.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 4 | | | Deep Space 1

The space probe Deep Space 1 was launched October 24, 1998, and it used a type of engine called an ion propulsion drive. The ion propulsion drive generated only a weak force (or thrust), but could do so for long periods of time using only small amounts of fuel. Suppose the probe, which has a mass of 474 kg, is traveling at an initial speed of 275 m/s. No forces act on it except the thrust of its engine. This external force is directed parallel to the displacement , which has a magnitude of (see Figure 6-7). Determine the final speed of the probe, assuming that its mass remains nearly constant.

Figure 6-7 The engine of Deep Space 1 generated a single force that pointed in the same direction as the displacement . The force performed positive work, causing the space probe to gain kinetic energy.

The physics of an ion propulsion drive.

Reasoning If we can determine the final kinetic energy of the space probe, we can determine its final speed, since kinetic energy is related to mass and speed according to Equation 6.2 and the mass of the probe is known. We will use the work–energy theorem , along with the definition of work, to find the final kinetic energy.
Knowns and Unknowns The following list summarizes the data for this problem:

Description | Symbol | Value | Comment
Explicit Data |  |  |
Mass | m | 474 kg |
Initial speed | v0 | 275 m/s |
Magnitude of force | F | 5.60 × 10−2 N |
Magnitude of displacement | s | 2.42 × 109 m |
Implicit Data |  |  |
Angle between force and displacement | θ | 0° | The force is parallel to the displacement.
Unknown Variable |  |  |
Final speed | vf | ? |

Modeling the Problem

| Step 1 Kinetic Energy
| An object of mass m and speed v has a kinetic energy KE given by Equation 6.2 as . Using the subscript “f” to denote the final kinetic energy and the final speed of the probe, we have that

Solving for vf gives Equation 1 at the right. The mass m is known but the final kinetic energy KEf is not, so we will turn to Step 2 to evaluate it.

| (1)

| Step 2 The Work–Energy Theorem
| The work–energy theorem relates the final kinetic energy KEf of the probe to its initial kinetic energy KE0 and the work W done by the force of the engine. According to Equation 6.3, this theorem is . Solving for KEf shows that

The initial kinetic energy can be expressed as , so the expression for the final kinetic energy becomes

This result can be substituted into Equation 1, as indicated at the right. Note from the data table that we know the mass m and the initial speed v0. The work W is not known and will be evaluated in Step 3.

| Step 3 Work
| The work W is that done by the net external force acting on the space probe. Since there is only the one force acting on the probe, it is the net force. The work done by this force is given by Equation 6.1 as

where F is the magnitude of the force, is the angle between the force and the displacement, and s is the magnitude of the displacement. All the variables on the right side of this equation are known, so we can substitute it into Equation 1 as shown in the right column.

Solution Algebraically combining the results of the three steps, we have
The final speed of the space probe is

Related Homework: Problems 18 and 20

In Example 4 only the force of the engine does work. If several external forces act on an object, they must be added together vectorially to give the net force. The work done by the net force can then be related to the change in the object’s kinetic energy by using the work–energy theorem, as in the next example.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 5 | | | Downhill Skiing

A 58-kg skier is coasting down a 25° slope, as Figure 6-8a shows. Near the top of the slope, her speed is 3.6 m/s. She accelerates down the slope because of the gravitational force, even though a kinetic frictional force of magnitude 71 N opposes her motion. Ignoring air resistance, determine the speed at a point that is displaced 57 m downhill.

Figure 6-8 (a) A skier coasting downhill. (b) The free-body diagram for the skier.

Reasoning The skier’s speed at a point 57 m downhill (her final speed) depends on her final kinetic energy. According to the work–energy theorem, her final kinetic energy, in turn, is related to her initial kinetic energy (which we can calculate directly) and the work done by the net external force that acts on her. The work can be evaluated directly from its definition.
Knowns and Unknowns The data for this problem are listed below:

Description | Symbol | Value | Comment
Explicit Data |  |  |
Mass | m | 58 kg |
Initial speed | v0 | 3.6 m/s |
Magnitude of kinetic frictional force | fk | 71 N |
Magnitude of skier’s displacement | s | 57 m |
Angle of slope above horizontal |  | 25° |
Implicit Data |  |  |
Angle between net force acting on skier and her displacement |  | 0° | The skier is accelerating down the slope, so the direction of the net force is parallel to her displacement.
Unknown Variable |  |  |
Final speed | vf | ? |

Modeling the Problem

| Step 1 Kinetic Energy
| The final speed vf of the skier is related to her final kinetic energy KEf and mass m by Equation 6.2:

Solving for vf gives Equation 1 at the right. Her mass is known, but her final kinetic energy is not, so we turn to Step 2 to evaluate it.

| (1)

| Step 2 The Work–Energy Theorem
| The final kinetic energy KEf of the skier is related to her initial kinetic energy KE0 and the work W done by the net external force acting on her by the work–energy theorem (Equation 6.3): . The initial kinetic energy can be written as , so the expression for the final kinetic energy becomes

This result can be substituted into Equation 1, as indicated at the right. The mass m and the initial speed v0 are known. The work W is not known, and will be evaluated in Steps 3 and 4.

| Step 3 Work
| The work W done by the net external force acting on the skier is given by Equation 6.1 as

where is the magnitude of the net force, is the angle between the net force and the displacement, and s is the magnitude of the displacement. This result can be substituted into Equation 2, as shown at the right. The variables and s are known (see the data table), and the net external force will be determined in Step 4.

| Step 4 The Net External Force
| Figure 6-8b is a free-body diagram for the skier and shows the three external forces acting on her: the gravitational force , the kinetic frictional force , and the normal force . The net external force along the y axis is zero, because there is no acceleration in that direction (the normal force balances the component mg cos 25° of the weight perpendicular to the slope). Using the data from the table of knowns and unknowns, we find that the net external force along the x axis is

All the variables on the right side of this equation are known, so we substitute this expression into Equation 3 for the net external force (see the right column).

Solution Algebraically combining the four steps, we arrive at the following expression for the final speed of the skier:
The final speed of the skier is

Related Homework: Problems 22 and 24

Check Your Understanding 2

A rocket is at rest on the launch pad. When the rocket is launched, its kinetic energy increases. Is the following statement true or false? “The amount by which the kinetic energy increases is equal to the work done by the force generated by the rocket’s engine.” (The answer is given at the end of the book.)
Background: The concepts of work, kinetic energy, and the work–energy theorem are pertinent to this question. Think about what forces determine the work that appears in the work–energy theorem.
For similar questions (including calculational counterparts), consult Self-Assessment Test 6.1. This test is described at the end of this section.

Need more practice?

Interactive LearningWare6.2
In screeching to a halt, a car leaves skid marks that are 65 m long. The coefficient of kinetic friction between the tires and the road is . How fast was the car going before the brakes were applied? Related Homework: Problem 18

Example 5 emphasizes that the work–energy theorem deals with the work done by the net external force. The work–energy theorem does not apply to the work done by an individual force, unless that force happens to be the only one present, in which case it is the net force. If the work done by the net force is positive, as in Example 5, the kinetic energy of the object increases. If the work done is negative, the kinetic energy decreases. If the work is zero, the kinetic energy remains the same. Conceptual Example 6 explores these ideas further.

| Conceptual Example 6 | | | Work and Kinetic Energy

Figure 6-9 illustrates a satellite moving about the earth in a circular orbit and in an elliptical orbit. The only external force that acts on the satellite is the gravitational force. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion.

Figure 6-9 (a) In a circular orbit, the gravitational force is always perpendicular to the displacement of the satellite and does no work. (b) In an elliptical orbit, there can be a component of the force along the displacement, and, consequently, work can be done.

Reasoning and Solution For the circular orbit in Figure 6-9a the gravitational force does no work on the satellite, since the force is perpendicular to the instantaneous displacement at all times. The gravitational force is the only external force acting on the satellite, so it is the net force. Thus, the work done by the net force is zero, and according to the work–energy theorem (Equation 6.3), the kinetic energy of the satellite (and, hence, its speed) remains the same everywhere on the orbit.
In contrast, the gravitational force does do work on a satellite in an elliptical orbit. For example, as the satellite moves toward the earth in the top part of Figure 6-9b, there is a component of the gravitational force that points in the same direction as the displacement. Consequently, the gravitational force does positive work during this part of the orbit, and the kinetic energy of the satellite increases. When the satellite moves away from the earth, as in the lower part of Figure 6-9b, the gravitational force has a component that points opposite to the displacement. Now, the gravitational force does negative work, and the kinetic energy of the satellite decreases.
Can you identify the two places on an elliptical orbit where the gravitational force does no work?
Related Homework: Problem 16

Test your understanding of the material in Sections 6.1 and 6.2:

· | Work
· | Kinetic Energy
· | The Work–Energy Theorem

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