6.7

Power

In many situations, the time it takes to do work is just as important as the amount of work that is done. Consider two automobiles that are identical in all respects (e.g., same mass), except that one has a “souped-up” engine. The car with the souped-up engine can go from 0 to 27 m /s (60 mph) in 4 seconds, while the other car requires 8 seconds to achieve the same speed. Each engine does work in accelerating its car, but one does it more quickly. Where cars are concerned, we associate the quicker performance with an engine that has a larger horsepower rating. A large horsepower rating means that the engine can do a large amount of work in a short time. In physics, the horsepower rating is just one way to measure an engine’s ability to generate power. The idea of power incorporates both the concepts of work and time, for power is work done per unit time.

| Definition of Average Power
| Average power is the average rate at which work W is done, and it is obtained by dividing W by the time t required to perform the work:

| (6.10a)

SI Unit of Power:

Check Your Understanding 4

Engine A has a greater power rating than engine B. Which one of the following statements correctly describes the abilities of these engines to do work? (a) Engines A and B can do the same amount of work in the same amount of time. (b) In the same amount of time, engine B can do more work than engine A. (c) Engines A and B can do the same amount of work, but engine A can do it more quickly. (The answer is given at the end of the book.)
Background: Work, time, and power are the key concepts. Remember that work and power are not the same thing.
For similar questions (including calculational counterparts), consult Self-Assessment Test 6.3. This test is described at the end of Section 6.9.

The definition of average power presented in Equation 6.10a involves the work that is done by the net force. However, the work–energy theorem relates this work to the change in the energy of the object (see, for example, Equations 6.3 and 6.8). Therefore, we can also define average power as the rate at which the energy is changing, or as the change in energy divided by the time during which the change occurs:

| (6.10b)

Since work, energy, and time are scalar quantities, power is also a scalar quantity. The unit in which power is expressed is that of work divided by time, or a joule per second in SI units. One joule per second is called a watt (W), in honor of James Watt (1736–1819), the developer of the steam engine. The unit of power in the BE system is the foot·pound per second (ft·lb/s), although the familiar horsepower (hp) unit is used frequently for specifying the power generated by electric motors and internal combustion engines:

Table 6-3 summarizes the units for power in the various systems of measurement.

Table 6-3 Units of Measurement for Power

System | Work | ÷ | Time | = | Power
SI | joule (J) |  | second (s) |  | watt (W)
CGS | erg |  | second (s) |  | erg per second (erg/s)
BE | foot·pound (ft lb) |  | second (s) |  | foot·pound per second (ft·lb/s)

Equation 6.10b provides the basis for understanding the production of power in the human body. In this context the “Change in energy” on the right-hand side of the equation refers to the energy produced by metabolic processes, which, in turn, is derived from the food we eat. Table 6-4 gives typical metabolic rates of energy production needed to sustain various activities. Running at 15 km /h (9.3 mi/h), for example, requires metabolic power sufficient to operate eighteen 75-watt light bulbs, and the metabolic power used in sleeping would operate a single 75-watt bulb.

Table 6-4 Human Metabolic Rates*

Activity | Rate (watts)
Running (15 km/h) | 1340 W
Skiing | 1050 W
Biking | 530 W
Walking (5 km/h) | 280 W
Sleeping | 77 W

a) |  | For a young 70-kg male.

The physics of human metabolism.

An alternative expression for power can be obtained from Equation 6.1, which indicates that the work W done when a constant net force of magnitude F points in the same direction as the displacement is . Dividing both sides of this equation by the time t it takes for the force to move the object through the distance s, we obtain

But W/t is the average power , and s/t is the average speed , so that

| (6.11)

The next example illustrates the use of Equation 6.11.

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 13 | | | The Power to Accelerate a Car

A car, starting from rest, accelerates in the direction (see Figure 6-21). It has a mass of and maintains an acceleration of for 5.00 s. Assume that a single horizontal force (not shown) accelerates the vehicle. Determine the average power generated by this force.

Figure 6-21 A car, starting from rest, accelerates in the +x direction for 5.00 s.

Reasoning The average power can be found by using the relation (Equation 6.11), provided that the magnitude F of the horizontal force and the average speed of the car can be determined. The force can be obtained from Newton’s second law as the product of the car’s mass and acceleration. The average speed can be determined by using the equations of kinematics.
Knowns and Unknowns The data for this problem are listed below:

Description | Symbol | Value | Comment
Explicit Data |  |  |
Acceleration | a | +4.60 m/s2 |
Mass | m | 1.10 × 103 kg |
Time | t | 5.00 s |
Implicit Data |  |  |
Initial velocity | v0 | 0 m/s | Car starts from rest.
Unknown Variable |  |  |
Average power |  | ? |

Modeling the Problem

| Step 1 Average Power
| The average power is the product of the magnitude F of the horizontal force acting on the car and the car’s average speed , as given at the right. At this point we do not have numerical values for either F or , but in Steps 2 and 3 we will show how to obtain them.

| (6.11)

| Step 2 Newton’s Second Law
| According to Newton’s second law, Equation 4.1, the net force is equal to the mass m of the car times its acceleration a. Since there is only one horizontal force acting on the car, it is the net force. Thus,

This expression also represents the magnitude F of the force, since a is a positive number. Values are known for both m and a, so we substitute this expression for F into Equation 6.11, as indicated at the right. In Step 3 an expression for will be found.

| Step 3 Equations of Kinematics
| Since the acceleration is constant, the equations of kinematics apply, and the average velocity of the car is one-half the sum of its initial velocity v0 and its final velocity v (Equation 2.6): . The final velocity is related to the initial velocity, the acceleration a, and the time t (all of which are known) by Equation 2.4: . Substituting this expression for v into the equation above for yields

This result, which represents the average velocity, is also equal to the average speed (which is the magnitude of the average velocity), since v0 is zero and a is positive. As indicated at the right, this expression for can be substituted into Equation 6.11.

Solution Algebraically combining the results of the three steps, we arrive at the following expression for the average power:
The average power generated by the net force that accelerates the car is

Related Homework: Problem 59

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