Study Unit

Mechanics

iii

Physical science has played an importantrole in the development of our civilization.Using the concepts of mechanics, heat,sound, chemistry, light, and electricity,our world has developed important newtechnologies that help us live better lives.

Many of the devices we take for granted, such as motorvehicles, computers, telephones, and electrical appliances,were invented and improved by applying physical scienceconcepts. The study of physical science can thereforeprovide you with an understanding of the basic scientificprinciples that are at the foundation of these technologicalinnovations. This understanding is essential to makinginformed decisions in all areas of our lives.

The first step toward the understanding of physical scienceis to learn the language of science. Science ideas arecommunicated by using specific science terms, and yourunderstanding of the material discussed will come easierwhen you become familiar with these terms.

For the study of physical science, you’ll also need to applysome basic math skills. In this study unit, mathematicalexamples and formulas are used to illustrate certain impor-tant science concepts, but the examples are kept as simpleas possible. That way, you can focus most of your attentionon the science ideas being presented rather than on the calculations.

The study of physical science is both challenging and reward-ing. It’s a challenge you’ll be glad you accepted, because asyou progress through the discussions of the various topics,you’ll realize the value of the knowledge you’re gaining.

In this study unit, you’ll begin your study of physicalscience by learning about mechanics. Mechanics includesthe study of energy and forces and their effect on objects.You’ll study mechanics first because this topic representsa general foundation for the other branches of physicalscience. You’ll learn about force, motion, energy, and power.You’ll also be introduced to the properties of liquids, gases,solids, and electrolytes.

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When you complete this study unit, you’ll be able to

• Calculate velocity and acceleration

• Recognize the relation between acceleration and force

• Apply Newton’s laws of motion

• Recognize the role of gravitational force

• Distinguish between potential and kinetic energy

• Calculate energy and power

• Determine pressure conditions within a liquid

• Apply Pascal’s and Archimedes’ principles

• Identify properties of gases

• Identify properties of solids

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MOTION 1

Bodies and Motion 2Units of Measurement 2Speed and Velocity 5Calculating Uniform Acceleration 13Falling Bodies 15

FORCE AND MOTION 20

What Affects the Motion of a Body? 20Newton’s First Law 21Newton’s Second Law 23Gravity 26Action and Reaction 28Vector and Scalar Quantities 29

MOTION IN A CURVED PATH 34

Centripetal Force 34Calculating Centripetal Force 38The Motion of Satellites 39Ptolemy’s Model of the Universe 40Copernicus’ Model of the Universe 42Kepler’s Laws 44The Universal Gravitational Law 45

ENERGY AND POWER 47

Work and Energy 47Kinetic Energy 49Potential Energy 50Mechanical Energy 52Resistance Due to Friction 53The Law of Energy Conservation 53Power 54Efficiency 56

PROPERTIES OF LIQUIDS 60

The Forces Exerted by Liquids 60Liquid Pressure 62Pressure at Any Point in a Liquid 65Pascal’s Principle 67The Hydraulic Press 71Pressure at the Free Surface of a Liquid 73Gages for Measuring Liquid Pressure 73

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Contentsvi

Archimedes’ Principle 75Specific Gravity 77The Meniscus, Capillarity, and Viscosity 79

PROPERTIES OF GASES AND SOLIDS 82

The Compressibility and Expansibility of Gases 82Atmospheric Pressure 84Barometers 85Atmospheric Pressure and Altitude 86Gage Pressure and Absolute Pressure 87Conductors and Insulators 88The Pressure of Solids 88

SELF-CHECK ANSWERS 93

APPENDIX 103

EXAMINATION 105

1

MOTION

Bodies can move in straight or curved lines with velocitiesthat may be uniform or accelerated. In this first section of your study unit, you’ll learn about the differences between speed and velocity, and then learn how to calculate acceleration.

The following list contains the definitions of some importantmotion terms. Take a few moments to familiarize yourselfwith the terms, and don’t hesitate to refer to the list as youwork through this part of your study unit.

• Body. Any distinct quantity of matter (an object).

• Motion. The change in position of a body.

• Speed. The distance traveled by a body in any directionin a unit of time.

• Velocity. The distance traveled by a body in a particulardirection (that is, a straight line) in a unit of time.

• Uniform velocity. A velocity at which a body covers thesame distance in each measured unit of time for whichit’s measured.

• Acceleration. Any change in velocity.

• Gravity. The force of attraction of the mass of the earth(or any other planet) for objects near its surface.

Mechanics

Mechanics

Bodies and Motion

In mechanics, any distinct quantity of matter is called abody. A body may be in the solid, liquid, or gaseous state,and it may have a certain shape, size, color, and mass. Inorder to describe a body completely and define what makes itunique among other bodies, it’s necessary to position thebody precisely in time and space. To do this, you’ll need todescribe how the body moves from place to place. The changein position of the body is called motion. It’s also important todetermine what makes the body move as it does. The causeof motion is a force.

Types of motion must be carefully defined so that each motioncan be consistently described and copied when necessary.Motions can be described as linear and curved (Figure 1).Linear motions are always in the same direction and follow a path in a straight line. Curved motions change direction atevery instant and follow a curved path.

Units of Measurement

Understanding science requires a basic knowledge of theunits that are used to make measurements. The basic con-cepts of length, mass, and time can be used to define mostphysical quantities. For example, speed is simply a length

2

A

B

A

B

Linear Motion

Curved Motion

FIGURE 1—The air dis-tance between City Aand City B is the short-est distance; it’s indicat-ed by a straight line. Anairplane can travel in astraight line through theair, and its motion is lin-ear. In contrast, a carmust follow the roadbetween the two cities,which is a winding path.The car’s motion iscurved.

Mechanics 3

divided by time. The volume of an object (the amount ofspace it occupies) can be calculated by multiplying thelength of the object times its width times its height.

It’s essential that the units used to describe basic quantitiesin science be standardized (that is, brought into conformitywith predetermined standards). It’s also necessary thatmeasured quantities be reproducible in a laboratory setting.Throughout history, many methods of measurements havebeen used to define fundamental units of measurement,but few have been consistent enough to be standardized.For example, in Egypt the unit of length was the cubit, whichwas defined as the length from elbow to the outstretchedfinger. Obviously, this length would be different for eachindividual person, and thus couldn’t be standardized. Asystem with consistent or standardized measuring unitswas necessary for accurate scientific study.

In 1799, the French adopted a decimal system of measure-ment called the metric system. They also set up standardsfor the basic units they defined. The basic units of measure-ment in the metric system are the meter (length), the liter(volume), and the gram (weight). The main advantage ofthe metric system is that all of its units are based on thenumber 10. Any metric quantity is obtained by multiplyingor dividing a base unit by ten. Thus, for example, 1 meterequals 10 decimeters, 1 decimeter equals 10 centimeters,and one centimeter equals 10 millimeters. Figure 2 showsa chart of basic metric units and their abbreviations.

Many people in the United States are familiar with theEnglish system of measurement (also called the customarysystem). The basic units of measurement in the Englishsystem are the foot (length), the pound (weight), and thequart (volume). English measurement units are commonlyseen in the United States on product labels, with metricequivalents printed alongside.

The problem with the English system is that it’s somewhatcomplicated—the relationships between units of the samequantity aren’t uniform. For example, 1 yard equals 3 feet,1 foot equals 12 inches, 1 pound equals 16 ounces, and1 gallon equals 4 quarts. Figure 3 shows a chart of basicEnglish units and their abbreviations.

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FIGURE 2—This chart liststhe basic units of the metricsystem of measurement,their abbreviations, and values.

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FIGURE 3—This chart liststhe basic units of theEnglish system of measure-ment, their abbreviations,and values.

Mechanics 5

Instead of having to memorize the many different conversionsof the English system, all measurements in the metric systemare based on multiples and products of 10. Thus, the metricsystem is much easier to use, particularly for scientific pur-poses. Basic measurement conversions for the English andmetric systems are shown in the Appendix at the back of thisstudy unit.

The modern international system of measurement is basedon the original metric system. This system is called theSysteme International (SI), and has been adopted worldwide.The seven base units of the SI metric system are the meter(length), the kilogram (mass), the second (time), the ampere(electric current), the kelvin (temperature), the candela (lightintensity), and the mole (amount of substance). These unitshave been standardized worldwide for accuracy in scientificmethods and research.

Thus, although the English system is still used in the UnitedStates for some casual purposes, the simpler and more pre-cise SI system is preferred for use in scientific and technicalapplications.

Speed and Velocity

The terms “speed” and “velocity” are often used interchange-ably; however, there are some important differences betweenthe two. Speed is the distance traveled by a body in anydirection in a unit of time. Velocity is the distance traveledby a body in a particular direction (that is, a straight line)in a unit of time.

It’s not surprising that direction is seldom considered whenmotion is discussed. We’re aware of speed because of speedlimits placed on car driving; however, since we must turnwhen the road turns, we’re rarely aware of the directions.The concept of maintaining accurate directions is often lostbecause, when traveling from city to city, we’re guided byroad signs. However, when you think of an airplane pilottrying to travel between two points in space, the importanceof direction as well as the amount of distance at a rate oftime becomes obvious. An airplane pilot must be consciousof where the airplane is in relation to its point of takeoff anddestination without guiding signs.

Mechanics

In problems dealing with motion along a straight line, or linear motion, the terms “speed” and “velocity” may be used interchangeably because direction remains unchanged.

Average Speed

Assuming that the motion is linear, all rules concerningspeed can be applied to velocity. The velocity of a body canbe determined by the following formula:

v = �st�

In this formula, v stands for velocity in meters per second,s stands for distance in meters, and t stands for time inseconds.

The velocity formula is also valid if the distance is measuredin kilometers and the time in hours; in this case, the velocityis measured in kilometers per hour. When the English orcustomary system is used, the distance is measured in miles,the time in hours, and the velocity in miles per hour.

There are other factors that must be considered in everydayapplication of the velocity formula. Let’s look at an exampleproblem. Suppose an automobile has traveled from City Ato City B, which are 420 kilometers apart. If it’s known thatthe trip took 6 hours, the velocity can be found by using theformula as follows:

v = �st�

v = �420

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v = 70 km/h (kilometers per hour)

Of course, the car didn’t travel at exactly 70 km/h at alltimes during the trip. (For example, there are intersectionswhere the car had to slow down or stop.) Instead, 70 km/his the average speed that the car traveled. In contrast, theinstantaneous speed is the real speed of the car at a particu-lar instant. A car’s instantaneous speed changes often duringa trip—the car may speed up to pass another vehicle or slowdown at intersections. In a car, an instrument called aspeedometer shows the car’s instantaneous speed, and anodometer registers the total distance traveled by the car.

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Mechanics 7

If the speed is changing frequently within an observeddistance, a better measure of the speed at any time canbe obtained by taking smaller and smaller intervals ofdistance and, therefore, smaller intervals of time. In thisprocedure of measurement, the only difficulty is the accuracywith which the smaller intervals can be measured. Mostbodies described in the following passage will have eitheruniform velocities or velocities that are changing in a waythat averages can be used.

Uniform Velocity

A body is said to be moving at uniform velocity if it moves ina fixed direction and covers equal distances in equal intervalsof time. The movement of such a body can be represented bygraphing points in a coordinate system; the result, whenplotted, will be a straight line.

Assume that a body travels through points A, B, C, and Dduring a period of 4 hours and moves 60 kilometers duringeach hour. The progress of this body is shown in the graphin Figure 4. If time is plotted on the horizontal axis of thecoordinate system and distance on the vertical axis, points A,B, C, and D can be located in the system shown. Becausethe distance covered in each time interval is the same, thepoints are located on a straight line.

240

180

120

60

1 2

Time, in Hours

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inK

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3 4

A (1, 60)

B (2, 120)

C (3, 180)

D (4, 240)

A 1

Time,h

Distance,km

60

120

180

240

2

3

4

B

C

D

FIGURE 4—Because the velocity is uniform, the progress of the body is illustratedas a straight line. It can be said that the distance is changing uniformly with thetime, or that the distance is a linear function of the time.

Mechanics

Read through the following example problems to practice theprinciples discussed in this section.

Example: An airplane travels between City A and City B, a distance of 750 kilometers (km). If the trip takes 2.5hours (h), what is the average speed of the airplane?

Solution: Use the formula for uniform velocity to calculatethe answer. Remember that v stands for the uniformvelocity (average speed), s stands for the distance covered,and t stands for the time it takes to get to the destination.Substitute the given values into the formula and solve.

v = �st�

v = �7

2

5

.

0

5

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v = 300 km/h

Answer: The average speed of the airplane is 300 km/h.

Example: A motorboat travels at a uniform speed of5 meters/second (m/s). How far does the boat travelin 10 minutes?

Solution: Use the formula for uniform velocity to calculatethe answer, and solve for distance (s) instead of velocity.The following variation of the velocity formula can beused to solve this problem:

s = v � t

In order to use the formula correctly, you must convert theminutes in this problem to seconds. There are 60 seconds inone minute, so multiply 60 times 10 to find the total numberof seconds in 10 minutes.

60 � 10 = 600 seconds

Now, substitute the values given in the problem into theformula and solve.

s = v � t

s = 5 m/s � 600 seconds

s = 3,000 meters

Answer: The boat travels 3,000 meters in 10 minutes.

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Mechanics 9

Example: How much time should be allowed for a 360-milecar trip if the car will be traveling at an average speed of45 miles per hour (mph)?

Solution: Use the formula for uniform velocity to calculatethe answer, and solve for time (t) instead of velocity.The formula t = s/v, a variation of the velocity formula,can be used to solve this problem. Substitute the valuesgiven in the problem into the formula and solve.

t = �sv

�

t = �3

4

6

5

0

m

m

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h

es�

t = 8 hours

Answer: The time allowed for the car trip should be 8 hours.

Acceleration

Acceleration occurs when the velocity of a body changes. Thevelocity may change in magnitude, in direction, or in both atthe same time. Any change in velocity is called acceleration.In the following discussion of acceleration, only linear motion(motion in a straight line) will be covered. The direction ofvelocity will therefore remain constant. Acceleration can beexpressed as follows:

acceleration = change in velocity ÷ time required

This relationship is shown in the following formula:

a = �v2 –

t

v1�

In this formula, a stands for acceleration, t stands for thetime interval, v2 stands for velocity at the end of the timeinterval t, and v1 stands for velocity at the beginning of thetime interval t.

Example: Consider a car whose velocity changes uniformlyfrom 22 meters per second (m/s) to 33 m/s over a timeinterval of 22 seconds (s). What is the acceleration of thecar?

Mechanics

Solution: Use the acceleration formula to find the answer.

a = �v2 –

t

v1�

Now, substitute the values given in the problem into theformula and solve. Note that the answer will be expressedin meters per second squared (m/s2).

a = 33 m/s – 22 m/s22 s

a = �1212�

a = 0.5 ms2

Answer: The acceleration of the car is 0.5 m/s2.

In general, acceleration is expressed as length units pertime unit squared. In this example, the car’s velocity changes0.5 meter per second every second, or 0.5 meter per secondsquared.

The meaning of acceleration is best expressed as a velocitychange during a unit of time. A graphical representation of avelocity change is shown is Figure 5. The graph gives pointsA, B, C, D, and E which a body reaches after traveling 1, 2,3, 4, and 5 seconds. If the points are plotted in a coordinatesystem that relates time and distance, as shown in the graph,the points form a curved line. The curve becomes steeperafter every consecutive second, which means that the velocityis greater in every second or that the body is moving with anacceleration.

When the velocity decreases in each consecutive time unit, abody is moving with a deceleration. The solution found in theacceleration formula will be negative. That is, a body movingwith a negative acceleration is slowing down or decelerating.

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Mechanics 11

Average Velocity

If the velocity changes with constant acceleration during atime interval, the average velocity may be calculated with thefollowing formula:

vav = �v1

2

+ v2�

In this formula, vav stands for the average velocity during atime interval, v1 stands for the original velocity, and v2

stands for the final velocity.

The distance s traveled by the body during a time interval t isthen

s = vav � t, or

s = �v1

2

+ v2� � t

This formula has a wide application in practical problems.

40

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30

20

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Time, in Seconds

Dis

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3 4 5

A (1, 3)

B (2, 12)

C (3, 27)

D (4, 48)

E (5, 75)

A 1

Time,s

Distance,m

3

12

27

48

75

2

3

4

5

B

C

D

E

FIGURE 5—This graph represents an accelerated motion. The distance covered in eachtime unit is greater than in the preceding time unit, and isn’t changing uniformly withtime. The increase in velocity indicates an acceleration of the body.

Mechanics

Example: Assume a body is traveling in a straight linewith an original velocity of 10 m/s. The body acceleratesuniformly and reaches its destination at a final velocityof 29.2 m/s. What is the average velocity of the body?

Solution: Use the average velocity formula to calculate theanswer. Substitute the values given in the problem intothe formula and solve.

vav = �v1

2

+ v2�

vav = 10 m/s + 29.2 m/s2

vav = �392.2�

vav = 19.6 m/s

Answer: The average velocity of the body is 19.6 m/s.

Example: If the body in the previous problem took 4 sec-onds to reach its destination, find the distance traveled.

Solution: Use the average velocity formula to calculatethe answer, and solve for distance (s) instead of averagevelocity. The following variation of the average velocityformula can be used to solve this problem:

s = vav � t

Now, substitute the values given in the problem into theformula and solve.

s = vav � t

s = 19.6 m/s � 4 s

s = 78.4 m

Answer: The body traveled a distance of 78.4 meters.

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Mechanics 13

Calculating Uniform Acceleration

As shown before, the following is the basic formula used tocalculate acceleration:

a = �v2 –

t

v1� (Formula 1)

The following variation of the acceleration formula can beused to calculate the final velocity (v2):

v2 = v1 + at (Formula 2)

In some problems, it’s necessary to find the time interval (t).The following variation of the acceleration formula can beused to calculate the time interval:

t = �v2

a

– v1� (Formula 3)

As you saw previously, the following variation of the accelera-tion formula can be used to calculate the distance covered bya body:

s = �v1

2

+ v2� � t (Formula 4)

If the value of v2 is substituted in Formula 4, the distancecan be expressed using the original velocity v1 and the timeinterval t:

s = v1t + �12

�at2 (Formula 5)

Now, let’s look at some example problems.

Example: A moving vehicle at the beginning of a measuredinterval has a velocity of 10 m/s. The vehicle then accel-erates uniformly with an acceleration of 2 m/s2. What isthe velocity of the vehicle after an interval of 4 seconds?

Solution: Use Formula 2 to solve this problem. Substitutethe given values into the formula, then solve.

v2 = v1 + at

v2 = 10 m/s + (2 m/s2 � 4 s)

v2 = 10 + 8

v2 = 18 m/s

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Answer: The velocity of the vehicle is 18 m/s after aninterval of 4 seconds.

Example: A body starts to move with a velocity of 10 m/sand has an acceleration of 5 m/s2. How far does the bodytravel in 8 seconds?

Solution: Because the original velocity, acceleration, andtime are given, you can calculate the distance by usingFormula 5. Substitute the given values into the formula,then solve. (Note: As you solve the problem, be sure toperform the operations inside parentheses first.)

s = v1t + �12

� at2

s = (10 m/s � 8 s) + �12

� � [5 m/s2 � (8 s)2]

s = 80 + �12

� � (5 � 64)

s = 80 + (�12

� � 320)

s = 80 + 160

s = 240 m

Answer: The body travels 240 meters in 8 seconds.

Example: For the body in the previous problem, what is itsvelocity at the end of 8 seconds?

Solution: Use Formula 2 to calculate the final velocity.Substitute the given values into the formula, then solve.

v2 = v1 + at

v2 = 10 m/s + (5 m/s2 � 8 s)

v2 = 10 + 40

v2 = 50 m/s

Answer: The velocity is 50 m/s at the end of 8 seconds.

Example: A car is traveling at a speed of 10 m/s. When thebrakes are applied, the car takes 2 seconds to stop. Whatis the acceleration?

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Mechanics 15

Solution: The acceleration you’re being asked to calculate inthis problem is actually a negative acceleration, or decel-eration. When the car stops, the final velocity (v2) is zero.You can use the final velocity formula to calculate theacceleration.

v2 = v1 + at

0 m/s = 10 m/s + (a � 2 s)

0 = 10 + 2a

Subtract 2a from both sides of the equation.

0 – 2a = 10 + 2a – 2a

–2a = 10

Divide both sides of the equation by –2 to solve for a.

–2a/–2 = 10/–2

a = –5 m/s2

Answer: The negative acceleration (deceleration) of the car is –5 m/s.

Falling Bodies

A free-falling body (such as a rock being dropped from thetop of a cliff) is an application of body movement with uni-form acceleration. Such a body is assumed to be falling inthe vicinity of earth and to have an acceleration due to theforce of the earth’s gravity. The gravity acceleration has thesymbol g in the following formulas and is measured in m/s2.

All formulas for calculating the final velocity, time, anddistance of a body moving with uniform acceleration canbe applied to falling bodies if the following conditions areobserved:

1. Assume that the gravity acceleration g is a constant(or unvarying rate of acceleration throughout the fall).

2. Ignore the resistance of air.

Mechanics

In acceleration formulas applied to the falling bodies, theacceleration (a) equals the gravity acceleration (g). You canuse the following formula to calculate the final velocity (v2):

v2 = v1 + gt (Formula 1)

You can use the following formula to calculate the distancetraveled (s):

s = v1t + �12

� gt2 (Formula 2)

Actually, the acceleration due to gravity is not exactly aconstant. Its magnitude depends on the distance of thebody from the center of the earth. For example, on a highmountain, the acceleration will be slightly different than atthe bottom of the sea. However, for all practical purposes,you should assume that the acceleration of gravity (g) is aconstant equal to 9.8 m/s2. In customary units, g is equalto 32 feet/second2.

When a body falls in air, the air resistance influences themovement of the body and exerts a slowed acceleration. Aclassic example that illustrates the effect of air resistance ona falling body is the comparison of the fall of a small piece oflead and that of a large feather of equal weight. The lead willfall faster because the air resistance acting on the surface ofthe feather has a far greater effect than it has on the muchsmaller lead surface.

Example: If a ball with an original velocity of zero is fallingfrom a tower and takes 4 seconds to hit the ground, howtall is the tower?

Solution: The height of the tower equals the distance traveledby the ball. Use Formula 2 to calculate the distance.Substitute the given values into the formula, then solve.

s = v1t + �12

� gt2

s = (0 m/s � 4 s) + �12

� � [9.8 m/s2 � (4 s)2]

s = 0 + �12

� � (9.8 � 16)

s = 0 + �12

� � 156.8

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Mechanics 17

s = 0 + 78.4

s = 78.4 meters

Answer: The tower is 78.4 meters tall.

Example: For the ball in the previous problem, with whatvelocity does the ball reach the ground?

Solution: Use Formula 1 to calculate the final velocity.Substitute the given values into the formula, then solve.

v2 = v1 + gt

v2 = 0 m/s + (9.8 m/s2 � 4 s)

v2 = 0 + 39.2

v2 = 39.2 m/s

Answer: The ball hits the ground at a velocity of 39.2 m/s.

Now, take a few moments to review what you’ve learned bycompleting Self-Check 1.

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Self-Check 1At the end of each section of Mechanics, you’ll be asked to pause and check your under-standing of what you have just read by completing a “Self-Check” exercise. Answeringthese questions will help you review what you’ve studied so far. Please complete Self-Check 1 now.

Questions 1–12: Fill in the blanks in the statements.

1. The basic unit of length in the SI measurement system is the _______.

2. The velocity of a moving body includes both the rate and the _______ of motion.

3. If distance is expressed in meters and time is expressed in seconds, the velocity of a body isexpressed in _______.

4. Any change in velocity is called _______.

5. A body that’s moving with a negative acceleration is said to be _______.

6. In mechanics, any distinct quantity of matter is called a _______.

7. In the SI system of measurement, the basic unit of mass is the _______.

8. The terms “speed” and “velocity” may be used interchangeably when motion is _______.

9. The average speed of a body can be determined by using the formula _______.

10. When a body has a uniform acceleration, its _______ changes uniformly.

11. The original velocity of a free-falling body is _______.

12. The formula used to calculate the final velocity of a free-falling body is _______.

(Continued)

Mechanics 19

Self-Check 1Questions 13–16: Solve each problem.

13. If the velocity of an object changes from 15 m/s to 50 m/s during a time interval of 4 s,what is the acceleration of the object?

__________________________________________________________________________

14. What is the average velocity of the object in the previous problem?

____________________________________________________________________________

15. If a coin is thrown with a starting velocity of 12 m/s down a dry well and hits bottom in 1.2 s,what is the depth of the well?

__________________________________________________________________________

16. A car starts from rest and accelerates uniformly at the rate of 1.5 m/s2. How long does it takethe car to reach a velocity of 18 m/s?

__________________________________________________________________________

Check your answers with those on page 93.

Mechanics

FORCE AND MOTION

Sir Isaac Newton was a great mathematician and scientistwho lived in England about 300 years ago. Although he mademany important discoveries and contributions to science,some of his most important work involved the mechanics of force and motion. Newton developed three basic theoriesrelating to inertia, the relation between force and mass, and the forces of action and reaction. These are known asNewton’s Laws of Motion. You’ll learn about Newton’s laws inthis section of your study unit. The difference between forceand mass is emphasized, and units of measure for mass andforce are explained.

The following are some important terms that you should befamiliar with.

• Force. Strength or energy used to cause a motion orchange; usually, a push or a pull.

• Inertia. A body’s resistance to motion.

• Mass. A body’s quantity of matter.

• Newton. The metric unit used to measure force; namedfor the scientist Isaac Newton.

What Affects the Motion of a Body?

In developing formulas for velocity and acceleration in theprevious chapter, no mention was made of what causes themotion of the body. Now, the questions arise: What causesmotion? Does any characteristic of the body help or hinderthe motion? What outside effects can change the motion?What contributes to the description of motion with sufficientprecision that the motion can be predicted?

The following discussions will give you some insight into theconditions that affect motion. For example, it’s easier to slidean object over a smooth surface than over a rough one. Theresistance offered by the surfaces affects the motion. Oil isused between surfaces of resistance, such as sliding pistonsand the inner cylinder surfaces in a car engine, to overcomethe resistance of the surfaces.

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Mechanics 21

The body itself may have an effect on the motion; it’s certain-ly easier to push a small stalled car than a large stalled car.Experience tells us that motion is always associated with apush or pull, which is called the force. There is always arelation between a motion and the force acting on the body.

Newton’s First Law

Consider a body sliding over a relatively smooth surface. Thesmoother the surface in contact, the easier it will be to movethe body. If a lubricant is introduced between the body andthe surface, the body will move even more freely. If wheels areput on it, the body might move even more freely. In theory,it might be possible to get the frictional resistance betweenthe body and the surface so small that, once the body startedmoving, there would be little or no tendency for it to stop.Under perfect conditions it might keep moving forever. Thatpossible result was observed by Galileo in his writings andlater expressed by Newton’s first law of motion: A body atrest or in motion at a constant speed tends to remain at rest or in motion at constant speed unless acted upon by someexternal force.

In the example of the sliding body, the friction was an exter-nal force that was minimized by lubrication between the bodyand the surface. In the absence of an external force the bodywould remain either at rest or in motion at constant speed.A body, whether moving or resting, resists any change due toan external force. But external forces do act upon bodies andcause them to start moving and to change their movement.

As mentioned earlier, it’s much easier to push a small stalledcar than a large stalled car. The first impression is to attributethe difference in difficulty to the difference in size of the cars.However, that’s not a precise way to figure the difference indifficulty. A better approach is to try to make more generalthe characteristic of a body to resist change in motion. Thecharacteristic is called inertia. The large car has more inertiathan the small car. Because of its greater inertia, it’s moredifficult both to put the larger car in motion and to slow itdown once it’s moving. It’s easily seen that the concept ofinertia is related to quantity of matter, otherwise known as

Mechanics

mass. Mass is defined as a measure of inertia or as a meas-ure of a body’s resistance to change in motion.

A common example of the effect of inertia is illustrated inFigure 6. A man standing on a bus that’s at rest is shown in Figure 6A. Because of inertia, the man’s body tends toremain at rest while the bus is standing still. When the busis moving forward and then slows down, inertia causes theman’s body to continue moving at the original speed. Theoverall effect is that of being pushed forward (Figure 6B).In Figure 6C, the bus moves forward. Because of inertia, theman’s body tends to remain at rest when the bus moves andspeeds up; the overall effect is that of being pushed back.

22

(A)

(B) (C)

FIGURE 6—This person is experiencing the effect of inertia while standing on a bus. Whenthe bus is at rest, he has no trouble maintaining his balance, as shown in 6A. After thebus is moving and then slows down, he is thrown toward the front of the bus (6B). Whenthe bus suddenly moves again, he is thrown off balance toward the back of the bus (6C).

Mechanics 23

Newton’s Second Law

A better insight into the meaning of mass and inertia canbe gained by studying Newton’s second law of motion, whichdescribes what happens to a body when external forces acton it. Most simply, Newton’s second law can be stated asfollows: When an external force acts on a body, that body isaccelerated with a magnitude proportional to the force andinversely proportional to the mass. In the form of an equation,this law can be expressed as follows:

a = �mF

�, or

F = ma

In this formula, a stands for acceleration in meters persecond squared, F stands for force in newtons, and mstands for mass in kilograms.

Newton’s second law relates mass and force to acceleration,which has already been defined. The equation of Newton’ssecond law is used to derive the unit of force. The metricunit of force is the newton. It’s the force which will acceleratea mass of one kilogram at a rate of one meter per secondsquared (remember, squared means the number is multipliedby itself). In other words, the mass of one kilogram under-goes a change in velocity of one meter per second everysecond if acted upon by a force of one newton. Mass is notonly a measure of inertia but is also thought of as a quantityof matter. As mentioned earlier, the metric unit of mass isthe kilogram.

The mass of a body is assumed to be a natural characteristicof the body. It doesn’t depend on location, shape, color, orsize. The greater the mass, the greater the force that’s neededto start it moving or to accelerate it.

Example: A force of 10 newtons (N) is applied to a ball witha mass of 0.5 kilograms (kg). What is the acceleration ofthe ball?

Solution: Use the formula for Newton’s second law to solvethe problem. Substitute the given values into the formula,then solve.

Mechanics

a = �mF

�

a = �01.50

kNg

�

a = 20 m/s2

Answer: The acceleration of the ball is 20 m/s2.

Example: Assuming the ball in the previous problem startsfrom rest, what is the velocity after 5 seconds?

Solution: Use velocity Formula 2 to calculate the velocity.Substitute the given values into the formula, then solve.(Remember that the value of v1 is 0.)

v2 = v1 + at

v2 = 0 m/s + (20 m/s2 � 5 s)

v2 = 0 + 100

v2 = 100 m/s

Answer: The velocity of the ball after 5 seconds is 100 m/s.

Example: How far does the ball in the previous problemtravel during the time interval of 5 seconds?

Solution: Use the distance formula to solve this problem.Substitute the given values into the formula, then solve.

s = v1t + �12

� at2

s = (0 m/s � 5 s) + �12

� [20 m/s2 � (5 s)2]

s = 0 + �12

� (20 � 25)

s = 0 + �12

� (500)

s = 0 + 250 meters

s = 250 meters

Answer: The ball travels 250 meters in 5 seconds.

Example: What force (in newtons) is required to accelerate abody with a mass of 10 kilograms at a rate of 6 m/s2?

Solution: Use a variation of the formula for Newton’s secondlaw to solve for force. Substitute the given values into theformula, then solve.

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Mechanics 25

F = ma

F = 10 kg � 6 m/s2

F = 60 N

Answer: A force of 60 N is required.

Example: A car with a mass of 1,200 kg is traveling at aspeed of 10 m/s. What force must be applied to stop thecar in 4 seconds?

Solution: The acceleration you’re being asked to calculateis a negative acceleration, or deceleration. When the carstops, the final velocity (v2) is zero. You can use the finalvelocity formula to calculate the acceleration.

v2 = v1 + at

0 m/s = 10 m/s + (a � 4 s)

0 = 10 + 4a

Subtract 4a from both sides of the equation.

0 – 4a = 10 + 4a – 4a

–4a = 10

Divide both sides of the equation by –4 to solve for a.

�––44a

� = �1–40�

a = –2.5 m/s2

The acceleration is negative, which means that the car isslowing down. Next, use the formula F = ma to calculate theforce needed to cause that acceleration.

F = ma

F = 1,200 kg � (–2.5 m/s2)

F = –3,000 N

Answer: A deceleration or braking force of –3,000 N must beapplied to the car to stop it in 4 seconds.

Mechanics

Gravity

In your study of accelerated motion, uniform accelerationwas discussed. Uniform acceleration is the accelerationdue to gravity. If the gravity acceleration g is substitutedfor general acceleration a in the equation of Newton’s secondlaw, the equation becomes

F = mg

In this formula, F stands for the force of gravity in newtons,m stands for mass in kilograms, and g stands for gravityacceleration (9.8 m/s2).

The force obtained by this equation is the force actingon a body and causing the earth’s gravity acceleration; it’scalled the force of gravity, or gravity. As mentioned before,the gravity acceleration varies over the surface of the earth,and the force of gravity varies proportionally to the gravityacceleration. On a high mountain peak (where the distancefrom the center of the earth is greater than it is at sea level),the force of gravity is slightly lower than at the bottom ofthe mountain. An even greater change in the force of gravitywould be experienced on the moon. Because the moon is lessmassive than the earth, the force of gravity on it is aboutone-sixth of the force of gravity on the earth.

Another important fact to keep in mind is that the directionof the force causing acceleration is in the same direction as the acceleration. For example, the gravity acceleration is inthe same direction as the force of gravity, that is, toward thecenter of the earth or, on the moon, toward the center of themoon.

In everyday measuring while using English units, it hasbecome common practice to use the term “weight” for boththe mass and force of gravity. The English unit of one poundis used loosely when either the mass or the force of gravitythat acts upon that mass is measured.

In a discussion of scientific problems such as problems ofmechanics, the term “weight” should either be avoided orused only to indicate the force of gravity that acts upon amass.

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Mechanics 27

When metric units are used, the confusion of mass and forceof gravity is easily avoided. Mass is measured in kilogramsand the force of gravity is measured in newtons. A quantityexpressed in kilograms, grams, or megagrams is the mass; aquantity expressed in newtons is a force—including the forceof gravity, or “weight.”

Example: An astronaut’s body has a mass of 70 kilograms.What is the force of gravity acting on his mass (that is,his “weight”) on earth?

Solution: Use the force formula. Substitute the known valuesinto the formula and solve. (If not indicated otherwise,assume that g = 9.8 m/s2.)

F = mg

F = 70 kg � 9.8 m/s2

F = 686 N

Answer: The force of gravity acting on the astronaut onearth is 686 newtons.

Example: What is the force of gravity (the “weight”) of thesame astronaut on the moon?

Solution: Use the force formula to solve this problem.Substitute the known values into the formula and solve.

F = mg

F = 70 kg � 9.8 m/s2

F = 686 N

Next, divide the answer by six, because the force of gravityon the moon is about one-sixth of the force of gravity on theearth.

F = 686 N ÷ 6

F = 114.3 N

Answer: The force of gravity acting on the astronaut on themoon is 114.3 newtons.

Example: What is the force of gravity acting on the mass ofa 3-kilogram package of potatoes?

Mechanics

Solution: Use the force formula. Substitute the known valuesinto the formula and solve.

F = mg

F = 3 kg � 9.8 m/s2

F = 29.4 N

Answer: The force of gravity acting on the package ofpotatoes is 29.4 newtons.

Example: The force of gravity acting on a girl’s mass onearth is 550 newtons. What is her mass?

Solution: You can use a variation of the force formula tosolve for mass. Substitute the known values into theformula and solve.

m = �Fg

�

m = �9.

585m0

/Ns2�

m = 56.123 kg, or 56 kg rounded

Answer: The girl’s mass is 56 kilograms.

Action and Reaction

Newton’s second law states that if an external force is exertedon a body, the body will be accelerated. Of course, if the bodyisn’t free to move, it will push back with a force equal to theapplied force. Under those conditions there will be no netexternal force acting on the body and the body won’t move.

For example, if you push downward on a desk, the deskmust push upward against your hand if there is no motion.Also, the harder you press, the harder the desk pushes backon your hand. This seemingly simple situation is the basisfor Newton’s third law, which can be stated in the followingway: Whenever one body exerts a force on a second body, thesecond body exerts an equal and opposite force on the firstbody. It’s usually stated as follows: Every action has an equaland opposite reaction.

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Mechanics 29

Force always occurs in pairs, never alone. If a hunter fires agun, the force exerted on the bullet by the exploding powderis matched by the recoiling force exerted on the gun and thehunter’s hand or shoulder.

Vector and Scalar Quantities

So far in this study unit, you’ve learned about the movementof bodies in a straight line. When a force acting on a bodywas considered, only motion in the same direction as oropposite to the force was studied. When the force and themotion of the body are not in the same opposite direction, adifferent procedure should be used to find the influence ofthe force on the body.

Assume that a block is sliding on a horizontal surface asshown in Figure 7. The force that causes this sliding is shownby the arrow F (force). As you can see, the force causes theblock to move horizontally to the right. However, the entireforce doesn’t contribute to the horizontal motion, because theforce also tends to lift the block, not just slide it along thesurface.

A vector is a quantity that has both a magnitude and adirection. Usually, a vector is represented on paper by astraight line with an arrowhead on one end. The lengthof the line is proportional to the magnitude of the quantityrepresented, and the line is drawn so that its directionmatches the direction of the quantity it represents.

Figure 8 shows a simple illustration of a vector. In the figure,a boat sails 300 miles east away from an island. The “300miles east” is a vector that can be illustrated with an arrow.The vector’s magnitude is 300 miles, and its direction is east.

FFIGURE 7—The velocityof the block moving onthe horizontal surfacedepends to a largeextent on the directionof the applied force. Themore nearly parallel tothe surface the force isapplied, the faster theblock will move.

Mechanics

In the field of mechanics, vectors can be used to representother quantities, such as forces, velocities, and accelerations.Some vectors that represent forces and accelerations areshown in Figure 9.

The vectors shown in Figure 9 are graphical representationsof physical quantities that have both magnitude and direc-tion. The vector in Figure 9A represents a force of 150 N ina northeasterly direction. This vector is three times as longas the vector shown in Figure 9B, which represents a forceof 50 N in a northeasterly direction. The vector in Figure 9Crepresents an acceleration of 30 m/s2. The vector’s magni-tude is indicated by its length, and its direction is expressedby the angle it encloses with a reference line.

30

50 N45°

150 N

45°

45°

(A) (B) (C)

FIGURE 9—These vectors are graphical representations of physical quan-tities that have both magnitude and direction.

East

Boat

Island

300 Miles

FIGURE 8—In this illustra-tion, a boat starts at theisland and then sails 300miles east. The vectortherefore has a magnitudeof 300 miles and an east-erly direction.

Mechanics 31

In mechanics, you’ll also deal with quantities called scalars.A scalar is a quantity consisting of a single real numberthat’s used to measured magnitude (size). Voltage, mass,time, temperature, energy, and speed measurements can bedescribed as scalar quantities. So, some examples of scalarquantities would be “20 degrees Fahrenheit,” “10 pounds,”“500 volts,” and “60 miles per hour.”

The difference between speed and velocity can thus beexpressed very simply: speed is a scalar quantity, andvelocity is a vector quantity. Therefore, the speed “60 milesper hour” is a scalar quantity, but if we add a direction andsay “60 miles per hour west” the quantity becomes a vector.

Now, take a few moments to review what you’ve learned bycompleting Self-Check 2.

Mechanics32

Self-Check 2Questions 1–8: Fill in the blanks in the statements.

1. The measure of inertia is the _______ of a body.

2. Newton’s second law defines the relation between _______ and _______.

3. The component that differentiates speed from velocity is _______.

4. Two examples of vector quantities are _________ and ___________.

5. Newton’s _______ law explains why a person seated in a car tends to fall forward when thecar suddenly stops.

6. Two examples of scalar quantities are ________ and _________.

7. A mass of 1 kg is accelerated 9.8 m/s2 by a force of _______ N.

8. A directed quantity is a quantity that has both _________ and _________.

Questions 9–18: Solve each problem.

9. What force (in newtons) is required to accelerate a body with a mass of 32 kilogramsat a rate of 12 m/s2?

__________________________________________________________________________

10. What force (in newtons) is required to accelerate a body with a mass of 7 kilogramsat a rate of 15 m/s2?

__________________________________________________________________________

11. If a body with a mass of 4 kg is moved by a force of 20 N, what is the rate of itsacceleration?

__________________________________________________________________________

(Continued)

Mechanics 33

Self-Check 212. A force of 500 N causes a body to move with an acceleration of 16 m/s2. What

is the mass of the body?

__________________________________________________________________________

13. If a suitcase has a mass of 20 kg, what is the force of gravity acting on it?

__________________________________________________________________________

14. The force of gravity on a boat is 3,000 N. What is the mass of the boat?

__________________________________________________________________________

15. On the moon, what would be the force of gravity acting on an object that has amass of 7 kg?

__________________________________________________________________________

16. What is the force of gravity acting on an astronaut on the moon if his mass is 80 kg?

__________________________________________________________________________

17. On earth, what is a child’s mass if the force of gravity on the child’s body is 100 N?

__________________________________________________________________________

18. A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force mustbe applied to stop the car in 3 seconds?

__________________________________________________________________________

Check your answers with those on page 94.

Mechanics

MOTION IN A CURVED PATH

When a body is moving in a curved path, different rulesapply to it than when it’s moving in a straight line. In thissection of your study unit, we’ll examine centripetal forceand gravitational force. You’ll also learn about the models ofplanetary movement developed by Ptolemy and Copernicus,as well as Kepler’s laws.

The following are some important terms you should familiarizeyourself with.

• Centripetal force. A center-seeking force related to acceleration.

• Frictional force. A naturally occurring force related totwo bodies in motion.

• Gravitational force. The force that provides the attractionbetween masses (such as planets) and keeps thosemasses in their orbits.

• Ellipse. An oval-shaped orbit of planets around the sun.

Velocity has been defined as a vector quantity that hasboth magnitude and direction. Earlier in this study unit, youlearned that any change in velocity results in an acceleratedvelocity. The acceleration may be due to a change of direc-tion, a change of magnitude, or to changes of both directionand magnitude.

In preceding discussions, only linear motions were consideredand the acceleration discussed was due to the change invelocity magnitude. Now the discussion will turn to motion in a curved path with a constant speed.

Centripetal Force

At first glance, it’s not obvious how a body traveling at a constant speed can be accelerated. Does a car rounding acurve experience an acceleration (Figure 10)? How can theacceleration exist? Does it have direction? If so, what is it?

These questions can be discussed by using a car on a curvedpath as an example. At each instant the velocity is different,

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Mechanics 35

as shown by the positions of velocity vectors in Figure 10. Allof the vectors are the same length because the magnitude ofthe velocity is constant, but each of the vectors has a differ-ent direction. Because the velocity changes direction, there’sacceleration. Whenever an acceleration is present, it must beassumed that a force causing the acceleration is also present.

In Figure 11, centripetal force (or center-seeking force) isacting on a car. The car is moving from the bottom of thefigure upward on a curved path. The force is directed towardthe center of the circular path of the car, causing the carto depart from the straight-line path and follow the circularpath. The vector directed toward the center of the path atthe bottom of Figure 11 indicates the centripetal force, whichis the frictional force of the road on the tires.

Icy Patch

Centripetal Force

FIGURE 11—In this illustra-tion, the car is moving fromthe bottom of the figureupward along the curvedroad. At the bottom of thefigure, the frictional forcepushing the car toward thecenter of the circular pathis large enough to keep thecar moving in the directionof the path. When the carstrikes the patch of ice atthe center of the figure, itskids and continues in thedirection it was headedbefore it hit the ice. Thevector drawn from the iceindicates the direction ofthe velocity when the cen-tripetal force ceases to actthrough loss of friction.

A

B C

FIGURE 10—This sketchshows three positions, A,B, and C, of a car that’srounding a curve at auniform magnitude ofvelocity. The vectors VA,VB, and VC represent thevelocities at points A, B,and C. The difference invectors proves the occur-rence of acceleration.

Mechanics

As the car moves, the direction changes every instant andthe car follows the road. As long as the car is traveling slowlyenough, it can usually round a curve without difficulty.However, if the car is going too fast, if the tires are worn,or if the road is wet or icy, a skid may result. When the carskids, as shown at the center of Figure 11, the centripetal(frictional) force necessary to negotiate the curve is lost,and the car moves in a straight-line direction—that is, ina straight line off the road.

When the roadbed is flat, the only force exerted towardthe center of the curve is the frictional force between thetires and the road. As the speed of the car increases, thatfrictional force is often not enough to cause the direction ofthe automobile to change and follow the curved roadbed.

As shown in Figure 12, the roadbed may be inclined orbanked from side to side. In this case, the normal forces onthe wheels have horizontal components that help the hori-zontal component of the friction to provide the centripetalforce that allows the car to travel in a curved path.

Another example of a body traveling in a curved path is amass that’s attached to a string and whirled in a circle ofconstant radius at a constant speed as shown in Figure 13.It’s immediately evident that the tension in the string acts asthe force that constrains the mass to move in the indicatedpath.

36

Center Seeking Force

Gravity

Friction

FIGURE 12—This a cross section of a banked roadbed. The angle of the bank incline is very pronounced at a racetrack; it allows cars to travel at a very high speed without skidding.

Mechanics 37

The center-seeking force is the centripetal force, and thattype of motion is called central-force motion. If the masskeeps whirling in a circular path at a higher and highervelocity, the needed tension becomes greater than the break-ing strength of the string, and the string will break. That typeof whirling mass can be used to make an effective slingshot(Figure 14).

FIGURE 13—When a per-son whirls a mass on theend of a string, the cen-tripetal force acts towardthe center of the circularpath and perpendicular tothe direction of the motionshown by vectors.

Path of Rock

FIGURE 14—In ancienttimes, a slingshot consistedsimply of a small leatherpad held by two leatherthongs. A stone would beplaced in the leather pad.The user would whirl theslingshot at an ever-increasing speed in a verti-cal plane, then release oneof the leather thongs. Thestone would then fly fromthe slingshot in a straightline.

Mechanics

Calculating Centripetal Force

The centripetal force acting on a mass that keeps moving in a circular path at a constant speed is directed toward the center of the circular path and can be expressed by thefollowing equation:

F = �mRv2�

In this formula, F stands for centripetal force in newtons,m stands for mass in kilograms, v stands for speed in metersper second (m/s), and R stands for the radius of the circularpath in meters.

Note that centripetal force does not act in the direction inwhich the body moves. It’s always directed toward the centerof the circular path of the body.

Example: A body has a mass of 40 kilograms and is movingwith a constant speed of 10 m/s in a circular path with aradius of 25 meters. What centripetal force is exerted onthe body?

Solution: Use the centripetal force equation to solve the problem. Substitute the known values into the formulaand solve.

F = �mRv2�

F = 40 kg � (10 m/s)2

25 meters

F = �40 �

25

100�

F = �402050

�

F = 160 N

Answer: The centripetal force on the body is 160 newtons.

Example: A car with a mass of 1,400 kilograms ismoving around a circular curve at a uniform velocityof 10 m/s. The curve has a radius of 50 meters. Whatis the centripetal force on the car?

Solution: Use the centripetal force formula. Substitute theknown values into the formula and solve.

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Mechanics 39

F = �mRv2�

F = 1,400 kg � (10 m/s)250 m

F = �1,40050

� 100�

F = �1405,0000�

F = 2,800 N

Answer: The centripetal force on the car is 2,800 N.

Example: A car with a mass of 1,100 kilograms is movingaround a circular curve at a uniform velocity of 15 m/s.The centripetal force on the car is 4,000 N. What is theradius of the curve?

Solution: You can use a variation of the centripetal forceformula to calculate the radius. Substitute the knownvalues into the formula and solve.

R = �m

F

v2�

R = 1,100 kg � (15 m/s)24,000 N

R = �1,1040,0

�

00225

�

R = �2447,0,50000

�

R = 61.875 m

Answer: The radius of the curve is 61.875 meters.

The Motion of Satellites

Another example of central-force motion is the motionof satellites around the earth (including the moon). Themoon travels around the earth in an almost circular path.The moon seems to travel at a constant speed, because eachorbit around the earth takes 27.3 days. Thus, there mustbe a centripetal force that keeps the moon in orbit. Thisforce is the gravitation force (Figure 15).

Mechanics

A similar force can be observed in man-made satellites.These satellites are commonly used for all sorts of communi-cation transmissions and are in constant orbit around theearth. Satellites are launched into orbit beyond the earth’satmosphere. Rockets are used to launch, place, and setthe speed for these satellites. Once satellites are set at aconstant speed at a predetermined distance from the earth,they can move freely in circular orbits without the applica-tion of any additional forces.

Even if a satellite’s path isn’t circular, it’s obvious that acentripetal force is necessary to keep it in orbit. That forceis the gravitational force mentioned earlier. To understandthe force more clearly, let’s examine some of the historicalmodels of the movement of celestial spheres.

Ptolemy’s Model of the Universe

Because human beings live on earth and have developedtheir observational and measurement techniques there, it’sonly natural that they try to describe the rest of the universein terms of the earth. In the first model of the universe pre-ferred by most people, the earth was stationary (stood still)and the rest of the planets, the sun, the moon, and the starsall moved with respect to the earth. That model was acceptedbecause by everyday observations (in ancient times) the earthappeared to be stationary.

40

Moon

Earth

FIGURE 15—The path of themoon around the earthreminds us of the motion ofa mass being whirled in acircle on a string, but there’sno string to exert the forcethat constrains the moon tofollow its path.

Mechanics 41

This model of a stationary earth was developed by theancient astronomer Ptolemy (pronounced “tall-a-mee”). Itwas widely accepted by the scientists of Ptolemy’s time andremained the most popular model for scientists for overone thousand years. In the Ptolemaic model (Figure 16), theearth stands still at the center of the universe and all otherbodies in the universe move around it in circular paths.When observed, the real planetary positions didn’t alwaysmatch their theoretical orbits. Therefore, to adjust for theactual observed positions, some of the planets were assumedto move in “secondary orbits.” These secondary orbits areindicated in Figure 16 by the smaller broken-line circles.

All of the planets and stars that were known to exist inPtolemy’s time were assumed to reside on the inside of avast rotating sphere that enclosed the moving planets.

Earth

Moon

Mercury

Venus

Sun

Mars

Jupiter

Saturn

Stars

FIGURE 16—In Ptolemy’s model, the earth was the center of the universe, and all the other planetswere assumed to move in circular orbits about the earth.

Mechanics

Ptolemy’s model did compare reasonably well with observa-tions made over the centuries, and could predict the futurepositions of the planets fairly well. There were no telescopesat that time, so all observations had to be made with thenaked eye. When later scientists developed better techniquesand more accurate viewing devices, they began to realize thatthere were serious differences between the true, observedplanetary positions and the positions predicted by Ptolemy’smodel.

Copernicus’ Model of the Universe

In the sixteenth century, Nicholas Copernicus, a native ofPoland, suggested a radical revision of the Ptolemaic model.He theorized that the earth rotates once a day on its axis,and at the same time orbits around the sun (which wouldbe assumed to be fixed in place). He further suggested thatthe other planets also orbit around the sun, but that themoon orbits around the earth. In the Copernican model, thestars were assumed to be at very great distances beyond theplanets and to be fixed in place. Figure 17 shows the orbitsof the planets in the Copernican model.

42

Moon

Venus

Sun

Saturn

Mercury

Earth

Mars

Jupiter

FIGURE 17—In the model of the universe suggested by Copernicus, the planets (including the earth)rotate around the sun in circular orbits.

Mechanics 43

The Copernican model does explain many observations. Therotation of the earth on its axis accounts for the apparentrising and setting of the stars. The orbital motion of theearth, as well as that of the moon, accounts for the apparentmotion of the sun and the moon with respect to the stars.

The Copernican model was much easier to understand andevaluate than the Ptolemaic model. When put to the test ofpredicting the future positions of the planets, however, it didonly slightly better than the Ptolemaic model, even thoughthe calculations were much simpler.

When the ideas of Copernicus were published, many longand bitter arguments resulted. The suggestion that the earthwas not the center of the universe went against the religiousteachings of the time and angered many government leaders.In many locations, Copernicus’ writings were banned andit was unlawful to express a belief in them. Both believersand nonbelievers made more attempts to make more exactastronomical measurements so that either or both modelscould be better tested. Remember, in the sixteenth century,optical instruments (such as the telescope) weren’t availableto make measurements.

Around that time, Tycho Brahe, a native of Denmark, builtan astronomical observatory near Copenhagen that was moreprecise than any other previously developed. Even thoughtelescopes were still in the future, techniques for measuringthe angular positions of the planets, as observed by the eye,had been made more accurate. Brahe’s instrumentation wassufficiently accurate to measure an angle to one-hundredthof a degree.

Tycho Brahe spent a lifetime making measurements and leftbehind a wealth of data to be used in checking the validity ofthe Copernican model. Brahe’s assistant, Johannes Kepler,was a believer in the theories of Copernicus, and he spentseveral years trying to corroborate the Copernican model byusing the data that Brahe and he had collected. After care-fully checking observations against predictions made withthe model, Kepler was forced to conclude that some errorsexisted in the model. The model is, however, correct inassuming that the earth rotates around the sun.

Mechanics

Kepler’s Laws

After very lengthy and exacting calculations, Kepler cameto the conclusion that the circular orbits of the Copernicanmodel couldn’t be allowed. Kepler then started to look atother possible geometric paths of the planets. It wasn’t longbefore he discovered that observations fit more closely if theassumed paths of the planets were elliptical (oval in shape).That led to a very careful study of some of the better-knownplanets and to Kepler’s discovery that the following lawsappear to govern the motion of planets:

1. The path of each of the various planets is an ellipse withthe sun at one focus (Figure 18).

2. As a planet moves in its elliptical path, a line drawn fromthe planet to the sun sweeps out equal areas in equaltimes.

3. The ratio of the square of the time required for a planetto complete its orbit to the cube of the planet’s averagedistance from the sun is a constant for all planets.

44

SunA B

EarthFIGURE 18—The path of theearth while travelingaround the sun is anellipse, not a circle. Thedeviation from the circularpath shown is exaggeratedto illustrate the principle.Shaded areas A and B,swept out in equal times,are equal.

Mechanics 45

The Universal Gravitational Law

Kepler’s laws are consistent with a law that governs theforces that exist between all masses. That law, called theuniversal gravitational law, was discovered by Isaac Newton.Newton’s universal law of gravitation states that every objectin the universe that has mass attracts every other object inthe universe that has mass. This force is proportional to theproduct of the two masses, and inversely proportional to thesquare of the distance between their centers. This relation-ship can be expressed with the following formula:

F = Gm1m2

R2

In this formula, F stands for the gravitational force in newtons,m1 and m2 stand for the masses of the bodies in kilograms,R stands for the distance between the bodies in meters, andthe letter G stands for the universal gravitational constant.(The gravitational constant is termed a “universal constant”because it’s thought to be the same in all places and at alltimes. The universal gravitational constant is expressed as avalue of 6.673 � 10–11.)

As you learned earlier, all objects possess inertia, andobjects with greater mass have greater inertia. So, whenyou consider two objects that are free to move (such as theearth and the moon), the object with the smaller mass (themoon) will do most of the moving because the earth has toomuch inertia to move any noticeable amount. If the force ofgravity didn’t act on the moon, the moon would move awayfrom the earth in a straight line because of its motion andinertia. Gravitational force is therefore a centripetal forcethat causes the moon to orbit the earth.

The gravitational force, defined by the equation of theuniversal gravitational law, provides the attraction betweenthe masses of celestial bodies and keeps the bodies in theirorbits in the universe.

Now, take a few moments to review what you’ve learned bycompleting Self-Check 3.

Mechanics46

Self-Check 3Questions 1–8: Fill in the blanks in the statements.

1. According to Kepler’s laws, the paths of planets are _______ curves.

2. The force exerted on the moon by the earth is called _______ force.

3. When a car moves around a curved path that’s not banked, the force that keeps it moving in a circular path comes from the _______ between the tires and the roadbed.

4. The value 6.673 � 10–11 is known as the _______ constant.

5. The universal gravitational law was discovered by the scientist _______.

6. A car racetrack is built with a steep inward slope to increase the _______ force.

7. If the centripetal force applied to a body moving in a circular path were suddenly discontinued, the body would immediately move in a _______.

8. Each orbit of the moon (at a constant speed) around the earth takes _______ days.

Questions 9–10: Solve each problem.

9. A car with a mass of 1,500 kilograms is moving around a circular curve at a uniform velocityof 16 m/s. The curve has a radius of 100 meters. What is the centripetal force on the car?

__________________________________________________________________________

10. A car with a mass of 1,200 kg is moving around a circular curve at a uniform velocity of 20m/s. The centripetal force on the car is 6,000 N. What is the radius of the curve?

__________________________________________________________________________

Check your answers with those on page 96.

Mechanics 47

ENERGY AND POWER

In this section of your study unit, you’ll learn about the relation of work and energy. Kinetic and potential energy will be studied and applied to various scientific problems.The difference between energy and power will be examined.Example problems in this section will illustrate the applica-tion of energy laws and methods of computing energy.

The following are some important terms that you shouldfamiliarize yourself with.

• Work. The product of a force and the distance throughwhich the force acts on a body.

• Energy. A measure of capacity to do work.

• Joule. The metric unit of work and energy.

• Kinetic energy. Energy in motion.

• Potential energy. Stored energy with the potential ofdoing work.

• Power. The rate at which work is done.

• Efficiency. The ratio of useful output power to totalinput power in a machine, expressed in percent.

Work and Energy

Work and energy are familiar concepts because everybodyhas done some work and at various times has felt “full ofenergy.” However, the term “work” as used in scientificdiscussions isn’t the same type of work we experience whenwe shovel snow, and the term “energy” has no relation toa sense of well-being.

Work is closely related to energy. In fact, work and energyare measured by the same units. Energy may be defined asa measure of capacity to do work. On the other hand, workis done when a change in energy occurs. Energy is a state,and work is a process occurring in time. In physical science,work is defined as the product of a force and the distancethrough which the force acts on a body. As an equation, it’sexpressed as follows:

W = Fs

Mechanics

In this formula, W stands for work in joules, F stands forforce in newtons, and s stands for distance in meters.

The metric unit of work and energy is the joule (rhymes with“tool”) which is derived as a product of newtons and metersand is sometimes indicated as a newton meter (N•m).

In the work equation given here, the force is assumed tobe in the same direction as the displacement of the body,or distance, if work is done. If no motion results from theapplication of a force, no work is done.

Assume that a block is pulled through a distance s by a forceF as shown in Figure 19. The block will move horizontally inthe direction of the force. In exerting the force on the blockand moving the block, work will be performed.

It’s worthwhile to remember here that the centripetal forcementioned earlier in the text does not perform work, becauseit’s not in the same direction as the velocity of the body.The centripetal force is perpendicular to the direction ofthe movement and doesn’t cause any change in the body’senergy; it only causes an accelerated movement of the body,or a change in the direction of velocity of the body.

It’s easy to see that work requires not only force but alsomotion of the body through a distance in the direction of theforce and a change in the body’s energy.

Example: What amount of work is done when a force of20 N is used to drag the block in Figure 19 a distanceof 10 meters across a floor?

Solution: Use the work formula to calculate the answer.

W = Fs

W = 20 N � 10 m

W = 200 J

Answer: The amount of work done is 200 joules.

48

s = 10 m

F = 20 N

FIGURE 19—In this illustra-tion, the block is movedthrough distance s by aforce F. The block doeswork measured in joules.The number of joules isdetermined by calculatingthe product of F and s.

Mechanics 49

Kinetic Energy

Kinetic energy is the energy of motion. Any object that hasmotion has kinetic energy. There are three basic types ofkinetic energy:

1. Vibrational energy, kinetic energy due to vibration

2. Rotational energy, kinetic energy due to rotating motion

3. Translational energy, kinetic energy due to motion fromone location to another

For the purposes of this study unit, we’ll be looking attranslational kinetic energy. The amount of kinetic energythat an object has depends on its mass and its speed. Theamount of kinetic energy in an object can be calculated byusing the following formula:

Ek = �m2v2�

In this formula, Ek stands for kinetic energy in joules,m stands for the mass of the object in kilograms, andv stands for the object’s velocity in meters per second.Note that kinetic energy values are always positive.

Example: Find the kinetic energy of an object that has amass of 4 kg and moves with a velocity of 5 m/s.

Solution: Use the kinetic energy formula. Substitute theknown values into the formula and solve.

Ek = �m2v2�

Ek = 4 kg � (5 m/s)22

Ek = �4 �2

25�

Ek = �1020

�

Ek = 50 J

Answer: The kinetic energy of the object is 50 joules.

Mechanics

Potential Energy

Potential energy is stored energy that has the potential ofdoing work. As a simple example, think of a bow and arrow.When the bow is laying on a table, it has no potential energy.However, if you pick up the bow and stretch back its elasticstring, it now has the potential ability to do work (that is, itnow has the ability to fire an arrow). Thus, the bow now haspotential energy. If you draw back the bow’s elastic string,place an arrow in it, and then release it, the string will havekinetic energy because the string will be in motion.

Another type of potential energy exists when an object islifted above the ground. Consider a box being lifted 5 meters,as shown in Figure 20. The force applied to the box is50 newtons, and it’s pulling the box vertically to a higherposition. The work performed can be calculated as follows:

W = Fs

W = 50 N � 5 m

W = 250 J

50

F = 50 N

h=

5m

FIGURE 20—In this illustra-tion, the block is movedthrough distance s by aforce F. The block doeswork measured in joules.The number of joules isdetermined by calculatingthe product of F and s.

Mechanics 51

Assume that the force applied is equal to the force of gravityacting on the box but in the opposite direction. There will beno acceleration, and the work done results only in a changeof position. The resulting potential energy acquired by thelifted box is called gravitational potential energy. It’s calledgravitational because the force is that necessary to overcomegravity, and it’s called potential because the box now has apotential of developing another form of energy. For example,if the box is allowed to fall from the high position to theground, it will develop kinetic energy. The ground is usuallythe reference level for the height of the body.

Gravitational potential energy is expressed by the followingformula:

Ep = mgh

In this formula, Ep stands for potential energy in joules,m stands for mass in kilograms, g stands for accelerationof gravity (9.8 m/s2), and h stands for height in meters.

Work doesn’t have to be performed for the body to havepotential energy. If the box is merely being held at a certainheight without any change in position or velocity, it will havea potential energy depending on its mass, gravity accelera-tion, and height. For example, for the box in Figure 20, theforce F = mg = 50 N, and the potential energy is calculatedas follows:

Ep = mgh

Ep = 50 N � 5 m

Ep = 250 J

Gravitational potential energy due to the position of a bodyis only one kind of potential energy. Another kind of energyis the energy in a storage battery and a dry cell, in whichchemical energy is the potential energy stored to be convertedto electrical energy. Another example of potential energy issteam in an enclosure under high pressure; it contains storedthermal energy (or heat energy) that can be transformed intokinetic energy, such as the motion of a turbine.

Example: A mass of 5 kilograms is lifted slowly through aheight of 10 meters. How much work is performed duringthe lifting?

Mechanics

Solution: Use the work formula to calculate the answer.Substitute the known values into the formula and solve.

W = mgs

W = 5 kg � 9.8 m/s2 � 10 m

W = 49 � 10

W = 490 J

Answer: The amount of work performed is 490 joules.

Mechanical Energy

Gravitational potential energy and kinetic energy are bothforms of mechanical energy. A body can have both kinds ofmechanical energy, and the two kinds can change. A veryinteresting example is a simple spring, like the type that’sused in a scale. The spring shown in Figure 21 is observedwhile varying forces are applied to it. In Figure 21A, no forceis applied to the spring. In Figure 21B, 20 N is applied, andthe spring is compressed 0.01 m. In Figure 21C, 40 N isapplied, and the spring is compressed 0.02 m. When a forceis applied, potential energy exists in the spring. When thespring is released, the spring has kinetic energy. Thus, workis performed while the spring is being compressed.

52

20 N 40 N

(A) (B) (C)

0.01 m

0.02 m

FIGURE 21—Experimentsshow that a spring isuniformly compressedwhen varying forces areapplied to it. A doubleforce will compress thespring a double amount.The work done storesenergy in the spring.

Mechanics 53

Resistance Due to Friction

The bodies considered up to this point have been ideal bodies,because any losses that might have resulted from frictionhave been neglected. If, however, a box slides across the floor,a significant effort will be required just to start it moving.This is because there is a resistance between the surface ofthe box and the floor. The resistance is due to friction, andit’s extremely difficult to predict because it varies from sur-face to surface and can also depend on such conditions astemperature and humidity.

Nevertheless, friction exists, and when work is done in arealistic situation, the work done against friction is usuallylost and can’t be reclaimed. The work lost because of frictionis usually in the form of heat. In the problems given in thistext, we will neglect frictional effects even though they’represent in all practical problems. That’s because taking frictioninto consideration would make our discussions unnecessarilycomplex and obscure the topics we’re trying to explore.

The Law of Energy Conservation

The law of energy conservation states that energy can’tbe created or destroyed—it can only be changed from oneform to another. An excellent example of this law can beseen in an automobile. In a car’s engine, fuel is burned. Theexpanding gases from the burning fuel then push againstmoving parts in the engine, which causes the car to move.Thus, chemical energy from the fuel is changed into heatenergy when the fuel is burned; and the heat energy isthen changed into mechanical energy to move the car. Asthis process occurs, no energy is created or destroyed—it’sjust changed from one form to another. The law of energyconservation can be expressed mathematically as

original energy = final energy

or with the following formula:

Epo + Eko = Epf + Ekf

Mechanics

In this formula, Epo stands for original potential energy,Eko stands for original kinetic energy, Epf stands for finalpotential energy, and Ekf stands for final kinetic energy.

The law of energy conservation is a universal law of physics.It applies to all forms of energy. When applied to mechanicalenergy, it means that the sum of potential and kinetic energybefore transformation equals the sum of potential and kineticenergy after transformation.

Each form of energy, such as heat, sound, light, electricalenergy, or nuclear energy, can be transformed to any otherform of energy, and in such a transformation the total energyis theoretically conserved. Of course, in practical applicationsthere are small losses of energy that aren’t considered here.

Power

When energy is used to perform work, it’s not only importantthat the loss of energy be as small as possible; it’s alsoessential that the work be done as quickly as possible.

Suppose that 1,000 bricks, each with a mass of 3 kilograms,must be carried up to a third floor 10 meters above theground. The work necessary to accomplish the task can be calculated as follows:

W = Fs

W = mgs

W = (1,000 � 3 kilograms) � 9.8 m/s2 � 10 meters

W = 3,000 � 9.8 � 10

W = 29,400 � 10

W = 294,000 J

Now, imagine that two workers are going to attempt tocomplete the brick-carrying task. One worker is strong andhealthy, but the other worker has recently suffered a backinjury. The healthy worker will be able to complete the jobmuch more quickly than the injured worker. So, when we’remaking work calculations, the rate at which the work is performed has to be considered.

54

Mechanics 55

Power is defined as the time rate at which work is performed.In the previous example, when two workers are performing anequal task, the worker who completes the job more quicklywill be using more power.

Power can be calculated with the following formula:

P = �Wt�

In this formula, P stands for power in watts, W stands forwork in joules, and t stands for time in seconds.

The metric unit of power is the watt. One watt of power isexerted when the work of one joule is performed during onesecond. The watt has been universally used as a unit forelectric power, but in metric it’s used for all types of powerincluding mechanical power.

In the English system of units, mechanical energy is meas-ured in foot-pounds. One foot-pound is the energy exertedwhen one pound of force moves a body one foot on a smoothsurface.

Historically, the horsepower has been a widely used Englishunit of power. Originally, one horsepower was defined as thetime rate of work that an average horse is capable of doing.Scientifically defined, one horsepower is the work of 550 foot-pounds per second.

1 horsepower (hp) = 746 watts (W)

1 watt (W) = 0.00134 horsepower (hp)

When metric units are used, the calculations are made mucheasier because the same unit is used for all kinds of power.

Example: How much horsepower is required to lift amass of 50 kilograms through a height of 5 metersin 2 seconds?

Solution: Use the power formula to calculate the answer.

P = �Wt�

P = �mtgs�

Mechanics

P = 50 kg � 9.8 m/s2 � 52s

P = �4902

� 5�

P = �2,4250�

P = 1,225 W

Now convert watts to horsepower. One watt equals 0.00134horsepower, so multiply 1,225 W by 0.00134 to convert thevalue to horsepower.

1,225 W � 0.00134 = 1.64 hp

Answer: 1.64 hp is required to lift the mass.

Efficiency

When energy is used to perform work in practical applica-tions, there’s always some loss of energy. Most of the lossis due to friction. However, the energy isn’t actually lost;it’s just changed to a form that’s not useful in a particularprocess. If the wheels of a machine get hot because of fric-tion, a part of the mechanical energy is transformed to heat.The heat of the wheels is dissipated to the air and is lost forthat particular process, even though it has been conservedin heating the air.

The ratio of useful output power to total input power in amachine, expressed in percent, is called the efficiency of amachine. Efficiency can be calculated by using the followingformula:

efficiency = �oiuntppuuttppoowweerr

� � 100

The efficiency of some machines may be very high, but it cannever be 100 percent. A machine with perpetual action (thatis, nonstop operation) isn’t possible in real practice. In reality,any machine will require a continued input of power to keepit running. In theoretical discussions of power, however, wecan neglect energy losses and assume that ideal conditions(conditions with 100 percent efficiency) exist.

56

Mechanics 57

Example: If a machine is supplied energy at a rate of2,000 W and it does useful work at a rate of 1,900 W,what is the efficiency of the machine?

Solution: Use the formula to calculate the efficiency ofthe machine.

efficiency = �oiuntppuutt

� � 100

efficiency = (1,900/2,000) � 100

efficiency = 0.95 � 100

efficiency = 95%

Answer: The efficiency of the machine is 95 percent.

Now, take a few moments to review what you’ve learned bycompleting Self-Check 4.

Mechanics58

Self-Check 4Questions 1–12: Fill in the blanks in the statements.

1. The SI basic unit for power is the _______.

2. A body has kinetic energy because of its _______ and _______.

3. For work to be done, energy must _______.

4. The definition used for work requires that the force and the distance through whichthe force moves lie in the _______ direction.

5. One horsepower is equal to _______ W.

6. When energy is conserved in a mechanical process, the sum of the _______ energyand the _______ energy is constant.

7. Work performed in a unit of time is called _______.

8. The SI basic unit of work is the _______.

9. An example of stored thermal energy is _______ under pressure.

10. A body that possesses energy because of its position has _______ energy.

11. When energy is stored in a compressed spring and then released, it usually showsup as _______ energy.

12. The work lost because of friction in a system usually shows up in the form of _______.

(Continued)

Mechanics 59

Self-Check 4Questions 13–18: Solve each problem.

13. What is the kinetic energy of an object that has a mass of 12 kilograms and moveswith a velocity of 10 m/s?

__________________________________________________________________________

14. An object with a mass of 50 kilograms is supported at a height of 3 meters abovethe ground. What is the potential energy of the object with respect to the ground?

__________________________________________________________________________

15. What is the kinetic energy, in joules, of an object that has a mass of 100 kilogramsand moves with a velocity of 10 m/s?

__________________________________________________________________________

16. An object with a mass of 125 kilograms is lifted through a height of 10 meters.How much work is done?

__________________________________________________________________________

17. Machine 1 has input power of 850 W and output power of 750 W. Machine 2 hasinput power of 950 W and output power of 875 W. Which one of the two machinesis more efficient?

__________________________________________________________________________

18. What is the efficiency of a motor that has input power of 400 W and provides 0.5 hp of mechanical power?

__________________________________________________________________________

Check your answers with those on page 97.

Mechanics

PROPERTIES OF LIQUIDS

In this section of your study unit, you’ll learn about the forcesthat are exerted within a liquid and the meaning of pressure.The principles of Pascal and Archimedes are explained. Thecapillarity and viscosity of liquids are also discussed.

The following are some important terms that you shouldfamiliarize yourself with.

• Pressure. The force exerted on a unit of area. Pressurecan be calculated with the formula P = F ÷ A, in whichP stands for pressure, F stands for force, and A standsfor area.

• Pascal. The basic unit of measurement for pressure.

• Free surface. The surface of a liquid subjected to atmospheric pressure.

• Manometer. An instrument used to measure pressure.

• Buoyant force. An upward force exerted on a body bya liquid.

• Specific gravity. The ratio of a weight of a body in airto the weight of an equal volume of water.

• Hydrometer. An instrument used to measure specificgravity.

The Forces Exerted by Liquids

If a tank with vertical sides and an open top is filled with aliquid, any area at the bottom of the tank must support theweight of the column of liquid that’s directly over the area.

This concept is illustrated in Figure 22. Area A in the figureis subjected to a force equal to the weight of the liquid in thecolumn above it, or

F = mg

60

Mechanics 61

In this formula, m stands for mass in kilograms and g standsfor the acceleration of gravity in meters per second squared(m/s2). The value of the acceleration of gravity is a constantvalue of 9.8 m/s2.

The volume of a column of liquid is obtained by finding theproduct of the base area and the height of the column, or

V = Ah

In this formula, V stands for volume in cubic meters, Astands for the area of the base surface in square meters, and h stands for the height of the column in meters.

The mass of the liquid column is determined by finding theproduct of the volume and the density, or

m = Vd

In this formula, m stands for mass in kilograms, V stands forvolume in cubic meters, and d stands for density in kilogramsper cubic meter.

The density of a body is defined as the mass of a unit ofvolume, here measured in kilograms per cubic meter (kg/m3). If the liquid is water, its density is 1,000 kg/m3. The known

A

FIGURE 22—Water in a tankexerts a force on the bot-tom surface of the tank. Theforce is evenly distributedover the surface area anddepends on the height ofthe water above the surfaceand the area of the surface.

Mechanics

densities of matter in solid, liquid, or gaseous states can befound in reference books. When you need to solve a problemthat involves a particular liquid, you simply look up the density for that liquid in such a book.

The force of gravity on the column of liquid, or the weight ofthat column, is found by the equation

F = Ahdg

Example: If the liquid in the tank in Figure 22 is water, theheight of the liquid column is 5 meters, and the area is 1 square meter (m2), calculate the force of gravity on thecolumn of water.

Solution: Use the formula F = Ahdg to calculate the force ofgravity. Substitute the known values into the formula and solve.

F = Ahdg

F = 1 m2 � 5 meters � 1,000 kg/m3 � 9.8 m/s2

F = 5 � 1,000 � 9.8

F = 5,000 � 9.8

F = 49,000 N

Answer: The force exerted by the liquid column on an areaat the bottom of the tank is 49,000 N.

Liquid Pressure

A quantity of high importance in the mechanics of liquids ispressure. Pressure is the force exerted on a unit of area, andcan be calculated with the following formula:

P = �AF

�

If the force is expressed in newtons and the area is expressedin square meters, the pressure is measured in newtons persquare meter, or pascals (abbreviated Pa). One pascal isthe pressure of one newton on one square meter (m2). Onekilopascal (kPa) is equal to 1,000 pascals.

62

Mechanics 63

To determine the pressure at a point in a liquid, the valueof the force exerted on the liquid column can be substitutedin the general formula for pressure as follows:

P = �AF

� = �AhAdg�

or when simplified,

P = hdg

In this formula, P stands for the pressure in kilopascals, h stands for the height of the liquid column in meters, d stands for the density of the liquid in kg/m3, and g standsfor the gravity acceleration.

Example: For the tank shown earlier in Figure 22, calculatethe pressure on area A at the bottom of the tank.

Solution: Use the pressure formula. Substitute the knownvalues into the formula and solve.

P = hdg

P = 5 m � 1,000 kg/m3 � 9.8 m/s2

P = 5,000 � 9.8

P = 49,000 Pa

Answer: The pressure on area A at the bottom of the tankis 49,000 Pa.

For measuring pressure in a liquid, the preferred unit is thekilopascal. (Again, 1 kilopascal is equal to 1,000 pascals.)Thus, the pressure at the bottom of the tank in Figure 23 is

P = 49,000 Pa, or 49 kPa

In the English system, pressure may be expressed in poundsper square foot or pounds per square inch.

At first thought, it seems inconsistent to measure the pres-sure at a point by a force “per square meter.” Such a unitseems to be associated with a relatively large area ratherthan a point. However, pressure is a point phenomenon.The dimension “per square meter” merely indicates theforce that would be exerted on a unit area if the pressurewere uniformly distributed over the area; that is, if thepressure at every point were the same as at the pointunder consideration.

Mechanics

Within the same container, if the bottom of the container isflat, the pressure will be the same at any point on the bottomsurface. The density of the liquid and the gravity acceleration(g = 9.8 m/s2) are both the same at any point. Therefore,the pressure depends only on the height of the liquid column.The upper liquid surface is horizontal and, if the bottomsurface is horizontal and even, the pressure will be evenlydistributed over the bottom area. At every point at thebottom of the tank the pressure is the same. The pressureat any point within a liquid depends on the height of theliquid above that point.

In some containers, such as the glass shown in Figure 23,the bottom surface is uneven. In such a container, there’s adifferent height of liquid above the surface at each point onthe bottom. In such a container, the pressure of the liquidvaries from point to point, depending on the height of the column of liquid above the point.

In the discussion of the pressure at the bottom of a tankcontaining a liquid, it was assumed that the force on thebottom of the container was due only to the pressure ofthe liquid. The fact that atmospheric pressure was actingdownward on the surface of the water was neglected. Thepressure due to the air above the liquid, or the atmosphere,is important however, and it should be added to the liquid

64

FIGURE 23—The bottomsurface of this containeris curved. At every point ofthis surface, the pressureis perpendicular to thesurface. If that weren’ttrue, the liquid would be inconstant motion.

Mechanics 65

pressure to get precise results. Atmospheric pressure willbe discussed later in this text. In the following problems,the atmospheric pressure isn’t considered.

Example: What is the water pressure at the bottom of a cupin which the water stands 0.03 meter high?

Solution: Use the pressure formula. Substitute the knownvalues into the formula and solve.

P = hdg

P = 0.03 meter � 1,000 kg/m3 � 9.8 m/s2

P = 30 � 9.8

P = 294 Pa, or 0.294 kPa

Answer: The pressure at the bottom of the cup is 0.294 kPa.

Example: If a total force exerted by the water in a containerwith a bottom area of 0.5 square meter is 600 newtons,what is the water pressure at the bottom of the container?

Solution: Because the total force is known, the generalpressure formula can be applied here. Substitute theknown values into the formula and solve.

P = �AF

�

P = �06.050mN2�

P = 1,200 Pa, or 1.2 kPa

Answer: The pressure at the bottom of the container is1.2 kPa.

Pressure at Any Point in a Liquid

When a liquid in a container is at rest, the direction of thepressure at any point on the surface of the container is atright angles, or normal, to the surface. Also, when an objectis immersed in a liquid, the liquid exerts a pressure on anysurface of the object that’s in contact with the liquid. The liquid pressure is everywhere normal to the surface of theimmersed object.

Mechanics

In Figure 24, two blocks are shown immersed in a liquid. Thetop surface of one body (labeled 1 in the figure) is horizontal.The pressure on that surface is vertical and can be found ifthe height of the liquid above a point on the surface is known.The same is true for a point at the bottom of the liquid. Youcan use the following formula to calculate the pressure:

P = hdg

Remember, the density of water is 1,000 kg/m3.

Example: If the tank shown in Figure 24 has a flat bottomand is filled with water to a height of 3 meters, calculatethe pressure at any point on the bottom of the tank.

Solution: Use the pressure formula. Substitute the knownvalues into the formula and solve.

P = hdg

P = 3 m � 1,000 kg/m3 � 9.8 m/s2

P = 3,000 � 9.8

P = 29,400 Pa, or 29.4 kPa

Answer: The pressure at any point on the bottom of the tankis 29.4 kPa.

66

2

1

A

h

FIGURE 24—The pres-sure on the blocksimmersed in liquid isperpendicular to thesurface of the blocks atevery point. Horizontalpressure at point A isequal to the horizontalpressure on the backside of the block. If thatweren’t true, the blockwould tip.

Mechanics 67

In the same liquid, the pressures are equal at all pointsthat are the same distance below the liquid surface. That’strue because h and d are the same for all such points. Thepressure at a particular point on a vertical or an inclinedsurface is the same as the pressure at any point on ahorizontal surface that’s at the same level below the liquidsurface as the point under consideration. For example, inFigure 24, the pressure at point A on the vertical surfacelabeled 2 in the figure is the same as the pressure at anypoint on the horizontal surface labeled 1.

Example: If the liquid in Figure 24 is water and if theheight (h) of the water above point A is 1 meter, calculatethe pressure at point A.

Solution: Use the pressure formula. Substitute the knownvalues into the formula and solve.

P = hdg

P = 1 m � 1,000 kg/m3 � 9.8 m/s2

P = 1,000 � 9.8

P = 9,800 Pa, or 9.8 kPa

Answer: Any point 1 meter below the surface is under thesame pressure, that is, 9.8 kPa.

Pascal’s Principle

In addition to the pressure due to the force of gravity, aconfined liquid may be subjected to an additional pressureby the application of an external force. For the confinedliquid shown in Figure 25, assume that a force F is appliedto the piston pressing on the upper surface of the liquid.Because the compressibility of liquids is very small, thepiston won’t move an appreciable amount but the forcedistributed over the area of the piston will set up a pressurein the liquid that will be distributed throughout the liquid.The feature of a confined liquid that allows the liquidto transmit undiminished pressure is known as Pascal’sprinciple. It can be stated as follows:

Whenever the pressure in a confined liquid is increased ordecreased at any point, the change in pressure is transmittedequally throughout the entire liquid.

Mechanics

In this statement it’s important to realize that the pressurerather than the force is transmitted. Particular care is requiredin designing systems that make use of Pascal’s principle.

To illustrate the application of Pascal’s principle, consider thetwo cylinders shown in Figure 26. One is a cylinder of a solidmaterial, concrete. Assume that it has a cross-sectional areaof 0.05m2. Further assume that the concrete cylinder iscompressed between the surfaces AB and CD by a force of10,000 N applied as shown. If that force is uniformly distrib-uted over the top of the cylinder, the pressure on the top ofthe cylinder can be calculated as follows:

P = �AF

�

P = �100.0,0500m2

N�

P = 200,000 Pa, or 200 kPa

The pressure is transmitted from the top of the cylinderundiminished to the bottom of the cylinder. Unless a forcelarge enough to cause the concrete to crumble is applied, thecylinder will retain its shape and there will be no need toapply vertical or sidewise forces to help it do so.

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Force

Container

Piston

Liquid

FIGURE 25—When a pis-ton exerts a force on theupper surface of the liq-uid, the pressure at thebottom doesn’t dependonly on the height of theliquid in the container; italso depends on theexternal force, which istransmitted undiminishedthrough the liquid.

Mechanics 69

The internal diameter of the liquid container in Figure 26 isequal to the diameter of the solid cylinder. The container ispartly filled with a liquid, and a force of 10,000 N is appliedto the surface of the liquid by means of a piston. The pistonfits in the container tightly enough to prevent the liquid fromleaking past it but loosely enough to permit it to move verti-cally in the container. The cross-sectional area of the insideof the container is 0.05m2, so that the pressure exerted bythe piston on the upper surface of the liquid is the sameas that applied to the top surface of the concrete cylinder,or 200,000 Pa (200 kPa). According to Pascal’s principle,the pressure applied to the upper surface of the liquid istransmitted with undiminished intensity through the liquidat the bottom of the container. There will be a small increasein the pressure at the bottom caused by the pressure of theliquid due to the force of gravity.

Some of the conditions that exist in the liquid-filled containerare entirely different from those that exist in the concretecylinder. As is to be expected, the liquid must be confinedby the walls of the container to prevent it from flowing away.More importantly, because the pressure is transmittedundiminished in all directions by the confined liquid, the

Piston

10,000 N10,000 N

Solid Cylinder

A

C

B

D

Liquid Container

FIGURE 26—The effects inthe systems using con-crete and liquid cylindersare identical. Both cylin-ders transmit the samepressure if their diame-ters are equal. It’s easyto see that the one usingthe liquid would be muchmore flexible; the pres-sure can be directed any-where by the proper pipedesign.

Mechanics

walls of the container must be made strong enough to resistrupture. The pressure exerted on the walls of the container isthe same as that exerted on the bottom; it equals the sum of200 kPa owing to the external force and the pressure exertedby the confined liquid. In practical applications, the internalpressure of the liquid isn’t usually considered.

Example: Assume that the surface of the piston shownin Figure 25 has an area of 0.01 m2 and that the pistonexerts a force of 2,000 newtons on the liquid. What isthe external pressure on the upper surface of the liquid?

Solution: Use the general pressure formula. Substitute theknown values into the formula and solve.

P = �AF

�

P = �02.,00100

mN2�

P = 200,000 Pa, or 200 kPa

Answer: The external pressure on the upper surface of theliquid is 200 kPa.

Example: Suppose that the surface of the piston shownin Figure 25 has an area of 0.05 m2 and that the pistoncarries a load of 500 kg. What will be the external pres-sure on the upper surface of the liquid?

Solution: Because F = mg, you can use the following varia-tion of the pressure formula to solve this problem.

P = �mAg�

P = 500 kg � 9.8 m/s2

0.05 m2

P = �40,.90050

�

P = 98,000 Pa, or 98 kPa

Answer: The external pressure on the upper surface of theliquid will be 98 kPa.

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Mechanics 71

Example: The cylinder in Figure 25 is filled with water0.4 meter high. What is the pressure at the bottom ofthe container if the internal pressure is considered?

Solution: Use the formula to calculate the internal pressure of the confined liquid at the bottom of the container.Substitute the known values into the formula and solve.

P = hdg

P = 0.4 m � 1,000 kg/m3 � 9.8 m/s2

P = 400 � 9.8

P = 3,920 Pa, or 3.92 kPa

The total pressure at the bottom is the sum of the externalpressure and the internal pressure of the confined liquid.This can be calculated as follows:

200 kPa + 3.92 kPa = 203.92 kPa

Answer: If the internal pressure is considered, the pressureat the bottom of the container is 203.92 kPa.

The Hydraulic Press

One type of device in which Pascal’s law is usefully applied isthe hydraulic press. A simplified diagram of a laboratoryhydraulic press is shown in Figure 27.

Two cylinders with different diameters are filled with oilso that the entire enclosed system is oil-filled with no airspaces. The pistons fit tightly enough that oil is preventedfrom leaking by them, but they’re still free to move verticallywith only slight resistance. The cylinders are connected by apipe through which oil can flow only from the smaller cylin-der to the larger one because of a check valve. (A check valveis a one-way valve that only allows flow in one direction.)

In the hydraulic press shown, the small piston is moved byoperating the hand lever. When the lever is lowered, thesmall piston is moved downward and oil is forced throughthe pipe from the small cylinder to the large cylinder. Thisaction lifts the large piston. By repeated strokes of the lever,the large piston can be raised to any desired height. Because

Mechanics

the same pressure is exercised over a large area, the uppersurface of the piston can apply a great force to any objectplaced in contact with it. Since the check valve prevents theoil from flowing back through the pipe, there is only a smallamount of oil in the small cylinder at any time. Additional oilfor the next stroke of the piston is permitted to enter thecylinder, as required, from the storage reservoir through a pipe and another check valve. After the large piston hasbeen raised to the desired height, the oil from the large cylinder can be returned to the reservoir through a pipe by opening another valve and the piston can return to itsoriginal position.

When the press is in use, the pressure P1 that’s appliedto the oil in the small cylinder by the piston is transmittedwith undiminished intensity to the oil in the large cylinderand to the bottom of the piston in that cylinder. The totalforce applied to the large piston by the oil is equal to thepressure P2 times the area of the bottom piston. The pressureis the same for every point on the surfaces of both pistons,and, therefore

P1 = P2

72

Large CylinderSmall Cylinder

SmallPiston

F

Lever

Reservoir

Check Valve Check Valve

Large Piston

FIGURE 27—In the hydraulic press, the pressure exerted by the small piston can be varied significantlyby varying the piston diameter or by using either a large force or a longer lever.

Mechanics 73

Pressure at the Free Surface of aLiquid

A surface liquid that’s subjected to atmospheric pressureis called a free surface. When a liquid is at rest, its freesurface must be horizontal. If a liquid has a free surface,the pressure at a point in the liquid can be calculated byusing the following formula:

P = hdg

In fresh water, which has a density of 1,000 kg/m3, thepressure at a point 3 meters below the surface is

P = hdg

P = 3 m � 1,000 kg/m3 � 9.8 m/s2

P = 3,000 � 9.8

P = 29,400 Pa, or 29.4 kPa

Gages for Measuring Liquid Pressure

A pressure gage is any device used to measure the pressureat a point in a liquid. One such type of gage, the manometer,is shown in Figure 28. The manometer is a U-shaped glasstube partially filled with a suitable liquid that’s heavy andpermits the measurement of a fairly high pressure with arelatively short tube.

The manometer in Figure 28 is being used to measurethe pressure of water at a certain point in a closed pipe.The leg nearer the pipe is called the closed leg, and theother leg is called the open leg. When the pressure is beingmeasured, the valve is opened and the water, indicated bythe light-colored shading, enters the closed leg of the tube.The blowoff valve is opened for a short time so that the airin the closed leg can be forced out by the water. The blowoffvalve is then closed and the water (light shading) is pusheddown on the column of mercury (dark shading) until it comesto rest in the closed leg of the tube at a and in the open leg

Mechanics

of the tube at b. At that time the pressure of the water in theclosed leg of the tube at a is balanced by the pressure dueto the height h of the column of mercury in the open leg ofthe tube. If the pressure in the pipe were greater, the waterwould exert more force on the mercury and the mercurywould rise to a higher level in the open leg of the tube.

A widely used type of pressure gage called the Bourdon gageis shown in Figure 29. The calibrated front face is represent-ed in Figure 29A, and the interior mechanism is shown inFigure 29C. The action depends on the motion of the end of the hollow metal tube whose cross section is shown inFigure 29B. An increase in pressure causes the tube to tend

74

Blow Off Valve

Valve

Pipe

a

b

h

FIGURE 28—The manome-ter can be made more orless sensitive by changingthe liquid in the opentube.

Mechanics 75

to straighten out, and the indicated linkages cause the point-er to rotate. If the scale is calibrated correctly, the pressureof the liquid can be read directly.

The Bourdon gage is widely used because it’s easily handledand read and is adaptable to a wide range of pressures. Formost purposes it gives a satisfactory degree of accuracy.

Archimedes’ Principle

The fact that some objects float in water and others sink tothe bottom has been the subject of study for many centuries.Archimedes, who lived in Greece between 287 and 212 B.C.,was a mathematician and inventor who was the first todiscover the principle underlying those phenomena. Theprinciple attributed to him can be simply stated in thefollowing way: When an object is immersed or floating ina liquid, it weighs less than it does in air by an amountequal to the weight of the liquid displaced by it.

Keep in mind that “weight,” as used here, means the force of gravity acting upon the object and the displaced liquid.According to that principle, a cork, when placed in wateras shown in Figure 30, sinks into the water until the weightof the displaced water is equal to the cork’s weight; then thecork floats. The cork floats in water because the water exerts

20 30

40

50

10

Scale

(A) (B) (C)

Tube

Tube End

Linkages

Pointer

FIGURE 29—Bourdon gages are quite accurate. They’re used to indicate steam pressure in afurnace.

Mechanics

an upward force on its submerged surfaces that’s sufficientto keep it afloat. The upward force, called the buoyant force,is equal to the weight of the cork. The cork floats becausethe buoyant force is equal to the weight of the cork.

Similarly, when ice equal in volume to the cork is placedin water, the ice floats as shown in Figure 30, but it sinksdeeper into the water than the cork does because it’s heavierthan the cork.

On the other hand, when concrete equal in volume to thecork is placed in the water, it sinks to the bottom, as shownin Figure 30, because it can’t displace enough water to weighas much as it does. However, the water does exert a buoyantforce on it equal to the weight of the water displaced by it.Because of the buoyant force the concrete appears to weighless in water than it does in air. If measurements are made,it’s found that the difference in the weight of the concreteblock in air and in water is exactly equal to the weight ofthe water displaced by the block.

Example: A concrete block has a volume of 0.05 m3 anda density of 3,000 kg/m3. What is the force of gravityacting on the block (the weight) in the air?

Solution: Substitute the known values into the formula andsolve.

F = Vdg

F = 0.05 m3 � 3,000 kg/m3 � 9.8 m/s2

76

Cork

Ice

Concrete

FIGURE 30—The cork andthe ice float in the liquidbecause they’re lighterthan the liquid in whichthey’re immersed. Theconcrete, which is heav-ier than the liquid, sinksto the bottom.

Mechanics 77

F = 150 � 9.8

F = 1,470 N

Answer: The force of gravity acting on the block in the air is1,470 N.

Example: What is the force of gravity acting on the concreteblock from the previous problem in water?

Solution: Start by calculating the weight of displaced water.

F = mg

F = 0.05 m3 � 1,000 kg/m3 � 9.8 m/s2

F = 50 � 9.8

F = 490 N

Subtract the weight of the displaced water from 1,470 N.

1,470 N – 490 N = 980 N

Answer: The force of gravity acting on the concrete block inwater is 980 N.

Specific Gravity

From Archimedes’ principle, it follows that the ratio of theweight of a body in air to the weight of an equal volume ofwater will determine whether a body sinks or floats. Thatratio is called the specific gravity of a body. If the specificgravity is greater than 1, the body will sink; if it’s lessthan 1, it will float.

Specific gravities of liquids are very often measured withhydrometers. A hydrometer is usually made from a graduated cylinder attached to a bulb that contains a heavy material (a material with a high specific gravity). (A graduated cylinderis a clear glass cylinder marked with measuring lines on itsside.) A typical hydrometer is shown in Figure 31.

A hydrometer can be used to determine the specific gravityof a liquid. When the hydrometer is placed in a vessel thatcontains a liquid to be tested, it will sink until the weightof the liquid it displaces is equal to its own weight. The depthto which the hydrometer sinks is a measure of the specificgravity of that liquid; it will sink deeper into a liquid thathas a lower specific gravity than it will into a liquid that hasa higher specific gravity.

Mechanics

The hydrometer is hollow except for the bulb, which is filledwith a heavy substance. The tube can be calibrated to readthe specific gravity directly, or to read some characteristicrelated to it. Depending on the range of specific gravities tobe measured, more or less heavy material can be added tothe bulb.

One of the uses of a hydrometer is to measure the conditionof the electrolyte solution in a storage battery of the typeused in a car. When a battery is fully charged, the amountof sulfuric acid mixed with the water is sufficient to give aspecific gravity of about 1.3. When the battery is dischargedto 50 percent, the specific gravity reading should be about1.2. Any specific gravity reading that’s less than 1.1 usuallyindicates a dead battery. A hydrometer used for checkingbatteries may be numerically calibrated to give actual valuesof specific gravity, or it may only give the conditions of thebattery, such as charged, weak, or dead.

78

Tube

Bulb

FIGURE 31—A hydrometer isbeing used here to measurethe specific gravity of a liquid.The shaded area in the figurerepresents the liquid beingmeasured. When the hydrometeris placed in the vessel, it willdisplace a weight of liquidequal to its own weight.

Mechanics 79

The Meniscus, Capillarity, andViscosity

To account for some of the phenomena observed in nature,it’s necessary to consider interactions between very smallparticles of matter such as molecules. That’s pointed outin Figure 32. In the illustration, a small-diameter open glasstube is partially immersed in a container of water. The sur-face of the water in the glass tube is higher than the waterlevel in the container. Further observation shows that thesurface of the water in the tube is shaped like the insideof a bowl (curved downward). If the same open glass tubeis inserted in a container of mercury (mercury has a specificgravity of 13.6), the surface of the mercury in the tube willbe lower than the surface of the mercury in the container.Also, the surface of the mercury in the tube will be shapedlike the outside of an umbrella (curved upward).

By observation of many combinations, it has been foundthat, if the force of attraction between the molecules of asolid and the liquid is greater than the force of attractionbetween the molecules of the liquid, the liquid is said towet the solid. The water wets the tube. However, if the forceof attraction between the molecules of the liquid is greater,the liquid doesn’t wet the solid. An example is mercury.

Meniscus in Water Meniscus in Mercury

FIGURE 32—A meniscusformed in water is con-cave (curved like theinside of a bowl), and ameniscus formed in mer-cury is convex (curvedlike the outside of anumbrella).

Mechanics80

The curved form assumed by the free surface of a liquidinside a tube or container is called a meniscus.

When the liquid wets the material of the tube and the liquidrises in the tube, that property of a liquid is called capillarity.Because of capillarity, oil flows upward through a wick, waterrises to the soil to feed the roots of plants, and liquids aredrawn up by a paper towel.

In a liquid, viscosity is a property that resists the free flow ofthe liquid. If a liquid will flow easily, its viscosity is low. Forexample, water has a relatively low viscosity and molasseshas a high viscosity. Viscosity acts like friction in that itopposes the motion of an object through a liquid.

Now, take a few moments to review what you’ve learned bycompleting Self-Check 5.

Mechanics 81

Self-Check 5Questions 1–8: Fill in the blanks in the statements.

1. The pressure at any point in a liquid container depends on the _______ of a liquidabove that point.

2. The specific gravity of a body can be measured with a _______.

3. Gage pressure (does/does not) include atmospheric pressure.

4. Pressure of a liquid within a container depends on the _______ of the water column.

5. The ratio of the weight of a body in air to the weight of an equal volume of water is called the_______ of the body.

6. The pressure at the flat bottom of a container full of liquid is the _______ at everypoint of the bottom area.

7. The basic SI unit of pressure is the _______.

8. A manometer is used to measure _______.

Questions 9–12: Solve each problem.

9. A concrete block has a volume of 0.10 m3 and a density of 2,000 kg/m3. What isthe weight of the block in water?

__________________________________________________________________________

10. The weight of one cubic meter of concrete in air is 30,000 newtons, and the weightof one cubic meter of water is 9,800 newtons. What is the weight of the concretewhen it’s submerged in water?

__________________________________________________________________________

11. A piston above a liquid in a closed container has an area of 0.75 m2, and the pistoncarries a load of 200 kg. What will be the external pressure on the upper surface ofthe liquid?

__________________________________________________________________________

12. In a tank full of water, the pressure on a surface 2 meters below the water level is1.5 kPa. What is the pressure on a surface 6 meters below the water level?

__________________________________________________________________________

Check your answers with those on page 100.

Mechanics

PROPERTIES OF GASES ANDSOLIDS

Gases have more compressibility and expansibility than liquids. Pressure of gaseous material is measured by abarometer. The significance of the atmospheric pressureis explained. The difference between absolute pressure andgage pressure is also examined. The conductivity of solidswill be studied and applied to various scientific concepts.

• Compressibility. The property of a gas that permits adecrease in volume without any change in mass.

• Expansibility. The property of a gas that permits anincrease in volume without any change in mass.

• Barometer. An instrument used for measuring atmospheric pressure.

• Conductivity. A measure of how easily heat and electriccurrent passes through a material.

• Insulation. The ability of a material to prevent flow ofheat or an electric current.

• Pressure. A value equal to force divided by area.

The Compressibility and Expansibilityof Gases

An important difference between a liquid and a gas is thatthe volume of a liquid is only slightly affected by a largeincrease or decrease in pressure whereas the volume of agas changes considerably under the same circumstances.The property of a gas that permits a decrease in volumewithout any change in mass is called compressibility. Allgases have high compressibility; that is, they can easily becompressed.

The property of a gas that permits an increase in volumewithout any change in mass is called expansibility. Gaseshave high expansibility; that is, they expand very easily.

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Mechanics 83

The conditions represented in Figure 33 demonstrate thecompressibility and expansibility of a gas. Assume that acylinder-shaped gas vessel contains a piston. The pistonfits the cylinder tightly enough to prevent gas from leakingaround it, but loosely enough to move up and down easilywhen a force is applied.

In Figure 33A, the piston is loaded so that it’s held in theindicated position when a certain mass of gas is admitted tothe cylinder. If the load on the piston is increased, the pistonwill move to the position indicated in Figure 33B. The gaswill be compressed as indicated by the darker shading. If theload on the piston is further increased, the piston will movedownward and the volume of the gas will be smaller and thecompression higher. This is indicated by the darkest shadein Figure 33C.

If the added loading is removed from the piston, the gas willexpand and the piston will return to its original position.

(A)

Vessel

Piston

(B) (C)

FIGURE 33—When an external force is applied to an enclosed volume of air, the air is compressed. Thevolume of the air becomes smaller, and the pressure in the enclosed air becomes higher in relation tothe increase of the applied force.

Mechanics

Atmospheric Pressure

The atmosphere is a layer of air surrounding the earth; itsthickness has been estimated at about 1,000 km (or approxi-mately 621 miles). Since air has a mass and a gravity force isexerted on that mass, the layer of air produces pressure. Asimple way to measure that pressure is to use a mercurybarometer. A barometer is shown in Figure 34.

In a simple mercury barometer, a glass tube of mercuryis placed in an open container of mercury. The air pressureon the open surface of mercury is sufficient to support thecolumn of mercury in the tube. In other words, the weight ofa column of air that’s about 1,000 kilometers high and thathas a certain cross-sectional area is the same as the weightof a column of mercury approximately 760 millimeters highand has the same cross-sectional area.

84

h

FIGURE 34—When mercuryis used for a barometer, theusable height of the columnis approximately 760 milli-meters. If water were usedinstead of mercury, theheight of the barometerwould be approximately 10meters.

Mechanics 85

At any one location, atmospheric pressure varies slightlyfrom day to day. For scientific work, standard atmosphericpressure is defined as a pressure equivalent to that producedby a column of mercury that’s 760 millimeters high at 0degrees C (Celsius). It’s usually expressed as 760 millimetersHg, because Hg is the chemical symbol for mercury. Thedensity of mercury at 0 degrees is 13,600 kg/m3.

The pressure exerted by a column of mercury 0.760 metershigh can be calculated by the same pressure formula as fora liquid, as follows:

P = hdg

P = 0.760 m � 13,600 kg/m3 � 9.8 m/s2

P = 10,336 � 9.8

P = 101,292.8 Pa, or 101.3 kPa

In practical applications, it may be assumed that theatmospheric pressure is approximately 100 kPa, or 760 mmHg. This is the value of normal atmospheric pressure. Thepressure is equivalent to atmospheric pressure of 14.7 psi(pounds per square inch) if expressed in English units.

Barometers

Two general types of barometers are used today: the liquidtype just described and the aneroid type. The liquid typebarometer is the standard type used in the laboratory andisn’t usually portable. An aneroid barometer consists of apartially evacuated corrugated cylinder (Figure 35). As theatmospheric pressure changes, the top of the cylinder movesup and down. The motion is magnified by the linkage systemand moves a pointer over a scale. The aneroid barometer isportable and is usually calibrated to agree with a standardmercury barometer.

Mechanics

Atmospheric Pressure and Altitude

Atmospheric pressure decreases with increasing altitudebecause the air column above the ground is decreasing.The pressure changes are shown by the graph in Figure 36.Pressure in kilopascals (kPa) is plotted against altitude inkilometers (km). At lower altitudes, atmospheric pressuredrops off at a relatively rapid pace, but at higher altitudes,it drops off much more slowly. As a matter of fact, atmos-pheric pressure at an altitude of 6 km drops to half the valueat sea level, yet atmosphere extends upward about another1,000 km.

86

Scale

Pointer

Linkage

Cylinder

FIGURE 35—The aneroidbarometer is widely usedbecause it’s fairly accurateand is of convenient size.Some barometers, alongwith being calibrated inmillimeters or inches ofmercury, will have legendssuch as “stormy” or “fair”to signify weather condi-tions.

Mechanics 87

Gage Pressure and Absolute Pressure

It’s the usual practice to assume that pressure applied tothe free surface of a liquid by the atmosphere is zero. Whenthe atmospheric pressure on a liquid is neglected, the liquidpressure is called a gage pressure. However, when a gasrather than a liquid is considered, atmospheric pressuremust usually be taken into account.

A pressure obtained by adding atmospheric pressure to thegage pressure is called absolute pressure. For instance, if the gage pressure of a gas is 50 kPa, the absolute pressure is 150 kPa (50 kPa + 100 kPa). On the other hand, if theabsolute pressure is 220 kPa, the gage pressure is 120 kPa(220 kPa – 100 kPa). An absolute pressure is usually greaterthan atmospheric pressure, but it may be equal to atmos-pheric pressure or be less than atmospheric pressure.

When the absolute pressure of a gas in a container isless than atmospheric pressure, a partial vacuum existsin the container. A partial vacuum can be produced byfirst enclosing a certain mass of air in a container and thenremoving some of the air. To produce what may be calleda perfect vacuum, it would be necessary to remove all theair from the container. Since that’s technically impossible,a perfect vacuum can’t be obtained.

0

100

90

80

70

60

50

40

30

20

10

2 4 6 8 10 12 14 16 18 20

Pre

ssu

re,in

Kil

opasc

als

Altitude, in Kilometers

FIGURE 36—This graphshows changes in the aver-age values of atmosphericpressure in the region onearth at a latitude of 40degrees north. To show thepressure changes for alti-tudes higher than 20 km, adifferent vertical scalewould have to be used.

Mechanics

Conductors and Insulators

Conduction is the transfer of heat through a substancefrom molecule to molecule as they collide. Solids are betterconductors of heat than liquids, because the molecules insolids are closer together than molecules in liquids. Silveris one metal that conducts heat very well. If you stir a hotliquid with a silver spoon, the spoon’s handle will becomehot very quickly. Heat from the liquid is transferred fromthe liquid to the metal through conduction. The heat istransferred from molecule to molecule until the entirespoon is hot. This is why metal cooking utensils haveplastic handles—to prevent you from burning your handwhen stirring hot liquids.

The term conductor is used to refer to substances thatare good conductors of electricity. Metals such as copper,aluminum, silver, and gold are examples of good electricalconductors.

In contrast, a solid that’s a poor conductor of heat or elec-tricity is called an insulator. Solids that contain lots of tinyair pockets, such as wood and cork, are good insulatorsbecause air is a good insulator. Materials such as rubber,plastic, and glass are also good insulators. This is whyrubber and plastic materials are used to coat the outsidesof electrical wires. A solid’s ability to provide heat insulationis also important in homebuilding, because good insulationmaterials can increase the efficiency of a home’s heating andcooling systems.

The Pressure of Solids

As you’ve learned, force acting on a unit of area of surface ispressure. Any solid object exerts pressure on the surface itrests on. The amount of pressure exerted by a solid relates toboth its mass and the area of its bottom surface. Look at theexample shown in Figure 37. Both objects in the illustrationweigh 20 kilograms. However, Object 1 has a smaller bottomsurface than Object 2. Which object will exert more pressureon the tabletop? You can determine the answer by applyingthe pressure formula P = �

AF

�.

88

Mechanics 89

Example: Object 1 in Figure 37 has a bottom surface area of 0.2 m2, and Object 2 has a bottom surface area of 0.8 m2. What is the pressure exerted by each object on the tabletop?

Solution: First, use the formula F = mg to calculate the forcefor each object.

F = mg

F = 20 kg � 9.8 m/s2

F = 196 N

Next, use the formula to calculate the pressure for Object 1.

P = �AF

�

P = �01.926mN2�

P = 980 Pa

Use the formula again to calculate the pressure for Object 2.

P = �AF

�

Object 1

20 kg 20 kg

Object 2FIGURE 37—This illustra-tion shows two objects.Both objects have a massof 20 kilograms, butObject 1 has a smallerbottom surface area thanObject 2. Thus, Object 1will exert more pressureon the tabletop.

Mechanics90

P = �01.986mN2�

P = 245 Pa

Answer: Object 1 exerts a pressure of 980 Pa, and Object 2exerts a pressure of 245 Pa. Thus, you can see that whenboth objects have the same mass, the object with thesmaller bottom surface area will exert more pressure onthe tabletop.

Now, take a few moments to review what you’ve learned bycompleting Self-Check 6.

Mechanics 91

Self-Check 6Questions 1–8: Indicate whether each statement is True or False.

_____ 1. Gases can be compressed and expanded more than liquids.

_____ 2. The pressure exerted by solid material depends directly on surface area and density.

_____ 3. If a mass of gas is confined in a closed container with a movable piston above the gas and a small downward force is applied to the piston, the volume of gas will be reduced considerably.

_____ 4. If the gage pressure of a gas is 70 kPa, the absolute pressure of the gas is 170 kPa.

_____ 5. The partially evacuated cylinder used in the aneroid-type barometer is not flexible.

_____ 6. A column of atmosphere of 1 m2 cross-sectional area and about 1,000 km high creates a pressure of approximately 1,000 kPa.

_____ 7. If the absolute pressure of a gas is 350 kPa, the gage pressure of the gas is 450 kPa.

_____ 8. A portable barometer is usually the aneroid type.

9. Two cube-shaped blocks are placed on a table. Both cubes have a mass of 3 kilograms. Cube1 has a bottom surface area of 0.9 m2, and Cube 2 has a bottom surface area of 0.3 m2.Which cube will exert more pressure on the tabletop?

__________________________________________________________________________

10. Two objects are placed on a table. Object 1 has a mass of 2 kilograms and a bottom surface area of 0.1 m2. Object 2 has a mass of 8 kilograms and a bottomsurface area of 0.4 m2. How much pressure does each object exert on the tabletop?

__________________________________________________________________________

Check your answers with those on page 101.

Mechanics92

NOTES

93

Self-Check 11. meter

2. direction

3. meters per second

4. acceleration

5. decelerating

6. body

7. kilogram

8. linear

9. v = s ÷ t

10. velocity

11. zero

12. v2 = v1 + gt

13. a = �v2 –

t

v1�

a = 50 m/s – 15 m/s4 s

a = �345�

a = 8.75 m/s2

14. vav = �v1

2

+ v2�

vav = 15 m/s + 50 m/s2

vav = 65 ÷ 2

vav = 32.5 m/s

15. s = v1t + �12

� at2

s = (12 m/s � 1.2 s) + �12

� � 9.8 m/s2 � (1.2 s)2

s = 14.4 + �12

� � 9.8 � (1.44)

s = 14.4 + �12

� � 14.112

An

sw

er

sA

ns

we

rs

Self-Check Answers94

s = 14.4 + 7.056

s = 21.456, or 21.5 m (rounded)

16. v2 = v1 + at

18 m/s = 0 m/s + 1.5 m/s2 � t

18 = 1.5 � t

Divide both sides of the equation by 1.5 to solve for t.

�11.85� = 1.5 � �

1t.5�

12 s = t

Self-Check 21. mass

2. mass, force

3. direction

4. force, velocity

5. first

6. mass, time

7. 9.8

8. magnitude, direction

9. F = ma

F = 32 kg � 12 m/s2

F = 384 N

10. F = ma

F = 7 kg � 15 m/s2

F = 105 N

11. a = �mF

�

a = �240kNg

�

a = 5 m/s2

Self-Check Answers 95

12. m = �Fa

�

m = �1

5

6

0

m

0

/

N

s2�

m = 31.25 kg

13. F = mg

F = 20 kg � 9.8 m/s2

F = 196 N

14. m = �Fg

�

m = �9

3

.

,

8

00

m

0

/

N

s2�

m = 306 kg

15. F = �m6g�

F = 7 kg � 9.8 m/s2

6

F = 7 � 1.6333

F = 11.43 N

16. F = �m6g�

F = 80 kg � 9.8 m/s2

6

F = �7864

�

F = 130.67 N

17. m = �Fg

�

m = �9.

1

8

0

m

0

/

N

s2�

m = 10.2 kg

Self-Check Answers96

18. Keep in mind that when the car stops, the final velocityis zero. Use the formula for final velocity to calculate theacceleration.

v2 = v1 + at

0 m/s = 30 m/s + (a � 3 s)

0 = 30 + 3a

Subtract 3a from both sides of the equation.

0 – 3a = 30 + 3a – 3a

–3a = 30

Divide both sides of the equation by –3 to solve for a.

�––33a

� = �3–30�

a = –10 m/s

The acceleration is negative, which means that the car isslowing down. The force needed to cause such accelera-tion can be calculated as follows:

F = ma

F = 1,500 kg � (–10 m/s2)

F = –15,000 N

Answer: A deceleration or braking force of –15,000 Nmust be applied to stop the car in 3 seconds.

Self-Check 31. elliptical

2. centripetal

3. friction

4. universal gravitational

5. Isaac Newton

6. centripetal

7. straight line

8. 27.3

Self-Check Answers 97

9. Use the centripetal force formula to calculate theanswer.

F = �mRv2�

F = 1,500 kg � (16 m/s)2

100 m

F = (1,500 � 256) ÷ 100

F = �38140,0000

�

F = 3,840 N

Answer: The centripetal force on the car is 3,840 newtons.

10. Use the centripetal force formula to calculate theanswer, and solve for the radius (R).

R = �mFv2�

R = 1,200 kg � (20 m/s)2

6,000 N

R = �1,2060,0

�

00400

�

R = �48

6

0

,0

,0

0

0

0

0�

R = 80 m

Answer: The radius of the curve is 80 meters.

Self-Check 41. watt

2. mass, velocity

3. change

4. same

Self-Check Answers98

5. 746

6. potential, kinetic

7. power

8. joule

9. steam

10. potential

11. kinetic

12. heat

13. Ek = �m2v2�

Ek = 12 kg � (10 m/s)2

2

Ek = �12 �2

100�

Ek = �1,2200�

Ek = 600 J

14. Ep = mgh

Ep = 50 kg � 9.8 m/s2 � 3 m

Ep = 490 � 3

Ep = 1,470 J

15. Ek = mv2/2

Ek = 100 kg � (10 m/s)2

2

Ek = �100 �2

100�

Ek = �10,2000�

Ek = 5,000 J

Self-Check Answers 99

16. W = mgs

W = 125 kg � 9.8 m/s2 � 10 m

W = 1,225 � 10

W = 12,250 J

17. Use the formula to calculate the efficiency of Machine 1.

efficiency = output/input � 100

efficiency = �785500

WW

� � 100

efficiency = 0.8823 � 100

efficiency = 88.23%

Then, calculate the efficiency of Machine 2.

efficiency = �oiuntppuutt

� � 100

efficiency = �897550

WW

� � 100

efficiency = 0.921 � 100

efficiency = 92.1%

Compare the efficiencies of the two machines.

Answer: Machine 2 is more efficient.

18. First, you need to convert horsepower to watts.

1 hp = 746 W, so 0.5 hp = 373 W

efficiency = �oiuntppuutt

� � 100

efficiency = �347030

WW

� � 100

efficiency = 0.9325 � 100

efficiency = 93.25%

Self-Check Answers100

Self-Check 51. height

2. hydrometer

3. does not

4. height

5. specific gravity

6. same

7. pascal

8. pressure

9. Calculate the weight of the block in the air.

F = Vdg

F = 0.10 m3 � 2,000 kg/m3 � 9.8 m/s2

F = 200 � 9.8

F = 1,960 N

Calculate the weight of displaced water.

F = mg

F = 0.10 m3 � 1,000 kg/m3 � 9.8 m/s2

F = 100 � 9.8

F = 980 N

Subtract the weight of the displaced water from 1,960 N.

1,960 N – 980 N = 980 N

Answer: The force of gravity acting on the concreteblock in water (the weight) is 980 N.

10. Subtract the weight of the water from the weight ofthe concrete.

30,000 N – 9,800 N = 20,200 N

Answer: The weight of the concrete when it’ssubmerged in water is 20,200 N.

Self-Check Answers 101

11. P = �mAg�

P = 200 kg � 9.8 m/s2

0.75 m2

P = 1,960 ÷ 0.75

P = 2,613 Pa, or 2.613 kPa

Answer: The external pressure on the upper surface ofthe liquid will be 2.613 kPa.

12. The height of the water column is 3 times higher at6 meters below the water level. Multiply the pressureat 2 meters by 3 to calculate the pressure at 6 meters.

P = 3 � 1.5 kPa

P = 4.5 kPa

Self-Check 61. True

2. False

3. True

4. True

5. False

6. False

7. False

8. True

9. Both cubes have the same mass. Thus, Cube 2 willexert more pressure on the tabletop because it has asmaller bottom surface area.

10. Use the formula F = mg to calculate the force forObject 1.

F = mg

F = 2 kg � 9.8 m/s2

F = 19.6 N

Self-Check Answers102

Next, use the formula P = �AF

� to calculate the pressurefor Object 1.

P = �AF

�

P = �0

1

.

9

1

.6

m

N2

�

P = 196 Pa

Use the formula F = mg to calculate the force forObject 2.

F = mg

F = 8 kg � 9.8 m/s2

F = 78.4 N

Use the formula P = �AF

� to calculate the pressure forObject 2.

P = �AF

�

P = �07.84.4

mN2�

P = 196 Pa

Answer: Both objects exert the same amount ofpressure on the tabletop.

103

ENGLISH TO METRIC CONVERSIONS

Length1 mile = 1.609 kilometers1 yard = 0.9144 meter1 foot = 30.48 centimeters1 inch = 2.54 centimeters

Area1 square inch = 0.000645 square meter1 square foot = 0.093 square meter1 cubic foot = 0.028 cubic meter

Liquid Volume1 gallon = 3.785 liters 1 quart = 0.946 liter 1 pint = 0.473 liter 1 fluid ounce = 29.573 milliliters

Weight1 pound = 0.454 kilogram1 ounce = 28.35 grams

Velocity, Force, Work, Pressure1 mile per hour = 1.6 kilometers per hour1 foot per second squared = 0.3 meter per second squared1 pound-force = 4.45 newtons1 foot-pound = 1.36 joules1 horsepower = 746 watts1 pound-force per square inch = 6,890 pascals

Ap

pe

nd

ixA

pp

en

dix

Appendix104

METRIC TO ENGLISH CONVERSIONS

Length1 kilometer = 0.62 mile 1 meter = 3.28 feet1 meter = 39.37 inches1 centimeter = 0.39 inch1 millimeter = 0.039 inch

Area1 square meter = 10.76 square feet1 cubic meter = 35.3 cubic feet

Liquid Volume1 liter = 1.057 quarts1 centiliter = 0.338 fluid ounce

Weight1 kilogram = 2.2046 pounds1 gram = 0.035 ounce

Velocity, Force, Work, Pressure1 kilometer per hour = 0.6 mile per hour1 meter per second squared = 3.28 feet per second squared1 newton = 0.225 pound-force1 joule = 0.74 foot-pound1 watt = 0.00134 horsepower1 pascal = 0.00145 pound-force per square inch

105

Mechanics

When you feel confident that you have mastered the material in this study unit, go to http://www.takeexamsonline.com andsubmit your answers online. If you don’t have access to theInternet, you can phone in or mail in your exam. Submit youranswers for this examination as soon as you complete it. Do notwait until another examination is ready.

Questions 1–25: Select the one best answer to each question.

1. In which of the following units is acceleration expressed?

A. NewtonsB. Foot-poundsC. KilogramsD. Meters per second squared

2. A stone falls from a ledge and takes 16 seconds to hit theground. The stone has an original velocity of 0 m/s. How tall isthe ledge?

A. 1,254.4 meters C. 156.8 metersB. 313.6 meters D. 78.4 meters

Exa

min

atio

nE

xam

ina

tion

EXAMINATION NUMBER:

00709102Whichever method you use in submitting your exam

answers to the school, you must use the number above.

For the quickest test results, go to http://www.takeexamsonline.com

Examination106

3. A tank with a flat bottom is filled with water to a height of 3.5 meters. What is the pressureat any point at the bottom of the tank? (You can ignore atmospheric pressure when calculat-ing your answer.)

A. 3.5 kPa C. 34.3 kPaB. 3,500 kPa D. 34,300 kPa

4. The ability of a material to transfer heat or electric current is called

A. insulation. C. conductivity.B. porosity. D. convection.

5. A steel block has a volume of 0.08 m3 and a density of 7,840 kg/m3. What is the force ofgravity acting on the block (the weight) in water?

A. 5,362.56 N C. 6,700.56 NB. 6,150.64 N D. 7,600.18 N

6. When an automotive battery is fully charged, the sulfuric acid and water mixture will havea specific gravity of about

A. 0.15. C. 1.3.B. 1.0. D. 2.5.

7. If an object weighs 39.2 N on Earth, what is its mass?

A. 8 N C. 4 kgB. 384.16 kg D. 78.4 N

8. If a small-diameter open glass tube is partially immersed in a vessel containing mercury,the surface of the mercury inside the tube will be

A. lower than the mercury in the vessel, and shaped like the outside of an umbrella.B. higher than the mercury in the vessel, and shaped like the outside of an umbrella.C. lower than the mercury in the vessel, and shaped like the inside of a bowl.D. higher than the mercury in the vessel, and shaped like the inside of a bowl.

9. How much time should be allowed for a 392-km car trip if the car will be traveling at 120 km/h?

A. 5.12 hours C. 2.72 hoursB. 3.27 hours D. 0.31 hours

10. A machine has an efficiency of 91%. If the energy supplied to the machine is 4,000 watts,what is the amount being used for work?

A. 4,091 watts C. 360 wattsB. 43.96 watts D. 3,640 watts

Examination 107

11. What is the kinetic energy of an object that has a mass of 20 kilograms and a speed of 2.8 m/s?

A. 56 J C. 17.2 JB. 22.8 J D. 78.4 J

12. If a gas has a gage pressure of 206 kPa, its absolute pressure is approximately

A. 2,060 kPa. C. 106 kPa.B. 306 kPa. D. 20.6 kPa.

13. According to Ptolemy’s model of the movement of celestial bodies,

A. the earth rotates around the sun.B. the sun is the center of the universe.C. the earth rotates around the moon.D. planets orbit in circular paths around the earth.

14. A car with a mass of 1,200 kilograms is moving around a circular curve at a uniform velocityof 20 meters per second. The centripetal force on the car is 6,000 newtons. What is theradius of the curve?

A. 160 meters C. 32 metersB. 80 meters D. 16 meters

15. What force is required to accelerate a body with a mass of 18 kilograms at a rateof 36 m/s2?

A. 2 N C. 648 NB. 54 N D. 6,350 N

16. Oil flows upward in the wick of a lantern because of the liquid property called

A. viscosity. C. meniscusity.B. capillarity. D. density.

17. A car travels from Boston to Hartford in 4 hours. The two cities are 240 kilometers apart.What was the average speed of the car during the trip?

A. 4 km/hour C. 60 km/hourB. 36 km/hour D. 960 km/hour

18. Which of the following is the term used to describe a body’s resistance to a change inmotion?

A. Inertia C. GravityB. Acceleration D. Mass

Examination108

19. If the speed of an object changes from 121 m/s to 98 m/s during a time interval of 12 s,what is the acceleration of the object?

A. -276 m/s2 C. -18.25 m/s2

B. -1.92 m/s2 D. -11 m/s2

20. An object has a mass of 180 kg on the moon. What is the force of gravity acting on theobject on the moon?

A. 1,764 N C. 10,584 NB. 294 N D. 684 N

21. In a practical machine, the power output is _______ the power input.

A. smaller than C. equal toB. larger than D. multiplied by

22. If a stone falls past a bird on a ledge moving at 4 m/s, how fast will the stone be movingjust before it hits the ground below 9 seconds later?

A. 92.2 m/s C. 352.8 m/sB. 48.2 m/s D. 36 m/s

23. The ratio of output power to input power, in percent, is called

A. conductivity. C. work.B. efficiency. D. horsepower.

24. An object with a mass of 21 kilograms is lifted through a distance of 7 meters. How muchwork is done?

A. 147 J C. 1,440.6 JB. 29.4 J D. 823.2 J

25. A car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocityof 25 meters per second. The curve has a radius of 80 meters. What is the centripetal forceon the car?

A. 625 N C. 15,625 NB. 703 N D. 20,250 N