WORK, ENERGY,

AND POWER

INTRODUCTIONIt wasn't until over 100years after Newton that the idea of energy became incorporated into physics,but today it permeates every branch of the subject.

It's difficult to give a precise definition of energy; there are different forms of energy because thereare different kinds of forces. There's gravitational energy (a meteor crashing into the earth), elasticenergy (a stretched rubber band), thermal energy (an oven), radiant energy (sunlight), electricalenergy (a lamp plugged into a wall socket), nuclear energy (nuclear power plants), and mass energy(the heart of Einstein's equation E =mc2). Energy can come into a system or leave it via variousinteractions that produce changes. One of the best definitions we know reads as follows: Force is theagent of change, energy is the measure of change, and work is the way of transferring energy fromone system to another. And one of the most important laws in physics (the Law of Conservation of

55

Energy, also known as the First Law 0'£ Thermodynamics) says that if you account for all its variousforms, the total amount of energy in a given processwill stay constant;that is, It willbe conserved. Forexample, electrical energy can be converted into light and heat (this is how a light bulb works),butthe amount of electrical energy coming in to the light bulb equals the total amount of light and heatgiven off. Energy cannot be created or destroyed; it can only be transferred (from one system toanother) or transformed (from one form to another).

WORKWhenyou lift a book from the floor,you exert a force on it, over a distance,and when you push a crateacrossa floor, you also exerta forceon it, over a distance. Theapplicationof forceover adistance, andthe resulting change in energy of the system that the forceacted on, give rise to the conceptofwork.When you hold a book in your hand, you exert a force on the book (normalforce) but, sincethe bookis at rest, the forcedoes not act through a distance,so you do no work on the book.Although you didwork on the book as you lifted it from the floor,onceit's at rest in your hand, you are no longerdoingwork on it. . .

De~nition. If a force F acts over a distance d, and F is parallel to d, then the work doneby Ris the product of force and distance: W=Fd.

Notice that, although work depends on two vectors (F and d), work itself is not a vector. Work isascalar quantity.

Example 4.1 You slowly lift a book of mass 2 kg at constant velocity adistance of 3 m. How much work did vou do on the book?

Solution. In this case,the forceyou exert must balancethe weight of the book (otherwisethe velocityof the book wouldn't be constant), so F =mg =(2kg)(9.8 m/s2) =20 N. Since this force is straightupward and the displacement of the book is also straight upward, F and d are parallel, so the workdone by your lifting force is W=Fd =(20 N)(3 m) = 60Nrn. The unit for work, the newton-meter(Nrn) is renamed a joule, and abbreviated J. SO the work done here is 60J.

The definitionabove takes care of cases in which F is parallel to the motion. If F is not parallel tothe motion, then the definition needs to be generalized. '

Definition~ If a force F acts over a distance d, and £} is the angle between F and d, thenthe work done by F is the product of the component of force in the directionof the motion and the distance: W=(F cos £})d. [Notice that you can. writethis definitionusing the dot product: W=F . d.]

Example 4.2 A 15 kg crate is moved along a horizontal floor by awarehouseworkerwho's pullingon It WIth a rope that makes a .300 anglewith the horizontal.The tension in the rope is 200 N and the crate slidesa distance of 10m. How much work is done on the crate by the worker?

56 CRACKING THE AP: PHYSICS EXAM

Solution. The figure below shows thatE, and d are not parallel. It's only the componentof the forceacting along the direction of motion, FT cos 0, that does work.

Therefore,

W= (Fr cos O)d = (200 N . cos 30°)(10 m) = 1730 J

Example 4.3 In the previous example. assume that the coefficient ofkinetic friction between tho? crate and the floor is 04.

(01) How much work is done by the normal force?tb) How much work 1'" done bv the friction force?

Solution.

(a) Clearly, the normal force is not parallel to the motion, so we use the general defini­tion of work. Since the angle between FN and d is 90° (by definition of normal) andcos 90° =0, the normal forcedoes zero work.

(b) The frictionforce, F, is also not parallel to the motion; it's antiparallel. That is, theangle between Ff and d is 180°. Sincecos 180° =-1, and since the strength of thenormal force is FN = Fw = mg = (15 kg)(9.8m/s2) = 147N, the work done by thefriction force is:

W= -FP= - Jll~ = -(0.4)(147 N)(lO m) = -590 J

The two previous examplesshow that work, which is a scalar quantity, may bepositive,negative,or zero. If the angle between F and d (0) is less than 90°, then thework is positive (because cos 0 is positive in this case); if 0 =90°, the work is zero(because cos 90° =0); and if 0 > 90°, then the work is negative (because cos 0 isnegative). Intuitively, if a forcehelps the motion, the work done by the force ispositive,but if the force opposes the motion; then the work done by the force isnegative. . .

WORK, ENERGY , AND POWER . 57

(a) How much work is done by gravity?(tI) How much work is done by the normal force?(c) How much work is done by friction?(d) What is the total work done?

E.,,,,ample 4.4 A box slides down an inclinedplane (incline angle =40")The mass of the block, m, is 3-5 kg, the coefficient of kinetic frictionbetween the box and the ramp. Jlv is 0.3, and the length of the ramp, d,isS m.

Solution.

(a) Recall that the force that's directlyresponsible for pulling the box down the planeisthe component of the gravitationalforce that's parallel to the ramp:FVi sin 8 =mg sin 8 (where 8 is the inclineangle). This component is parallel to themotion, so the work done by gravity is

Wbygravily =(mg sin 8)d =(35 kg)(9.8 N/kg)(sin 40°)(8 m) =1760 J

Notice that the work done by gravity is positive,as we would expectit to be, sincegravity is helping the motion.Also,be carefulwith the angle 8. The generaldefini­tion of work reads W=(F cos 8 )d,where 8 is the angle between F and d. However,the angle between Fwand d is not 40° here, so the work done by gravity is not(mg cos400)d. The angle 8 used in the calculation above is the inclineangle.

(b) Since the normal force is perpendicular to the motion, the work done by this force iszero.

(c) The strength of the normal force is Fwcos 8 (where 8 is the inclineangle), so thestrength of the frictionforce is F, = J.llN =J.llwcos 8 = J.lkmg cos 8. Since Ff isantiparallel to d, the cosineof the angle between these vectors (180°) is -I, so thework done by frictionis

Wbyfriclion =-Fp. =-(J.lkmg cos 8)(d) =-(0.3)(35 kg)(9.8 N/kg)(cos 40°)(8 m) =-630 J

Noticethat the work done by friction is negative,as we expect it to be, sincefrictionis opposing the motion.

58 • CRACKING THE AP: PHYSICS EXAM

(d) The total work done is foUnd simply by adding the values of the work done by eachof the forces acting on the box:

WtDtal =~W=Wbygravity + Wbyno~force + Wbyfnction =1760 + 0 + (-630) =1130 J

WORK DONE BY A VARIABLE FORCEHa force remains constant over the distance through which it acts, then the work done by the forceis simply the product of forceand distance.However, if the forcedoes not remainconstant, then thework done by the forceis given by a definiteintegral.Focusingonly on displacements that are alonga straight line (call it the x axis), let F be-a force whose component in the x directionvaries withpositionaccordingto the equation F =F(x). Then the work done by this force as it actsfrompositionx =Xl to position X =x2 is equal to

w=jX2F(X) dxXl

Ha graph ofF vs. x is given, then the work done by F as it acts from x =Xl to X =x2 is equal to thearea bounded by the graph of F, the X axis, and the vertical lines x =Xl and x =x2:

F

//;,---~~. : \

I 7f

I----If-------:~--x

area =work done by F

Example 4.5 Theforce exertedby a sprmg when it's displaced by x fromitsnaturallength isgivenby theequation f(~) =-h" where k is a positiveconstant. What is the work done by a spring us it pushes out fromt = -x 2 to " = -Xl (where x:! > ")?

Solution. Sincethe force is variable,we calculate the following definite integral:

WORK, ENERGY, AND POWER . 59

Anothersolutionwould involve sketching a graph ofF(x) =~kx and calculating theareaunder thegraph from x = -x2 to X = - xl.

F

F(x) =-kx'r-- -- -- - kx;

'"-"k-----X

Here, the region is a trapezoid with area A =t (basel +base.) x height, so

W =A =t(kx2 + kx l ) ( X2 - Xl)

=tk(x2 +XI)(X2 - Xl )

=tk(X~-xnKINETIC ENERGYConsider an object at rest (va =0), and imagine that a steady force is exerted on it, causing it to

.accelerate. Let'sbernorespecific; let theobject's massbe m,and letFbe the force actingon the object,pushing it in a straight line. The object's acceleration is a = Flm, so after the object has traveled adistance A s under the actionof this force, its final speed, v, is given by Big Five #5:

Butthe quantity FLis is the work done by the force, so W=tmv2. Thework done on the object hastransferredenergy to it, in the amount tmo'. Theenergy an object possesses by virtue of its motionis therefore defined as tmv2 and is called kinetic energy:

K = lrmJ22

THE WORK-ENERGY THEOREMKinetic energyis expressed in joules just likework, sincein the casewe just lookedat, W=K. In fact,the derivationabovecanbe extendedto an object with a nonzeroinitialspeed,and the sameanalysiswill show that the total work done on an object-or, equivalently, the work done by the net force­will equal its change in kinetic energy; this is known as the work-energy theorem:

Wtotal = Li K

Notice that kinetic energy,like work, is a scalarquantity.

60 CRACKING THE AP: PHYSICS EXAM

Example 4.6 \l\'11,lt is the kinetic energy of a baseball (mass=0 15kg)movingwith a speed of 36m/&?

Solution. From the definition,

Example 4.7 A tennisball(mass=0.06 kg)is hit straight upward withanirutial speed of 45 m/s. How high would it go if air resistance werenegligible?

Solution. This could be done using the BigFive,but let's try to solve it using the concepts of workand energy. As the ball travels upward, gravity acts on it by doing negative work. [The work isnegative because'gravity is opposing the upward motion. Fw and d are in opposite directions, so(} =180°, which tellsus that W=(Fw cos (J)d =-Fwd.] At the moment the ball reaches its highestpoint,its speed is 0, so its kineticenergy is also O. The work-energy theorem then says

Example 4.8 Consider the box sliding down the inclined plane inExample 4.4.If it starts fromrest at the top of the ramp, with \'11 hat speeddoes it reach the bottom?

Solution. Itwas calculatedin Example 4.4that Wtota\ =1130J.Accordingto the work-energytheorem,

Wtotal =M< => Wtotal =~f - K, =Kf=tmv2 => v=~2Wtotal = 2(1130 J) =8 m I s

. m 35kg

Example4.9 Apoolcue strikinga stationary billiardball (mass =025 kg)gIv~ the ball a speed of 15 m/s. If the rorce(If the cue OIl the ball was20N; over what distancedid this force act:'

Solution. The kinetic energy of the ball as it leaves the cue is

K =tmv'- = t(0~25 kg)(l.5 m/s)2 =0.28 J

The work Wdone by the cue gave the ball this kinetic energy, so

K 0.28 JW=M< => W=.l<t => Fd === K => d=- =--=0.014 m =1.4em

F . 20N

WORK, ENERGY, AND POWER . 61

POTENTiAL ENERGYKinetic energyis the energy an object has by virtue ofits motion. Potential energyis independentofmotion; it arisesfrom the object's position (or the system'sconfiguration). Forexample, a ballat theedge of a tabletophas energy,whichcouldbe transformed into kinetic energyif it falls off. Anarrowin an archer'spulled backbow has energythat couldbe transformedintokinetic energyif the archerreleases the arrow. Bothof theseexamples illustrate the conceptof potential energy, the energy anobject or system has by virtue of its position or configuration. In each ease, work was done on theobject to put it in the givenconfiguration (theballwas liftedto the 'tabletop, thebowstringwaspulledback), and sincework is the meansof transferringenergy, these thingshave stored energy that can beretrieved, as kinetic energy. This is potential energy, denoted by li.

Because there are differenttypes of forces, there are different types of potential energy. Theballat the edge of the tabletopprovidesan example of gravitational potential energy, U av' whichis theenergystoredby virtueofan object's positionin a gravitational field. ThisenergywoJid be convertedto kinetic energy as gravity pulled the ball down to the floor. For now, let's concentrate on gravita-tional potential energy. '

Assumetheballhas a massmof2kg,and that the tabletopish =1.5 m abovethe floor. Howmuchwork did gravitydo as the ballwasliftedfromthe floor to the table? Thestrengthof the gravitationalforce on the ball is Fw =mg=(2 kg)(9.8 N/kg) =20N. TheforceFw points downward, and the ball'smotion was upward, so the work done by gravity during the ball's ascentwas

Wbygravity =-F"h =-mgh =-(20 N)(l.5 m) =-30 JSo someone performed +30 Jof work to raise the ball from the floor to the tabletop. That energyis.now stored and, if the ballwas givena push to send it over the edge,by the time the ball reached thefloor it would acquire a kinetic energy of 30 J. We therefore say that the change in the ball'sgravitational potential energy in movingfrom the floor to the table was +30 J. That is,

AUgrav =-Wbygravity

Notice that potential energy,like work (and kinetic energy), is expressed in joules.In general, if an object ofmassmis raiseda heighth(which is smallenoughthatgstaysessentially

constantover this altitude change), then the increase in the object's gravitational potentialenergy,is

AU =mghgray

An important fact that makes the above equation possible is that the work done by gravity as theobject is raised does not depend on the path taken by the object. The ball could be lifted straightupward, or in somecurvy path; it would make no difference. Gravityis said to a conservative force ,becauseof this property.

If we decide on a reference level to call h =0, then we can say that the gravitational potentialenergyofan object ofmassmat a heighthis U v =mgh. In order to use this lastequation,it's essentialthat we choose a reference levelforheight.Forexample,considera passengerin an airplanereadinga book. If the book is 1 m above the floor of the plane then, to the passenger, the gravitationalpotentialenergyof the bookis mgh, where h =1 m. However, to someone on the ground lookingup,the floor of the plane may be, say, 9000 m above the ground. So, to this person, the gravitationalpotential energy of the book is mgH, where H =9001 m. What both would agree on, though, is thatthe difference in potential energy between the floor of the plane and the position of the book ismg x (l m), since the airplane passengerwould calculate the difference as mg x (1 m - 0 m), whilethe person on the ground would calculate it as mg x (9001 m - 9000 m). Differences, or changes, inpotential energy are unambiguous, but values of potential energy are relative.

62 CRACKING THE AP: PHYSICS EXAM

)

)

Example 4.10 A stuntwoman (mass » 60 kg) scalesa 40-metE:'r tall rockface, What is her gravitational potential energy trelative to the ground)?

Solution. Calling the ground h =0, we find

U~av =mgh =(60 kg)(9.8 N/kg)(40 m) =23,500 J

Example 4.11 If the stuntwoman in the previous example were to Jumpoff the chff, what would be her final speed as she landed OIl a large,air­filled cushion lying on the ground?

Solution. The gravitational potential energy would be transformed into kinetic energy. So

U ----' K U" 2 !2.U 2(23,500 J) =28 mls-.-, => ~ t mv => v =~---;;- = 60kg

CONSERVATION OF MECHANICAL ENERGYWe have seen energy in its two basic forms: Kinetic energy (K) and potential energy (U). The sum ofan object's kinetic and potential energies is called its mechanical energy, E:

E=K+U

<Notice that because U is relative, so is E.) Assuming that no nonconservative forces (friction, forexample) act on an object or system while it undergoes some change, then mechanical energy isconserved. That is, the initial mechanical energy, E

i, is equal to the final mechanical energy, Ep or

Ki + Ui =Kf + U,

This is the simplest form of the Law of Conservation of Total Energy, which we mentioned at thebeginning of this section.

Example 4.12 A ball of mass 2 kg is gentlv pushed off the edge of atabletop that is 1..':; m above the floor. Find the speed of the ball as itstrikes the floor.

Solution. Ignoring the friction' due to the air, we can apply Conservation of Mechanical Energy.Calling the floor our h =0 reference level, we write ;

Ki +Ui =Kf -u,O+mgh = tmv2 +0

v =~2gh

= ~r-2(-9.-8m-Is-2)-O-.s-m-)

=5.4mls

WORK, ENERGY, AND POWER . 63

Noticethat the ball's potential energy decreased,while its kineticenergy increased. This is the basicidea behind conservation of mechanical energy: One form of energy decreases while the otherincreases.

Example 4.13 A box is projected up a long ramp (incline angle with thehorizontal== 20°) with an initialspeed of 8 m/6 Ifthe surfaceof the rampis verysmooth tessentially frictionless), how highup the ramp will thebox go? What distance along the ramp will it slide?

Solution. Because friction is negligible, we can apply Conservationof Mechanical Energy. Callingthe bottom of the ramp our h == 0 reference level,we write

Ki +u'i =Kf +Uf

tmv~ + O = O + mgh1 2

h= IVag

1(8 m/s)2_ 2 .

- 9.8m/s2

=3.3 m

Since the incline angle is (J == 20°, the distance d it slides up the ramp is found in this way:

h =dsinO

'd =~= 3.3m = 9.6 msinO sin 20°

Example4.14 Askydiverjumps froma hoveringhelicopterthat'~ 3000 mabove the ground If air resistancecan be Ignored, how fast will he befalling when his altitude is 2000 m?

Solution. Ignoring air resistance, we can apply Conservation of Mechanical Energy. Calling theground our h =0 reference level, we write

t; +Uj = Kf -u,O+mgH =t mv2 +mgh

v =~2g(H-h)

=~2(9.8 m / 52)(3000 m - 2000 m)

== l 40m / s

64 • CRACKING THE AP: PHYSICS EXAM

(That's over 300 mph! This shows that air resistance does playa role, even before the parachute isopened.) .

Theequation Kj +Uj =K,+ Utholds ifno nonconservativeforces are doing work.However, if work

is done by such forces during the processunder investigation,then the equationneedsto be modifiedto account for this work as follows:

Example 4.15 Wtle E, 'oyote {mas!'> -= 40kg) falls off a 50-mpter-highcliff.Un the""ay dOlVT1; the torce(If cHI resistance b<t~ an average strength<If 100 N Find the': speed with,",hich he crashes into the ground

Solution., The forceofair resistanceopposes the downward motion, so it does negativework on thecoyoteas he falls: Wr =-F)to Calling the ground h =0, we find that

Xi -u, +Wr =Kf+Uf

. o+mgh +(-Frh) =tmv2 +0

v =~2h(g - Fr 1m) = ~2(50)(9.8 -100 I 40) = 27 m / s

Example 4.16 A skier "tarts from rest at the top of a 20°Incline and skisin a straight line to the bottom of the slope, a distanced (measuredalongthe slope)of 400 m If the coefficient of kinetic friction between the skisand the snow IS 0.2, calculate the skier's speed at the bottom of the run.

Solution. Thestrength of the frictionforceon the skier isF,= IIIN= Ilk(mg cos (J), so the work doneby friction is -Ftd =-Ilk(mg cos (J)·d. The vertical height of the slope above the bottom of the run(which we designate the h = 0 level) is h = d sin (}. Therefore, Conservation of Mechanical Energy(including the negative work done by friction) gives

K, +U i +Wfriction =Kf +U,

O+mgh+(-pkmgcos(J ·d) =t mv2+0

mg(dsin0) + (-Pkmgcos(} ·d) =tmv2

gd(sin(J - Ilk cosO) =t v2

V=~r-2-gd"'-(-si-n-O---Ilk-C-OS 0)

= ~2(9.8)(400)[sin20° - (0.2)cos 20°]

= 35 m/s

WORK, ENERGY, AND POWER . 65

[C] POTENTIAL ENERGY CURVESThe behavior of a system can be analyzed if we are given a graph of its potential energy, u<X) as a

function of position and its mechanical energy, E. SinceK + U =E, we have t mv2 +U(x) =E, which

can be solved for v, the velocity at position x:

2v=± -[E-U(x)]m

Forexample, consider the following potential energy curve:

E

U = U(x) .

The graph shows how the potential energy, U, varies with position, x. A particular value ofthe totalenergy, E =EO' is also shown. Motion of an object whose potential energy is given byU(x) and whichhas a mechanical energy of Eo is confined to the region -xo~ x ~ xo' because only in this range is Eo ~U(x). At each position x in this range, the kinetic energy, K =Eo·- U(x), is positive, However, if x> Xo(or if x < -xo)' then U(x) > Eo' which is physically impossible because the difference Eo - U(x), whichshould gIve K, is negative.

This' particular energy curve, with U(x) =.tk~, describes one of the most important physicalsystems: a simple harmonic oscillator. The force felt by the oscillator can be recovered from the 'potential energy curve. Recall that, in the case of gravitational potential energy, we definedA Ugrav =-Wby av' In general, AU =-W. If we account for a variable force of the form F =F(x), whichdoes the wor' W, then over a small displacement Ax, we have A U(x) = - W = -F(x) Ax, soF(x) =- AU(x) j Ax. In the limit as Ax ~ 0, this last equation becomes

dUF(x) = - -

dx

Therefore, in this case, we find F(x) =-(djdx)( tkX2) =-kx, which specifies a linear restoring force, aprerequisite for simple harmonic motion. This equation, F(x) =-kx, is called Hooke's Law and isobeyed by ideal springs (see Example 4.5).

With this result, we can appreciate the oscillatory nature of the system whose energy curve issketched above. If x is positive (and not greater than xo)' then U(x) is increasing, so dU j dx is positive,which tells us that F is negative . So the oscillator feels a force-and an acceleration-in the negativedirection, which pulls it back through the origin (x =0), If x is negative (and not less than -xo)' thenU(x) is decreasing, so dll]dx is negative, which tells us that Eis positive. So the oscillator feels aforce-and an acceleration-in the positive direction, which pushes it back through the origin (x =0).

66 • CRACKING THE AP: PHYSICS EXAM

Furthermore, the difference between Eo and U, which is K, decreases as x approaches Xo(or as xapproaches -x), dropping to zero at these points. Thefactthat K decreases to zeroat ±xotellsus thatthe oscillator's speed decreases to zero as it approaches these endpoints,beforechanging directionand heading back toward the origin-where its kinetic energy and speed are maximized-foranother oscillation. Bylookingat the energycurve with these observations in mind,you can almostsee the oscillator moving back and forth betweenthe barriersat x =±xo'

The origin is a point at which U(x) has a minimum, so the tangent line to the curveat this pointis horizontal; the slope is zero. Since F =~Uldx, the force F is 0 at this point [Which we alsoknowfromthe equationF(x) =-lex]; thismeans that this is a point ofequilibrium. If theoscillator is pushedfrom this equilibrium point in either direction, the force "F(x)' will attempt .to restore it tox =0,so this is a.point of stable equilibrium. However, a point where the U(x)curvehas a maximumis also a point of equilibrium, but it's an Unstable one,because if the systemwere movedfrom thispoint in either direction, the force would accelerate it away from the equilibrium position.

Consider the following potential energy curve: '

E

Point C is a position of stable equilibrium and Eisa point of unstable equilibrium. Since U(x) isdecreasing at Points A and F, F(x) is positive, accelerating the system in the positive x direction.PointsBand D mark the barriers of oscillation if the systemhas a mechanical energy Eo of O.

POWERSimply put, power is the rate at whichwork getsdone (orenergygets transferred, whichis the samething). Supposeyou and I each do 1000 Jof work,but I do the'work in 2 minutes whileyou do it inI minute. We both did the same amount of work, but you did it more quickly; you were morepowerful. Here's the definition of power:

WorkPower=-- -in symbols~

timew

p=-t

Theunit ofpoweris the jouleper second(I/s), whichis renamedthe watt, and symbolized W (notto be confused with the symbolfor work, W). Onewatt is I jouleper second:1W =1JI s. Herein theUnitedStates, whichstilluses olderunits likeinches, feet, yards,miles, ounces, pounds,and soforth,you stillhear ofpower ratingsexpressed in horsepower(particularly of engines). Onehorsepowerisdefined as, well, the power output of a large horse. Horses can pull a ISO-pound weightat a speed

WORK, ENERGY, AND POWER 61

of2 t mph for quite a while. L~t's assume that Fand d are parallel, so that W=Fd; then the definitionp =Wit becomesP =Fdlt, which is Ft: Therefore,

P =Fv => 1 horsepower (hp) =(150 Ib)(2 t mph)

Now for some unit conversions:

150lb=150 lbX 4.~~ N =667.5 N

2 1 h _ 2t mi 1609 m . 1 hr - 1 117 I-mp ---x X---. m s2 hr l mi 3600s

Therefore,

1 hp = (667.5 N)(1.117 m/s) = 746 W

Bycontrast, a human in good physicalconditioncan do work at a steady rate ofabout 75W(about1/10 that ofa horse!) butcan attain power levelsas much as twicethis much forshort periods oftime.

Example 4.1.7 A mover pushes a large crate (mass in =- 75 kg) from theinside of the truck tL' the hack end ta distance (If 6 m). exerting a steadv,push of 300 N If he moves the crate this distance III 20 s, what is hispower output during this lime?

Solution. The work done on the crate by the mover is W=Fd =(300 N)(6 m) =1800 J. If this muchwork is-done in 20 s, then the power delivered is P =Wit =(1800 J)/(20 s) =90W.

Example4.18 What must be the power output of an elevatormotor thatcan lift d total mass of 1000 kg and give the elevator a constant speed of80 rn/~?

Solution. The equation P =Fv, with F =mg, yields

P= mgv =(1000 kg)(9.8 NIkg)(8.0 mls) =78,000 W =78kW

68 • CRACKING THE AP: PHYSICS EXAM

CHAPTER 4 REVIEW QUESTIONS

SECTION I: MULTIPLE CHOICE

1. A force F of strength 20 N acts on anobject of mass 3 kg as it moves a distanceof 4 m. If F is perpendicular to the 4 mdisplacement, the work it does is equal to

(A) 0 J(B) 60J(C) 80 J(D) 600 J(E) 2400 J

2. Under the influence of a force, an object ofmass 4 kg accelerates from 3 mls to 6 mlsin 8 s. How much work was done on theobject during this time?

(A) 27 J(B) 54 J(C) 72 J(D) 96 J(E) Cannot be determined from the

information given

3. A box of mass m slides down a frictionlessinclined plane of length L and verticalheight h. What is the change in its gravita­tional potential energy?

(A) -mgL(B) - mgh(C) -mgLlh(D) -mghlL(E) -mghL "

4. An object of mass m is traveling at con­stant speed v in a circular path of radius r.How much work is done by the centrip­etal force during one-half of a revolution?

(A) 1tmv2

(B) 21tm&(C) 0(D) 1tmv2r

(E) 21tm&r

5. While a person lifts a book of mass 2 kgfrom the floor to a tabletop, 1.5 m abovethe floor, how much work does th egravitational force do on the book?

(A) -30 J(B) -15 J(C) 0 J(D) 15 J(E) 30 J

6. A block of mass 3.5 kg slides down africtionless inclined plane of length 6 mthat makes an angle of 30° with th ehorizontal. If the block is released fromrest at the top of the incline, what is itsspeed at the bottom?

(A) 4.9 m/s(B) 5.2 mls(C) 6.4 mls(D) 7.7m/s(E) 9.1 mls

7. A block of mass 3.5 kg slides down aninclined plane of length 6 m that makes anangle of 60° with the horizontal. Thecoefficient of kinetic friction between the "block and the incline is 0.3. If the block isreleased from rest at the top of the incline,what is its speed at the bottom?

(A) 4.9 mls(B) 5.2 mls(C) 6.4 mls(D) 7.7 mls(E) 9.2 mls

8. "As a rock of mass 4 kg drops from theedge of a 40-meter-high cliff, it experi­ences air "resis tance, whose averagestrength during the descent is 20 N. Atwhat speed will the rock hit the ground?

(A) 8 mls(B) 10 mls(C) 12 mls(D) 16 mls(E) 20 mls

WORK, ENERGY, AND POWER . 69

9. An astronaut drops a rock from the top ofa crater on the Moon. When the rock ishalfway down to the bottom of the crater,its speed is what fraction of its finalimpact speed?

(A) 1/4.j2(B) 1/4(C) 1/2.j2(D) 1/2(E) 1/.j2

70 • CRACKING THE AP: PHYSICS EXAM

10. A force of 200 N is required to keep anobject sliding at a constant speed of 2 m/sacross a rough floor. How much power isbeing expended to maintain this motion?

(A) SOW(B) 100 W(C) 200 W(D) 400 W(E) Cannot be determined from the

information given

1

}

SEalON II: FREE RESPONSE

1. A box of mass m is'released from rest at Point A, the top of a long, frictionless slide. Point A isat height H above the level of Points Band C. Although the slide is frictionless, the horizontalsurface from Point B to C is not. The coefficient of kinetic friction between the box and thissurface is Ilk' and the horizontal distance between Points Band Cis x.

A

B

x

C

(a) Find the speed of the box when its height above Point B is tH.

(b) Find the speed of the box when it reaches Point B.

(c) Determine the value of Ilkso that the box comes to rest at Point C.

(d) Now assume that Points B'and C were not on the same horizontal level. In particular,assume that the surface from B to C had a uniform upward slope so that Point C werestill at a horizontal distance of x fromB but now at a vertical height of y above B.Answer the question posed in part (c).

(e) If the slide were not frictionless, determine the work done by friction as the box movedfrom Point A to Point B if the speed of the box as it reached Point B were half the speedcalculated in part (b). .

WORK, ENERGY AND POWER 11

2. The diagram below shows a roller-coaster ride which contains a circular loop of radius r . Acar (mass m) begins at rest from Point A and moves down the frictionless track from A to Bwhere it then enters the vertical loop (also frictionless), traveling once around the circle fromBto C to D to E and back to B, after which it travels along the flat portion of the track from Bto F (which is not frictionless).

c

F

(a) Find the centripetal acceleration of the car when it is at Point C-

(b) Determine the speed of the car when its position relative to Point B-is specified by theangle (J shown in the diagram.

(c) What is the minimum cut-off speed V c that the car must have as it enters the loop atPoint B to make it around the loop? '

(d) What is the minimum height H necessary to ensure that the car makes it around theloop?

(e) If H = 6r and the coefficient of friction between the car and the flat portion of the trackfrom B to F is 0.5, how far along this flat portion of the track will the car travel beforecoming to rest at Point F?

12 • CRACKING THE AP: PHYSICS EXAM

3. [C] A particle of mass m =3 kg has the potential energy function

U(x) =3(x - 1) - (z - 3)3

where x is measured-in meters and U in joules. The following graph is a sketch of this poten­tial energy function.

Energy

IIIIII

E1 - - - l- - - --IIIIII

The energies indicated on the vertical axis are evenly spaced: that is, E3

- E2=E

2- E

1" The

energy EI is equal to U(xI) , and the energy E

3is equal to U(x),

(a) Determine the numerical values of Xl and x3

•

(b) Describe the motion of the particle if its total energy is E2

,

(c) What is the particle's speed at x =Xl if its total energy, E, equals 58 J?

(d) Sketch the graph of the particle's acceleration as a function of x. Be sure to indicate Xl

and x3

on your graph,

(e) The particle is released from rest at X = tXI' Find its speed as it passes through X =Xl'

WORK, ENERGY, AND POWER 73

Therefore, to find the maximum speed at which static friction can continue to provide thenecessary force, we write

(c) Ignoring friction, the forces acting on the car are gravity (Fw/ downward) and the normalforce from the road (Fwwhich is now tilted toward the center of curvature of the road):

(d) Because of the banking of the tum, the normal force is tilted toward the center of curvatureof the road. The component of FN toward the center can provide the centripetal force, makingreliance on friction unnecessary.

SOLUTIONS TO THE CHAPTER REVIEW QUESTIONS . 443

There's no vertical acceleration, so FN cos 8 =Fw =mg, so FN =mg/ cos 8. The component ofFN toward the center of curvature of the turn, FN sin 8, provides the centripetal force:

. mv2

FNsm6 = ­r

2mg sin8 =mv

cos8 rv2

gtan8 = ­r

28 =tan-1E....

gr

CHAPTER 4 REVIEW QUESTIONS

SECTION I: MULTIPLE CHOICE1. (A) Since the force F is perpendicular to the displacement, the work it does is zero.

2. (B) By the work-energy theorem,

W=M< =tm(v2-v~) = t(4 kg)[(6m/ S)2 - (3 m/ s)2]=54 J

3. (B) Since the box (mass m) falls through a vertical distance of h, its gravitational potentialenergy decreases by mgh. The length of the ramp is irrelevant here.

4. (C) Since the centripetal force always points along a radius toward the center of the circle, andthe velocity of the object is always tangent to the circle (and thus perpendicular to the radius),the work done by the centripetal force is zero. Alternatively, since the object's speed remainsconstant, the work-energy theorem tells us that no work is being performed.

5. (A) The gravitational force points downward while the book's displacement is upward. there­fore, the work done by gravity is -mgh =--(2 kg)(9.8 N/kg)(l.5 m) =-30 J.

6. (D) The work done by gravity as the block slides down the inclined plane is

W=(mg sin 8)(L) =[(3.5 kg)(9.8 N/kg) sin 30°](6 m) =103 J

Now, applying the work--energy theorem, we find that

W =M<=tm(v2-v~)=tmv2 => v =~2mW = 2(103J) = 7.7 m / s3.5 kg

444 • CRACKING THE AP PHYSICS EXAM

7. (E) Apply Conservation of Mechanical Energy (including the negative work done by F, theforce of slidingfriction):

K-u, +Wf = Kt +Uf

0+mgh-~L=tmv2+0

mg(LsinO) - tpmgcosO)L =tmv2

t' =~r-2g-L-(s-in-8---Jl-c-os-O-)

=~2(9.8 m I s2)(6 m)[sin60° - (0.3)cos60°]

=9.2 m / s

8. (E) Apply Conservation of Mechanical Energy (including the negative work done by Fr' theforce of air resistance):

Kj +Ui +Wr =K, +Uf

0+mgh -Frh=tmv2 +0

V= 2h(rng-Fr)m

=~2(4O m)[(4 kg)(9.8 N I kg)- 20 N]-l kg

=20m/s

9. (E) Because the rockhas lost half of its gravitational potentialenergy, its kinetic energy at thehalfwaypoint is halfof its kinetic energyat impact. Since K is proportional to rfl, if Kalhalfwaypoinl

is equal to t Kat impact' then the rock's speed at the halfwaypoint is ~112 =1l...fi its speed atimpact.

10. (D) Usingthe equationP =Fv, we find that P =(200 N)(2 m/s) =400 W.

SEalON II: FREE RESPONSE1. (a) ApplyingConservation of Energy,

KA +UA =Kat H / 2 +Uat H / 2

O+mgH =t mv2+mg(tH)

tmgH=tmv2

v =~gH

SOLUTIONS TO THE CHAPTER REVIEW QUESTIONS 445

(b) ApplyingConservation of Energyagain,

KA +UA == KB +UB

O+mgH == tmv~ +0

VB == ~2gH

(c) Bythe work-energytheorem, we want the work doneby friction to be equal (butopposite)to the kinetic energyof the box at Point B:

W == M< =tm(v~ - v~) =-tmv~ =-tm(~2gH)2 =-mgH

Therefore,

W =-mgH ~ - Ffx=-mgH ~ - f.lkmgx =-mgH ~ f.lk =H / x

(d) Apply Conservation of Energy (including the negative work done by friction as the boxslides up the ramp from B to C):

~cL~~ ~

~ iY~ ,

------- -- - - - -- - - - - - - -- -- - - - - - - - -~:B x

KB +UB +Wf = Kc +Uc

tm(~2gH)2 +O-FfL =O+mgy

mgH +O-FfL=O+mgy

mg(H - Y) - (f.lkmgcos8)(L) =0

H-y H-yf.lk = Lcos8 x

(e) The result ofpart (b) reads VB =~2gH. Therefore, by Conservation of Mechanical Energy

(with the work done by the frictional force on the slide included),we get

KA +UA +Wf = Ki +UB

1 (1 )2O+mgH+Wf =2"m 2"vB +0

mgH +Wf =tm(t~2gHt

mgH +Wf = i mgH

Wf=-imgH

446 • CRACKING THE AP PHYSICS EXAM

12. (a) The centripetal acceleration of the car at Point C is given by the equation a=v2 [r, where

Vcis the speed of the car at C. To find v2' we apply Conservation of Energy:

KA +UA =Kc +UcO H _ 1 2+mg - "2 mvc +mgr

mg(H - r)=tmv2

1'2 =2g(H -r)

Therefore,

v2 2g(H -r)a - - - --'='--'--_":"C - r - r

(b) In terms of 8, the car's height above the bottom of the track (point B) is given by the

equation h =r + (-r cos 8),

r(-cos 0) D

) E c

so we get

B

KA +UA = K+U

0+ mgH =tmv2 +mg(r - rcos9)

mg[H - r(l - cos8)] =tmv2

v =~r-2----:g['----H---r--'--(I---co-s-8""""')1

(c) When the car reaches Point D, the forces acting on the car are its weight, Fw' and thedownward normal force, FN' from the track. Thus, the net force, Fw + FN' provides thecentripetal force. In order for the car to maintain contact with the track, FN must not vanish.Therefore, the cut-off speed for ensuring that the car makes it safely around the track is thespeed at which FN just becomes zero; any greater speed would imply that the car would makeit around. Thus,

v2F +0 =m cut-off

W rrrF::r=

=> Vcut-off =~ ---;;- =-V gr

SOLUTIONS TO THE CHAPTER REVIEW QUESTIONS . 447

(d) Using the cut-off speed calculated in part (c), we now apply Conservation of MechanicalEnergy:

KA +UA =KD+UD

0+ mgH =tmv;ut-off+mg(2r)

mgH =tm(gr) +mg(2r)

=tmgr

H = t r

(e) First, we calculate the car's kinetic energyat PointB; then, we determinethe distance x thecar must travel from Bto F for the work done by friction to eliminate this kinetic energy. So,applying Conservation of Mechanical Energy, we find

KA +UA = KB +UB

0+ mg(6r) =tmv~ +0

6 - 1 2mgr -2:mvB

Now, by the work-energy theorem,

F. 1 2~ - fX= -2: mvB

-pmgx =-6mgr

6r 6rx =-=-=12r

J.l 0.5

3. (a) Thepointwherex = Xl is a local minimumofthe function U(x), and the pointwherex = x3is

a local maximum. Therefore, at eachoftheselocations, the derivative ofU(x) mustbe equalto O.

dU = 0 ~ 3-3(x-3)2 = 0 ~ (X-3)2 = 1 ~ x =2or4dx

Therefore, Xl =2 m and x3=4 m.

(b) If the particle's total energyis E2 =t (E1 +E3) =t [U(x1) + U(x3) ] =t [U(2) + U(4)] =t (4J+8J) =6J, thentheparticle canoscillate betweenthepointsmarkedaand bin thefigureon thenextpage.

448 • CRACKING THE AP PHYSICS EXAM

1

,)

Energy

)

At pointsaand b,the object has no kinetic energy(since U=E2 at thesepoints),so these are theturning points at which the object momentarily comes to rest beforebeing accelerated backtoward x =Xl (wherethe potentialenergyis minimized). The particlecannotbe found at x <aor withinthe intervalb<x<C,sinceU> E2 in theseintervals(and thiswould implya negativeK,which is impossible). If x> c, then sinceU decreases, K increases: the particle would moveoff with increasing speed.

(c) Since K + U = E,we have K = E- U. Therefore,

K(x1) =E- U(x1)

=E-[3(x-1)- (x-3)3lx:2

=58J-[(3-(-1)] J=54J

Therefore,

K=tmv; => VI = ~2K = 2(5k4

J) =6 m/sm 3 g

(d) Since F(x) = -dUIdx = -[3 - 3(x - 3)2] = 3(x- 3)2 - 3, dividing by m givesa:

a(x) = F(x) [3(x _3)2-3] J= (x-3i - 1 (in ml 52)m 3kg

SOLUTIONS TO THE CHAPTER REVIEW QUESTIONS . 449

Acceleration

ornls21---.-~r---r-~:"'-.,-----.--x

-1 rnls2m_m u __

Xl =2

(e) At x =tXl' the particle's energy is £3 = U(tXl) = U(1) = [3(x-1)-(x-3)3] =8 J.=1

This is the particle's total energybecause it has no initial kinetic energy (it is released from rest).As the particle passes through X =Xl' its potential energy decreases to

Therefore, since its total energy is 8 J, the particle's kinetic energy at X = Xl must beK =E - U =8 J- 4 J=4 J. This implies that its speed is

K =tmv2 ~ v = ~2K =~2(4 n= 1.6 rn I s

m 3kg

450 • CRACKING THE AP PHYSICS EXAM