Mechanics Potential Energy Conservative and Nonconservative Forces

Home , Conservative force

Lana Sheridan

De Anza College

Oct 31, 2018 Last time

• work done by a varying force

• kinetic energy

• the work-kinetic energy theorem

• power Overview

• concept of potential energy

• conservative and nonconservative forces

• potential energy deﬁnition

• some kinds of potential energy

• potential energy diagrams WALKMC08_0131536311.QXD 12/8/05 17:47 Page 205

8–1 CONSERVATIVE AND NONCONSERVATIVE FORCES 205

8–1 Conservative and Nonconservative Forces In physics, we classify forces according to whether they are conservative or nonconservative. The key distinction is that when a conservative force acts, the work it does is stored in the form of energy that can be released at a later time. In this section, we sharpen this distinction and explore some examples of conservative and nonconservative forces. Perhaps the simplest case of a conservative force is gravity. Imagine lifting a box of mass m from the ﬂoor to a height h, as in Figure 8–1. To lift the box with constant speed, the force you must exert against gravity is mg. Since the upward distance is h, the work you do on the box is W = mgh. If you now release the box andWork allow it to Done drop back Lifting to the ﬂoor, a gravity Box does the same work, W = mgh, and in the process gives the box an equivalent amount of kinetic energy. Work done by person (applied force) W = Fd cos(0◦) = mgh. Work done by person = mgh Work done by gravity = mgh ▲ FIGURE 8–1 Work against gravity Lifting a box against gravity with constant speed takes a work mgh. When the box is released, gravity does the same work on the box as it falls. Gravity is a conservative force.

ContrastWhen this box with falls, the force this energyof kinetic becomes friction, which kinetic is nonconservative. energy. To slide a box of mass m across the ﬂoor with constant speed, as shown in Figure 8–2, Wnet = mgh = ∆K. you must exert a force of magnitude mkN = mkmg. After sliding the box a distance d, the work you have done is W = mkmgd. In this case, when you release the box it simply stays put—friction does no work on it after you let go. Thus, the work done by a nonconservative force cannot be recovered later as kinetic energy; instead, it is converted to other forms of energy, such as a slight warming of the ﬂoor and box in our example. The differences between conservative and nonconservative forces are even more apparent if we consider moving an object around a closed path. Consider, for example, the path shown in Figure 8–3. If we move a box of mass m along this path, the total work done by gravity is the sum of the work done on each segment of the path; that is Wtotal = WAB + WBC + WCD + WDA. The work done by gravity from A to B and from C to D is zero, since the force is at right angles to the displacement on these segments. Thus WAB = WCD = 0. On the segment from B to C, gravity does negative work (displacement and force are in opposite directions),

Work = ␮kmgd

␮kN

8–1 CONSERVATIVE AND NONCONSERVATIVE FORCES 205

8–1 Conservative and Nonconservative Forces In physics, we classify forces according to whether they are conservative or nonconservative. The key distinction is that when a conservative force acts, the work it does is stored in the form of energy that can be released at a later time. In this section, we sharpen this distinction and explore some examples of conservative and nonconservative forces. Perhaps the simplest case of a conservative force is gravity. Imagine lifting a box of mass m from the ﬂoor to a height h, as in Figure 8–1. To lift the box with constant speed, the force you must exert against gravity is mg. Since the upward distance is h, the work you do on the box is W = mgh. If you now release the box and allow it to drop back to the ﬂoor, gravity does the same work, W = mgh, and in the process gives the box an equivalent amount of kinetic energy. Potential Energy

Work done by person = mgh Work done by gravity = mgh ▲ FIGURE 8–1 Work against gravity Lifting a box against gravity with constant speed takes a work mgh. When the box is released, gravity does the same work on the box as it falls. Gravity is a conservative force.

Contrast this with the force of kinetic friction, which is nonconservative. To slide a boxWhen of mass the m boxacross is inthe theﬂoor air, with it constant has the speed, “potential” as shown toin Figure have 8–2 kinetic, you mustenergy. exert a force of magnitude mkN = mkmg. After sliding the box a distance d, the work you have done is W = mkmgd. In this case, when you release the box it simply stays put—friction does no work on it after you let go. Thus, the work done byThe a nonconservative man put in work force liftingcannot it,be recovered as long aslater the as boxkinetic is energy; held in in- the air, stead, itthis is converted energy isto stored.other forms of energy, such as a slight warming of the ﬂoor and box in our example. The differences between conservative and nonconservative forces are even more apparent if we consider moving an object around a closed path. Consider, for example, the path shown in Figure 8–3. If we move a box of mass m along this path, the total work done by gravity is the sum of the work done on each segment of the path; that is Wtotal = WAB + WBC + WCD + WDA. The work done by gravity from A to B and from C to D is zero, since the force is at right angles to the displacement on these segments. Thus WAB = WCD = 0. On the segment from B to C, gravity does negative work (displacement and force are in opposite directions),

Work = ␮kmgd

␮kN

This illustrates that there is another type of energy that it makes intuitive sense to assign in some systems.

That is a kind of energy that results from the conﬁguration of the system, the potential energy. WALKMC08_0131536311.QXD 12/8/05 17:47 Page 205

8–1 CONSERVATIVE AND NONCONSERVATIVE FORCES 205

8–1 Conservative and Nonconservative Forces In physics, we classify forces according to whether they are conservative or nonconservative. The key distinction is that when a conservative force acts, the work it does is stored in the form of energy that can be released at a later time. In this section, we sharpen this distinction and explore some examples of conservative and nonconservative forces. Perhaps the simplest case of a conservative force is gravity. Imagine lifting a box of mass m from the ﬂoor to a height h, as in Figure 8–1. To lift the box with constant speed, the force you must exert against gravity is mg. Since the upward distance is h, the work you do on the box is W = mgh. If you now release the box and allow it to drop back to the ﬂoor, gravity does the same work, W = mgh, and in the process gives the box an equivalent amount of kinetic energy. Work Done Lifting a Box

◦ ▲ FIGURE 8–1 Work against gravity Work done by person (appliedWork done force) by personWapp = mgh= Fd cos(0 ) = Womghrk. done by gravity = mgh Lifting a box against gravity with constant speed takes a work mgh. When the box is released, gravity does the same work on the box as it falls. Gravity is a conservative force.

◦ WorkContrast done by gravitythis withWg the= Fd forcecos( 180of kinetic) = − mghfriction,. which is nonconservative. To slide a box of mass m across the ﬂoor with constant speed, as shown in Figure 8–2, you must exert a force of magnitude mkN = mkmg. After sliding the box a distance d, the work you have done is W = mkmgd. In this case, when you release the box it simply stays put—friction does no work on it after you let go. Thus, the work done by a nonconservative force cannot be recovered later as kinetic energy; instead, it is converted to other forms of energy, such as a slight warming of the ﬂoor and box in our example. The differences between conservative and nonconservative forces are even more apparent if we consider moving an object around a closed path. Consider, for example, the path shown in Figure 8–3. If we move a box of mass m along this path, the total work done by gravity is the sum of the work done on each segment of the path; that is Wtotal = WAB + WBC + WCD + WDA. The work done by gravity from A to B and from C to D is zero, since the force is at right angles to the displacement on these segments. Thus WAB = WCD = 0. On the segment from B to C, gravity does negative work (displacement and force are in opposite directions),

Work = ␮kmgd

␮kN

▲ FIGURE 8–2 Work against friction mg Pushing a box with constant speed against friction takes a work mkmgd. ␮kN When the box is released, it quickly d N comes to rest and friction does no further work. Friction is a nonconservative force. WALKMC08_0131536311.QXD 12/8/05 17:47 Page 206 Conservative and Nonconservative Forces The work done by gravity when raising and lowering an object 206 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY around a closed path is zero.

▲ FIGURE 8–3 Work done by gravity on W = 0 Side View a closed path is zero Gravity does no work on the two horizontal segments of the path. On the two D C vertical segments, the amounts of work done are equal in magnitude but opposite in sign. Therefore, the total work mg done by gravity on this—or any—closed path is zero. W = mgh W = –mgh h

mg mg W = 0

The path takenbut it doesn’tdoes positive matter; work fromif D itto A comes (displacement back and to force the are start, in the same the work done isdirection). zero. Hence, WBC =-mgh and WDA = mgh. As a result, the total work done by gravity is zero: Forces (like gravity) that behave this way are called conservative Wtotal = 0 + -mgh + 0 + mgh = 0 forces. With friction, the results are quite1 different.2 If we push the box around the closed horizontal path shown in Figure 8–4, the total work done by friction does not vanish. In fact, friction does the negative work W =-fkd =-mkmgd on each segment. Therefore, the total work done by kinetic friction is

Wtotal = -mkmgd + -mkmgd + -mkmgd + -mkmgd =-4 mkmgd

These results1 lead 2to the1 following2 deﬁnition1 of2 a conservative1 2 force:

Conservative Force: Deﬁnition 1 A conservative force is a force that does zero total work on any closed path.

▲ FIGURE 8–4 Work done by friction W = – µ mgd Top View on a closed path is nonzero k The work done by friction when an object moves through a distance d is D C - mkmgd. Thus, the total work done by fk friction on a closed path is nonzero. In this case, it is equal to - 4 mkmgd.

fk W = – µ kmgd d f W = – µ kmgd k

fk AB

W = – µ kmgd

d WALKMC08_0131536311.QXD 12/8/05 17:47 Page 206

206 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

mg mg W = 0

but it does positive work from D to A (displacement and force are in the same direction). Hence, WBC =-mgh and WDA = mgh. As a result, the total work done by gravity is zero:

Wtotal = 0 + -mgh + 0 + mgh = 0

With friction, the results are quite1 different.2 If we push the box around the closed horizontal path shown in Figure 8–4, the total work done by friction does not vanish. In fact, friction does the negative work W =-fkd =-mkmgd on each segment. Therefore, the total work done by kinetic friction is

Wtotal = -mkmgd + -mkmgd + -mkmgd + -mkmgd =-4 mkmgd

ConservativeThese and results1 Nonconservative lead 2to the1 following2 deﬁnition1 of2 a conservative Forces1 2 force: The work doneConservative by friction Force: Deﬁnition when 1 pushing an object around a closed A conservative force is a force that does zero total work on any closed path. path is not zero.

fk W = – µ kmgd d f W = – µ kmgd k

fk AB

W = – µ kmgd

Forces (like friction) where the work done over a closed path is not zero are called nonconservative forces. WALKMC08_0131536311.QXD 12/8/05 17:47 Page 205

8–1 CONSERVATIVE AND NONCONSERVATIVE FORCES 205

8–1 Conservative and Nonconservative Forces In physics, we classify forces according to whether they are conservative or nonconservative. The key distinction is that when a conservative force acts, the work it does is stored in the form of energy that can be released at a later time. In this section, we sharpen this distinction and explore some examples of conservative and nonconservative forces. Perhaps the simplest case of a conservative force is gravity. Imagine lifting a box of mass m from the ﬂoor to a height h, as in Figure 8–1. To lift the box with constant speed, the force you must exert against gravity is mg. Since the upward distance is h, the work you do on the box is W = mgh. If you now release the box and allow it to drop back to the ﬂoor, gravity does the same work, W = mgh, and in the process gives the box an equivalent amount of kinetic energy.

Contrast this with the force of kinetic friction, which is nonconservative. To slide a box of mass m across the ﬂoor with constant speed, as shown in Figure 8–2, you must exert a force of magnitude mkN = mkmg. After sliding the box a distance d, the work you have done is W = mkmgd. In this case, when you release the box it simply stays put—friction does no work on it after you let go. Thus, the work done by a nonconservative force cannot be recovered later as kinetic energy; instead, it is converted to other forms of energy, such as a slight warming of the ﬂoor and box in our example. The differences between conservative and nonconservative forces are even moreNonconservative apparent if we consider moving Forces: an object Friction around a closed path. Consider, for example, the path shown in Figure 8–3. If we move a box of mass m along this path, Thethe total work work done done byby gravity kinetic is frictionthe sum of is the always work done negative. on each segment of the path; that is Wtotal = WAB + WBC + WCD + WDA. The work done by gravity fromKinetic A to B and friction from C points to D is inzero, the since opposite the forcedirection is at right angles to the to the velocity dis- / placement on these segments. Thus WAB = WCD = 0. On the segment from B to C, gravityinstantaneous does negative displacement. work (displacement and force are in opposite directions),

Work = ␮kmgd

␮kN

Wfric = −fk d = −µk Nd where d is the distance the object moves along the surface. Nonconservative Forces: Friction

Air resistance is another nonconservative force.

When a force does negative work on a system, energy is transferred out of the system.

In the case of kinetic friction, this energy increases the temperature of the two surfaces that rub on each other, and may also leave as sound waves.

This energy is lost to the system, but not to the universe. WALKMC08_0131536311.QXD 12/8/05 17:47 Page 205

8–1 CONSERVATIVE AND NONCONSERVATIVE FORCES 205

8–1 Conservative and Nonconservative Forces In physics, we classify forces according to whether they are conservative or nonconservative. The key distinction is that when a conservative force acts, the work it does is stored in the form of energy that can be released at a later time. In this section, we sharpen this distinction and explore some examples of conservative and nonconservative forces. Perhaps the simplest case of a conservative force is gravity. Imagine lifting a box of mass m from the ﬂoor to a height h, as in Figure 8–1. To lift the box with constant speed, the force you must exert against gravity is mg. Since the upward distance is h, the work you do on the box is W = mgh. If you now release the box and allow it to drop back to the ﬂoor, gravity does the same work, W = mgh, and in Conservativethe process gives the box Forces: an equivalent Work amountDone of kinetic Liftingenergy. a Box Work done by person (applied force) W = Fd cos(0◦) = mgh. Work done by person = mgh Work done by gravity = mgh ▲ FIGURE 8–1 Work against gravity Lifting a box against gravity with constant speed takes a work mgh. When the box is released, gravity does the same work on the box as it falls. Gravity is a conservative force.

ContrastWhen this box with falls, the force this energyof kinetic becomes friction, which kinetic is nonconservative. energy. To slide a box of mass m across the ﬂoor with constant speed, as shown in Figure 8–2, Wnet = mgh = ∆K. you must exert a force of magnitude mkN = mkmg. After sliding the box a distance d, the work you have done is W = mkmgd. In this case, when you release the box it simplyThe stays man put—friction put in work doeslifting no work it, on as it longafter you as the let go. box Thus, is held the work in the air, done bythis a nonconservative energy is stored. force cannot be recovered later as kinetic energy; instead, it is converted to other forms of energy, such as a slight warming of the ﬂoor and box in our example. The differences between conservative and nonconservative forces are even more apparent if we consider moving an object around a closed path. Consider, for example, the path shown in Figure 8–3. If we move a box of mass m along this path, the total work done by gravity is the sum of the work done on each segment of the path; that is Wtotal = WAB + WBC + WCD + WDA. The work done by gravity from A to B and from C to D is zero, since the force is at right angles to the displacement on these segments. Thus WAB = WCD = 0. On the segment from B to C, gravity does negative work (displacement and force are in opposite directions),

Work = ␮kmgd

␮kN

▲ FIGURE 8–2 Work against friction mg Pushing a box with constant speed against friction takes a work mkmgd. ␮kN When the box is released, it quickly d N comes to rest and friction does no further work. Friction is a nonconservative force. For any conservative force acting on an object, we can say that the object has some amount of stored energy that depends on its conﬁguration.

Potential energy energy that system has as a result of its conﬁguration. Is always the result of the eﬀect of a conservative force.

Conservative Forces: Potential Energy

Any box that has been lifted a height h has had the same work done on it: mgh.

The path the box took to get to that height doesn’t matter.

This is because gravity is a conservative force. Potential energy energy that system has as a result of its conﬁguration. Is always the result of the eﬀect of a conservative force.

Conservative Forces: Potential Energy

Any box that has been lifted a height h has had the same work done on it: mgh.

The path the box took to get to that height doesn’t matter.

This is because gravity is a conservative force.

For any conservative force acting on an object, we can say that the object has some amount of stored energy that depends on its conﬁguration. Conservative Forces: Potential Energy

Any box that has been lifted a height h has had the same work done on it: mgh.

The path the box took to get to that height doesn’t matter.

This is because gravity is a conservative force.

For any conservative force acting on an object, we can say that the object has some amount of stored energy that depends on its conﬁguration.

Potential energy energy that system has as a result of its configuration. Is always the result of the effect of a conservative force. The “configuration” of the system refers to how close the box is to center of the Earth.

To have a potential energy, we must include the Earth in the system and make the weight of the box an internal force.

Any time a potential energy is introduced, the source of the conservative force becomes part of the system.

Potential Energy

Potential energy energy that system has as a result of its conﬁguration. Is always the result of the eﬀect of a conservative force.

What is the “conﬁguration” of the box? Any time a potential energy is introduced, the source of the conservative force becomes part of the system.

Potential Energy

Potential energy energy that system has as a result of its conﬁguration. Is always the result of the eﬀect of a conservative force.

What is the “conﬁguration” of the box?

The “conﬁguration” of the system refers to how close the box is to center of the Earth.

To have a potential energy, we must include the Earth in the system and make the weight of the box an internal force. Potential Energy

Potential energy energy that system has as a result of its conﬁguration. Is always the result of the eﬀect of a conservative force.

What is the “conﬁguration” of the box?

The “conﬁguration” of the system refers to how close the box is to center of the Earth.

To have a potential energy, we must include the Earth in the system and make the weight of the box an internal force.

Any time a potential energy is introduced, the source of the conservative force becomes part of the system. Potential Energy

Potential energy energy that system has as a result of its conﬁguration. Is always the result of the eﬀect of a conservative force.

∆U = −Wcons

Only conservative forces can have associated potential energies!

If a nonconservative force acts, any work done to displace the system (at constant velocity) leaves the system again as heat and sound.

That energy isn’t stored ⇒ no potential energy. Gravitational Potential Energy

The change of potential energy when lifting an object of mass m near the Earth’s surface:

∆U = mg(∆h)

If we choose the convention that U = 0 at the Earth’s surface, then an object (mass m) at a height h has gravitational potential energy: U = mgh

The Earth (which creates the gravitational force on the box) is part of our system description. (a) U = (mg)y = (400 N)(2 m) = 800J (b) U = (mg)y = (400 N)(2 m)(1 − cos 30◦) = 107J (c) U = 0.

Problems 207

34. A 4.00-kg particle is subject to a net force that varies on the object during this time interval? (c) What is the W with position as shown in Figure P7.15. The particle speed of the particle at t 5 2.00 s? starts from rest at x 5 0. What is its speed at (a) x 5 5.00 m, (b) x 5 10.0 m, and (c) x 5 15.0 m? Section 7.6 Potential Energy of a System 35. A 2 100-kg pile driver is used to drive a steel I-beam into 40. A 1 000-kg roller coaster car is initially at the top of a M the ground. The pile driver falls 5.00 m before coming rise, at point Ꭽ. It then moves 135 ft, at an angle of 40.08 into contact with the top of the beam, and it drives the below the horizontal, to a lower point Ꭾ. (a) Choose beam 12.0 cm farther into the ground before coming the car at point Ꭾ to be the zero configuration for to rest. Using energy considerations, calculate the aver- gravitational potential energy of the roller coaster– age force the beam exerts on the pile driver while the Earth system. Find the potential energy of the system pile driver is brought to rest. when the car is at points Ꭽ and Ꭾ, and the change 36. Review. In an electron microscope, there is an electron in potential energy as the car moves between these AMT gun that contains two charged metallic plates 2.80 cm points. (b) Repeat part (a), setting the zero configura- Ꭽ apart. An electric force accelerates each electron in tion with the car at point . the beam from rest to 9.60% of the speed of light over 41. A 0.20-kg stone is held 1.3 m above the top edge of a this distance. (a) Determine the kinetic energy of the water well and then dropped into it. The well has a electron as it leaves the electron gun. Electrons carry depth of 5.0 m. Relative to the configuration with the this energy to a phosphorescent viewing screen where stone at the top edge of the well, what is the gravita- the microscope’s image is formed, making it glow. For tional potential energy of the stone–Earth system an electron passing between the plates in the electronGravitational (a) before Potential the stone is Energyreleased and (b) when it reaches gun, determine (b) the magnitude of the constant the bottom of the well? (c) What is the change in gravi- electric force acting on the electron, (c) the accelera- tational potential energy of the system from release to tion of the electron, and (d) the time interval the elec- reaching the bottom of the well? tron spends between the plates. 42. A 400-N child is in a swing that is attached to a pair 37. Review. You can think of the work–kinetic energy the- W of ropes 2.00 m long. Find the gravitational potential GP orem as a second theory of motion, parallel to New- energy of the child–Earth system relative to the child’s Q/C ton’s laws in describing how outside influences affect lowest position when (a) the ropes are horizontal, the motion of an object. In this problem, solve parts (b) the ropes make a 30.08 angle with the vertical, and (a), (b), and (c) separately from parts (d) and (e) so (c) the child is at the bottom of the circular arc. you can compare the predictions of the two theories. A 15.0-g bullet is accelerated from rest to a speed of Section 7.7 Conservative and Nonconservative Forces 780 m/s in a rifle barrel of length 72.0 cm. (a) Find 43. A 4.00-kg particle moves y (m) the kinetic energy of the bullet as it leaves the bar- M from the origin to posi- rel. (b) Use the work–kinetic energy theorem to find tion Ꭿ, having coordi- Ꭿ Q/C Ꭾ (5.00, 5.00) the net work that is done on the bullet. (c) Use your nates x 5 5.00 m and y 5 result to part (b) to find the magnitude of the average 5.00 m (Fig. P7.43). One net force that acted on the bullet while it was in the force on the particle is barrel. (d) Now model the bullet as a particle under 1Problemthe gravitationalfrom Serway & Jewett, force 9th ed, page 207. constant acceleration. Find the constant acceleration acting in the negative y x (m) of a bullet that starts from rest and gains a speed of direction. Using Equa- O Ꭽ 780 m/s over a distance of 72.0 cm. (e) Modeling the tion 7.3, calculate the bullet as a particle under a net force, find the net Figure P7.43 work done by the gravi- Problems 43 through 46. force that acted on it during its acceleration. (f) What tational force on the conclusion can you draw from comparing your results particle as it goes from O of parts (c) and (e)? to Ꭿ along (a) the purple path, (b) the red path, and 38. Review. A 7.80-g bullet moving at 575 m/s strikes the (c) the blue path. (d) Your results should all be identi- hand of a superhero, causing the hand to move 5.50 cm cal. Why? in the direction of the bullet’s velocity before stopping. 44. (a) Suppose a constant force acts on an object. The (a) Use work and energy considerations to find the force does not vary with time or with the position or average force that stops the bullet. (b) Assuming the the velocity of the object. Start with the general defini- force is constant, determine how much time elapses tion for work done by a force between the moment the bullet strikes the hand and f S the moment it stops moving. 5 ? S W 3 F d r 39. Review. A 5.75-kg object passes through the origin i at time t 5 0 such that its x component of velocity is and show that the force is conservative. (b) As a spe- S 5.00 m/s and its y component of velocity is 23.00 m/s. cial case, suppose the force F 5 3 i^ 1 4j^ N acts on a (a) What is the kinetic energy of the object at this time? particle that moves from O to Ꭿ in Figure P7.43. Cal- S (b) At a later time t 5 2.00 s, the particle is located at culate the work done by F on the1 particle2 as it moves x 5 8.50 m and y 5 5.00 m. What constant force acted along each one of the three paths shown in the figure (b) U = (mg)y = (400 N)(2 m)(1 − cos 30◦) = 107J (c) U = 0.

Problems 207

34. A 4.00-kg particle is subject to a net force that varies on the object during this time interval? (c) What is the W with position as shown in Figure P7.15. The particle speed of the particle at t 5 2.00 s? starts from rest at x 5 0. What is its speed at (a) x 5 5.00 m, (b) x 5 10.0 m, and (c) x 5 15.0 m? Section 7.6 Potential Energy of a System 35. A 2 100-kg pile driver is used to drive a steel I-beam into 40. A 1 000-kg roller coaster car is initially at the top of a M the ground. The pile driver falls 5.00 m before coming rise, at point Ꭽ. It then moves 135 ft, at an angle of 40.08 into contact with the top of the beam, and it drives the below the horizontal, to a lower point Ꭾ. (a) Choose beam 12.0 cm farther into the ground before coming the car at point Ꭾ to be the zero configuration for to rest. Using energy considerations, calculate the aver- gravitational potential energy of the roller coaster– age force the beam exerts on the pile driver while the Earth system. Find the potential energy of the system pile driver is brought to rest. when the car is at points Ꭽ and Ꭾ, and the change 36. Review. In an electron microscope, there is an electron in potential energy as the car moves between these AMT gun that contains two charged metallic plates 2.80 cm points. (b) Repeat part (a), setting the zero configura- Ꭽ apart. An electric force accelerates each electron in tion with the car at point . the beam from rest to 9.60% of the speed of light over 41. A 0.20-kg stone is held 1.3 m above the top edge of a this distance. (a) Determine the kinetic energy of the water well and then dropped into it. The well has a electron as it leaves the electron gun. Electrons carry depth of 5.0 m. Relative to the configuration with the this energy to a phosphorescent viewing screen where stone at the top edge of the well, what is the gravita- the microscope’s image is formed, making it glow. For tional potential energy of the stone–Earth system an electron passing between the plates in the electronGravitational (a) before Potential the stone is Energyreleased and (b) when it reaches gun, determine (b) the magnitude of the constant the bottom of the well? (c) What is the change in gravi- electric force acting on the electron, (c) the accelera- tational potential energy of the system from release to tion of the electron, and (d) the time interval the elec- reaching the bottom of the well? tron spends between the plates. 42. A 400-N child is in a swing that is attached to a pair 37. Review. You can think of the work–kinetic energy the- W of ropes 2.00 m long. Find the gravitational potential GP orem as a second theory of motion, parallel to New- energy of the child–Earth system relative to the child’s Q/C ton’s laws in describing how outside influences affect lowest position when (a) the ropes are horizontal, the motion of an object. In this problem, solve parts (b) the ropes make a 30.08 angle with the vertical, and (a), (b), and (c) separately from parts (d) and (e) so (c) the child is at the bottom of the circular arc. you can compare the predictions of the two theories. A 15.0-g bullet is accelerated from rest to a speed of (a)SectionU = (mg7.7 )Conservativey = (400 N)( 2and m) N =onconservative800J Forces 780 m/s in a rifle barrel of length 72.0 cm. (a) Find 43. A 4.00-kg particle moves y (m) the kinetic energy of the bullet as it leaves the bar- M from the origin to posi- rel. (b) Use the work–kinetic energy theorem to find tion Ꭿ, having coordi- Ꭿ Q/C Ꭾ (5.00, 5.00) the net work that is done on the bullet. (c) Use your nates x 5 5.00 m and y 5 result to part (b) to find the magnitude of the average 5.00 m (Fig. P7.43). One net force that acted on the bullet while it was in the force on the particle is barrel. (d) Now model the bullet as a particle under 1Problemthe gravitationalfrom Serway & Jewett, force 9th ed, page 207. constant acceleration. Find the constant acceleration acting in the negative y x (m) of a bullet that starts from rest and gains a speed of direction. Using Equa- O Ꭽ 780 m/s over a distance of 72.0 cm. (e) Modeling the tion 7.3, calculate the bullet as a particle under a net force, find the net Figure P7.43 work done by the gravi- Problems 43 through 46. force that acted on it during its acceleration. (f) What tational force on the conclusion can you draw from comparing your results particle as it goes from O of parts (c) and (e)? to Ꭿ along (a) the purple path, (b) the red path, and 38. Review. A 7.80-g bullet moving at 575 m/s strikes the (c) the blue path. (d) Your results should all be identi- hand of a superhero, causing the hand to move 5.50 cm cal. Why? in the direction of the bullet’s velocity before stopping. 44. (a) Suppose a constant force acts on an object. The (a) Use work and energy considerations to find the force does not vary with time or with the position or average force that stops the bullet. (b) Assuming the the velocity of the object. Start with the general defini- force is constant, determine how much time elapses tion for work done by a force between the moment the bullet strikes the hand and f S the moment it stops moving. 5 ? S W 3 F d r 39. Review. A 5.75-kg object passes through the origin i at time t 5 0 such that its x component of velocity is and show that the force is conservative. (b) As a spe- S 5.00 m/s and its y component of velocity is 23.00 m/s. cial case, suppose the force F 5 3 i^ 1 4j^ N acts on a (a) What is the kinetic energy of the object at this time? particle that moves from O to Ꭿ in Figure P7.43. Cal- S (b) At a later time t 5 2.00 s, the particle is located at culate the work done by F on the1 particle2 as it moves x 5 8.50 m and y 5 5.00 m. What constant force acted along each one of the three paths shown in the figure (c) U = 0.

Problems 207

34. A 4.00-kg particle is subject to a net force that varies on the object during this time interval? (c) What is the W with position as shown in Figure P7.15. The particle speed of the particle at t 5 2.00 s? starts from rest at x 5 0. What is its speed at (a) x 5 5.00 m, (b) x 5 10.0 m, and (c) x 5 15.0 m? Section 7.6 Potential Energy of a System 35. A 2 100-kg pile driver is used to drive a steel I-beam into 40. A 1 000-kg roller coaster car is initially at the top of a M the ground. The pile driver falls 5.00 m before coming rise, at point Ꭽ. It then moves 135 ft, at an angle of 40.08 into contact with the top of the beam, and it drives the below the horizontal, to a lower point Ꭾ. (a) Choose beam 12.0 cm farther into the ground before coming the car at point Ꭾ to be the zero configuration for to rest. Using energy considerations, calculate the aver- gravitational potential energy of the roller coaster– age force the beam exerts on the pile driver while the Earth system. Find the potential energy of the system pile driver is brought to rest. when the car is at points Ꭽ and Ꭾ, and the change 36. Review. In an electron microscope, there is an electron in potential energy as the car moves between these AMT gun that contains two charged metallic plates 2.80 cm points. (b) Repeat part (a), setting the zero configura- Ꭽ apart. An electric force accelerates each electron in tion with the car at point . the beam from rest to 9.60% of the speed of light over 41. A 0.20-kg stone is held 1.3 m above the top edge of a this distance. (a) Determine the kinetic energy of the water well and then dropped into it. The well has a electron as it leaves the electron gun. Electrons carry depth of 5.0 m. Relative to the configuration with the this energy to a phosphorescent viewing screen where stone at the top edge of the well, what is the gravita- the microscope’s image is formed, making it glow. For tional potential energy of the stone–Earth system an electron passing between the plates in the electronGravitational (a) before Potential the stone is Energyreleased and (b) when it reaches gun, determine (b) the magnitude of the constant the bottom of the well? (c) What is the change in gravi- electric force acting on the electron, (c) the accelera- tational potential energy of the system from release to tion of the electron, and (d) the time interval the elec- reaching the bottom of the well? tron spends between the plates. 42. A 400-N child is in a swing that is attached to a pair 37. Review. You can think of the work–kinetic energy the- W of ropes 2.00 m long. Find the gravitational potential GP orem as a second theory of motion, parallel to New- energy of the child–Earth system relative to the child’s Q/C ton’s laws in describing how outside influences affect lowest position when (a) the ropes are horizontal, the motion of an object. In this problem, solve parts (b) the ropes make a 30.08 angle with the vertical, and (a), (b), and (c) separately from parts (d) and (e) so (c) the child is at the bottom of the circular arc. you can compare the predictions of the two theories. A 15.0-g bullet is accelerated from rest to a speed of (a)SectionU = (mg7.7 )Conservativey = (400 N)( 2and m) N =onconservative800J Forces 780 m/s in a rifle barrel of length 72.0 cm. (a) Find 43. A 4.00-kg particle moves y (m) the kinetic energy of the bullet as it leaves the bar- (b) U = (mg)y = (400 N)(2 m)(1 − cos 30◦) = 107J M from the origin to posi- rel. (b) Use the work–kinetic energy theorem to find tion Ꭿ, having coordi- Ꭿ Q/C Ꭾ (5.00, 5.00) the net work that is done on the bullet. (c) Use your nates x 5 5.00 m and y 5 result to part (b) to find the magnitude of the average 5.00 m (Fig. P7.43). One net force that acted on the bullet while it was in the force on the particle is barrel. (d) Now model the bullet as a particle under 1Problemthe gravitationalfrom Serway & Jewett, force 9th ed, page 207. constant acceleration. Find the constant acceleration acting in the negative y x (m) of a bullet that starts from rest and gains a speed of direction. Using Equa- O Ꭽ 780 m/s over a distance of 72.0 cm. (e) Modeling the tion 7.3, calculate the bullet as a particle under a net force, find the net Figure P7.43 work done by the gravi- Problems 43 through 46. force that acted on it during its acceleration. (f) What tational force on the conclusion can you draw from comparing your results particle as it goes from O of parts (c) and (e)? to Ꭿ along (a) the purple path, (b) the red path, and 38. Review. A 7.80-g bullet moving at 575 m/s strikes the (c) the blue path. (d) Your results should all be identi- hand of a superhero, causing the hand to move 5.50 cm cal. Why? in the direction of the bullet’s velocity before stopping. 44. (a) Suppose a constant force acts on an object. The (a) Use work and energy considerations to find the force does not vary with time or with the position or average force that stops the bullet. (b) Assuming the the velocity of the object. Start with the general defini- force is constant, determine how much time elapses tion for work done by a force between the moment the bullet strikes the hand and f S the moment it stops moving. 5 ? S W 3 F d r 39. Review. A 5.75-kg object passes through the origin i at time t 5 0 such that its x component of velocity is and show that the force is conservative. (b) As a spe- S 5.00 m/s and its y component of velocity is 23.00 m/s. cial case, suppose the force F 5 3 i^ 1 4j^ N acts on a (a) What is the kinetic energy of the object at this time? particle that moves from O to Ꭿ in Figure P7.43. Cal- S (b) At a later time t 5 2.00 s, the particle is located at culate the work done by F on the1 particle2 as it moves x 5 8.50 m and y 5 5.00 m. What constant force acted along each one of the three paths shown in the figure Problems 207

34. A 4.00-kg particle is subject to a net force that varies on the object during this time interval? (c) What is the W with position as shown in Figure P7.15. The particle speed of the particle at t 5 2.00 s? starts from rest at x 5 0. What is its speed at (a) x 5 5.00 m, (b) x 5 10.0 m, and (c) x 5 15.0 m? Section 7.6 Potential Energy of a System 35. A 2 100-kg pile driver is used to drive a steel I-beam into 40. A 1 000-kg roller coaster car is initially at the top of a M the ground. The pile driver falls 5.00 m before coming rise, at point Ꭽ. It then moves 135 ft, at an angle of 40.08 into contact with the top of the beam, and it drives the below the horizontal, to a lower point Ꭾ. (a) Choose beam 12.0 cm farther into the ground before coming the car at point Ꭾ to be the zero configuration for to rest. Using energy considerations, calculate the aver- gravitational potential energy of the roller coaster– age force the beam exerts on the pile driver while the Earth system. Find the potential energy of the system pile driver is brought to rest. when the car is at points Ꭽ and Ꭾ, and the change 36. Review. In an electron microscope, there is an electron in potential energy as the car moves between these AMT gun that contains two charged metallic plates 2.80 cm points. (b) Repeat part (a), setting the zero configura- Ꭽ apart. An electric force accelerates each electron in tion with the car at point . the beam from rest to 9.60% of the speed of light over 41. A 0.20-kg stone is held 1.3 m above the top edge of a this distance. (a) Determine the kinetic energy of the water well and then dropped into it. The well has a electron as it leaves the electron gun. Electrons carry depth of 5.0 m. Relative to the configuration with the this energy to a phosphorescent viewing screen where stone at the top edge of the well, what is the gravita- the microscope’s image is formed, making it glow. For tional potential energy of the stone–Earth system an electron passing between the plates in the electronGravitational (a) before Potential the stone is Energyreleased and (b) when it reaches gun, determine (b) the magnitude of the constant the bottom of the well? (c) What is the change in gravi- electric force acting on the electron, (c) the accelera- tational potential energy of the system from release to tion of the electron, and (d) the time interval the elec- reaching the bottom of the well? tron spends between the plates. 42. A 400-N child is in a swing that is attached to a pair 37. Review. You can think of the work–kinetic energy the- W of ropes 2.00 m long. Find the gravitational potential GP orem as a second theory of motion, parallel to New- energy of the child–Earth system relative to the child’s Q/C ton’s laws in describing how outside influences affect lowest position when (a) the ropes are horizontal, the motion of an object. In this problem, solve parts (b) the ropes make a 30.08 angle with the vertical, and (a), (b), and (c) separately from parts (d) and (e) so (c) the child is at the bottom of the circular arc. you can compare the predictions of the two theories. A 15.0-g bullet is accelerated from rest to a speed of (a)SectionU = (mg7.7 )Conservativey = (400 N)( 2and m) N =onconservative800J Forces 780 m/s in a rifle barrel of length 72.0 cm. (a) Find 43. A 4.00-kg particle moves y (m) the kinetic energy of the bullet as it leaves the bar- (b) U = (mg)y = (400 N)(2 m)(1 − cos 30◦) = 107J M from the origin to posi- rel. (b) Use the work–kinetic energy theorem to find (c) Ution= 0. Ꭿ, having coordi- Ꭿ Q/C Ꭾ (5.00, 5.00) the net work that is done on the bullet. (c) Use your nates x 5 5.00 m and y 5 result to part (b) to find the magnitude of the average 5.00 m (Fig. P7.43). One net force that acted on the bullet while it was in the force on the particle is barrel. (d) Now model the bullet as a particle under 1Problemthe gravitationalfrom Serway & Jewett, force 9th ed, page 207. constant acceleration. Find the constant acceleration acting in the negative y x (m) of a bullet that starts from rest and gains a speed of direction. Using Equa- O Ꭽ 780 m/s over a distance of 72.0 cm. (e) Modeling the tion 7.3, calculate the bullet as a particle under a net force, find the net Figure P7.43 work done by the gravi- Problems 43 through 46. force that acted on it during its acceleration. (f) What tational force on the conclusion can you draw from comparing your results particle as it goes from O of parts (c) and (e)? to Ꭿ along (a) the purple path, (b) the red path, and 38. Review. A 7.80-g bullet moving at 575 m/s strikes the (c) the blue path. (d) Your results should all be identi- hand of a superhero, causing the hand to move 5.50 cm cal. Why? in the direction of the bullet’s velocity before stopping. 44. (a) Suppose a constant force acts on an object. The (a) Use work and energy considerations to find the force does not vary with time or with the position or average force that stops the bullet. (b) Assuming the the velocity of the object. Start with the general defini- force is constant, determine how much time elapses tion for work done by a force between the moment the bullet strikes the hand and f S the moment it stops moving. 5 ? S W 3 F d r 39. Review. A 5.75-kg object passes through the origin i at time t 5 0 such that its x component of velocity is and show that the force is conservative. (b) As a spe- S 5.00 m/s and its y component of velocity is 23.00 m/s. cial case, suppose the force F 5 3 i^ 1 4j^ N acts on a (a) What is the kinetic energy of the object at this time? particle that moves from O to Ꭿ in Figure P7.43. Cal- S (b) At a later time t 5 2.00 s, the particle is located at culate the work done by F on the1 particle2 as it moves x 5 8.50 m and y 5 5.00 m. What constant force acted along each one of the three paths shown in the figure There is also spring potential energy!

Choosing U = 0 when the spring is at its natural length (relaxed):

1 U = kx2 2

(The spring must be part of our system.)

Spring Force: Another Conservative Force The spring force is also a conservative force.

If we stretch a spring, we can say that the spring stores the energy.

That energy is converted to kinetic energy when the end of the spring is released. Spring Force: Another Conservative Force The spring force is also a conservative force.

If we stretch a spring, we can say that the spring stores the energy.

That energy is converted to kinetic energy when the end of the spring is released.

There is also spring potential energy!

Choosing U = 0 when the spring is at its natural length (relaxed):

1 U = kx2 2

(The spring must be part of our system.) 200 Chapter 7 Energy of a System

U 1 2 s energy function for a block–spring system, given by Us 5 2kx . This function is U ϭ Ϫ1 kx 2 Potential Energy Diagramss 2 plotted versus in Figure 7.20a, where is the position of the block. The force E x x Fs Potential energy can be plotted as a function ofexerted position. by theeg. spring on the block is related to Us through Equation 7.29: 200 Chapter 7 Energy of a System dU potential energy of a spring: F 52 s 52kx s dx x Ϫ x max 0U x max As we saw in Quick Quiz 7.8, the x component of the force is equal to 1the2 nega- s energy function for a block–spring system, given by Us 5 2kx . This function is U ϭ Ϫ1 kx 2 tive of the slope of the U-versus-x curve. When the block is placed at rest at the sa 2 plotted versus x in Figure 7.20a, where x is the position of the block. The force F E equilibrium position of the spring (x 5 0), where Fs 5 0, it will remain there unless s someexerted external by forcethe spring Fext acts on on the it. Ifblock this external is related force to Ustretchess through the Equationspring from 7.29: equilibrium, x is positive and the slope dU/dx is positive; therefore, the force Fs exerted by the spring is negative and the block acceleratesdUs back toward x 5 0 when The restoring force exerted by the Fs 52 52kx spring always acts toward x ϭ 0, released. If the external force compresses the spring,dx x is negative and the slope is x the position of stable equilibrium. negative; therefore, F is positive and again the mass accelerates toward x 5 0 upon Ϫx max 0 x max As we saw in Quicks Quiz 7.8, the x component of the force is equal to the nega- release. S tive of the slope of the U-versus-x curve. When the block is placed at rest at the a F From this analysis, we conclude that the x 5 0 position for a block–spring sys- s equilibrium position of the spring (x 5 0), where F 5 0, it will remain there unless tem is one of stable equilibrium. That is, any movement aways from this position m resultssome in externala force directed force Fbackext acts toward on xit. 5 If 0. this In general, external configurations force stretches of a systhe- spring from temequilibrium, in stable equilibrium x is positive correspond and theto those slope for dU which/dx isU (positive;x) for the therefore,system is a the force Fs minimum.exerted by the spring is negative and the block accelerates back toward x 5 0 when The restoring forcex ϭ 0 exertedx max by the springb always acts toward x ϭ 0, released.If the block If in the Figure external 7.20 is forcemoved compresses to an initial positionthe spring, xmax andx is thennegative released and the slope is 1 2 the position of stable equilibrium. fromnegative; rest, its total therefore, energy initiallyF is positive is the potential and again energy the 2 masskx max storedaccelerates in the spring. toward x 5 0 upon Figure 7.20 (a) Potential energy s Fsp = −kx As the block starts to move, the system acquires kinetic energy and loses potential as a function of x for the friction- release. S energy. The block oscillates (moves back and forth) between the two points x 5 1 less block–spring system shown in From this analysis, we conclude that the x 5 0 position for a block–spring sys- Figure from Serway & Jewett, 9th ed. Fs 2x and x 5 1x , called the turning points. In fact, because no energy is trans- (b). For a given energy E of the sys- temmax is one ofmax stable equilibrium. That is, any movement away from this position tem, the block oscillates between formed to internal energy due to friction, the block oscillates between 2xmax and the turning points,m which have the 5 1xresultsmax forever. in a(We force will directeddiscuss these back oscillations toward furtherx 0. inIn Chapter general, 15.) configurations of a sys- 5 6 coordinates x xmax. temAnother in stablesimple mechanicalequilibrium system correspond with a configuration to those for of stablewhich equilibrium U(x) for theis system is a a ball rolling about in the bottom of a bowl. Anytime the ball is displaced from its x ϭ 0 x minimum. max lowest position, it tends to return to that position when released. bPitfall Prevention 7.10 If the block in Figure 7.20 is moved to an initial position xmax and then released Now consider a particle moving along the x axis under the influence1 2of a conser- Energy Diagrams A common from rest, its total energy initially is the potential energy kx stored in the spring. vative force F , where the U-versus-x curve is as shown in Figure 7.21.2 Oncemax again, Figuremistake 7.20 is to think(a) Potential that potential energy As the blockx starts to move, the system acquires kinetic energy and loses potential as energya function on the of graph x for inthe an friction- energy Fx 5 0 at x 5 0, and so the particle is in equilibrium at this point. This position, lessdiagram block–spring represents system the height shown of in however,energy. is The one ofblock unstable oscillates equilibrium (moves for back the followingand forth) reason. between Suppose the thetwo points x 5 some object. For example, that (b). For a given energy E of the sys- particle2xmax is anddisplaced x 5 1tox maxthe, rightcalled (x .the 0). turning Because points. the slope In fact,is negative because for xno . energy0, is trans- is not the case in Figure 7.20, tem, the block oscillates between F 5 2dU/dx is positive and the particle accelerates away from x 5 0. If instead the where the block is only moving x formed to internal energy due to friction, the block oscillates between 2xmax and the turning points, which have the particle is at x 5 0 and is displaced to the left (x , 0), the force is negative because horizontally. 1xmax forever. (We will discuss these oscillations further in Chapter 15.) coordinates x 5 6x . , max the slopeAnother is positive simple for x mechanical 0 and the particle system again with acceleratesa configuration away from of thestable equi equilibrium- is librium position. The position 5 0 in this situation is one of unstable equilibrium a ball rolling about in thex bottom of a bowl. Anytime the ball is displaced from its because for any displacement from this point, the force pushes the particle farther Pitfall Prevention 7.10 awaylowest from position, equilibrium it tendsand toward to return a position to that of positionlower potential when energy.released. A pencil Now consider a particle moving along the x axis under the influence of a conser- Energy Diagrams A common balanced on its point is in a position of unstable equilibrium. If the pencil is dis- mistake is to think that potential placedvative slightly force from Fx, itswhere absolutely the U vertical-versus- positionx curve and is isas then shown released, in Figure it will surely 7.21. Once again, energy on the graph in an energy fallF over.x 5 0In at general, x 5 0, configurations and so the particleof a system is inin unstable equilibrium equilibrium at this correspond point. This position, diagram represents the height of to those however, for which is one U(x ) of for unstable the system equilibrium is a maximum. for the following reason. Suppose the some object. For example, that particleFinally, a is configuration displaced to called the rightneutral ( x equilibrium . 0). Because arises the when slope U is is constant negative for x . 0, over some region. Small displacements of an object from a position in this region is not the case in Figure 7.20, F 5 2dU/dx is positive and the particle accelerates away from x 5 0. If instead the where the block is only moving producex neither restoring nor disrupting forces. A ball lying on a flat, horizontal horizontally. surfaceparticle is an is example at x 5 of0 andan object is displaced in neutral to equilibrium. the left (x , 0), the force is negative because the slope is positive for x , 0 and the particle again accelerates away from the equilibrium position. The position x 5 0 in this situationU is one of unstable equilibrium A plot of U versus becauseFigure for any 7.21 displacement from Positivethis point, slope theNegative force slope pushes the particle farther away fromx for equilibriuma particle that has and a position toward a xposition Ͻ 0 of lowerx Ͼ 0 potential energy. A pencil of unstable equilibrium located balancedat xon 5 0. its For point any finite is displace-in a position of unstable equilibrium. If the pencil is displaced slightlyment of the from particle, its theabsolutely force on vertical position and is then released, it will surely the particle is directed away from x fall over.x 5In 0. general, configurations of a system0 in unstable equilibrium correspond to those for which U(x) for the system is a maximum. Finally, a configuration called neutral equilibrium arises when U is constant over some region. Small displacements of an object from a position in this region produce neither restoring nor disrupting forces. A ball lying on a flat, horizontal surface is an example of an object in neutral equilibrium.

U A plot of U versus Figure 7.21 Positive slope Negative slope x for a particle that has a position x Ͻ 0 x Ͼ 0 of unstable equilibrium located at x 5 0. For any finite displacement of the particle, the force on the particle is directed away from x x 5 0. 0 Potential Energy Diagrams Recall that the work done by a force is the area under the force-displacement curve.

The work done by the spring relates to the change in the spring potential: Wsp = −∆Usp

So the area is also equal to −∆Usp. CHAPTER 2 | KINEMATICS 65 Potential Energy Diagrams Comparison: dx Slope of x-t curve = v.(v = dt ) Area under v-t graph = ∆x. CHAPTER 2 | KINEMATICS 65

Slope of U-x curve = −F . Area under F -x graph = −∆U. dU (F =200 − dx ) Chapter 7 Energy of a System

1 2 Us energy function for a block–spring system, given by U 5 kx . This function is 1 2 s 2 Us ϭ Ϫ2 kx Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly plotted versus x in Figure 7.20a, where x is the position of the block. The force Fs with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that E the graph shows some horizontal motion—the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down. exerted by the spring on the block is related to Us through Equation 7.29: Discussion The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since and are both dUs positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both and Fs 52 52kx are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at dx 1.5 s), its velocity is zero, but its acceleration is still . Its acceleration is for the whole trip—while it is moving up and x while it is moving down. Note that the values for are the positions (or displacements) of the rock, not the total distances traveled. Finally, note Ϫx max 0 x max As we saw in Quick Quiz 7.8, the x component of the force is equal to the nega- that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall Figurewhile arcing2.40 Vertical up as position, well as vertical down, velocity as we, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Noticetive that velocityof changesthe linearly slope of the U-versus-x curve. When the block is placed at rest at the with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that will discuss in more detail later. the graph shows some horizontal motion—the shapea of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down. equilibrium position of the spring (x 5 0), where Fs 5 0, it will remain there unless Making Connections: Take-Home Experiment—Reaction Time Discussion some external force Fext acts on it. If this external force stretches the spring from A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between yourThe thumb interpretation and index offinger these, separated results is byimportant. At 1.00 s the rock is above its starting point and heading upward, since and are both about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedlypositive. At 2.00, and s, try the to rock catch is itstill above its starting point, but the negative velocity means it is moving downward.equilibrium, At 3.00 s, both and x is positive and the slope dU/dx is positive; therefore, the force Fs between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brakeare was negative, twice thismeaning reaction the time? rock is below its starting point and continuing to move downward. Notice that when theexerted rock is at its highest by point the (at spring is negative and the block accelerates back toward x 5 0 when 1.5 s), its velocity is zero, but its accelerationThe is still restoring. Itsforce acceleration exerted is by thefor the whole trip—while it is moving up and while it is moving down. Note that the valuesspring for are always the positions acts (or displacements) toward xof theϭ rock,0, not the total released.distances traveled. Finally If , thenote external force compresses the spring, x is negative and the slope is that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts trainingthe in the position famous Vomit of Comet, stable for example, equilibrium. experience free-fall while arcingnegative; up as well as down, therefore, as we F is positive and again the mass accelerates toward x 5 0 upon will discuss in more detail later. s release. S Making Connections: Take-Home Experiment—Reaction Time Fs From this analysis, we conclude that the x 5 0 position for a block–spring sys- A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedlytem, and is try toone catch it of stable equilibrium. That is, any movement away from this position between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot mto go from the gas pedal to the brake wasresults twice this reaction in time? a force directed back toward x 5 0. In general, configurations of a system in stable equilibrium correspond to those for which U(x) for the system is a minimum. x ϭ 0 x max b If the block in Figure 7.20 is moved to an initial position xmax and then released 1 2 from rest, its total energy initially is the potential energy 2kx max stored in the spring. Figure 7.20 (a) Potential energy As the block starts to move, the system acquires kinetic energy and loses potential as a function of x for the friction- less block–spring system shown in energy. The block oscillates (moves back and forth) between the two points x 5 (b). For a given energy E of the sys- 2xmax and x 5 1xmax, called the turning points. In fact, because no energy is trans- tem, the block oscillates between formed to internal energy due to friction, the block oscillates between 2xmax and the turning points, which have the 1xmax forever. (We will discuss these oscillations further in Chapter 15.) 5 6 coordinates x xmax. Another simple mechanical system with a configuration of stable equilibrium is a ball rolling about in the bottom of a bowl. Anytime the ball is displaced from its Pitfall Prevention 7.10 lowest position, it tends to return to that position when released. Now consider a particle moving along the x axis under the influence of a conser- Energy Diagrams A common mistake is to think that potential vative force Fx, where the U-versus-x curve is as shown in Figure 7.21. Once again, energy on the graph in an energy Fx 5 0 at x 5 0, and so the particle is in equilibrium at this point. This position, diagram represents the height of however, is one of unstable equilibrium for the following reason. Suppose the some object. For example, that particle is displaced to the right (x . 0). Because the slope is negative for x . 0, is not the case in Figure 7.20, F 5 2dU/dx is positive and the particle accelerates away from x 5 0. If instead the where the block is only moving x horizontally. particle is at x 5 0 and is displaced to the left (x , 0), the force is negative because the slope is positive for x , 0 and the particle again accelerates away from the equilibrium position. The position x 5 0 in this situation is one of unstable equilibrium because for any displacement from this point, the force pushes the particle farther away from equilibrium and toward a position of lower potential energy. A pencil balanced on its point is in a position of unstable equilibrium. If the pencil is displaced slightly from its absolutely vertical position and is then released, it will surely fall over. In general, configurations of a system in unstable equilibrium correspond to those for which U(x) for the system is a maximum. Finally, a configuration called neutral equilibrium arises when U is constant over some region. Small displacements of an object from a position in this region produce neither restoring nor disrupting forces. A ball lying on a flat, horizontal surface is an example of an object in neutral equilibrium.

U A plot of U versus Figure 7.21 Positive slope Negative slope x for a particle that has a position x Ͻ 0 x Ͼ 0 of unstable equilibrium located at x 5 0. For any finite displacement of the particle, the force on the particle is directed away from x x 5 0. 0 Potential Energy, Conservative Force, & Equilibrium The value of200 a conservative force FCathapter a particular 7 Energy point can of be a System found as the slope of the potential energy curve:

U 1 2 s energy function for a block–spring system, given by Us 5 2kx . This function is U ϭ Ϫ1 kx 2 s 2 plotted versus in Figure 7.20a, where is the position of the block. The force E x x Fs exerted by the spring on the block is related to Us through Equation 7.29: dU F 52 s 52kx s dx x Ϫx max 0 x max As we saw in Quick Quiz 7.8, the x component of the force is equal to the negative of the slope of the U-versus-x curve. When the block is placed at rest at the a equilibrium position of the spring (x 5 0), where F 5 0, it will remain there unless Fsp = −(slope of U(x)) = −kx s If F is the only force acting on the particle, stationary pointssome external force Fext acts on it. If this external force stretches the spring from (slope = 0) are force equilibrium points. equilibrium, x is positive and the slope dU/dx is positive; therefore, the force Fs exerted by the spring is negative and the block accelerates back toward x 5 0 when The restoring force exerted by the spring always acts toward x ϭ 0, released. If the external force compresses the spring, x is negative and the slope is the position of stable equilibrium. negative; therefore, Fs is positive and again the mass accelerates toward x 5 0 upon release. S Fs From this analysis, we conclude that the x 5 0 position for a block–spring system is one of stable equilibrium. That is, any movement away from this position m results in a force directed back toward x 5 0. In general, configurations of a system in stable equilibrium correspond to those for which U(x) for the system is a minimum. x ϭ 0 x max b If the block in Figure 7.20 is moved to an initial position xmax and then released 1 2 from rest, its total energy initially is the potential energy 2kx max stored in the spring. Figure 7.20 (a) Potential energy As the block starts to move, the system acquires kinetic energy and loses potential as a function of x for the friction- less block–spring system shown in energy. The block oscillates (moves back and forth) between the two points x 5 (b). For a given energy E of the sys- 2xmax and x 5 1xmax, called the turning points. In fact, because no energy is trans- tem, the block oscillates between formed to internal energy due to friction, the block oscillates between 2xmax and the turning points, which have the 1xmax forever. (We will discuss these oscillations further in Chapter 15.) 5 6 coordinates x xmax. Another simple mechanical system with a configuration of stable equilibrium is a ball rolling about in the bottom of a bowl. Anytime the ball is displaced from its Pitfall Prevention 7.10 lowest position, it tends to return to that position when released. Now consider a particle moving along the x axis under the influence of a conser- Energy Diagrams A common mistake is to think that potential vative force Fx, where the U-versus-x curve is as shown in Figure 7.21. Once again, energy on the graph in an energy Fx 5 0 at x 5 0, and so the particle is in equilibrium at this point. This position, diagram represents the height of however, is one of unstable equilibrium for the following reason. Suppose the some object. For example, that particle is displaced to the right (x . 0). Because the slope is negative for x . 0, is not the case in Figure 7.20, F 5 2dU/dx is positive and the particle accelerates away from x 5 0. If instead the where the block is only moving x horizontally. particle is at x 5 0 and is displaced to the left (x , 0), the force is negative because the slope is positive for x , 0 and the particle again accelerates away from the equilibrium position. The position x 5 0 in this situation is one of unstable equilibrium because for any displacement from this point, the force pushes the particle farther away from equilibrium and toward a position of lower potential energy. A pencil balanced on its point is in a position of unstable equilibrium. If the pencil is displaced slightly from its absolutely vertical position and is then released, it will surely fall over. In general, configurations of a system in unstable equilibrium correspond to those for which U(x) for the system is a maximum. Finally, a configuration called neutral equilibrium arises when U is constant over some region. Small displacements of an object from a position in this region produce neither restoring nor disrupting forces. A ball lying on a flat, horizontal surface is an example of an object in neutral equilibrium.

U A plot of U versus Figure 7.21 Positive slope Negative slope x for a particle that has a position x Ͻ 0 x Ͼ 0 of unstable equilibrium located at x 5 0. For any finite displacement of the particle, the force on the particle is directed away from x x 5 0. 0 Energy Diagrams and Equilibrium 200 Chapter 7 Energy of a System System is in equilibrium when Fnet = Fsp = 0.

U 1 2 s energy function for a block–spring system, given by Us 5 2kx . This function is U ϭ Ϫ1 kx 2 s 2 plotted versus in Figure 7.20a, where is the position of the block. The force E x x Fs exerted by the spring on the block is related to Us through Equation 7.29: dU F 52 s 52kx s dx x Ϫx max 0 x max As we saw in Quick Quiz 7.8, the x component of the force is equal to the negative of the slope of the U-versus-x curve. When the block is placed at rest at the a equilibrium position of the spring (x 5 0), where F 5 0, it will remain there unless In this case, the force is always back toward the x = 0 point, so s some external force F acts on it. If this external force stretches the spring from this is a stable equilibrium. ext equilibrium, x is positive and the slope dU/dx is positive; therefore, the force Fs Examples: exerted by the spring is negative and the block accelerates back toward x 5 0 when The restoring force exerted by the • spring force spring always acts toward x ϭ 0, released. If the external force compresses the spring, x is negative and the slope is • ball inside athe bowl position of stable equilibrium. negative; therefore, Fs is positive and again the mass accelerates toward x 5 0 upon release. S Fs From this analysis, we conclude that the x 5 0 position for a block–spring system is one of stable equilibrium. That is, any movement away from this position m results in a force directed back toward x 5 0. In general, configurations of a system in stable equilibrium correspond to those for which U(x) for the system is a minimum. x ϭ 0 x max b If the block in Figure 7.20 is moved to an initial position xmax and then released 1 2 from rest, its total energy initially is the potential energy 2kx max stored in the spring. Figure 7.20 (a) Potential energy As the block starts to move, the system acquires kinetic energy and loses potential as a function of x for the friction- less block–spring system shown in energy. The block oscillates (moves back and forth) between the two points x 5 (b). For a given energy E of the sys- 2xmax and x 5 1xmax, called the turning points. In fact, because no energy is trans- tem, the block oscillates between formed to internal energy due to friction, the block oscillates between 2xmax and the turning points, which have the 1xmax forever. (We will discuss these oscillations further in Chapter 15.) 5 6 coordinates x xmax. Another simple mechanical system with a configuration of stable equilibrium is a ball rolling about in the bottom of a bowl. Anytime the ball is displaced from its Pitfall Prevention 7.10 lowest position, it tends to return to that position when released. Now consider a particle moving along the x axis under the influence of a conser- Energy Diagrams A common mistake is to think that potential vative force Fx, where the U-versus-x curve is as shown in Figure 7.21. Once again, energy on the graph in an energy Fx 5 0 at x 5 0, and so the particle is in equilibrium at this point. This position, diagram represents the height of however, is one of unstable equilibrium for the following reason. Suppose the some object. For example, that particle is displaced to the right (x . 0). Because the slope is negative for x . 0, is not the case in Figure 7.20, F 5 2dU/dx is positive and the particle accelerates away from x 5 0. If instead the where the block is only moving x horizontally. particle is at x 5 0 and is displaced to the left (x , 0), the force is negative because the slope is positive for x , 0 and the particle again accelerates away from the equilibrium position. The position x 5 0 in this situation is one of unstable equilibrium because for any displacement from this point, the force pushes the particle farther away from equilibrium and toward a position of lower potential energy. A pencil balanced on its point is in a position of unstable equilibrium. If the pencil is displaced slightly from its absolutely vertical position and is then released, it will surely fall over. In general, configurations of a system in unstable equilibrium correspond to those for which U(x) for the system is a maximum. Finally, a configuration called neutral equilibrium arises when U is constant over some region. Small displacements of an object from a position in this region produce neither restoring nor disrupting forces. A ball lying on a flat, horizontal surface is an example of an object in neutral equilibrium.

U 1 2 s energy function for a block–spring system, given by Us 5 2kx . This function is U ϭ Ϫ1 kx 2 s 2 plotted versus in Figure 7.20a, where is the position of the block. The force E x x Fs exerted by the spring on the block is related to Us through Equation 7.29: dU F 52 s 52kx s dx x Ϫx max 0 x max As we saw in Quick Quiz 7.8, the x component of the force is equal to the negative of the slope of the U-versus-x curve. When the block is placed at rest at the a equilibrium position of the spring (x 5 0), where Fs 5 0, it will remain there unless some external force Fext acts on it. If this external force stretches the spring from equilibrium, x is positive and the slope dU/dx is positive; therefore, the force Fs exerted by the spring is negative and the block accelerates back toward x 5 0 when The restoring force exerted by the spring always acts toward x ϭ 0, released. If the external force compresses the spring, x is negative and the slope is the position of stable equilibrium. negative; therefore, Fs is positive and again the mass accelerates toward x 5 0 upon release. S Fs From this analysis, we conclude that the x 5 0 position for a block–spring system is one of stable equilibrium. That is, any movement away from this position m results in a force directed back toward x 5 0. In general, configurations of a system in stable equilibrium correspond to those for which U(x) for the system is a minimum. x ϭ 0 x max b If the block in Figure 7.20 is moved to an initial position xmax and then released 1 2 from rest, its total energy initially is the potential energy 2kx max stored in the spring. Figure 7.20 (a) Potential energy As the block starts to move, the system acquires kinetic energy and loses potential as a function of x for the friction- less block–spring system shown in energy. The block oscillates (moves back and forth) between the two points x 5 (b). For a given energy E of the sys- 2xmax and x 5 1xmax, called the turning points. In fact, because no energy is trans- tem, the block oscillates between formed to internal energy due to friction, the block oscillates between 2xmax and the turning points, which have the 1xmax forever. (We will discuss these oscillations further in Chapter 15.) 5 6 coordinates x xmax. Another simple mechanical system with a configuration of stable equilibrium is a ball rolling about in the bottom of a bowl. Anytime the ball is displaced from its Pitfall Prevention 7.10 lowest position, it tends to return to that position when released. Now consider a particle moving along the x axis under the influence of a conser- Energy Diagrams A common mistake is to think that potential vative force Fx, where the U-versus-x curve is as shown in Figure 7.21. Once again, energy on the graph in an energy Fx 5 0 at x 5 0, and so the particle is in equilibrium at this point. This position, diagram represents the height of however, is one of unstable equilibrium for the following reason. Suppose the some object. For example, that particle is displaced to the right (x . 0). Because the slope is negative for x . 0, is not the case in Figure 7.20, F 5 2dU/dx is positive and the particle accelerates away from x 5 0. If instead the where the block is only moving x horizontally. particle is at x 5 0 and is displaced to the left (x , 0), the force is negative because the slope is positive for x , 0 and the particle again accelerates away from the equilibrium position. The position x 5 0 in this situation is one of unstable equilibrium because for any displacement from this point, the force pushes the particle farther away from equilibrium and toward a position of lower potential energy. A pencil balanced on its point is in a position of unstable equilibrium. If the pencil is displaced slightly from its absolutely vertical position and is then released, it will surely fall over. In general, configurations of a system in unstable equilibrium correspond to those for which U(x) for the system is a maximum. Finally, a configuration called neutral equilibrium arises when U is constant over some region.Energy Small displacements Diagrams of an and object Equilibrium from a position in this region produce neither restoring nor disrupting forces. A ball lying on a flat, horizontal surface is an example of an object in neutral equilibrium.

In this case, the force is always away from the x = 0 point, so this is a unstable equilibrium. Examples: • ball on upside-down a bowl Neutral Equilibrium

A system can also be in neutral equilibrium.

In this case, no forces act, even when the system is displaced left or right.

Example: • ball on a ﬂat surface Summary

• potential energy • conservative and nonconservative forces

Quiz given out tomorrow. Homework • Ch 8 Probs: 1, 3, 5, 7