### 4th Edition Errata (earlier editions below)

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Ch. 1 p. 19: In the paragraph labeled “Vector Velocity” the phrase “over a 5 month period” should be “over a 5 year period.”

Ch. 1 Fig. 1.49 (p. 29): In the WileyPLUS/VitalSource version (but in no other version), the exponents in the first column should all be positive, not negative.

Ch. 1 answer to checkpoint 6 (p. 44): Part (a) is .

Ch. 1 answer to checkpoint 9 (p. 17): The x component in part (a) should be -0.940.

Ch. 1 answer to checkpoint 15 (p. 44): Part (a) is kg m/s.

Ch. 1 prob. P53 (p. 41 and A-1): Answer is kg m/s.

Ch. 1 prob. P61 (p. 42 and A-1): Answer is kg m/s.

Ch. 1 prob. P63 (p. 42 and A-1): Answer is kg m/s. Also, the problem statement states that you don’t need to calculate the speed, which is true for calculating gamma but not for calculating the momentum.

Ch. 1 prob. P65 (p. 42 and A-1): Answer is 35.4.

Ch. 2 prob. P27 (p. 84 and A-1): (a) 37.5 mi/h; (b) 45 mi/h; (c) A graph of *v*_{x} vs. *t* is not a straight line.

Ch. 2 prob.P33 (p. 85 and A-1): The pot fell from rest at 4.5 m above the top of the window.

Ch. 2 p. 53: In the example solution. in two places the units of should be :

Ch. 2 pp. 58-59: In Time Step 1, m. In Time Step 2 the calculation of the future position should read as follows:

Ch. 2 prob. P21 (p. 83 and A-1): Answers are

Ch. 2 prob. P41 (p. 86 and A-1): Answer is 13 cm.

Ch. 2 P. 58 and P. 59: In Time Step 1 and P. 59 the first calculation of the force now should be N, not N/kg. The first calculation of r future (P. 58) should be <4, 4.43, 0> m. On P.59, just before Time Step 2, the new r should be <4, 4.43, 0>. Then in Time Step 2 (P. 59) the new r should be <8, 5.30, 0> m.

Ch 2 P. 60, first paragraph: “One case in which an iterative solution is possible is the case of constant net force.” should be “One case in which an analytical solution is possible is the case of constant net force.”

Ch. 2 P. 61, end of Figure 2.26: the “+” should be replaced with “and”.

Ch.2 P.62, near bottom: 0.0.146 m should be 0.146 m.

Ch. 3 prob. P13d (p. 124 and A-1): Answer is N.

Ch. 3 prob. P27b (p. 124 and A-1): Answer to part(b) is .

Ch. 3 prob. P37e (p.125 and A-1): Answer to part (e) is N.

Ch. 3 answer to checkpoint 8 (p. 129): Units are m/s.

Ch. 4, section 4.14, subsection “A Constant-Density Model of the Atmosphere” (p. 158), first sentence: ()

Ch. 4 answer to checkpoint 7 (p. 172): Part (a) is N.

Ch. 4 prob. P33a (p. A-1): The answer to part (a) is .

Ch. 4 prob. P45 (p.167): The final position is m.

Ch. 5 prob. P14 (p. 208): The coefficient of friction between the floor and the bottom block should be 0.5.

Ch. 5 (p. 184): In the second equation in the Solution the derivative should be .

Ch. 5 prob. P7 (p. 207): (a) You should show 3 arrows representing the forces acting on the box; one pointing down for the gravitational force, one pointing to the left for the force exerted by Rope 1, and one pointing up and to the right for the force exerted by Rope 2. (b) Zero.

Ch. 5 prob. P9 (p. 207): (a) < 195.34, 163.91, 0 > N;

(b) < -195.34, 326.09, 0 > N

Ch. 15 prob. P15 (p. 208): (a) 2.95

(b) 50.2

(c) 14.8 N

Ch. 5 prob. P19a (p. 208): The answer to part (a) is N

Ch. 5 prob. P27 (p. 209):

(a) <0, 0, 0>

(b) 0.975 radially inward

(c) 0.975 N, radially inward

(d) The saddle of the wooden horse

Ch. 5 prob. P33e (p. 209, p. A-2 ): The answer to part (e) is .

Ch. 5 prob. P37b (p. 210 and A-2): Answer to part (b) is 871 kg m/s/s.

Ch. 5 prob. P39 (p. 210 and A-2): Answer is kg.

Ch. 5 prob. P51 (p.212 and A-2): Speed is 1.64 m/s, period is 1.99 s.

Ch. 5 checkpoint 4 answers (p. 214): For points A and D, the answers to part (b) are nonzero, and the answers to part (c) are zero.

Ch. 6 Figure 6.42 (p. 252): The labels “1” and “2” should be interchanged.

Ch. 6 checkpoint 4 answers (p. 283): (a) negative, decreases, (b) positive, increases, (c) negative, decreases, (d) positive, increases, (e) zero, stays the same

Ch. 6 prob. P13 (p. 225 and A-2): (d) is 3.

Ch. 6 prob. P29 (p. 277 and A-2): Answer is 66 J.

Ch. 6 prob. P54 (p. 279 and A-2): (a) 9.8 J, 6.3 m/s; (b) 19.6 J, 6.3 m/s

Ch. 6 problems P52, P58, P59, P60, and P61 are in section 6.8 of the problems but belong in section 6.10 because there are references to escape speed. Also, P66 refers to making a graph and belongs in section 6.10.

Ch. 6 prob. P64 (p. 280): If you use the approximate value G = 6.7e-11 the orbit is indeed a bound state (total energy is negative), but if you use a more accurate value for G the total energy is positive, and the orbit is not bound (and not an ellipse).

Ch. 6 prob. P73 (p. 281): The coordinates of *B* should be , not .

Ch. 7 p. 301, next to last paragraph: The last part of the sentence should say, “the energy of an *open* system need not be constant”. (The energy of a *closed* system will necessarily stay constant.)

Ch. 7 p. 313: To correctly detect when the oscillation of a spring-mass system subject to sliding friction should stop, it isn’t adequate to watch for the velocity to become very small, because that also occurs at the end points of the oscillation. The correct criterion for stopping is that the velocity is very small AND the spring force is less than the sliding friction force.

Ch. 7 prob. P25 (p. 318): (a) *K* = 1.1 eV, *U* = -1.3 eV, *K*+*U* = -0.2 eV.

(b) 0.2 eV. (c) .

Ch. 7 prob.P29 (p. 318): *F* = 172 N; internal energy change = -516 J.

Ch. 7 summary (p. 316): The first sentence at the top right of the page should say, “Sliding friction force can be modeled in terms of the ball-and-spring model of solids.” (A more detailed treatment of friction is found in Ch. 9.)

Ch. 7 prob. P30 is in section 7.5 of the problems but belongs in section 7.4, since it does not involve *Q*.

Ch. 8 prob. P9 (p. 344): Answer is 0.9 eV.

Ch. 9 prob. P13 (p. 377): This problem belongs in section 9.2 because it involves moment of inertia. The answer is 24.6 J.

Ch. 9 prob. P43, Figure 9.63: The answer to part (b) is

The figure may be hard to understand. Here is an improved figure:

Ch. 10 p. 411 (Q13 Background): The first fission reactor was constructed in 1942, not 1941.

Ch. 10 p. 411-412: Problems P18 and P20 involve 1D collisions and so belong with problems from earlier sections of Ch. 10.

Ch. 10 p. 414: In the program skeleton the electric charges for the gold nucleus (79e) and the alpha particle (2e) are reversed. The initial value of the alpha particle momentum was calculated from the kinetic energy of the 10 MeV alpha particles used by Rutherford.

Ch. 11 p. 421, line 7: “The vector dot product” should be “The vector cross product”.

Ch. 11 prob. P21 (p. 412): The answer to part (b) is < 2.82, 0, 17.6 > m/s.

Ch. 11 prob. P35 (p. 462): (a) +0.308 ; (b) +0.308 ; (c) +0.0717 rad/s

Ch. 11 prob. P59 (p. 466): -2.33 rad/s

Ch. 11 prob. P60 (p. 466): Delete the symbols **(a)** and **(b)**.

Ch. 12 p. 489, 2nd paragraph: The specific heat of iron is 0.45 J/K/g.

Ch. 12 prob. P25 (p. 508): (c) J/K

Ch. 13 p. 521, example entitled “Source Location”: The absolute value of must be used:

Ch. 13 P33 (p. 541): The charge is negative, so the field is <225,0,0> N/C.

Ch. 13 P37 (p. 542): The start should read “At location <0, 0, 0> in a room…”

Ch.13 P45c (p. 542): The answer to part (c) is ,

Ch 13 P61 (p. 544): (a) left end.

Ch. 13 P62 (p. 544): The electric field units should be N/C.

Ch. 13 answer to checkpoint 8 (p. 545): The second answer is missing, C, which is 2e.

Ch. 14 EXP6 part g (p. 575): The field is

Ch. 14 P47 (p. 583): The correct answer is c.

Ch. 14 P57 (p. 584): The answer to part (b) is N/C.

Ch. 15 p. 604: It is not clearly stated that here the analysis is confined to the field between the plates; the analysis of the field outside the plates is carried out on p. 605. Also, in the calculation on p. 605 of , the sign of should be reversed.

Ch. 15 p. 605, Step 4: In both cases the denominator should be inverted. The fractions should be written , which yields

Ch. 15 P39 (p. 620): The answer to part (a) in the back of the book is backwards; the left side of the metal foil is charged positive, and the right side is charged negative.

Ch 15 P51 (p. 621): At r = 3 cm the field is zero.

Ch. 15 P62 (p. 624): in part b the units of polarizability are C·m/(N/C).

Ch. 15 P65 (p. 624), second line: Missing charge units (C).

Ch. 16, top of p. 641: The expression for is missing an overall minus sign.

Ch. 16, top of p. 653, 4th and 6th lines: pos=(…) should be pos=vector(…).

Ch 16 P49 (p. 666): (a) Change to *q; *(c)

Ch. 16 Summary (p. 661): Field is *negative* gradient of potential; the correct field components should have minus signs:

Ch. 17 p. 689: In the VPython code fragment, just after the statement **while t** **tmax:** insert the statement **rate(30)** .

Ch. 17 p. 690: In the VPython code fragment, delete the statement **rate(1000)**, which is not needed, since no display is made in this loop. Also, due to changes in VPython, the calculations of dl and loc should be written like this:

dl = wire.point(i+1).pos – wire.point(i).pos

loc = wire.point(i).pos + dl/2

Ch. 17 prob. P23b (p. 709): Answer to part b for magnetic field is 0.166 T.

Ch. 17 prob. P25 (p. 709): (a) electrons/s. (b) To the left. (c) m/s

Ch. 17 prob. P28 (p. 709): “long bulb” should be “light bulb”.

Ch. 17 prob. P37 (p. 710): (b)

(c) +z.

Ch. 18 by Figure 18.48 (p. 736): In the 2nd full paragraph, the electrons pile up at location 4. In the 3rd full paragraph, + charge accumulates at location 3.

Ch. 18 prob. P33(p. 758): The answer to the second question is electrons/s.

Ch. 19 (p. 766): In the second paragraph of the subsection “Discharging a Capacitor” the second sentence should read “We’ll think about it step by step, iteratively, choosing a time step that is **short** compared to the time required to discharge the capacitor (several seconds), but **long** compared to the time it takes surface charge to arrange itself on the wire and battery (nanoseconds).”

Ch. 19 prob. P63 (p. 800): (b) 10.4 degrees; (d) and have the same length; (g) J.

Ch. 19 prob. P77d (p. 802): The answer is 3.78 V.

Ch. 20 p. 819, Fig. 20.25: The caption should say “out of the page.”

Ch. 20 p. 820, Fig. 20.26: The caption should say “with an upward Coulomb electric field.”

Ch. 20 p. 822, Fig. 20.31: The magnetic field should point out of the page, as in the previous diagrams.

Ch. 20 prob. P33 (p. 856): The answer to part (a) is that the direction is -z (into the page) and the deflection angle is 4.49 degrees; in part (b) the force is < 0, -8e-18, 0 > N.

Ch. 20 p. 860, Fig. 20.118: The upper voltmeter reading should be -0.25 V.

Ch. 20 prob. P71 (p. 863): The units are Nm.

Ch. 20 prob. P72 (p. 863): The magnet is a cylinder with diameter 1 cm and length 3 cm.

Ch. 21 first paragraph on p. 885: The first sentence should read like this: If and are short enough that the angle between the two radial lines is small, then the length gets larger proportional to , but decreases proportional to , so the magnitude of the product is the same at both locations.

Ch. 21 prob. P23 (p. 900 and A-5): Answer for direction is “into the page.”

Ch. 21 prob. P30 (p. 901): In part (d) the location inside the cube should be <0.032, 0, 0> m.

Ch. 22 prob. P28 (p. 936): In two places in the first paragraph, “H” should be “Hz” (hertz; cycles per second).

Ch. 22 problems P40, P41, P42 (p. 938): These problems are associated with Section 22.7, not 22.6.

Ch. 23 p. 950: In the Example, the electric field should be V/m.

Ch. 23 p. 957, Fig. 23.37: The colors of the electric and magnetic field vectors were reversed; here is the correct figure:

### Errata for previous editions

Second edition volume 1 (2007)