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Unformatted text preview: Chapter 7 Potential Energy and Energy
Conservation 7.1: From Eq. (7.2),
mgy = (800 kg)(9.80 m/52)(440 m) : 3.45 x 10‘5 J = 3.45 MJ. 7.2: a) For constant speed, the net force is zero, so the required force is the sack’s weight,
(5.00 kg)(9.80 m/s2) = 49 N. b) The lifting force acts in the same direction as the sack’s
motion, so the work is equal to the weight times the distance, (49.00 N)(15.0 m) = 735 J;
this work becomes potential energy. Note that the result is independent of the speed, and
that an extra ﬁgure was kept in part (b) to avoid roundoff error. 7.3: In Eq. (7.7), taking K1 = 0 (as in Example 6.4) and U2 = 0, K2 = U1 + Wother.
Friction does negative work —fy, so K2 = mgy — fy; solving for the speed '02, 2(mg  f); _ 2((200 kg)(9.80 11052) .3 60"N)(3.00 m) _ _ m 7, 55 m/s. 7.4: a) The rope makes an angle of arcsin(§:%WE«) = 30° with the vertical. The needed
horizontal force is then m tan9 z (120 kg)(9.80 m/s2)tan 30° = 679 N, or 6.8 ><102 N
to two ﬁgures. b) In moving the bag, the rope does no work, so the worker does an
amount of work equal to the change in potential energy, (120 kg)(9.80 m/s2)(6.0 m)(1 
cos 30°) 2: 0.95 x 103 J. Note that this is not the product of the result of part (a) and
the horizontal displacement; the force needed to keep the bag in equilibrium varies as the angle is changed. 7.5: a) In the absence of air resistance, Eq. (7.5) is applicable. With yl — y2 = 22.0 m,
solving for 122 gives v2 = 11% + 2g(y2 — y1) = \/(12.0Mm/s)2 + m) = 24.0 m/s. b) The result of part (a), and any application of Eq. (7.5), depends only on the
magnitude of the velocities, not the directions, so the speed is again 24.0 m/s. 0) The
ball thrown upward would be in the air fora longer time and would be slowed more by air resistance. 7.6: a) (Denote the top of the ramp as point 2.) In Eq. (7.7), Kg = 0, Wothe, : —(35 N) x
(2.5 In): —87.5 {Land taking U1 = 0 and U2 = mgyg = (12 kg)(9.80 m/32)(2.5 Insin 30°) = 147 J, 111 = ﬂwﬂiﬁmﬁn m 6.25 m/s, or 6.3 m/s to two ﬁgures. Or, the work done by friction and the change in potential energy are both proportional to the distance the
crate moves up the ramp, and so the initial speed is proportional to the square root of the distance up the ramp; (5.0 m 6.25. m/s.
179 180 Chapter 7 b) In part a), we calculated Wothe, and U2. Using Eq. (7.7), Kg = %(12 kg)(11.0 m/s)2—
87.5 J — 147 J“: 491.5 J 2 491.5 J) _ _ 9.05 m/s. 7.7: As in Example 7.7, K2 = 0, U2 = 94 J, and U3 = 0. The work done by friction is —(35 N)(1.6 m) = —56 J, and so K3 = 38 J, and v3 = ﬂag: = 2.5 m/s. 7.8: The speed is v and the kinetic energy is 4K. The work done by friction is pro
portional to the normal force, and hence the mass, and so each term in Eq. (7.7) is
proportional to the total mass of the crate, and the speed at the bottom is the same for
any mass. The kinetic energy is proportional to the mass, and for the same speed but four
times the mass, the kinetic energy is quadrupled. 7.9: In Eq. (7.7), K1 = 0, Wother is given as —0.22 J, and taking U2 = 0, K2 = ng —
0.22 J, so 0.22 J
= . .8 2 . — WWW = 2. .
'02 0 m/s )(0 50 In) 0.20 kg) 8 m/s 7.10: (a) The ﬂea leaves the ground with an upward velocity of 1.3 m/s and then is in
freefall with acceleration 9.8 m/s2 downward. The maximum height it reaches is therefore
(v3 — vgy)/2(«—g) = 9.0 cm. The distance it travels in the ﬁrst 1.25 ms can be ignored. (b) W = KE 2 $771112
= $910 x 10‘6g)(1300m/S)2 =1.8ergs = 1.8 x 10‘7J 7.11: Take 3; = 0 at point A. Let point 1 be A and point 2 be B.
K1 + U1 + Wother = K2 + U2
U1 = 0, U2 = mg(2R) = 28, 224 J, Wother = W;
K1 = ém’uf = 37,500 J, K2 = §mv§ = 3840 J
The workenergy relation then gives Wf = K2 + U2 — K1 = —5400 J. 7.12: Tarzan is lower than his original height by a distance l(cos 30 — cos 45), so his
speed is 'v = \/2gl(cos 30° — cos 45°) = 7.9 m/s, 3. bit quick for conversation. 7.13: a.) The force is applied parallel to the ramp, and hence parallel to the oven's
motion, and so W = F3 = (110 N)(8.0 m) = 880 J. b) Because the applied force ? is
parallel to the ramp, the normal force is just that needed to balance the component of the
weight perpendicular to the ramp, n := '11) cos a, andso the. friction force is fk = pkmg cos a Potential Energy and Energy Conservation 181 and the work done by friction is
Wf z —,ukmg cosas = —(0.25)(10.0 kg)(9.80 m/sz) cos 37°(8.0 m) = —157 J, keeping an extra ﬁgure. 0) mgs since 2 (10.0 kg)(9.80 m/82)(8.0 m) sin 37° = 472 J,
again keeping an extra ﬁgure. (1) 880 J — 472 J — 157 J = 251 J. e) In the direction up
the ramp, the net force is F — mg sino: —— grka cosa
= 110 N —— (10.0 kg)(9.80 m/sz)(sin 37° + (0.25) cos 37°)
= 31.46 N, so the acceleration is (31.46 N) / (10.0 kg) = 3.15 m/s2. The speed after moving up the ramp is v : J2EE = ﬁngﬁ m/s2)(8.0 m) = 7.09 m/s, and the kinetic energy is (1/2)mv2 =
252 J. (In the above, numerical results of speciﬁc parts may differ in the third place if
extra ﬁgures are not kept in the intermediate calculations.) 7.14: a) At the top of the swing, when the kinetic energy is zero, the potential energy
(with respect to the bottom of the circlular arc) is mgl(1 — cos 9), where l is the length
of the string and 6 is the angle the string makes with the vertical. At the bottom of the
swing, this potential energy has become kinetic energy, so mgl(1 — cos 6) = émv2, or v = x/Zgl(1— 0036) = «279.80 m/s2)(0.80m)(1W:cosi5°) = 2.1 m/s. b) At 45° from
the vertical, the speed is zero, and there is no radial acceleration; the tension is equal
to the radial component of the weight, or my cost? = (0.12 kg)(9.80 m/s2)cos45° =
0.83 N. c) At the bottom of the circle, the tension is the sum of the weight and the radial acceleration, mg + mug/l = mg(1+ 2(1 «— cos45°)) = 1.86 N, or 1.9 N to two ﬁgures. Note that this method does not use the intermediate calculation
of 'u. 7.15: Of the many ways to ﬁnd energy in a spring in terms of the force and the distance,
one way (which avoids the intermediate calculation of the spring constant) is to note that
the energy is the product of the average force and the distance compressed or extended.
a) (1 / 2)(800 N)(0.200 m) = 80.0 J. b) The potential energy is proportional to the square
of the compression or extension; (80.0 J)(0.050 m/0.200 1n)2 = 5.0 J. 7.16: U = %ky2, where y is the vertical distance the spring is stretched when the weight w = mg is suspended. y = TEE, and k = E, where a: and F are the quantities that “calibrate” the spring. Combining, U see? _ 1,.te20kg)<9r§9erem>ﬁ = 360,. = 2 F/a: ” 2 (720N/0.150m) 182 Chapter 7 7.17: a) Solving Eq. (7.9) for ac, w = 2—,? = m 0.063 m.
b) Denote the initial height of the book as h and the maximum compression of the
spring by m. The ﬁnal and initial kinetic energies are zero, and the book is initially a height a: + h above the point where the spring is maximally compressed. Equating initial
and ﬁnal potential energies, @0032 = mg(:c + h). This is a quadratic in :v, the solution to which is
Fm M). .351
11: mg (1.20 kg)(9.80 m/s2) [1 i 2(1600 N/m)(0.80 m) (1.20 kg) (9.80 m/sz) (1600 N/m)
= 0.116 m, w0.101 m. The second (negative) root is not unphysical, but repreSents an extension rather than a.
compression of the spring. To two ﬁgures, the compression is 0.12 In. 7.18: a) In going from rest in the slingshot’s pocket to rest at the maximum height, the
potential energy stored in the rubber band is converted to gravitational potential energy;
U = mgy = (10 x 10‘3 kg)(9.80 m/s2)(22.0 m) = 2.16.]. b) Because gravitational potential may is proportional to mass, the larger pebble
rises only 8.8 m. c) The lack of air resistance and no deformation of the rubber band are two possible
assumptions. 7.19: The initial kinetic energy and the kinetic energy of the brick at its greatest height
are both zero. Equating initial and ﬁnal potential energies, £10032 = mgh, where h is the greatest height. Solving for h, 2 2
= £13m (1800 N/m)(0.15W119)”:1'7 m‘ h 2mg ‘ 2(120 kg)(9.80 m/sz) 7.20: As in Example 7.8, K; = 0 and U1 = 0.0250 J. For '02 = 0.20 m/s, K2 = 0.0040 J,
so U; = 0.0210 J = ékﬁ, so x = i1 / %£I% = :l:0.092 m. In the absence of friction, the
glider will go through the equilibrium position and pass through :1: = —0.092 m with the
same speed, on the opposite side of the equilibrium position. 7.21: a) In this situation, U2 = 0 when m = 0, so K2 = 0.0250 J and 02 = sagging =
0.500 m/s. b) If '02 = 2.50 m/s, K2 = (1/2)(0.200 kg)(2.50 m/a)2 = 0.625 J = U1, so Potential Energy and Energy Conservation 183 $1 = WWEﬁ/j; m 0.500 m. Or, because the speed is 5 times that of part (a), the kinetic energy is 25 times that of part (a), and the initial extension is 5 x 0.100 m = 0.500 m.
7.22: a) The work done by friction is Wo,.,,., s» —,ukmgA;r: = —(0.05)(0.200 kg)(9.80 m/S2)(0.020 m) = —0.00196 J, so K2 = 0.00704. J and v2 = = 0.27 m/s. b) In this case mmer = —0.0098 J, so K2 = 0.0250 J — 0.0098 J = 0.0152 J, and 112 2 JW = 0.39 m/s. c) In this case, K2 = 0, U2 = 0, so U1 + Wothe, = 0 =0.0250 J — 0140.200 kg)
(9.80 m/s2) x (0.100 In), or in, = 0.13.
7.23: a) In this case, K1 = 625,000 J as before, Won,“ = —~17,000 J and U2 = (1/2)ky§ + mgyz
= (1/2)(1.41 x 105 N/m)(——1.00 m)2 + (2000 kg)(9.80 m/s2)(—1.00 m)
= 50, 900 J. The kinetic energy is then K2 = 625,000 3 “50,900 J — 17,000 J = 557,100 J, corresponding
to a speed 02 = 23.6 m/s. b) The elevator is moving down, so the friction force is up
(tending to stop the elevator, which is the idea). The net upward force is then —mg +
f —— k2: = —(2000 kg)(9.80 m/s2) + 17,000 N — (1.41 x 105 N/m)(—1.00 m) = 138,400 N,
for an upward acceleration of 69.2 m/sz. 7.24: From ﬁlm2 = %m02, the relations between m, o, k and :1: are kxz = mvz, k1: = 5mg. Dividing the ﬁrst by the second gives 2: = 35%, and substituting this into the second gives k = 25%;”, so a) 85 b), (2.50 III/S)?‘
= mm = 0.128
9”” 5(9.80 m/s2) 1‘"
_ £1.15” Elﬁn3.93002 _ s
k _ 25 (250 III/s), _ 4.46 x10 N/m. 7.25: a) Gravity does negative work, —(0.75 kg) (9.80 m/s2)(16 m) = —118 J. b) Grav—
ity does 118 J of positive work. c) Zero d) Conservative; gravity does no net work on any complete round trip. 184 Chapter 7 7.26: a) 8: b) —(0.050 kg)(9.80 m/s2)(5.0 m) = —2.5 J. (a) (b)
c) Gravity is conservative, as the work done to go from one point to another is path—independent.
7.27: a) The displacement is in the ydirection, and since 75) has no ycomponent1 the work is zero. P2 3:2 2
b)/ EMF? = —12/ $261.76 = 13%13» (x3 — .133) = —0.104 J.
P1 .151 c) The negative of the answer to part (b), 0.104 m3 d) The work is independent of
path, and the force is conservative. The corresponding potential energy is U = ngméﬂﬂi’: m
(4 N / m2)x3. 7.28: a) From (0, 0) to (O, L), a: = 0 and so 7‘? = 0, and the work is zero. From (0, L)
o (L, L), if and d? are perpendicular, so ? ml? = 0, and the net work along this
path is zero. b) From (0, 0) to (L, O), ? Hit—1’ = 0. From (L, 0) to (L, L), the work is
that found in the example, W; = 0L2, so the total work along the path is CL”. C) Along
the diagonal path, a: = y, and so ?  d? = Cy dy; integrating from 0 to L gives (It
is not a coincidence that this is the average to the answers to parts (a) and d) The
work depends on path, and the ﬁeld is not conservative.
7.29: a) When the book moves to the left, the friction force is to the right, and the work is "(1.2 N)(3.0 m) = “3.6 J. b) The friction force is now to the left, and the work is
again —3.6 J. c) —7.2 J. (1) The net work done by friction for the round trip is not zero, and friction is not a conservative force. 7.30: The friction force has magnitude iika = (0.20)(30.0 kg)(9.80 m/s2 = 58.8 N.
a) For each part of the move, friction does —(58.8 N)(10.6 m) = —623 J, so the total work done by friction is —1.2 kN. b) “(58.8 N)(15.0 m) = ~882 N.
0) Potential Energy and Energy Conservation 185 _ The net work done by friction depends on the path, so friction is not a conservative
force. 7.31: The magnitude of the friction force on the book is ‘ [14ka = (0.25)(1.5 kg)(9.80 m/s2) = 3.68 N. a.) The work done during each part of the motion is the same, and the total work
done is —2(3.68 N)(8.0 m) = —59 J (rounding to two places). b) The magnitude of the
.. displacement is x/é(8.0 In), so the work done by friction is —\/§(8.0 m)(3.68 N) = ~42 N.
c) The work is the same both coming and going, and the total work done is the same
as in part (a), —59 J. (1) The work required to go from one point to another is not
path independent, and the work required for a round trip is not zero, so friction is not a
conservative force. 7.32: a.) — b) —%k(m§ — The total work is zero; the spring force is
conservative c) From (1:1 to $3, W = —%k(a:§  From $3 to 3:2, W = —%k(m§ — The net work is —%k(:r§ — This is the same as the result of part (a). 7.33: From Eq. (7.17), the force is _ dU_ d 1 _ 606
Fm— die—Cedx(.r6)m :37. The minus Sign means that the force is attractive.
7.34: From Eq. (7.15), F3, = +g~ = “401123 = —(4.8 J/m4)$3, and so Fm(0.800 m) = —(4.8 J /m4)(—0.80 m)3 = 2.46 N.
7.35: 2—: = 2163: + k’y, 9% = 2169 + We and g—g = 0, so from Eq. (7.19), ? = ... (2kg: + k’y) E— (2193; + Hz) 5. 7.36: From Eq. (7.19), F =  %g~ iwgég f, since U has no z—dependence. 39% = 3% and 0U_2o:
a—y—Ty ,SO 186 Chapter 7 7.37: a . . . . . , k»Whiammwiﬁw.wmi.»nnw.mmmwiwb 0.6 "Malawi 1.2 "L?" 1.6 0.6 0. 1 l. 1.4 1.6
r r b) Setting F, = 0 and solving for 1" gives rm,“ = (2a/b)1/6. This is the minimum of
potential energy, so the equilibrium is stable. 0) U (T'min) = M £3;
_ a b
— i537 5,5176% _ “(Engligimlﬁ
ab2 52 b2 21a? ' 22; M13“
To separate the particles means to remove them to zero potential energy, and requires the
negative of this, or E0 = (32/411. (:1) The expressions for E0 and rmin in terms of a and b
are I)2 6 _2a 2 E 1rmin _ b '
Mulitplying the ﬁrst by the second and solving for I) gives b = 2Eor2ﬁn, and substituting
this into the ﬁrst and solving for :1 gives a = Eur12 Using the given numbers, min ‘ E0 a = (1.54 x 10“18 J)(1.13 x 1010 m)12 = 6.68 ><10_138Jm12
b = 2(1.54 ><1(T18 J)(1.13 x 1010 m)6 = 6.41 x 10—73Jm6. (Note: the numerical value for a might not be within the range of standard calculators, and the powers of ten n'iay have to be handled seperately.) 7 .38: 3.) Considering only forces in the mdirection, Fa, = and so the force is zero d3 1
when the slope of the U vs :5 graph is zero, at points b and d. b) Point I) is at a potential
minimum; to move it away from h would require an input of energy, so this point is stable.
c) Moving away from point (i involves a decreaso of potential energy, hence an increase in kinetic energy, and the marble tends to move further away, and so d is an unstable point. 7.39: a) At constant speed, the upward force of the three ropes must balance the force,
so the tension in each is. onethird of the man’s weight. The tension in the rope is the force Potential Energy and Energy Conservation 187 he exerts, or (70.0 kg)(9.80 m/s2)/3 = 229 N. b) The man has risen 1.20 In, and so the
increase in his potential energy is (70.0 kg)(9.80 m/sz)(1.20 m) = 823 J. In moving up a
given distance, the total length of the rope between the pulleys and the platform changes
by three times this distance, so the length of rope that passes through the man’s hands is 3 X 1.20 m = 3.60 In, and (229 N)(3.6 m) = 824 J.
7.40: First ﬁnd the acceleration: 2 _ 2 2
v vo “ _ S.§;.99.§Z§L = 3.75 m/sx2 a 3 2(93 —— x0) _ 2(1.20m) Then1 choosing motion in the direction of the more massive block as positive: Fnet=Mg—mg= (M+m)e=Ma.+ma
Mtga) =m(9+a)
M g + a (9.80 + 3.75) m/s2
_ = w_..._ a WWW m 2.24
m g — a (9.80 — 3.75) m/s2 M = 2.24m SinceM + m = 15.0kg:
2.24m + m = 15.0 kg
m = 4.63 kg
M = 15.0 kg — 4.63 kg = 10.4 kg 7.41: 6.) K1 + U1 + ill/13th.,r = K2 + U2
U1 = U2 2 IQ = 0
WWW = W; = —pkmgs, with s = 280 ft = 85.3 In
The workenergy expression gives %mvf  pkmgs 7— 0 v1 = Wye = 22.4 m/s = 50 mph; the driver was speeding. b) 15 mph over speed limit so $150 ticket. 7.42: a) Equating the potential energy stored in the spring to the block’s kinetic energy,
ékmz = §mv2, or / k 400 N/m
'0 — an: — «%EE~(0.220 m) —— 3.11 m/s. b) Using energy methods directly, the initial potential energy of the spring is the ﬁnal gravitational potential energy, %kx2 = mgL sin 6, or you}? _ %(400.N/m)(0.220 m)2 _ 0 821
mgsina * (2.00 kg)(.9.80 m/s2)sin37.0° “‘ ‘ m' 188 Chapter 7 7.43: The initial and ﬁnal kinetic energies are both zero, so the work done by the spring
is the negative of the work done by friction, or éksz = ,ukmgl, where l is the distance the block moves. Solving for ,uk, = (1mm:2 (1/2)(100 N/m)(0.20 m)2 W = mWwW_WWW—nwm = 0.41.
mg! (0.50 kg)(9.80 m/s2)(1.00 m) Mk 7.44: Work done by friction against the crate brings it to a halt: fka 2 potential energy of compressed spring
360 J
fk * aim The friction force working over a 2.00m distance does work fkm = (—64.29 N)(2.00 m) =
—128.6 J. The kinetic energy of the crate at this point is thus 360J — 128.6J = 231.4 J, and its speed is found from = 64.29N 2 = 231.4J
2(231.4 J)
2 = WWW:W m 2 2
v BUDkg 9.256111 /s
'u = 3.04 m/s 7.45: a) mgh = (0.650 kg)(9.80 m/s2)(2.50 m) = 15.9 J b) The second height is 0.75(2.50 m) = 1.875 m, so second mgh = 11.9 J; loses
15.9 J  11.9 J = 4.0 J on ﬁrst bounce. This energy is converted to thermal energy. c) The third height is 0.750.875 1n) = 1.40 m, so third mgh = 8.9 J; loses
11.9 J — 8.9 J = 3.0 J on second bounce. 7.46: a) U A — U3 = mg(h — 2R) = §mvﬁ. From previous considerations, the speed at
the top must be at least Thus, mg(h — 2R) > éng, or h > gs. b) UA  U0 = (350)ng = Kc, so so = ,/(5.00)gs = JEBYJOXQBO m/s2)(20.0 m) = 31.3 m/s. The radial moderation is and = 3%; x 49.0 m/sz. The tangential direction is down, the
normal force at point C is horizontal, there is no friction, so the only downward force is gravity, and at“, = g = 9.80 m/sz.
7.47: a) Use workenergy relation to ﬁnd the kinetic energy of the wood as it enters the
rough bottom: U1 = K2 gives K2 : mgyl = 78.4 J. Potential Energy and Energy Conservation 189 Now apply workenergy relation to the motion along the rough bottom:
K1 + U1 + Wattler = K2 + U2 Wother = Wf = *ukmgs, K; = U1 = U2 = 0; K1 = 78.4 J 78.4 J  pkmgs = 0; solving for 3 gives 3 = 20.0 In. The wood stops after traveling 20.0 m along the rough bottom. b) Friction does —78.4 J of work. 7.48: (a) KEBottom + Wf = PETbP
% mug — ,ukmg cos 9d = mgh
d = h/ sinB
1 2 h
500  ,ukg cos 0m —~ gh
1 2 2 cos 40° __ 2
2(15m/s) (0.20)(9.8 m/s )Sin40oh — (9.8 m/s )h
h = 9.3m (b) Compare maximum static friction force to the weight component down the plane. f5 = namg cos 6 = (0.75)(28 kg) (9.8 m/s2) cos 40"
= 158 N
mg sing = (28 kg)(9.8 m/s2)(sin 40") = 176 N > f3 so the rock will slide down. (c) Use same procedure as (a), with h = 9.3 m PEpr + Wf = KEBottom h 1
mgh  ,ukmg cos = 5 mo: v3 = J2EE — 2nkghcos 0/ sinﬁ = 11.8m/s 7.49: a) K1 + U1 + Wothm = K2 + U2
Let point 1 be point A and point 2 be point B. Take 3; = 0 at point B.
mgyl + émvi" = §mv§, with h = 20.0 m and v1 = 10.0 m/s 02 = + 29h = 22.2 m/s b) Use K1 + U1 + Wm,er = K2 + U2, with point 1 at B and point 2 where the spring
has its maximum compression 9:. U1 : U2 = K2 = 0; K1 = ém’uf with 01 = 22.2 m/s
Wothe, = Wf + We. = ,uk'mgs — yea2, with s = 100 m + a: 190 Chapter 7 The workenergy relation gives K1 + Wothe, = 0. L
2 Putting in the numerical values gives :02 + 29.42: — 750 = 0. The positive root to this
equation is :c = 16.4 In. moii — pkmgs — ﬁlm? 2 0 b) When the spring is compressed w = 15.4 In the force it exerts on the stone is
F81 = kg: = 32.8 N. The maximum possible static friction force is max f8 = JW9 = (0.80)(15.0 kg)(9.80 m/s") = 118 N. The spring force is less than the maximum possible static friction force so the stone remains at rest. 7.50: First get speed at the top of the hill for the block to clear the pit.
1 y 2 g
1
20 m = E(93 m/s2)t2
t = 2.03
40111
“UToPt = In —) vTop : = m/S Energy conservation: K’JE'ﬁl'Br.)ttom = PETop + KETop 1 1
—m'vlza = mgh + — mug. 2 0., = Mme;
= WEE/s)? + 2(9.8 m/s2)(70 m)
= 42 m/s 7.51: K1 + U1 + Wother = K2 + U2 Point 1 is where he steps off the platform and point 2 is where he is stopped by the
cord. Let y = 0 at point 2. y1 = 41.0 m. Wower = —%kw2, Where :1: = 11.0 In is the amount
the cord is stretched at point 2. The cord does negative work. K1 = K2 = U2 : 0, so mgyl —%k:1:2 = O and k = 631 N/m.
Now apply F 2 km to the test pulls:
F = k2: so 3 = F/k = 0.602 m. 7.52: For the skier to be moving at no more than 30.0 m/s; his kinetic energy at the
bottom of the ramp can be no bigger than
11333 (85.0 kg)(30.0 m/s)2 2 = WWWm 38,250.] Potential Energy and Energy Consemctz'an 191 Fi‘iction does —4000 J of work on him during his run, which means his combined PE
and KB at the top of the ramp must be no more than 38, 250 J + 4000 J = 42, 250 J. His .KE at the top is
mi)2 (85.0 kg)(2.0m/s)2 2 2 His PE at the top should thus be no more than 42, 250 J — 170 J = 42, 080 J, which gives
.. a height above the bottom of the ramp of 42
h ._— = = 50.5 In mg (85.0 kg)(9.80 m/sz) = 170J 7.53: The net work done during the trip down the barrel is the sum of the energy stored
in the spring, the (negative) work done by friction and the (negative) work done by gravity.
Using ékmz = %(F2/k), the performer’s kinetic energy at the top of the barrel is 1 (4400 N)2 _ Mumwuw _ _ 2 I 3
K — 21100 N/m (40 N)(40 m) (60 kg)(9.80 m/s )(2.5 m) 7.17 x 10 J, and his speed is ¢gzeyg§ﬁgﬁvil = 15.5 m/s.
7.54: To be at equilibrium at the bottom, with the spring compressed a distance :60, the
spring force must balance the component of the weight down the ramp plus the largest value of the static friction, or k:ch = w sing + f. The workenergy theorem requires that
the energy stored in the spring is equal to the sum of the work done by friction, the work done by gravity and the initial kinetic energy, or 1 1
51033 = (wsinﬂ — f)L + Emvz, where L is the total length traveled down the ramp and v is the speed at the top of the
ramp. With the given parameters, %k:r3 = 248 J and kmo = 1.10 x 103 N. Solving for I:
gives k = 2440 N /m. 7.55: The potential energy has decreased by (12.0 kg)(9.80 m/s2)(2.00 m) — (4.0 kg) x
(9.80 m/s2)(2.00 m) = 156.8 J. The kinetic energy of the masses is then %(m1 + 11%)?)2 = (8.0 kg)v2 = 156.8 J, so the common speed is v = W = 4.43 m/s, or 4.4 m/s to two ﬁgures.
7.56: a) The energy stored may be found directly from 1
5kg; = K1 + wother — mgyz = 625,000 J — 51,000 J — (—5s,soo J) = 6.33 x 105 J. b) Denote the upward distance from point 2 by h. The kinetic energy at point 2 and
at the height h are both zero, so the energy foundin part (a) is equal to the negative
of the work done by gravity and friction, —(mg + f)h = —((2000 kg)(9.80 m/s2) + 192 Chapter 7 17,000 N)h = (36,600 N)h, so h = gggﬁgﬁ = 17.3 In. c) The net work done on
the elevator between the highest point of the rebound and the point where it next reaches
the spring is (mg — f )(h — 3.00 m) = 3.72 x 104 J. Note that on the way down, friction does negative work. The speed of the elevator is then ‘/ m 6.10 m/s. (1)
When the elevator next comes to rest, the total work done by the spring, friction, and
gravity must be the negative of the kinetic energy K3 found in part (c), or K3 = 3.72 x 104 J = —(mg—f)x3 + gag = —(2,600 N)2>3 + (7.03 x 104 N/mng. (In this calculation, the value of A: was recalculated to obtain better precision.) This
is a quadratic in .123, the positive solution to which is w _ _.....__..______1.. m
3  2 (7.03 x" 104 N/m)
x [2.60 x 103 N + Jﬁ’f’i’é‘ﬁ‘ x 103 N)2""¥ZE71073;”§?T04"N/fffjﬁIWfi’ﬁTﬁ] = 0.746 m, corresponding to a force of 1.05 x 105 N and a stored energy of 3.91 x 104 J. It should be
noted that different ways of rounding the numbers in the intermediate calculations may
give different answers. 7.57: The two design conditions are expressed algebraically as log == f + my = 3.66 X 104N
(the condition that the elevator remains at rest when the spring is compressed at distance
change in energy is the work Wm},er = — f y). Eliminating y in favor of k by y = 5535,591va
leads to 1: (3.66 x__1__04 N)2 a (1.70 x 104 N)(3.6§m>5 104 N) 2 k" k
1. 6 4 , 4
= 62.5 x 104 J + N 10 N) This is actually not hard to solve for k = 919 N/m, and the corresponding :3 is 39.8 m.
This is a very weak spring constant, and would require a space below the operating range
of the elevator about four ﬂoors deep, which is not reasonable. b) At the lowest point, the
spring exerts an upward force of magnitude f + mg. Just before the elevator stops, however,
the friction force is also directed upward, so the net force is (f + mg) + f — mg = 2f, and
the upward acceleration is if = 17.0 m/s2. 7.58: One mass rises while the other falls, so thenet loss of potential energy is
(0.5000 kg — 0.2000 kg)(9.80 m/sﬂ)(0.400 m) = 1.176 J. This is the sum of the kinetic energies of the animals. Ifthe animals are equidistant from the center, they have the same Speed, so the kinetic energy of the combination is %mtotvg, Potential Energy and Energy Conservation 193 and ...mm mm = 1.83 m/s. 7.59: a) The kinetic energy of the potato is the work done by gravity (or the potential energy lost), §mv2 = mgl, or v = x/2gl = m/sz)(2.50 m) = 7.00 m/s. 2
b) T—mg=mvl—=2mg, so T = 3mg = 3(0.100 kg)(9.80 m/sz) = 2.94 N.
7.60: a) The change in total energy is the work done by the air, 2 _ (1/2)((18.6 m/s)’’ — (30.0 m/s)2
_ (0145 kg) ( —(40.0 III/5F) + (9.80 m/s2)(53.6 m)) (K2 + U2)  (K1 + U1) = m (l (v;  113+ are) n ——80.0 J. b) Similarly, _ (1/2)((11.9 m/s)2 + (~23? m/s)2
(K3 + U3) — (K2 + U2) _ (0145 kg) ( —(18.6 m/s)2) — (9.80 In/s2)(53.6 m)) = —31.3 J. c) The ball is moving slower on the way down, and does not go as far (in the :c
direction), and so the work done by the air is smaller in magnitude. 7.61: a) For a friction force f, the total work done sliding down the pole is mgd — fd.
This is given as being equal to mgh, and solving for f gives f mg = mg (1 __ _ When h = d, f = 0, as expected, and when h = 0, f = mg; there is no net force
on the ﬁreman. b) (75 kg)(9.80 In/s2)(1 M 1'0 m) 2 441 N. c) The net work done is 2.5 111
(mg — f )(d — y), and this must be equal to ﬁmvz. Using the above expression for f, gm? = (mg — f)(d — y) =m9 (tiy) =mgh(l~z~), 194 Chapter 7 from which 0 = \/ Qgh(1 — 3; When 3; = 0, 'v = x/QEE, which is the original condition.
When y = d, v = O; the ﬁreman is at the top of the pole. 7.62: a) The skier’s kinetic energy at the bottom can be found from the potential energy at the top minus the work done by friction, K1 = mgh — WF = (60.0 kg) 9.8 N
10,500 J, or K1 : 38,200J — 10,500.] = 27,720 J. Then '01 = £1? = {M53341 =
30.4 m/s. b) K2 = K1 — (WF + WA) = 27,720 J # (pkmgd+ fwd), K2 = 27,720 J — [(.2)(588 N)x
(82 m) + (160 N)(82 m)], or K2 = 27,720 J — 22,763 J = 4957 J. Then, _ 2”}? _ 5174538“ N
’02— “"1"??— c) Use the WorkEnergy Theorem to ﬁnd the force. W = AKE, F = KE/d =
(4957J)/(2.5 m) = 1983N m 2000 N. 7.63: The skier is subject to both gravity and a normal force; it is the normal force that
causes her to go in a circle, and when she leaves the hill, the normal force vanishes. The
vanishing of the normal force is the condition that determines when she will leave the
hill. As the normal force approaches zero, the neCessary (inward) radial force is the radial
component of gravity, or meal/R = mg cos a, where R is the radius of the snowball. The
speed is found from conservation of energy; at an angle a, she has descended a vertical
distance R(1 — cos a), so §mv2 = ng(1 — cos a), or v2 = 2gR(1 ~— cos (1). Using this in
the previous relation gives 2(1 —cos a) = cos a, or a = arccos(§) = 48.2°. This result does
not depend on the skier’s mass, the radius of the snowball, or g. 7.64: If the speed of the rock at the top is m, then conservation of energy gives the speed 1)., from %m'ug = gmvf + mg(2R), R being the radius of the circle, and so v3 = v? + 49R.
2 The tension at the top and bottom are found from Tt + my = m—ga and Th — mg = 5551*, so
Tb — Tt = %(v§ — of) + 2mg = 6mg = 610.
7.65: a) The magnitude of the work done by friction is the kinetic energy of the package at point B, or ,ukmgL = %mvf3, or a was _ <1/2)(4.80 m/sr .__
#k * “7r” “ 7mm  "392 b) Wother = KB — UA = $(0200 kg) (4.80 111/3)? ~— (0.200 kg) (9.30 m/s2)(1.60 m) = —0.832 Equivalently, since KA : KB = 0, UA + WAB + W30 = 0, 01' wAB = 0, — W30 = mg(—(1.60 m) — (0.300)(—3.00 m)) = “0.832 J. Potential Energy and Energy Conservation 195 7.66: Denote the distance the truck moves up the ramp by m. K1 = %mv§, U; =
mgL Sinai, K2 = 0, U2 = mgr sinﬁ and Won“... = —,urmg.r cos ,8. From Wow... = (K2 +
U2) — (K1 + U1), and solving for w, K1 + mgL sin a (ea/29) + L sina x .WWWWMWNMWW = “mmmw.—MW.MM = mg(sin )3 + it, cos 6) sin 3 + u. cos ,6 '
7.67: 8.) Taking U(0) = 0, U(a:) = —— [03 Fa. da: = 33:2 + gxa = (30.0 N/m):r2 + (6.00 N/m2)$3. b) K2 = U1 — U2
= ((30.0 N/m)(1.00 m)2 + (6.00 N/m2)(1.00 m)3)
— ((30.0 N/m)(0.50 m)2 + (6.00 N/m2)(0.50 m)3)
= 27.75 J, and so v2 = = 7.85 m/s. 7.68: The force increases both the gravitational potential energy of the block and the
potential energy of the spring. If the block is moved slowly, the kinetic energy can be
taken as constant, so the work done by the force is the increase in potential energy, AU = mge sinél + %k(a0)2.
7.69: With U2 = 0, K1 = 0, K2 2 %mv§ = U1 = ﬁler” + mgh, and solving for v2, ’02 3 if“ + 29h = + 2(9.80 m/s2)(1.20 m) = 7.01 m/s. 7.70: a) In this problem, use of algebra avoids the intermediate calculation of the spring
constant k. If the original height is h and the maximum compression of the spring is d,
then mg(h + d) = %kd2. The speed needed is when the spring is compressed ‘5', and from
conservation of energy, mg(h + d/ 2) — yew/2)2 = émvz. Substituting for k: in terms of h+d, d mg(h+d)___1 2
mg(h+2) 4 ~2mv, which simpliﬁes to 3 1
2— _ _
v 29(4h+4d). Insertion of numerical values gives *0 = 6.14 m/s. b) If the spring is compressed a. distance
3:, é—kw? z mgm, or x = Using the. expression from part (a) that gives 19 in terms of 196 Chapter 7 h and (1, d2 w (12% = 0.0210 In. 2mg(h :3) : "Hid x = (2mg) 7.71: The ﬁrst condition, that the maximum height above the release point is h, is
expressed as yea:2 = mgh. The magnitude of the acceleration is largest when the spring
is compressed to a distance :r; at this point the net upward force is its: — mg = ma, so the
second condition is expressed as :r = (m/k) (g+a). 3.) Substituting the second expression
into the ﬁrst gives m(g + (1)2 1 m 2 2_ _ WWW
§k(—) (9+0) #771911, or 19— 29h . k b) Substituting this into the expression for x gives 1‘. = E211“. 7.72: Following the hint, the force constant k is found from w = mg : kd, or k = "—53.
When the ﬁsh falls from rest, its gravitational potential energy decreases by mgy; this
becomes the potential energy of the spring, which is ﬁfty2 = {15,33}? Equating these, 7.73: With K1 = 0,
U1 = K2 + U2 — Wother
= émvg + mgL sincx — (—mgLnk cos a) = m (cg/2 + gL(sina + it}, cos (1))
= (1.50 kg) ((7.00 m/s)2/2 + (9.80 m/s2)(6.00 m)(sin 300° + (0.50) cos30.0°))
= 119 J. 7.74: a) From either energy or force considerations, the speed before the block hits the
spring is v = 29L(si”ﬁ”é”:mﬁfé"t§§j = ﬂgﬁmﬁvin/ﬁXAOOmein 531° — (0.20) cos 531°)
= 7.30m/s. b) This does require energy considerations; the combined work done by gravity and
friction is mg(L+d) (sin 6 — pk cos 6), and the potential energy of the spring is %kd2, where
d is the maximum compression of the spring. This is a quadratic in d, which can be written
as k
2 m... WWMMWWW... m .....—. . z
d 2mg(sin 6 — ,uk cos 6') d L 0' Potential Energy and Energy Conservation 197 The factor multiplying d2 is 4.504 mil, and use of the quadratic formula gives d = 1.06 m.
c) The easy thing to do here is to recognize that the presence of the spring determines d,
but at the end of the motion the spring has no potential energy, and the distance below
the starting point is determined solely by how much energy has been lost to friction. If
the block ends up a distance y below the starting point, then the block has moved a
distance L + all down the incline and L + d — y up the incline. The magnitude of the
friction force is the same in both directions, itka cos 9, and so the work done by friction
is —,u.k(2L + 2d — y)mg cos 8. This must be equal to the change in gravitational potential
energy, which is mgy sin 6. Equating these and solving for 3; gives 211k = L m WW «w— W_— :3: d W.....m,.m“W,_'
y ( +d) (L+ tan9+uk Using the value of d found in part (b) and the given values for pk and 9 gives y = 1.32 m. 7.75: a) KB = Wother — U3 = (20.0 N)(0.25 m)  (1/2)(40.0 N/m)(.25 m)2 = 3.75 J,
so 113 = = 3.87 m/s, or 3.9 m/s to two ﬁgures. b) At this point (point 0),
KC = 0, and so Uc = Wothe, and :rc = —@%%7% = —0.50 In (the minus sign denotes a
displacement to the left in Fig. (7.65)), w ich is 0.10 m from the wall. 7.76: The kinetic energy K ’ after moving up the ramp the distance 3 will be the energy
initially stored in the spring, plus the (negative) work done by gravity and friction, or 1
K’ 2 aka; — mg(sina + pk cosa)s. Minimizing the speed is equivalent to minimizing K ’ , and differentiating the above ex
pression with respect to a and setting 9% = 0 gives 0 = —mgs(cosa — ,uk sin 0), M or tana = i, oz = arctan Pushing the box straight up (a = 90°) maximizes the
vertical displacement h, but not 5 = h/ sin a. 7.77: Let 3:1 = 0.18 m, 312 = 0.71 m. The spring constants (assumed identical) are then
known in terms of the unknown weight w, 4km = w. The speed of the brother at a given
height h above the point of maximum compression is then found from 1
(4k)w§ = l E v2 + mgh,
or 2 2 g 2
v. = e12); 1 2,. = g _ 2;.) , w 198 Chapter 7 so 1: = \/(A9:§0mm7s‘2)((077“1 m)2/(0.18 m) — = 3.13 m/s, or 3.1 m/s to two
ﬁgures. b) Setting '0 = 0 and solving for h, 2 2
= 2’”? a ﬁ = 1.40 in,
mg 22:1 h or 1.4 m to two ﬁgures. 0) No; the distance :01 will be different, and the ratio 335 =
law7515113 : 3:1(1+ will be different. Note that on a. small planet, with lower 9, $1 :31 will be smaller and h will be larger. 7.78: a) am = 032.23/dt2 : —w3m, F... = max = #rmugm
ay = dzy/dt2 = —wgy = —w[2,y, Fy = may = —mw3y
1
b) U z — [/Fwdar+/Fydy] =mw§ [/mdm+/ydy] = Emwﬂxg +y2)
C) 'vau 7— dx/dt = ——x0w0 sin wot = ‘mow0(y/y0) 1),, == dy/dt = +yowo cos wot = +yowo(a:/2:g) (i) When :1: = :00 and y = 0, '03 = 0 and 11,, = mm; 1 1 1 1
K 1* §m(vg + 12:) 2: imygwg, U = Ewgmmg andE = K + U = Emwﬂccg + yg)
(ii) When it = 0 and y = yo, 1),; = —a:owo and Du = 0
1 2 2 1 2 2 1 2 2 2
K = Ewomxo, U = Emwoyo andE = K + U = Emwomo + yo) Note that the total energy is the same.
7.79: a) The mechanical energy increase of the car is
K2 — K1 = %(1500 kg)(37 m/s)2 =1.027 x106 J. Let a be the number of gallons of gasoline consumed.
a(1.3 x 108 J)(0.15) = 1.027 x 106 J
oz = 0.053 gallons b) (1.00 gallons) /a = 19 accelerations
7.80: (a) Stored energy = mgh m (pl/)gh: pA(1m)gh = (1000 kg/m3)(3.0 x 106 m2)(1 m) (9.8 3;) (150m)
= 4.4 x 1012 J. Potential Energy and Energy Conservation 199 (b) 90% of the stored energy is converted to electrical energy, so (0.90)(mgh) = 1000 kW h
(0.90)pVgh = 1000 kW h (1000 kW h) 363.3%) V = (mewaxas mysj
= 2.7 x 103 m3
Change in level of the lake:
AAh = Vwater
Ah = K — 2'7 5—1229: = 9.0 x 10—4m A — 3.0'x'106 m2
7.81: The potential energy of a horizontal layer of thickness dy, area A, and height y
is dU = (dm)gy. Let p be the density of water.
dm = pdV = pAdy, so dU = pAgy dy.
The total potential energy U is
U = f0” dU = pAg johydy = #219152.
A=3.0><106 In2 andh= 150m,soU=3.3><1014 J=9.2>< 107 kWh. 7.82: a) Yes; rather than considering arbitrary paths, consider that 7’: " l5; 0%)] i b) No; consider the same path as in Example 7.13 (the ﬁeld is not the same). For this
force, 17 = 0 along Leg 1, .1“?  d? = 0 along legs 2 and 4, but 7‘7  d7 aé 0 along Leg 3. 7.83: 8.) Along this line, a: = y, so T"  d? = —ay3 dy, and (I 92
/ Fydy a m— (y; w y?) = —50.6 J.
In 4 +—b b) Along the ﬁrst leg, dy = 0 and so F  d I = 0. Along the second leg, :1: = 3.00 In,
so Fy = —(7.50 N/m2)y2, and f” Fydy = —(7.5/3 N/mz) (y; — y?) = —67.5 J. m c) The work done depends on the path, and the force is not conservative. 200 Chapter 7 7.84: a) b) (1): as = 0 along this leg, so .17" = 0 and W = 0. (2): Along this leg, y = 1.50 m, so
‘1?  d? = (3.00 Minna, and W = (1.50 N/m)((1.50 m)2—0) = 3.38 J (3) id? = 0,
so W = 0 (4) y = 0, so 7!? = 0 and W = 0. The work done in moving around the closed
path is 3.38 J. c) The work done in moving around a closed path is not zero, and the
force is not conservative.
7.85: a) For the given proposed potential U (31:), 4% = k:c + F, so this is a possible
potential function. For this potential, U (0) = F2/2k, not zero. Setting the zero of
potential is equivalent to adding a constant to the potential; any additive constant will
not change the derivative, and will correspond to the same force. b) At equilibrium, the
force is zero; solving —k:c + F = 0 for 2‘: gives 9:0 = F/k. U(:co) m “Fa/k, and this is a
minimum of U , and hence a stable point. 6) d) No; Fm = 0 at only one point, and this is a stable point. e) The extreme values
of :1: correspond to zero velocity, hence zero kinetic energy, so U (xi) = E, where art
are the extreme points of the motion. Rather than solve a quadratic, note that U = %k(m  F/k)2 — Fz/k, so U(xi) = E becomes Potential Energy and Energy Conservation 201 f) The maximum kinetic energy occurs when U is a minimum, the point .130 =
F/k found in part At this point K = E ~— U = (Fz/k) — (—Fg/k) = ZFQ/k, so
u = 2F/ 7.86: a) The slope of the U vs. 3: curve is negative at point A, so F3 is positive
(Eq. (7.17)). b) The slope of the curve at point B is positive, so the force is negative. 0) The
kinetic energy is a maximum when the potential energy is a minimum, and that ﬁgures to
be at around 0.75 m. d) The curve at point C looks pretty close to ﬂat, so the force is zero.
e) The object had zero kinetic energy at point A, and in order to reach a point with more
potential energy than U (A), the kinetic energy would need to be negative. Kinetic energy
is never negative, so the object can never be at any point Where the potential energy is
larger than U On the graph, that looks to be at about 2.2 m. f) The point of minimum
potential (found in part is a stable point, as is the relative minimum near 1.9 m. g) The
only potential maximum, and hence the only point of unstable equilibrium, is at point C. 7.87: a) Eliminating [3' in favor of a and :30 (,6 = a/cco), U(,)=a_ﬁme§s§ea:%[(n)2_(n)]. :1: 33 U($0) = fgﬂ — 1) = 0. U(:I:) is positive for a: < 230 and negative for x > £60 and 16 must
be taken as positive). 202 Chapter 7 The proton moves in the positive mdirection, speeding up until it reaches a maximum
speed (see part (c)), and then slows down, although it never stops. The minus sign in the
square root in the expression for v(a:) indicates that the particle will be found only in the
region where U < 0, that is, a: > :30. c) The maximum speed corresponds to the maximum kinetic energy, and hence the minimum potential energy. This minimum occurs when % = 0, or ﬁnal—aeseﬂa dm 2:0 2: a: C! which has the solution :1: = 23:0. U (2%) = ~Zgg, so 1) = ‘ /W~. d) The maximum speed
0 O
dU occurs at a point where a; 2 0, and from Eq. (7.15), the force at this point is zero. e) :31 = and mat) = v(a:) = genius; = mg) m (W m 23)] = —— — 2/9). The particle is conﬁned to the region where U(:1:) < U(a:1). he maximum speed still occurs at at = 22:0, but now the particle will oscillate between
$1 and some minimum value (see part f) Note that U — U (3:1) can be written as 0: $0 2 $0 2 a {130 1 $0 2 —2 <—) —(—>+~ =—2 (—>w <—>——, 320 a: co 9 9:0 as 3 3: 3
which is zero (and hence the kinetic energy is zero) at a: z 33:0 = :51 and a: = £230.
Thus1 when the particle is released from 330, it goes on to inﬁnity, and doesn’t reach any maximum distance. When released from $1, it oscillates between gxo and 3500. ...
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