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Unformatted text preview: Chapter 6 Work and Kinetic Energy 6.1: a) (2.40 N)(1.5 m) = 3.60 J b) (—0.600 N)(1.50 1n) = —0.900 J c) 3.60 J — 0.720 .1 = 2.70 J. 6.2: a) “Pulling slowly” can be taken to mean that the bucket rises at constant speed, so
the tension in the rope may be taken to be the bucket’s weight. In pulling a given length
of rope, from Eq. (6.1), W = F3 = mgs = (6.75 kg)(9.80 m/sz)(4.00 m) = 264.6 J. b) Gravity is directed opposite to the direction of the bucket's motion, so Eq. (6.2)
gives the negative of the result of part (a), or —265 J. c) The net work done on the
bucket is zero. 6.3: (25.0 N)(12.0 m) = 300 J. 6.4: a) The friction force to be overcome is
f = mm. m itka = (0.25)(30.0 kg)(9.80 111/32) = 73.5 N, or 74 N to two ﬁgures. b) From Eq. (6.1), F3 = (73.5 N)(4.5 m) = 331 J. The work is positive, since the
worker is pushing in the same direction as the crate’s motion. c) Since f and s are oppositely directed, Eq. (6.2) gives —fs = —(73.5 N)(4.5 m) m —331 J. (1) Both the normal force and gravity act perpendicular to the direction of motion,
so neither force does work. e) The net work done is zero. 6.5: a) See Exercise 5.37. The needed force is F — w. “my... 9.3.5.290 REW ‘ c0505 — pk sii'IE ‘ cos 30° — (0.25) sin 30° = 99.2 N, keeping extra ﬁgures. b) F3 coed) : (99.2 N)(4.50 In) cos 30° = 386.5 J, again keeping
an extra ﬁgure. c) The normal force is mg + Fsin 42, and so the work done by friction is
—(4.50 m)(0.25)((30 kg)(9.80 m/s2) + (99.2 N) sin 30°) = —386.5 J. (1) Both the normal
force and gravity act perpendicular to the direction of motion, so neither force does work.
e) The net work done is zero. 6.6: From Eq. (6.2), F's cosgb = (180 N)(300 m) cos 15.0° = 5.22 x 104 J. 155 156 Chapter 6 6.7: 2Fscosqb = 2(1.80 x 106 N)(O.75 ><103 m) cos 14° = 2.62 x 109 J, or 2.6 x 109 J to
two places. 6.8: The work you do is: _. F .3 = ((30 N): — (40 mi)  ((—90 m)2  (30 H03“)
= (30 N)(—9.0 m) + (—40 N)(—3.0 m)
= _270 N.m + 120N111 = —150J 6.9: a) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is —mg(y2 —— yl). When y1 = y2, me == 0. b) Tension does no work.
(ii) Let I be the length of the string. ng = —mg(y2 — y1) = —mg(2l) = —25.1 J The displacement is upward and the gravity force is downward, so it does negative
work. 6.10: a) From Eq. (6.6), 2
$0600 kg) ((50.0 km/h) = 1.54 x105 J. b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the
speed of any object increases the kinetic energy by a factor of four. 611: F01“ the T'Rex: K = gigﬂ? = 4.32 x 103 J. The person's velocity would be '0 = 2(4.32 x 103 J)/70 kg = 11.1 m/s, or about 40 km/h.
6.12: (3.) Estimate: 1) % lm/s (walking) 1) a: 2m/s (running) K: m m 70 kg
Walking: KE = ém’u2 = %(70kg)(1m/s)2 = 35J
Running: KE = ;—(70 kg)(2 m/s)2 = 140.]
(b) Estimate: 1) m 60 mph = 88 ft/s z 30 m/s
m z 2000 kg
1 KE 2 5(2000 kg)(30 m/s)2 = 9 x 105J
(c) K E = ngdty = mgh Estimate h. a: 2 m
KE = (1 kg)(9.8m/sz)(2 m) z 20 J
6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier.
Wtot : K2 — K1
K1 z émvg, K2 = 0 Work is done by gravity and friction, so Wm = ng + Wf. Work and Kinetic Energy 157 Wm = —mg(y2 — in) = m9h
W; = —fs = (,ukmgcosot)(h/ sin a) = —,ukmgh/ tana Substituting these expressions into the worleenergy theorem and solving for v0 gives v0 = mdi'tmmj 6.14: (a) W = AKE —mgh = émvfz — émvﬁ
‘00 = x/U? +297i WWWW
= ‘/ (25.0 m/s)2 + 2(9.80m/82)(15.0 m)
= 30.3 m/s 0”) W = AKE
1 2 1 2
—mgh = émvf w Emvo
h _ sired _ QBEELiliﬂ 2g 2(9.80 111/52)
= 46.8 m 6.15: a) parallel to incline: force component 2 my sin a, down incline; displacement =
h/ sina, down incline W" = (my sin a)(h/ sin a) = mgh perpendicular to incline: no displacement in this
direction, so W_L = 0.
Wm = W" + W; = mgh, same as falling height h. b) Wm = K2 — K1 gives mgh = aims2 and v = £575, same as if had been dropped
from height h. The work done by gravity depends only on the vertical displacement of the
object. When the slope angle is small, there is a small force component in the direction of
the displacement but a large displacement in this direction. When the slope angle is large,
the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller. c) h = 15.0 m, so 1) = V295 = 17.1s 6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the
work done by friction, by a factor of four. With the stopping force given as being inde
pendent of speed, the distance must also increase by a factor of four. 158 Chapter 6 6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so W = (1/2)m'u2 = (1/2)(0.145 kg)(32.0 111/3)? = 74.2 J. 6.18: As the example explains, the boats have the same kinetic energy K at the ﬁnish
line, so (1/3771waa = (1/2)mgv§, or, with m3 = 2mA, v3 = 21%. 3.) Solving for the ratio
of the speeds, 12A / v3 = b) The boats are said to start from rest, so the elapsed time
is the distance divided by the average speed. The ratio of the average speeds is the same
as the ratio of the ﬁnal speeds, so the ratio of the elapsed times is tB/tA = vA/vg = 6.19: a) From Eq. (6.5), K2 2 113/16, and from Eq. (6.6), W = —(15/16)K1. b) No;
kinetic energies depend on the magnitudes of velocities only. 6.20: From Equations (6.1), (6.5) and (6.6), and solving for F, 2 95 2 in»:  v?) : iEEQPngOO wetnegnefz = 32 0 N F W
s s (2.50 m) M = W_Mﬁﬁ;ﬁﬂﬂﬂﬁﬁl 2 16 8 cm 0.21: = .—
3 F (40.0 N) 6.22: a) If there is no work done by friction, the ﬁnal kinetic energy is the work done
by the applied force, and solving for the speed, _ 2W _ §f§_ 2(36.0 N)(1.20 m) _
y _ m _ m _ (4.30 kg) — 4.48 m/s. b) The net work is F3 — fks = (F  nkmg)s, so v = MW...WWW. m “3230.0 N (0.30)(4.30 kg) (9.80 m/sz))(1.20 m) (4.30 kg)
= 3.61 m/s. (Note that even though the coefﬁcient of friction is known to only two places, the difference
of the forces is still known to three places.) 6.23: a) On the way up, gravity is opposed to the direction of motion, and so W =
—mgs = ——(0.145 kg)(9.80 m/s2)(20.0 m) = “28.4 J. W 2(—.28..4_J)
b = 2 _ = 2 m...“
) 02 v1 + 2m (25.0 m/s) + (0.145 kg) = 15.26 m/s. Work and Kinetic Energy 159 c) No; in the absence of air resistance, the ball will have the same speed on the way
down as on the way up. On the way down, gravity will have done both negative and
positive work on the ball, but the net work will be the same. 6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1)
gives W = F3 = mgs = (4.80 kg)(9.80 m/s2)(25.0 m) = 1176 J. b) Since the melon is released from rest, K1 = 0, and Eq. (6.6) gives 6.25: a) Combining Equations (6.5) and (6.6) and solving for v2 algebraically, v2 = \/v;" + 2&5 = \/ (4.00 m/s)2 + 39100 NIQQMEQ = 4.96 m/s. (7.00 kg) Keeping extra ﬁgures in the intermediate calculations, the acceleration is
a = (10.0 kgm/sg)/(7.00 kg) = 1.429 m/s2. From Eq. (2.13), with appropriate change in
notation, v3 = vf + 20.3 = (4.00 m/s)2 + 2(1429 m/s2)(3.0 In), giving the same result. 6.26: The normal force does no work. The work—energy theorem, along with Eq. (6.5),
gives 1) = ([35 = 3g = = m where h = Lsin6 is the vertical distance the block has dropped, and 9 is the angle the
plane makes with the horizontal. Using the given numbers, '0 = \/§(9.80 m/s2)(0.75 iii)”§ihw'§éf§”5 = 2.97 m/s. 6.27: a) The friction force is nkmg, which is directed against the car‘s motion, so the
net work done is —,ukmgs. The change in kinetic energy is AK = K1 = —(1/2)mv§,
and so 5 = 413/an9. b) Fromthe result of part (a), the stopping distance is pro
portional to the square of the. initial speed,.and so foran initial speed of 60 km/h,
3 = (91.2 In)(60.0/80.0)2 = 51.3 In. (This method avoids the intermediate calculation
of ,uk, which in this case is about 0.279.) 6.28: The intermediate calculation .of thespring'. constant may be avoided by using
Eq. (6.9) to see that the work is proportional to. the square of the extension; the work needed to compress the spring 4.00 cm is (12.0 J) = 21.3 J. 160 Chapter 6 6.29: a) The magnitude of the force is proportional to the magnitude of the extension
or compression; (160 N)(0.015 m/0.050 m) = 48 N, (160 N)(0.020 m/0.050 m) = 64 N. b) There are many equivalent ways to do the necessary algebra. One way is to note 1 1 0 N
that to stretch the spring the original 0.050 In requires 5 (0.050 1n)2 = 4 J, so that stretching 0.015 m requires (4 J)(0.015/0.050)2 = 0.360 J and compressing 0.020 111
requires (4 J )(0.020/ 0.050)”! = 0.64 J. Another is to ﬁnd the spring constant k = (160 N)+
(0.050 1n) = 3.20 X 103 N/m, from which (1/2)(3.20 x 103 N/m)(0.015 m)2 = 0.360 J and
(1/2)(3.20 x103 N/m)(0.020 m)2 = 0.64 J. 6.30: The work can be found by ﬁnding the area under the graph, being careful of the
sign of the force. The area under each triangle is 1/2 base x height. a) 1/2 (8 m)(10 N) = 40 J.
b) 1/2 (4 m)(10 N) = +20 J.
c) 1/2 (12 m)(10 N): 60 J.
6.31: Use the Work—Energy Theorem and the results of Problem 6.30. (2)(4O J)
= *,.—~.«uu—mmw~— = 2' I
a) v \/ 10 kg 83 m/s
b) At a; = 12 In, the 40 Joules of kinetic energy will have been increased by 20 J, so  (336331
11 —‘/ 10 kgw m 3.46 m/s. 6.32: The work you do with your changing force is 6.9 N 6.9 6.9
F(:r)d:1: =/ (—20.0N)d:r —] 3.0—wdx
o o 0 III = (—20.0N)x39  (3.0%)62/2) 39 = —138Nm— 71.4Nm = ——209.4J or —209J The work is negative because the cow continues to advance as you" vainly attempt to push her backward.
6.33: Wm == K2 — K1 K1 2 ém'ug, K2 = 0 Work is done by the spring force. WM 2 —%k:r2, where a: is the amount the spring
is compressed. 1 —§k:r2 =w~2«mv§ and :L' = 'vM/m/k = 8.5 cm Work and Kinetic Energy 161 6.34: a) The average force is (80.0 J) / (0.200 m) = 400 N, and the force needed to hold
. the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance
quadruples the work so an extra 240 J of work must be done. The maximum force is quadrupled, 1600 N. Both parts may of course be done by solving for the spring constant k = 2(80.0 J) +
(0.200 m)2 = 4.00 X 103 N / m, giving the same results. 6.35: a) The static friction force would need to be equal in magnitude to the spring force, psmg : kd or a, = tig£436mhﬁéﬁ§g§§7§j = 1.76, which is quite large. (Keeping extra
ﬁgures in the intermediate calculation for d gives a different answer.) b) In Example 6.6, the relation 1 1
,ukmgd + ikd2 = Emu? was obtained, and d was found in terms of the known initial speed '01. In this case, the
condition on d is that the static friction force at maximum extension just balances the spring force, or kd = aging. Solving for of and substituting, k
of = —d2 + 2gdakd m _ m92 — T (P: + 2ﬂsﬂk) = (nggygﬁglﬁﬁ) ((0.60)2 + 2(0.60)(0.47)), from which 01 = 0.67 m/s. 6.36: a) The spring is pushing on the block in its direction of motion, so the work is
positive, and equal to the work done in compressing the spring. From either Eq. (6.9) or
Eq. (6.10), W = %k:r2 = %(200 N/m)(0.025 m)2 = 0.06 J. b) The workenergy theorem gives _ Jam new m
1)... “. (4; 0.18 /S. 6.37: The work done in any interval is the area under the curve, easily calculated when
the areas are unions of triangles and rectangles. a) The area under the trapezoid is
4.0 Nm : 4.0 J. b) No force is applied in this interval, so the work done is zero.
c) The area of the triangle is 1.0 N m x 1.0 J, andsince the curve is below the axis
(Fa, < 0), the work is negative, or 1.0 J. (I) The net work is the sum of the results of
parts (a), (b) and (c), 3.0 J. (c) +1.0 J — 2.0 J = —1.0 J. 162 Chapter 6' 6.38: a) K = 4.0 J, so 1) = yams”; = \/2(4.0 J)/(2.0 kg) = 2.00 m/s. b) No work is
done between :r = 3.0 m and a: = 4.0 m, so the speed is the same, 2.00 m/s. 0) K m 3.0 J, so 1) = = = 1.73 m/s.
6.39: a) The spring does positive work on the sled and rider; (1 / 2)lcal:2 = (1 / 2)mv2, or v = = (0.375 m)\/(4000 N/m)/(70 kg) = 2.83 m/s. b) The net work done by
the spring is (1/2)]c — mg), so the ﬁnal speed is v = \/ = ¢i§ﬂ§lﬁ ((0.375 m)2 — (0.200 m)2) = 2.40 m/s. 6.40: a) From Eq. (6.14), with all = qub, P: 90
W = Fcosqbdl = 2wa cosqbdgﬁ = 2szin 90.
0 P1 In an equivalent geometric treatment, when T" is horizontal, d? = Fdx, and the
total work is F 2 210 times the horizontal distance, in this case (see Fig. 6.20(a)) Rsin 60, giving the same result. b) The ratio of the forces is ﬂﬁw = 2 cot 60. 210R sin 60 Sill 60 60
c) __.._.......,,......m...............m = M..........m...,. = _
wR(1— c0500) 2 (1 — 008%) 23% 2 ' 6.41: a) The initial and ﬁnal (at the maximum distance) kinetic energy is zero, so the
positive work done by the spring, (1 / 2)k:c2, must be the Opposite of the negative work
done by gravity, —mgL sin 6, or x _ \Fﬁgtsme _ x/EEHBQOOinkszXLS‘O‘Inlein 400° _ ........W........ WWW..” .mmmmmm WWMMMM mm. mm _ 5_7 cm_ It At this point the glider is no longer in contact with the spring. b) The intermediate
calculation of the initial compression can be avoided by considering that between the point 0.80 m from the launch to the maximum distance, gravity does a negative amount of work given by —(0.090() kg)(9.80 m/52)(1.80 m — 0.80 m) sin 400° = —0.567 J, and so
the kinetic energy of the glider at this point is 0.567 J. At this point the glider is no longer
in contact with the spring. 6.42: The initial and ﬁnal kinetic energies of the brick are both zero, so the net work done
on the brick by the spring and gravity is zero, so ( 1 / 2)kd2 «— mgh = 0, or d = x/2mgh/ =
@030 kg)(9.80 m/s2)(3.6 m)/(450 N/m) = 0.53 m. The spring will provide an upward
force while the spring and the brick are in contact. Whenthis force goes to zero, the spring
is at its uncompressed length. Work and Kinetic Energy 163 6.43: Energy = (p0wer)(time) = (100 W)(3600 s) = 3.6 x 105 J K: ém‘u2 sou: \/2K/m= 1005 form: 70 kg. 6.44: Set time to stop: 2F = ma :11.ka = ma
0. = ,ukg = (0.200)(9.80m/sz) = 1.96m/s2 ’U = ’00 + at
0 = 8.00 m/s — (1.96 m/sz)t
t = 4.088
p = _ $323
1(20.0 kg)(8.00 111/82)
—— 3...”...WTW =
— 4.08 S 157W 6.45: The total power is (165 N)(9.00 m/s) = 1.485 x 103 W, so the power per rider is
742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long periods). 19
6.46: a) ﬁgﬂgmﬂﬂ = 3.2 x1011 W. 3.2 x 1011 W
b WW. 3 . .
) 2.6 x 108 folks I 2 kW/pemon
11
c) WWWr = 8.0 x 108 m2 = 800 km2. (0.40)1.0 x 103 W/m2 6.47: The power is P = Fv. F is the weight, mg, so P = (700 kg)(9.8 III/S2) (2.5 m/s) =
17.15 kW. So, 17.15 kW/75 kW. = 0.23, or about 23% of the engine power is used in
climbing. 6.48: a) The number per minute would be the average power divided by the work (mgh)
required to lift one box, ‘ (Montage/13% _ 141 /s
(30 kg)(9.80 m/s2)(0.90 m) — ' ’
or 84.6 /min. b) Similarly:
MQQQmW = 0.378 /s, (30 kg)(9.80 m/s”)(0.90 m) or 22.7 /min. 164 Chapter 6 6.49: The total mass that can be raised is Sigi‘LEBJSZEEXVm/lﬁllﬁﬁfl m 2435 kg (9.80 m/s2)(20.0 In) so the maximum number of passengers is m = 28.
6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19), _ Wh _ (3800 N)(2.80 m) _ 3 _
P— t _ (4.003) "2.66x10 W—3.57hp. 651. F = (We... 2 ntzgisesggenzemp) = 8. x we N_ 'u (65 km/h)((1 km/h)/(3.6 m/Eji 6.52: Here, Eq. (6.19) is the most direct. Gravity is doing negative work, so the rope
must do positive work to lift the skiers. The force ? is gravity, and F = N mg, where N
is the number of skiers on the rope. The power is then P = (ng)(v) cosqﬁ
1 m/s = (50)(70 kg)(9.80 m/s2)( 12.0 km/h) (gmémﬁmﬁ) cos(90.0° — 15.0°) = 2.96 x 104 W. Note that Eq. (1.18) uses 05 as the angle between the force and velocity vectors; in this
case, the force is vertical, but the angle 15.0° is measured from the horizontal, so 9!) 2
900° — 15.0° is used. 6.53: a) In terms of the acceleration a and the time t since the force was. applied, the
speed is 'u = at and the force is me, so the power is P = Fv = (ma)(at) = mazt. b) The
power at a given time is proportional to the square of the acceleration, tripling the ac
celeration would mean increasing the power by a factor of nine. 0) If the magnitude of
the net force is the same, the acceleration will be the same, and the needed power is
proportional to the time. At t = 15.0 s, the needed power is three times that at 5.0 s, or
108 W. 6.54: (1K (1 1 2
is“ a a (5” )
= mi: @
dt
= mm = mav
= F?) = P 6.55: Work done in each stroke is W = F3 and PM. = W/t = IOOFS/t
t= 1.00 s, F = 2mg and s = 0.010 m. P... = 0.20 W. Work and Kinetic Energy 165 6.56:
dm )
M, mm“. ““““ P _‘/" “W “* “ﬁrm”; L KE: ~;(dm)v2
M
dmzfdx
V213
_ T
L1 M 2m: 2
1 M) (Ml/L 2
=— —— Mm— wdm
2(AI/l 4T22 0153 2
_,1 7T __ 2 2 2
~2(L)(T2)(3)*3WML/T 5 revolutions in 3 seconds —> T = 3/ 5 8 KB = §w2(12.0kg)(2.00 m)2/(3/58)2 = 877.1 6.57: a) (140 N)(3.80 m) = 532 J b) (20.0 kg)(9.80 m/82)(3.80 m)(— sin 25°) = —315 J
c) The normal force does no work. d) Wf 2': _fk3 = ﬂukns = —,u.kmgs (3089 = —(0.30)(20.0 kg)(9.80 m/s2)(3.80 m) cos 25° = —203 J e) 532 J — 315 J — 203 J = 15 J (14.7 J to three ﬁgures).
f) The result of part (6) is the kinetic energy at the top of the ramp, so the speed is v = ME = = 1.21 m/s. 6.58: The work per unit mass is = gh.
a) The man does work, (9.8 N/kg)(0.4 m) = 3.92 J /kg.
b) (3.92 J/kg)/(70 J/kg) x 100 = 5.6%. c) The child does work, (9.8 N/kg)(0.2 m) = 1.96 'J/kg. (1.96 J/kg)/(70 J/kg) x 100 =
2.8%. d) If both the man and the child can. do. work at the rate of 70 J /kg, and if the child
only needs to use 1.96 J /kg instead of 3.92 J /kg, the child should be able to do more pull
ups. 166 Chapter 6 6.59: a) Moving a distance L along the ramp, 5“, = L, sout = L sin a, so I M A = min. b) If AMA = IMA, (Font/Fin) = (sin/sent) and so (Font)(sout) — (Em)(sin), or
want = in 0)
d) E z lilies}. ,__ 1F sutllmsmeeil = Essiﬁs 2 £942
I/Vin (Finxsin) Sin/Soul; w — /s (7.35 ><103 J)
6. : = — = = —»~w ‘WW« 2 41.7 k .
60 a) m g g (9.80 m/s2)(18.0 m) g
Wn 8.25 X 103 J
b) n — ‘5‘ m 18.63"” ‘ 458 N
e) The weight is my = % = 408 N, so the acceleration is the net force divided by
the ass gay—408 me — 1 2 m/s2
m ’ 41.7 kg m“ ‘ '
6.61: a) 1 2 1 21rR 2 1 27r(6.66 x 106 In) 2 12
_ = _ W E __ 4 WW”.W..W.WWWW = 2. 1 _
27m} 2m ( T ) 2(86’ 00 kg) ((90.1 min)(60 s/min) 59 x 0 J b) (1/2)mv2 = (1/2)(86,400 kg) ((1.00 m)/(3.00 3))2 = 4.80 x 103 J. 6.62: 3) WI = .__fk3 = —p,kmg (:0st
: (0.31)(5.00 kg)(9.80 m/s2) cos 12.0°(1.50 m) = w22.3 J (keeping an extra ﬁgure) b) (5.00 kg)(9.80 m/sz)sin12.0°(1.50 m) = 15.3 J. c) The
normal force does no work. (:1) 15.3 J — 22.3 J = —7.0 J. e) K2 = K1 + W = (1/2)(5.00 kg)(2.2 m/s)2 — 7.0 J = 5.1 J, and so 112 = = 1.4 m/s.
6.63: See Problem 6.62: The work done is negative, and is proportional to the distance
3 that the package slides along the ramp, W = mg(sin 6 —— ,uk cos 9)3. Setting this equal to
the (negative) change. in kinetic energy and solving for 3 gives (1/2)m'vi vi mg(sin 0 — ,uk cos 6)  M?g(sin6 — pk Work and Kinetic Energy 167 (2.2 111/ s)2
M5130§5W3im53 — W As a check of the result of Problem 6.62, (2.2 m/s)\/1: (1.57m)/(2.6 In) 2 1.4 m/s.
6.64: a) From Eq. (6.7), The force is given to be attractive, so Fm < 0, and It must be positive. If :02 > m1, f; < i,
and W < 0. b) Taking “slowly” to be constant speed, the net force on the object is zero,
so the force applied by the hand is opposite F2, and the work done is negative of that found
in part (a), or k i w A , which is positive if 2:; > :31. c) The answers have the same 31 532 magnitude but opposite signs; this is to be expected, in that the net work done is zero. 6.65: F = mg(RE/1r)‘z
2 RE 2
W n —f Fds = —/ dr = —ngE(—(1/r):E) = ngE
1 00 = 2.6 m. Wtot = K2 —K1,K1 = 0
This gives K2 = ngE = 1.25 x 1012 J K2 = §mv§ so '02 = ﬁKg/“Et = 11,000 m/s 6.66: Let x be the distance past P. Mk = 0.100 + A33 when x = 12.5 m,,uk = 0.600
A : 0.500/12.5 m = 0.0400/m (a)
W=AKE2Wf=KEf—KEi
— / ,ukmgda: = 0 — émv?
g/ (0.100 + Am)da: = 50?
0 9 [(0.100)z:f + flag—g] = (9.80 We) [(0.100):..—f + (0.0400/m):—?] = $450 III/5);, Solve for m : xf = 5.11m
(b) m. = 0.100 + (0.0400/m)(5.11 m) = 0.304 168 Chapter 6‘ (C) Wf = K13f — KEi
—p.kmg:1: = 0 — émv?
2
2: = reg/2111,51 = mm = 10.3m 2(0.100)(9.80 111/52) 6.67: a) amﬁ = (4.00 N/m3)(1.00 1n)3 = 4.00 N.
b) 052:3 = (4.00 N/m3)(2.00 m)3 = 32.0 N. c) Equation 6.7 gives the work needed to
move an object against the force; the work done by the force is the negative of this, “fan amsda: = mg!— (374 —x“)
1131 4 2 ' With 331 = 33,, = 1.00 m and :02 = 22;, = 2.00 In, W = —15.0 J, this work is negative.
6.68: From Eq. (6.7),with 2:1 = 0, $2 $2 2 3 k 2 b 3 C 4
W: de= (kw—bx +cx)d:r=—x ——a:2+—a:2
0 0 2 3 4
= (50.0 N/m);tg — (233 N/In2):1:g + (3000 N/m3)xg. a) When 11:2 = 0.050 In, W = 0.115 J, or 0.12 J to two ﬁgures. b) When :02 = —0.050 In,
W = 0.173 J, or 0.17 J to two ﬁgures. 0) It’s easier to stretch the spring; the quadratic
—b;r2 term is always in the —mdirection, and so the needed force, and hence the needed
work, will be less when x2 > O. 6.69: a) T = mam = m% = (0.120 kg)§9§9ﬂ[§)j = 0.147 N, or 0.15 N to two ﬁgures. (0.40 m)
b) At the later radius and speed, the tension is (0.120 kg) 9153343» = 9.41 N, or 9.4 N to two ﬁgures. 0) The surface is frictionless and horizontal, so the net work is the work
done by the cord. For a massless and frictionless cord, this is the same as the work
done by the person, and is equal to the change in the block’s kinetic energy, K2 —— K1 =
(1/2)m(v§ — of) = (1/2)(0.120 kg) ((2.80 III/s)2 — (0.70 m/s)2) = 0.441 J. Note that in
this case, the tension cannot be perpendicular to the block’s velocity at all times; the cord
is in the radial direction, and for the radius to change, the block must have some nonzero
component of velocity in the radial direction. 6.70: a) This is similar to Problem 6.64, but here a > 0 (the force is repulsive), and
3:; < 031, so the work done is again negative; W = (— m me) = (2.12 x10"26 Nm2((0.200.m'1) — (1.25 x 109 1111))
= —2.65 x 10—17 J. Work and Kinetic Energy 169 L
3:1 Note that 1:1 is so large compared to 3:2 that the term is negligible. Then, using Eq. (6.13)) and solving for v2, 2W W‘Wﬂmwéwffzfss XTF" J) = 2 ._ m 5 2 ._.W.WW = _ 5 I
'02 v1 + m \/(:3.00 x10 m/s) + (1.67 x10_27 kg) 2 41 x 10 m/s b) With K2 = 0, W = —K1_ Using W = .._._9L —26 I 2
$2 a __ 3.3m I 232 X 1010 m. = "IE “ mof = (1.67 x 1037” kg)(3.00 ><105 m/s.)2
c) The repulsive force has done no net work, so the kinetic energy and hence the speed of
the proton have their original values, and the speed is 3.00 x 105 m/s. 6.71: The velocity and acceleration as functions of time are
da: 2
v(t) = d—t = 2at + 3/315 , a.(t) = 20 + 6,6t
a) v(t = 4.00 s) = 2(0.20 m/s2)(4.00 s) + 3(0.02 m/s3)(4.00 s)2 = 2.56 m/s.
b) ma = (6.00 kg)(2(0.20 111/32) + 6(0.02 m/s3)(4.00 s) = 5.28 N.
c) W = K2 — K1 = K2 = (1/2)(6.00 kg)(2.56 m/s)2 = 19.7 J.
6.72: In Eq. (6.14), dl = (12: and d) m: 31.0" is constant, and so
92 1:2
W2] Fcosqﬁdl=f Fcosqbdw
P1 3: 1
1.50 m = (5.00 N/mz) cos31.0°f 1‘2 dx = 3.39 J. 1.00 m The ﬁnal speed of the object is then o2 = of + 271:: = “(4.00 m/s)2 + =2 6.57 m/s.
6.73: 3.) K2 — K1 = (1/2)m (v3 — of)
= (1/2)(80.0 kg)((1.50 m/s)2 — (5.00 m/s)2) = —910 J. b) The work done by gravity is —mgh = —(80.0 kg)(9.80 m/s2)(5.20 m) =
—4.08 x 103 J, so the work done by the rider is —910 J  (—4.08 X 103 J) = 3.17 X 103 J.
°° b b °° b
. W = — d = w . = m .
674 a) m at" 9’ (—(n ——1)):r“‘1 .. (n— 1):]:3‘1 170 Chapter 6‘ Note that for this part, for n > 1, :51‘" —> 0 as x —> 00. b) When 0 < n < 1, the
improper integral must be used, — ' ﬂ  1 _ n — 1
W — legnoo (n — 1) ($2 $0 ) ’ and because the exponent on the 1:"; — 1 is positive, the limit does not exist, and the integral diverges. This is interpreted as the force F doing an inﬁnite amount of work, even though F —> 0 as :52 ——> oo.
6.75: Setting the (negative) work done by the spring to the needed (negative) change
in kinetic energy, %km2 2 %mvg, and solving for the spring constant, __ meg (1200 kg)(0.65 rn/s)2 k _ mm = wmWWWWWW. = 1.03 105 N .
$2 (0.070 m)2 X /m 6.76: a) Equatng the work done by the spring to the gain in kinetic energy, %kx3 = 1 2
k 400 N/m
=‘/—— = w .000 = .93 .
“ mm“ \/0.0300 kg(0 m) 6 “1/3 amp , so
b) Wm must now include friction, so imp2 = Wtot = %k:r3 — fro, where f is the magnitude
of the friction force. Then, _ _ 2 _ _
v . mxo m 0:0 = WWWW . w , = 4. _ \/0.0300 kg(0 06 In) (0.0300 kg)(0 06 m) 90 m/s 0) The greatest speed occurs when the acceleration (and the net force) are zero, or k1: = f, :c = k = 3%,?wa = 0.0150 111. To ﬁnd the speed, the net work is Wm = %k ($3 — :62) —~ f (:60 — 1:), so the maximum speed is mm = (/5 e3  x2) —' a (mo — as) m m
 2(000 N)
 (/(00300 kg) ((0060 m) (00150 m) ) (00300 kg) (0.060 in 0.0150 m)
= 5.20 m/s, which is larger than the result of part (b) but smaller than the result of part (a). 6.77: Denote the initial compression of the spring by x and the distance from the ini
tial position by L. Then, the work done by the spring 'is %ka:2 and the work done by
friction is makmgw + L); this form takes into account the fact. that while the spring is
compressed, the frictional force is still present (see Problem 6.76). The initial and ﬁnal Work and Kinetic Energy 171 kinetic energies are both zero, so the net work done is zero, and yen? = pkmgﬁr + L).
Solving for L, _ (1/3196 1 _ HEREQWQEBQEX _ _ m
L _ MERE: m _ (0.30)(2.50 kg)(9.80 m/S2) ' (0250 m) _ 0‘813 ’ or 0.81 m to two ﬁgures. Thus the book moves .81 m + .25 m = 1.06 m, or about 1.1 m. 6.78: The work done by gravity is Wg = —mgL sinf) (negative since the cat is moving
up), and the work done by the applied force is FL, where F is the magnitude of the
applied force. The total work is Wtot = (100 N)(2.00 m) — (7.00 kg)(9.80 m/s2)(2.00 m)sin30° = 131.4 J. The cat’s initial kinetic energy is §mvf : %(7.00 kg)(2.40 m/s)2 = 20.2 J, and _ ._ @EEEFEEE _
112 —\/ — \/ (7’00   H 6.58 m/s. 6.79: In terms of the bumper compression :1: and the initial speed 110, the necessary
relations are 1 1
Ekxz z Emvg, k3: < 5mg. Combining to eliminate it and then 3:, the two inequalties are 2 2
x > v— and k < 253%”.
59 v 8.) Using the given numbers, (20.0 m/s)2 _
m > 5(9.80 111/32) " 8'16 m’
(1700 kg)(9.80 m/S2)2 2 WWWMWW =1. 4 .
k < 5 (20.0 III/SP 02x10 N/m b) A distance of 8 m is not commonly available as space in which to stop a car. 6.80: The students do positive work, and the force that they exert makes an angle of
300° with the direction of motion. Gravity does negative work, and is at an angle of
600" with the Chair’s motion, so the total work done is Wtot = ((600 N) cos 300° —
(85.0 kg)(9.80 m/s2)cos60.0°)(2.50 m) = 257.8 J, and so the speed at the top of the
ramp IS 172 Chapter 6‘ Note that extra ﬁgures were kept in the intermediate calculation to avoid roundof‘rr error. 6.81: a) At maximum compression, the spring (and hence the block) is not moving, so the block has no kinetic energy. Therefore, the work done by the block is equal to its initial kinetic energy, and the maximum compression is found from $ch2 : émvz, or . k
X = “331; = “Egan546.00 m/s) = 0.600 m. b) Solving for v in terms of a known X, H: 500 N/m 6.82: The total work done is the sum of that done by gravity (on the hanging block) and
that done by friction (on the block on the table). The work done by gravity is (6.00 kg)gh
and the work done by friction is —p,k(8.00 kg) gh, so Wm, = (6.00 kg — (0.25)(8.00 kg)(9.80 m/s2)(1.50 m) = 58.8 J. This work increases the kinetic energy of both blocks; 1
Wtot = §(m1 + 7712)!)2, SO 6.83: See Problem 6.82. Gravity does positive work, while friction does negative work.
Setting the net (negative) work equal to the (negative) change in kinetic energy, 1
(m1 — #km2)9h = —§(m1 + m2)v2, and solving for pk gives .mamWWWW ‘ ume.‘WWWHm _ (6.00 kg) + (1/2)(14.00 kg)(0.900 m/s)2/((9.§gm/s2)(2.00 m)) = 0.79. 6.84: The arrow will acquire the energy that was used in drawing the bow (z'.e., the
work done by the archer), which wilI be the area under the curve that represents the force
as a function of distance. One possible way of estimating this work is to approximate the
F 03. :1: curve as a parabola which goes to zero at .13.: 0 and m = 3:0, and has a maximum Work and Kinetic Energy 173 of F0 at x = $29, so that F(:r) = 4—:ng$($g w This may seem like a crude approximation
to the ﬁgure, but it has the ultimate advantage of being easy to integrate; 1” 4F "’0 4F 2 3 2
/ Fdw = «39/ ($013 — 32) do: = (a:qu — Eg) = —F0x0.
0 $0 0 530 With F0 = 200 N and 11:0 2 0.75 In, W = 100 J. The speed of the arrow is then % «m = 89 m/s. Other ways of ﬁnding the area under the curve in Fig. (628) should give similar results. 6.85: fk = 0.251019 so Wf = WM = —(0.25mg)s, where .3 is the length of the rough
patch. Wtot = K2 — K1
K1 = gmvg, K2 = %mv§ = %m(0.45v§) = 0.2025(%mv3) The workenergy relation gives (0.25mg)s = (0.2025 — 1)%mv3 The mass divides out and solving gives 3 = 1.5 m. 6.86: Your friend’s average acceleration is o — on 6.00 m/s 2
= = .WWW, = 2.00
a t 3.005 m/S
Since there are no other horizontal forces acting, the force you exert on her is given by
Fnet = me = (65.0 kg)(2.00 III/82) = 130N Her average velocity during your pull is 3.00 m/s, and the distance she travels is thus
9.00 m. The work you do is Fa: = (130 N)(9.00 m) = 1170 J, and the average power is
therefore 1170 J / 3.005 = 390 W. The work can also be calculated as the change in the kinetic energy.
6.87: a) (800 kg)(9.80 m/s2)(14.0 m) = 1.098 x 105 J, or 1.10 x 105 J to three ﬁgures. b) (1/2)(800 kg)(18.0 m/s)2 2 1.30 x 105 J.
5 5
C) L10><10 J+1.30 £10th=399 kW.
60 s
6.88: P = Fv = moo
= m(2a + 6ﬁt)(2at + 3ﬁt2) = m(4a2t + 180m? + 1862.133)
= (0.96 N/s)t + (0.43 N/sz)t2 + (0.043 N/s3)t3. At t = 4.00 s, the power output is 13.5 W. 6.89: Let t equal the number of seconds she walksevery day. Then, (280 J /s)t +
(100 J/s)(86400 s — t) = 1.1 ><107 J. Solvingfor t, t = 13,111 3 = 3.6 hours. 174 Chapter 6 6.90: a) The hummingbird produces energy at a rate of 0.7 J /s to 1.75 3/3. At
10 beats/s, the bird must expend between 0.07 J / beat and 0.175 J /beat. b) The steady output of the athlete is 500 W/ 70 kg = 7 W/kg, which is below the
10 W/ kg neceSSary to stay aloft. Though the athlete can expend 1400 W/70 kg = 20 W/ kg
for short periods of time, no humanpowered aircraft could stay aloft for very long. Movies
of early attempts at humanpowered ﬂight bear out this observation. 6.91: From the chain rule, P = %W = %(mgh) = EELS—git, for ideal eﬁciency. Expressing
the mass rate in terms of the volume rate and solving gives
(2000 x 106 W) m3
. . . . ~ ~ = 1.30 103 —.
(0.92)(9.80 m/s2)(17o m)(1000 kg/m ) X s 6.92: a) The power P is related to the speed by Pt = K = émv21 so 1) = a. b) a dv_d 35:: sewn/eggs L. = E _ E m m dt zﬂ 2mt
W 2P We 0) :c—zr0=/vdt= Ejdﬁdt: Wgtgr— it?
m m 3 9m 6.93: a) (7500 x 10’3 kg3)(1.05 X 103 kg/m3)(9.80 m/s2)(1.63 m) = 1.26 X 105 J.
b) (1.26 x105 J)/(86,400 s)=1.46 W.
6.94: a) The number of cars is the total power available divided by the power needed per car, 13.4 x 106 W = 177, (2.8 x 103 N)(27 m/s) rounding down to the nearest integer. b) To accelerate a total mass M at an acceleration a and speed '0, the extra power
needed is M cw. To climb a hill of angle a, the extra power needed is M g sinav. These
will be nearly the same if a N gsin a; if gsin (1 ~ g tana ~ 0.10 m/sz, the power is about
the same as that needed to accelerate at 0.10 m/s2. c) (1.10 X 106 kg)(9.80 m/s2)(0.010)(27 m/s) = 2.9 MW. d) The power per car
needed is that used in part (a), plus that found in part (c) with M being the mass of a
single car. The total number of cars is then 2'9 X 106 . .. .. _ 35
(2.8 x 103 N + (8.2 x 104 kg)(9.80 m/s2)(0.010))'("27.m/s) ‘ ’ rounding to the nearest integer.
6.95: a) P0 = Fv = (53 x 103 N)(45 m/s) = 2.4.MW.
b) P1 = me?) = (9.1 x 105 kg)(1.5 m/sz)(45 m/s) = 61 MW. I Work and Kinetic Energy 175 c) Approximating sin a by tan a, and using the component of gravity down the incline as my sin a, P2 = (mg sin adv = (9.1 x 105 kg)(9.80 m/s2)(0.015)(45 m/s) = 6.0 MW.
6.96: 8.) Along this path, 3,; is constant, and the displacement is parallel to the forca, so
W = ayfzdx = (2.50 N/m2)(3.00 m)§3%9924n): = 15.0 J.
b) Since the force has no ycomponent, no work is done moving in the ydirection.
c) Along this path, 3; varies with position along the path, given by y 2 1.53m, so
E, = a(1.5m):v = 1.504022, and 2. 3
W = ij d3 = 1.50:]:1:2 dz: = 1.5(2.50 N/m2)£«w%mww = 10.0 J. 6.97: a) P = F0 = (an + Fair)”
= ((0.0045)(62.0 kg)(9.80 m/s”)
+ (1/2)(1.00)(0.463 m2)(1.2 kg/m3)(12.0 m/s)2)(12.0 m/s)
= 513 W. b) ((0.0030)(59.0 kg)(9.80 111/52)
+ (1/2)(0.88)(0.366 m2)(1.2 kg/m3)(12.0 m/s)2)(12.0 m/s)
= 355 w. c) ((0.0030)(59.0 kg)(9.80 m/s2)
+ (1/2)(0.88)(0.366 m2)(1.2 kg/In3)(6.0 m/s)2)(6.0 m/s)
= 52 w. 3
P = 1.63 x 103 N_ E = (60.0 lkm/h)((1 m/s)/(3.8 km/h)) b) The speed is lowered by a factor of onehalf, and the resisting force is lowered by a
factor of (0.65 + 0.35/4), and so the power at the lower speed is 6.98: a) F (28.0 kW)(0.50)(0.65 + 0.35/4) = 10.3 kW = 13.8 hp.
0) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65 + 0.35 x 4) = 114.8 kW = 154 hp. 6.99: a (800 hP)(745 W/hP) _
) t‘aﬁfa‘mﬁnfm7imm0  358 N b) The extra power needed is 60.0 km/h .
35mm sm(arctan(1/10)) _ 29.3 kW _ 39.2 hp, m/s mgv" = (1800 kg)(9.80 111/82) 1 76 Chapter 6 so the total power is 47.2 hp. (Note: If the sine of the angle is approximated by the tangent,
the third place will be different.) c) Similarly, 60.0 km/h
3.6 35% m/s mgr)" = (1800 kg)(9.80 m/sz) sin(arctan(0.010)) = 2.94 kW 2 3.94 hp. This is the rate at which work is done on the car by gravity. The engine must do work on
the car at a rate of 4.06 hp. cl) In this case, approximating the sine of the slope by the
tangent is appropriate, and the grade is u (8.00 hp)(746 W/hp) mm m 0 0203
(1300 kg)(9.80 m/s2)(6o.0 km/h)((1 m/s)/(3.6 km/h)) ' ’ very close to a 2% grade.
6.100: Use the Werk—Energy Theorem, W = AK E, and integrate to ﬁnd the work. 1 a:
AKE = 0 — Emu}; and W =/ (—mg sin a — ,urng cos a)da:.
0
Then,
‘” . . A2:2
W = —mg (sma + As: cos a)d:c, W = —mg am as: + Wéw cos a .
0
Set W = AKE.
1 2 . A232
——2mv0 = —mg[sm as: l T cosa] . To eliminate 2:, note that the box comes to a rest when the force of static friction balances
the component of the weight directed down the plane. So, mg sina = Am mg cos a; solve
this for cc and substitute into the previous equation. x_ sina
m A meal.
Then,
sina 2
. A m
lvz " + sine: sm aw ~+~ WW<A 605a) cosa
2 0 g A0080 2 ' , . 3 sinza , _
and upon canceling factors and collectlng terms, «)3 =  v. 01' the box W111 remain
osa
3g sin2 a
stationar whenever 112 > y 0 _ A cosa 6.101: a) Denote the position of a piece of the spring by l; l = 0 is the ﬁxed point and
l = L is the moving end of the spring. Then the velocity of the point corresponding to l, Work and Kinetic Enemy 177 denoted u, is u(l) = "0% (when the spring is moving, 1 will be a. function of time, and so u
.. is an implicit function of time). The mass of a piece of length oil is dm = % di, and so 1 2 11141222 if: . . and My? L2 M112
K—de— 2L3/0 z (ii—WEN b) ;—k:t2 = §mu2, so 1) = i/(k/m) = «(3200 N']i”ﬁj”7(626§“3“il§)"(2.50 x 102 m) = 6.1 m/s.
0) With the mass of the spring included, the work that the spring does goes into the kinetic
energies of both the ball and the spring, so ﬂea:2 = §mv2 + %Mv2. Solving for v, v = when!“ 3; m \/WW§§393§1§9MW(2.50 x 102 m) = 3.9 m/s.
m + M/3 (0.053 kg) + (0.243 kg)/3 d) Algebraically, _ Moments“ 3 ' d
2m'u (1+M/3 ) 040J an
1 (1/2)k:1:2
_ M 2 WWWWWW m
6 v (1 + 3m/M) 0 60 J 6.102: In both cases, a given amount of fuel represents a. given amount of work W0 that
the engine does in moving the plane forward against the resisting force. In terms of the range R and the (presumed) constant speed '0, W0=RF=R(G'U2+£). 112
.In terms of the time of ﬂight T, R = vT, so Wo=vTF=T(av3+%). 3.) Rather than solve for R as a function of v, differentiate the ﬁrst of these relations with
" respect to 1), setting %9 m 0 to obtain %F + 13% = 0. For the maximum range, dag—2 = D, so % = 0. Performing the differentiation, % = 20w — 2,6/v3 = 0, which is solved for i a 1/4 _ 3.5 x 106 Nom2/s2 1’4 _ _ 178 Chapter 6 b) Similarly, the maximum time is found by setting %(Fv) = 0; performing the
differentiation, 3cm2 — {3/02 = 0, which is solved for _ 6 1/4_ 3.5x 105 N~m2/S2 1/4_ _
v — (3a — 3(030 N'S2/m2) ~ 25 m/s — 90 km/h. 6.103: a) The walk will take oneﬁfth of an hour, 12 min. From the graph, the oxygen
consumption rate appears to be about 12 cm3/kgmin, and so the total energy is (12 em3/kgmin)(70 kg)(12 min)(20 J/cma) = 2.0 x 105 J. b) The run will take 6 min. Using an estimation of the rate from the graph of about
33 cma/kgmin gives an energy consumption of about 2.8 X 105 J. c) The run takes
4 min, and with an estimated rate of about 50 cm3/kgmin, the energy used is about
2.8 x 105 J. d) Walking is the most efﬁcient way to go. In general, the point where the
slope of the line from the origin to the point on the graph is the smallest is the most efﬁcient speed; about 5 km/ h. 6.104: From ?=m?, E, = max, Fy = may and Fz = maz. The generalization of
Eq. (6.11) is then dvx dvy d'uz
an, = 123—, ay = v,,—— (1,, = v; dz The total work is then (22.y2.22)
Wtot=f ( £1.91. 21) ":2 dv 5'2 d'v 2’ dv
=m vx‘m5d3+/ 'v de+/ 112—de)
(1;, d3: m y dy 2, dz “:2 “1,12 1’22
m / 1130111,, + f “0,, 011),, + f 1),, deg
1’31 vyl ":1 1
_ 2 2 2 2 2 2
— Em (1&2 _ 2:1 + “1,2 _ vyl + “:2  “21)
1 1
2 2
= —mv  —mv .
2 2 2 1 ...
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This note was uploaded on 04/11/2009 for the course ENG 100 taught by Professor Delwiche during the Spring '08 term at UC Davis.
 Spring '08
 Delwiche

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