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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 + ...
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 Spring '08
 Delwiche
 Energy, Force, Friction, Kinetic Energy, Work

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