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Unformatted text preview: Work 1 123 K3+Ug3+Us3+AEm =K2+U82+U52+Wm 01+0J+%k(x3—x2)2+0J=%mv§+01+OJ+OJ Using 122 = 9.39 m/s, k = 500 Mm and m = 5.0 kg,’ the above equation yields (x3 — x2) = Ax = 93.9 cm. ((1) The initial energy: mgy0 = (5.0 kg)(9.8 m/s2)(5.0 m) = 245 J. The energy transformed to thermal energy
during each passage is fk(x2 — x,) = ,ukmgoc2 —x,) = (0.25)(5.0 kg)(9.8 m/s2 )(2.0 In) = 24.5 J
The number of passages is equal to 245 J / 24.5 J or 10. 11.54. Model: Assume an ideal spring, so Hooke’s law is obeyed. Treat the physics student as a particle and apply the law of conservation of energy. Our system is comprised of the spring, the student, and the ground. We
also use the model of kinetic friction. Visualize: We place the origin of the coordinate system on the ground directly below the end of the compressed
spring that is in contact with the student. Known
m x0=0m v0=0mls
xl—x0=0.50m k=80,000N/m
y0=yl=10m m=100kg
uk=0.15 v2=0mls Find a
V1 As = yzlsm 30° . Solve: (a) The energy conservation equation is l —2—mv,2 + mgyl +%k(xl —xe)2 +0 I = émvﬁ +mgy0 +%k(xI xo)2 +0 J =y0 = 10 m, x, =xe, v0 = 0 m/s, k = 80,000 N/m, m = 100 kg, and (x, ~xo) =05 m,
I 2 1 2 k
2—mvl = 5kg] —x0) => v1: 71:0!!! —x0) = 14.14 m/s v, = 0 m/s and y2 = (As)sin30", the above equation is simpliﬁed to
mg(As)sin30° + 111an = mgyo + gm, — xn )2 (15111 the free—body diagram for the physics student, we see that n = wcos30°. Thus, the conservation of energy As(mgsin30° + pkmgcos30°) = mgyo + ékﬂ:l — x0 )2
= 100 kg, k = 80,000 N/m, (x, — x0) = 0.50 m, yo = 10 m, and ,uk = 0.15, we get I
mgyo + _k(xi " x0 )2
As ‘ 2 __ H. = 32.1 m
mg(sm30° + Mk cos30°) ll—24 Chapter 11 Assess: y2 = (As)sin30° = 16.05 m, which is greater than y0 = 10 m. The higher value is due to the transforma—
tion of the spring energy into gravitational potential energy. 11.55. Model: Treat the block as a particle, use the model of kinetic friction, and apply the energy conservation law. The block and the incline comprise our system.
Visualize: We place the origin of the coordinate system directly below the block’s starting position at the same level as the horizontal surface. On the horizontal surface the model of kinetic friction applies. Known x0=0m X2—x1=L
m=h n=h=0m
v0=v3=0mls I‘k Find
V1 3’3 Solve: (a) For the ﬁrst incline, the conservation of energy equation gives
Kl +Ugl +AEm =K0 +Ug0 +Wm :—m1/l2+0J+OJ=OJ+mgyo +0 J: v1=1l2gyo =1’23h (b) The work of friction creates thermal energy. Applying once again the conservation of energy equation, we have . l 1
K3 + Ug3 +AE1h = K1+ U3l + Wm Emvf +mgy3 +j;(xz —x,) = Emv? +mgyl +Wm Using v3 = 0 m/s, yI = 0 m, Wm = 0 L}; =ykmg, v1 = «'12 h, and (x2 x,) '= L,‘ we get
1
mgy3 + /.tk‘mgL = —2m(2gh) = y3 = h — ukL
Assess: For pk = 0, y3 = h which is predicted by the law of the conservation of energy.
11.56. Model: Assume an ideal spring, so Hooke’s law is obeyed.
Visualize: ( WWWW Solve: For a conservative force the work done on a particle as it moves from an initial to a ﬁnal positiOn is
independent of the path. We will show that WA_,C_m = A4,, for the spring force. Work done by a spring force F = 40: is given by
W = dex = —f kxdx
This means
x k 2 2 k 2 2 x k 2 2
WM, = — kxdx = —§(x3 — xA), WMC = — kxdx = —E(xc — xA), and W93 =  kxdx = ——2—(x,, ~xc)
xA «VA Xc
Adding the last two: 1 1—30 Chapter 11 From the horizontal motion, 1
x2 = x1+ v1,‘(t2 —1f,)+5ax(t2 —t,)2 5.0 In =>(x1 +5.0 m) = x1 + v1(0.639 s) +0 in => 1),: 0 639 = 7.825 m/s
. s Having found the v1 that will take the skier to the other side of the tank, we now use the energy equation to ﬁnd the minimum speed v0. We have K1+Um = Ko +Ugo => é—mvf +mgyl =§mv§ +mgy0 v0 = 1M + 2g(yl — yo) = ‘[(7.825 m/s)2 + 2(9.8 m/sz )(2.0 m) = 10.0 m/s 11.73. Model: Assume the spring to be ideal that obeys Hooke’s law, model the block as a particlehand apply the model of kinetic friction. Visualize: We placed the origin of the coordinate system on the free end of the compressed spring which is in
contact with the block. Because the horizontal surface at the bottom of the ramp is frictionless, the spring energy
appears as kinetic energy of the block until the block begins to climb up the incline. .. 960’]: V1 xo’ Yovo x6 of the block as it leaves the spring, we don’t need to. We can use energy
y of the spring to the energy of the block as it begins projectile motion
ate the increase in thermal energy. The energy equation is Solve: Although we could find the speed v, conservation to relate the initial potential energ
at point 2. However, friction requires us to calcul K2 + U1;2 + AE‘,l = K0 + U + Wext => gmv22+ mgy2 + kas = sik(xo — are)2 The distance along the slope is As = yzlsin45°. The friction force is ﬂ = [1.31, and we can see from the freebody diagram that n = mgcos45°. Thus V2 1/2
[If(2:0 — re)2 — 23y2 — zukgyz cos 45°/ sin45°]
m
1/2 10:25:? (0.15 m)2 — 2(9.8 m/s2 )(2.0 m) — 2(0.20)(9.8 m/s2 )(2.o m)cos 45°/ sin45° = 8.091 m/s Having found the velocity v2, we can now find (x3 II — x2) = d using the kinematic equations of projectile motion:
y3 = y2 + v2},(t3 — t2)+::azy(t3 —— t2)2
2.0 m = 2.0 m + (v2 sim45°)(t3  t2) + %(—9.8 m/s2)(t3 — t2)2
==> (t3 —t2) = 0 s and 1.168 s Finally;
1
x3 = x2 + v2x(t3 —t2)+2—a2x(t3 t2)2 d = (x3 — x2) = (v2 cos45°)(1.l68 s) + 0 m = 6.68 m *— Work 1 131 11.74. Model: Assume
of kinetic friction. The ball,
Visualize: i" an ideal spring that obeys Hooke’s law, the particle model for the ball, and the model the spring, and the barrel comprise our system. K,+Us,+AEm =K0+Uso+Wm l l l
Emv,2 +§k(xl ~xo)2 +(fk)(xl —x0) = 5m; +§k(xe xe)2 + F(Jcl ~xo) $0.0 kg)(2.0 m/s)2 + %(3000 N / m)(0.3 In)z + ,uk The friction force is j: = ,ukmg. With /.tk = 0.30 and m = 1.0 kg, we can solve this equation to obtain F = 460 N.
(b) Using the energy equation for our system once again, we have (mg)(0.30 m) = 0 J + O J + F (0.30 m) K2 +Us2 +AEuI = K1+Usl+ Wm f é—mvzz +Ek(xe ~xe)2 +(f;{)(x2 —xl) =émv,2 +:—k(x1 —Jc0)2 +0 I gm; + ,ukmg(x2 —x,) = 0 J+%k(x, —x0)2 %(1 .0 kg)»; + (0.30)(1.0 kg)(9.8 m/s2 )(15 m) = $6000 N/m)(0.30 m)2 (0.5 kg)v22 +4.41 J = 135 J => v2 = 16.2 m/s ‘.75. Solve: (a) U(r)
0 1 1—32 Chapter 11 (c) Due to the absence of nonconservative forces in our system of two particles, the mechanical energy is conserved.
v  = 0
11 v2; = 0
Before .
ri=1.0>< 1014 m
m,=8.0x1030kg m2=2.0x103°kg
R,=11.0x108m R2=7.0x103m
Aﬁer rf=R1+R2= 18.0x103 m The equations of energy and momentum conservation are 1 1 Gm m 1 l Gm m
K+U =K+U —mv2+——mv2+———l—2— =— v2+—mv2+———‘—2— 
r gf x gi 211i 2 22f rf 2min 2 221 rj _‘
l 2 1 2 l 1
.. ’Z'mivir +E'm2V2r = Gmlmz — "‘—
1 "r 7‘1
m1
pf = pi ==> mlv" + mzvzf = 0 kg m/s => v2f = ——,—n—v1f
' 2
Substituting this expression for v2‘ into the energy equation, we get
2
1 l m 1 1 2Gm l 1
—mivlzf +“m2 “LVN = GmlmZ _"'_ =9 v12: = z ""—
2 2 m2 rf ri (1+ml/m2) I; I; With G = 6.67 x10‘“ N ~ m(kg)‘2, r,= R1+ R2=18x108m, 4: 1.0 x mm, m]: 8.0 x 103°kg, and m: 2.0 x10”
kg, the above equation can be simpliﬁed to yield 30
v” =1,72x105 m/s, and v2, = "'31 — W ><(1.72><105 m/s) =6.89><105 m/s m2 V" ' 2.0 x1030 kg
The speed of the heavier star is 172,000 m/s. That of the lighter star is 689,000 m/s. Newton ’5 Theory of Gravity 1225 (h) The previous result shows there is a linear relationship between x and y, hence there is a linear relationship
between log 14 and log v. The graph of a linear relationship is a straight line, so the graph of log vversuslog u will (c) The slope of the straight line represented by the equation y = (q/p)x + logC/ p is (1/17. Thus, the slope of the log
vversuslog u graph will be q/p. (d) From Newton’s theory, the period T and radius r of an orbit around the sun are related by 47:2 3 — r GM This equation is of the form T P = Cr 4, with p = 2, q = 3, and C = 41t2/GM. If the theory is correct, we expect a graph
of log Tversus~log r to be a straight line with slope q/p = 3/2 = 1.500. The experimental measurements of actual
planets yield a straight line graph whose slope is 1.500 to four signiﬁcant figures. Note that the graph has nothing
to do with theory——it is simply a graph of measured values. But the fact that the shape and slope of the graph agree
precisely with the prediction of Newton’s theory is strong evrdence for its correctness. ' ' (e) The predicted y—intercept of the graph is log C/p, and the experimentally determined value is 9.264. Equating
these, we can solve for M. Because the planets all orbit the sun, the mass we are ﬁnding is M = Mm. T2: 1 l 4752 471.2 l
—10 C: —I = ~9.264 : =10—18528 =
2 g 2 0 GMSM GMSM 1013.523
2
=> M,.,, = 4%.10'8528 = 1.996 ><1030 kg
1 The tabulated value, to three signiﬁcant ﬁgures, is Msun = 1.99 x 103° kg. We have used the orbits of the planets to
“weigh the sun!” r2 (2r) r r T 1' 4r T2
2 2 2 .3 1/2
£(M+ﬂz_)=47t2r =>T= 4n! 1
r 4 T G (M+ "1/4) 2.59. Model: Model the earth and moon as spherical masses.
Ive: The mechanical energy of the earthmoon system is E —w _ _GM,M,,. __(6.67><10““ Nmzlkg2)(598><102‘ @036an kg) AEmm = —PAt = ~(l.0 ><10l3 J/s)(100 yearsx 3.16 x107 s/year) = —3.16 ><1022 J dEmh __ GMeMm
dr 21;:
N 2r; _ 2(3.84x108 m)’ AE —3.16 ><1022
GMeMm m" (6.67x10‘” N mz/kg2)(5.98 x1024 kg)(7.36xio22 kg)( ‘0 1232 Chapter 12 12.73. Model: Assume the solar system is a point particle.
Solve: (a) The radius of the orbit of the solar system in the galaxy is 25,000 light years. This means r = 25000 light years = 2500(3 x108)(365)(24)(3600) m = 2.36 x 1020 m 20
T = .21”: = (2”)(2‘36 x 10 m) = 6.64 x 1015 s = 2.05 x108 years v (2.30 ><105 m/s) 9
5 X10 3”“ = 24.4 orbits.
2.05 X 10 years
(c) Applying Newton’s second law yields (b) The number of orbits = GM m 2 Z 5 2 20
“.2... s. = mﬁv 8me r = (2.30x10 mils? (2.362x102 m) = 1.87X104, kg
r r G 6.67x10 Nm /kg ((1) The number of stars in the center of the galaxy is 1.87 x 10‘“ kg
1.99 x10” kg = 9.4 x 10m 12.74. Model: Model the earth as a spherical mass and the shuttle and payload as point masses. We’ll assume mpnyload << mshimle'
Visualize: Freebody diagram
of payload
/ Shuttle: rs = Re + 300 km Payload: rp = Re + 290 km M, = 100 kg 1':
Re = 6.37 x 106 m
Me = 5.98 x 1024 kg
ism Solve: (a) The payload, in steady state, is undergoing uniform circular motion. This means that the net force is
directed toward the center of the earth. There is no tangential force component, since such a force would cause the
payload to speed up or slow down and the motion wouldn’t be uniform. Since the net force is due to gravity and
tension, and the gravitational force is radial, the tension force cannot have any tangential component. Thus the rope is radially outward at angle 0°.
(b) To move with the shuttle, the payload’s period T9 in an orbit of radius rp must exactly match the period T5 of the shuttle in an orbit of radius r5. The shuttle’s period around the earth is given by T: = (47:2/ GMQIS , where we’ve used the assumption mp << ms to infer that the rope’s tension will be too small to have any inﬂuence on the shuttle’s
motion. Newton’s second law for the payload’s motion is GMm mv2 m 2m 47;2 e P P P P (Elm)r= 2 _T=mpar; =..._... .— =mr—2.—
r r r T H T
P P P P P Using TP = T5 and the above expression for T 5, this becomes GMem GMe GMem r3
r: P_T=mprp r53 = If) Fur:T
3 3
GM . GM
#7,: emp 1 5, = amp 1_ Re+290km =4.04N
If, I; r: Re+300km Assess: The fact that the tension is so small justiﬁes our assumption that it will have no effect on the motion of
the shuttle. ‘ i Newton ’s Theory of Gravity 12—33 12.75. Model: spherical mass. Momentum is conserv
Visualize: Solve: For the given orbit, 7;, = Re +1><IO6 m = 7.37 ><106 m. The speed of a satellite in this orbit is GM v0 = = = 7357 m/s
'6 The two satellites collide, stick together,
[ the perfectly inelastic collision is (400 kg + 100 kg)v, and move with velocity v,. The equation for momentum conservation for = (400 kg)(7357 m/s)  (100 kg)(7357 m/s)
==> vl = 4,414 m/s 1 GM (500 kg) 1 2 GM (500 kg)
~500k 4414 / 2— e =—500k — e
2( gx m S) 7.37x10“ m 2( gm "2
, we can simplify the energy equation to
v22 — (2.452 X 10" m/s)v2 + (8.876 x107 mzlsz) = O m2/s2 => v2 = 20,107 m/s and 4,414 m/s
'Iocity of 20,107 m/s for v2 yields ' ﬁchanical energy is conserved. Also, the an
. Please refer to Figure CP12.76. (a) Angular momentum is L = mrvsinﬁ. The angle [3 = 90°
tum requires This is an isolated system,
erved. at points 1 and 2, so conservation of angular r
/_ I I_ __2_ I
mnv, — mrzvz :5 V — V2 12—34 Chapter 12 Theenergy conservation equation is l , GMm 1 , GM , , 1 1
—m(v2)2  = —m(v,)2 — => (v2 2 (v,)2 = 20 ~———
r2 2 1 r2 71
Using the angular momentum result for v1’ gives
2 2 r 2 r — r r2 — r2 r — r
(v;) (v;) ——2— =ZGM ' 2 =>(v;)2 ‘ 22 =2G ‘ 2
"1 ’16 'i ’i’i
d vl’ = :2—
"1
(b) For the circular orbit,
—u . 2 2 24
v‘ = _(_;_Il:l_ ___ (6.67x10 N 1;! /kg )(5.958X10 kg) :77” m/S M
rI (6.37X10 m+3x10 m) For the elliptical orbit,
rl = Re +300km=6.37X106 m+3><105 m:6.67><106 m r2 = Re +35900 km=6.37x105 m+3.59x107 m=4.23x107 m 2GM(r2/rl)
(n + 6) = v: = 10,160 m/s V1 (c) From the workkinetic energy theorem, W = AK = %mvl’2 — émv? = 2.17 X 1010 J ((1) v; = M512
"1 + "2
Using the same values of rl and r2 as in (b), v; = 1600 m/s. For the circular orbit,
v2 = 91—M— =307O m/s
r2 (e) W = émv2 —%mv;2 = 3.43x109 J 2
(f) The total work done is 2.513 X 10‘0 J. This is the same as in Example 12.6, but here we’ve learned how the work ‘ has to be divided between the two bums. ...
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 Winter '08
 Susskind,L
 Physics, mechanics

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