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Unformatted text preview: CHAPTER FIVE SOLUTIONS 5.5 (a) The force of gravity is rce of gravity and the
direction of motion is 9 = 90“ — 30° = 60°, and so the work done by
gravity is given as Wg = 49(2.5) c0360” = 6 J
(b) The normal force exerted
N= mgcosSO" = 42.4 N, fk = hkN= O.436(42.4 N) = 18.5 N. This force is
. Therefore the on the block by the incline is so the friction force is directed opposite to th
work it does is Wf= fks cost) = (18.5 N)(2.5 m)(cos180°), g
(c) Since the normal ‘ ' e displacement (i.e. 9: 180°) ME
5.6 PX =16.Oc0325”= 14.5 N,
F), = 16.0sin25“ = 6.76 N gives N= 31.3 N
(a) W}? = (16.0c0325°)(2.20) = 31.9]
(b) WN= (31.3 cos 90°)(2.20) =0
(0
Wgravity = (24.5COS90°)(2.2) = 0
(d) Wnet = WF + WN+ Wgravity = 319 J m
5.7 (a) The workenergy theorem, W= AKE, gives 5000J= 21 (2500 kg)va  0, or vr = 2 m/s. W= (Fcoss)s=F(25.0 m) = 5000 J, so F: 200 N.
——________ , we have %(7.00)(3.00)2 = % (0.00245)v2, so v = 160 m/s.
—»—___,_:_m 5.9 (a) We use the workenergy theorem to nnd the work. 1 W: AKE= E meZ — %vaZ = 0  % (70.0 kg)(4.0 m/s)2 = —560 J.
(b) W= (Fcoss)s = (fc03180")s = —fs.
But, the frictional force I“ is found as,
f= ukN = (0.70)(70.0 kg)(9.80 111/52): 480 N. Thus, W= fs = —(480 N)s = —S60 J, which gives, 3 = 1.2 m.
________ 5.10 (a) KEA =%va2 =% (0.60 kg)(2.0 m/s)2 = 1.2 J. (b) KEB =%mv132 = 7.51= %(0.60 kngZ. Thus, (c) Work = AXE: KEB  KEA = 7.5 J— 1.21= 6.3 J. (b) 5.8 Using KE2%mV2 1 Net work — AKE= Emez ~ %mvi2, or J 57 5.18 5.19 5.20 5.21 5.22 5.23 CHAPTER FIVE SOLUTIONS magnitude of this distance to be 32.14 m. Because this distance is
now below the zero reference level, it is a negative number. Thus, P53 = (1000 kg)(9.80 m/sz)( 32.14 m] =  3.15 x 105 J, and
APE= PEA  P153 = 0  (3.15 x 105 J) = 3.15 x 105 J. (a) Relative to the ceiling, y = l 111. Thus, PE= mgy= (2.00 kg)(9.80 m/sZ)(1.00 m) = 19.6 J
(b) Relative to the ﬂoor, y= 2.00 In, so PE: mgy = (2.00 kg)(9.80 m/52)(2.00 m) = 39.2 J
(c) Relative to the height of the ball, y = O, and PE =0. Choosing the gravitational potential energy
to be Zero at the bottom of the swing, the total
energy (E=KE+ PEg)0f the pendulum at the
top of the swing is, Etop =0+ mgU.  Loose) = mg]. (1  c059).
At the bottom of the swing, we have Ebouom=0+%mv2= %mv. Using conservation of energy, %mv2 =mgL (1 , c059), or V=V2gL (1  c039)=\/2(9.80)(2.0)(1  C0525°) =l.9m/s. Let in be the mass of the ball, R be the radius of the circle, F be the 30 N force. The work—energy theorem yields (choosing y = 0 at the bottom of the circle) mg(2R) + % mvtopz + pitcher's work = %mvbottom2, or 2ng+ %mvmp2 +F(JER) = é’mvbotmmz, so that vbonom=‘\/4gR + vmpl + Z’gR =\/4(9.80)(0.6) + (15)2 + 2" 35250'6 01‘ Vbottom = I'D/S. (a) We take the zero level of potential energy at the lowest point of the
arc. When the string is held horizontal initially, the initial position
is 2 m above the zero level. Thus, PE= mgy~— wy = (40 N)(2 m) = 80 j. (b) From the sketch, we see that at ‘
an angle of 30° the ball is
(2.0 m)(l  cos30“) above the
lowest point of the arc. Thus,
PE: (40 N)(2.0 m)(1  c0330“) = 1 l J. (c) The zero level has been selected
at the lowest point the arc.
Therefore, PE = O at this point. Using conservation of energy, we
have %(50)(10)2 =% (50)(1.O)2 + (50)(9.80)H, so that H.= 5.1 In. There are no non—coriservative forces that do any work. As a result, we
use conservation of mechanical energy, with the zero level for 59 CHAPTER FIVE SOLUTIONS 1 1
= Fthrusts ‘ Fresistances = i me2 ‘ E mViZ. = (7.5 x 104  4.0 x 104)(500) = % (1.5 x 104) (sz  (60V), or Vf = 77 m/s. 5.35 The initial vertical height of the car above the zero reference level at
the base of the hill is yi = (5.0 m)sin20c = 1.71 m.
The energy lost through friction is
W11C = —fs = —(4000 N)(5.0 m) = —2 x104 3.
We now use,
Wnc=%me2  %mn2 + mgyr  ngi
1 2 x 104J=§ (2100 kg)v2  0 + 0  (2100 kg)g(1.71 m), or v = 3.8 m/s. 5.36 (a) The initial vertical height of the child above the zero level for
gravitational energy at the bottom of the arc is
(2.00 m)(1  c0530“) = 0.268 m.
In the absence of friction, we use conservation of mechanical energy as %mVi2 + mgyi = % £1in2 + ngf, 01‘ 0 + mg(0.268 m) z %mv2 + O. This gives, v = 2.29 m/s. (b) In the presence of friction, we use 1 1
Wnc=§me2  'Z‘mVjZ + mgyf  mgyl. W11C :% (25.0 kg)(2.00 m/s)2  0 + 0 — (25.0 kg)g(0.268 m) = ~15.6 J. 5.37 (a) Let L = 3 rn be the length of the ramp, and m = 10 kg be the mass of
the block, and 9 = 30° be the angle of the incline. Also let [1k be the
coefﬁcient of kinetic friction between block and floor and D = S m.
We may start with conservation of energy along the ramp and write, mglsin6+ O = %mvbonom2, so that Vbottom =VZgIsin6 =‘JZ(9.80)(3) sin30 = 5.42 m/s. (b) Using energy considerations, we have 0  mgL sine:  01(ng SO that HR = f3 sin6= sin30° = 0.3. (c) All the initial potential energy is lost due to friction, which is
ngi a mgL sine: (10.0)(9.80)(3.00)si1130° = 147 j. 5.38 “35+ KBinitial = Wnc + (PE+ KBfinal Migy1i+ Mzgyzi + M3gy3i + KEii + mm + K1931 =Wnc + M1sr1r+ Mzgyzf
+ Mggygf + KElf+ KEzf+ Kng
1 0 = 30(4) + (5.0(4.0) + 10.0(0) + 15.0(—4.0))9.80 + E (5.0 + 10.0 + 15.0) V2
which gives v = 4.26 m/s 5.39We shall take the zero level for PE at the base of the hill. The skier starts
at a vertical height above this level of
no = (200 m)sin 10.50 = 36.45 m. 62 ...
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This homework help was uploaded on 02/05/2008 for the course PHYS 111 taught by Professor Shapiro during the Fall '01 term at Wesleyan.
 Fall '01
 Shapiro
 Physics

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