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Unformatted text preview: 1445. An automobile having a mass of 2 Mg travels
up 3 7° slope at a constant speed of v = 100 km/h. If
mechanical friction and wind resistance are neglected, determine the power developed by the engine if the auto—
mobile has an efﬁciency 6 = 0.65. Equation of Motion: The force F which is required to maintain the
car’s constant speed up the slope must be determined first. ' + Z Er. = max: F — 2(103)(9.81)sin 7° 2 2(103)(0) F :2 2391.08 N ' . l00(10") m ( l h
P : ‘ e d 1h .‘ *: ———— =
ower Here,l.he spec of eczema, [ h I x 3 5 27.78 m/s. The power output can be obtained using Eq. l4l0. P = F  v :. 239] .08(27.78) : 66.418(103) W = 66.4]8 kW Using Eq. 14l l, the required power input from the engine to provide
the above power output is power output power input:
a w=102kw Ans 0.65 1450. Acar .__has a mass m and acceleratcs' along 'a
horizontal _s__t'raight_road from rest Suchfthat'the 'pQWer is
always 'a 'cgnstaht'tamount P. Determine how far itmUSt
travel to’ reaChh speedofv. Power : Sine: the: power ouput is constant. than the traction forceFvatics.
with 1.). Applying Eq'. 14 10. w: have; P = F W
P
P 2 F1) F= ~—
1:
Equation (pfMarion :
+ P P
.+ 2F = ;  = = ......_
_ , 11‘anr U ma a mu
. . . _ va'u
K Humane: : Applymg cquatmn d5 = —. we. have
a . s " mt)2 mu’
d: = d =
ID In P v _: 7 3P Ans 1477. The SIb collar is released from rest at A and
travels along the smooth guide. Dotormjhe itsirspeed when
its center reaches point C and the normal force: it exerts 15% + VA = Tc + Vc on the ‘rod at this point. The spring hasan :gnstrotohed
length of 12 in.. and point C is locatedjust before the and 0 + 0 +}[2(12m£’)2 + 5(2) : 1(_5_)v2 + lumm REV + (E9): 12 2
2 12 12 12 " ‘1 of the curvod portion of the rod. 12 2 32.2 2 12
v = 12.556 ft/s =12.6ft/s Ans
$15.
* _ . 5 (12.556)!
nZFu — m ; N + Fsm50.19¢4° = _
a" C ‘ 32.2( 1 ) c
) F5
_ . 12 10 12 “a
E—ks. Ii: 12 *2 _2__= 5°”
2( )[ (12) + {12) 121 7.241011:
Thus, NC = Ans 1486. The rollercoaster car has a speed of 15 ft/s when
it is at the crest of a vertical parabolic track. Determine
the car’s velocity and the normal force it exerts on the track when it reaches point B. Neglect friction and the T
mass of the wheels. The total weight of the car and the "ﬁlm,
passengers is 350 lb. y=_2;3(mooo—2) MMLVMMKVA s
dy—_l __2 9_tanl(2_ o — — — . —  )—63.43
dx 100 x=2oo 200“
d_zy___1_
dxz ' 100
DammatA
T4+VA =Tb+ls 3501b 1 350 15, 0 1 350 1 30200 (333°
aim)‘ ’+ 71%)“) ‘ 5‘ ’
v,=114.48=n4ru3 Ans Q dy I? , N
[1%)] 1+(—2)2 ’ a
P=T= l =1118.0fl
5 I‘m 350
+ 2F. =ma..; 350 63.43°—N =(_._)
/ c°s B 32.2 (114.48)2
1118.0 N, = 29.1 lb Ans *1492. The 2lb collar has a speed of 5 ft/s at A. The
attached spring has an unstretched length of 2 ft and a J'
stiffness of k = 10 lb/ﬂ. If the collar moves over the
smooth rod, determine its speed when it reaches point B,
the normal force of the rod on the collar, and the rate of
decrease in its speed. Damm aLB : H+M=E+ﬁ ‘2 11 2 :13. 21103—220
ﬂmjrs) +§(10)(4.5 2) +2045) 2(32_2](va) +2( )( ) + vs = 34.060 ftis = 34.1 ftfs Am 1
F45»?
§=ma=1i =4, 2“:
dx x=3
2 /g 2 N 7/.5_7°
9=—71.57° Err {3 J
(3’13" N
WT ’ /V 1“ __ _3 1 I
,9a df; =Q‘LI—iilllgsmzsn ,1 El . a n__ 2 (34'060)2 yzr; =ma,., —N+10c0518.43 +2cos7l.57 31.623 ) N=7.84lb Ans
. D . D 2
+39»: mm; 2sm71.57 —105m18.43 =[3—2—2Ja. a: = —20.4 Ms2 Ann 156. A man hits the SO—g golf ball such that it leaves
the tee at an angle of 40° with the horizontal and strikes
the ground at the same elevation a distance of 20 m away.
Determine the impulse of the club C on the ball. Neglect
the impulse caused by the ball's weight while the club is
striking the ball. (in (+ T) (V) S); = (so); + (VOL!
20 = 0 + v cos40°(t) 1
s = so +vut+§act2 1
0 = o + v sin40°(t) ~ E(9131)? t:1.85 s v = 14.115 m/s rrtvl + Xi’th : mv2 o + J'th : (0.05)(14.115) dez = 0,706 N‘s 4:840“ Ans *1528. Block A weighs 10 lb and block 8 weighs 3 lb.
If B is moving downward with a vetocity (#3)] =3 ft/s
at ! = 0, determine the velocity of A when I = 1 s; The
coefﬁcient of kinetic friction between the horizontal plane and block A is ,u,‘ = 0.15. :A + 255 = I
vA = 2v, ((1) mvl + SJ.th = mu: —ﬂ2)3 T1 151.0 ' m
32_2< u— 0+0. ( ig—ﬁivm (+l) mv1+EIth= va T T i I. 3 (vnh 1011)
322(3) + 3(1) — 2T(1)— 572( 2 ) ET 1.5!)» T —32.2T— 10002 = 11.70 ion 3a, —64.4T+ 1.5m)2 = —105.6
T = 1.50 lb (VA): = —6.00 ft/s = 6.00 ft/s —) Ans 1537. A 40lb box slides from rest down the smooth
ramp onto the surface of a ZOlb cart. Determine the speed of the box at the instant it stops sliding on the cart;
If someone ties the cart to the ramp at B, determine the
horizontal impulse the box will exert at C in order to stop
its motion. Neglect friction and the size of the box. Datum atB : n+m=u+e (£%wa+o 0+40<15):% v3 = 31.08 ft/s Box and cart: 4. (H) va1 =Emv2 0+(%)(31.08):(42;:9)V1 v2 = 20.7 ft/s Ans '6) mv1,+£Jth=mvz (§%J(31.0‘8)~1F 41:0
#3 dez=38.61b s Ans 1542. The man M weighs 150 lb and jumps onto the
boat 3 which has a weight of 200 lb. If he has a horizontal
component of velocity relative to the boat of 3 ft/s, just
before he enters the boat, and the boat is traveling
v5 = 2 ft/s away from the pier when he makes the jump.
determine the resulting velocity of the man and boat. +
(a) VM = VB + VM/B
vM = 2+3 vM = 5ft/s (:9 val = Em v2 150 200
322(5) + 32.2(2 (V3); = 3.29 ft/S ) 350 3 3720392 Ans 1554. Blocks A and B have masses of40 kg and 60 kg,
respectively.They are placed on a smooth surface and the
spring cannected between them is stretched 2 m. If they
are released from rest, determine the speeds of both
blocks the instant the spring becomes unstretched. (L) 21an = 2va 0+0=40vA “60% n+w=n+% 1 1 l
0 + 508mm: = 5010va + imam? VA 2 3.29 mls Ans v5 = 2.19 m/s Ans ...
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This note was uploaded on 02/13/2010 for the course ME 324 taught by Professor Neptune during the Spring '08 term at University of Texas.
 Spring '08
 Neptune

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