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Unformatted text preview: 146. When a 7kg projectile is fired from a cannon
barrel that has a length of 2 m, the explosive force exerted
on the projectile, while it is in the barrel, varies in the 15
manner shown. Determine the approximate muzzle velo— city of the projectile at the instant it leaves the barrel. Neglect the effects of friction inside the barrel and assume 10
the barrel is horizontal. The work done is measured as the area under the force—displacement
curve. This area is approximately 31.5 squares. Since each square has
an area or 2.5(10°)(0.21 5 (1n) 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.3 2.0" T1+£U1—2 = T: I .
0+1(31.51t2.5)(10“)to.211= 5mm) v2 = 2121 m/s = 2.12 km/s (approx) Ans _ 147. Design considerations for the bumper B on the
‘ 5Mg train car require use of a nonlinear spring having the
loaddeﬂection characteristics shown in the graph. Select
the prOper value of k so that the maximum deﬂection
of the spring is limited to 0.2 111 when the car, traveling
at 4 m/s, strikes the rigid stop. Neglect the mass of the
.car wheels. s (m) (111
(5000M)2 ~ / kszds = 0 t) 51103)t9.81) N F 15.0 MN/m2 148. The crate, which has a mass of 100 kg, is sub
ted to the action of the two forces. If it is originally at
L'determine the distance it slides in order to attain a
dof 6 m/s. The coefﬁcient of kinetic friction between
rate and the surface is 111. = 0.2. t t g 800 N
»" 0fM0twn: Stnee the crate slides, the friction force devel
,een the crate and its contact surface is F; = MN = 0.2 N. W); _\A
“g E‘l 13'7. We have
M ~ma ~ N 3  a
."v +100 3 — SOOSln3O —100(9.81)=10010)
N=1321N 5:0” “fogs: an: Energy: The horizontal components of forces
hereas :1 leh‘ in the direction of displacement do pg“;
wsince ilc‘ rtction torce Ff = 020321) = 264.2 N does T. + Z U1...2 = T:
> a] renacts in the opposttc direction to that of displace
mes and ton N,.lhc vemcal components of the 800 N o 4 I
the weight 01" the crate do not displace hence 0 + 800 cos 30 (s) + 100 s — 264.21' = 5(100)(62) nice the crate is ori aye ginally at rest. T. = 0. Applying S = rn Ans 189 1414. Determine the velocity of the 20kg block A
after it is released from rest and moves 2 m down the
plane. Block B has a mass of 10 kg and the coefﬁcient of
kinetic friction between the plane and block A is pk = 0.2.
Also, what is the tension in the cord? Block A kw, =0; NA 20(9.81)cos60°=0 NA =98.l N
System:
20 I? 8/)N
7i +£Ul2 = 7i 0
60 7'
T
(0+0)+ 20(9.8l)(sm60°)20A2(98. 1)(2) 10(9.81)(2)= alcohhéao); 0, Z NA
v=2,638 =2,64 m/s Ans “4
N50. blockA :
BlockB: “NagoN
1; +211] 2 = 1;
TI +EUI 1 = Ti
0+20(9.81)(sin60°)(2 T2 — 1
0+T(2)10(9r81)(l)= illl0)(24638)2 ) ( ) 0'2(98‘”(2)‘ ZOOM638V
T= 115 N Ans
Ans
1415. Block A has a weight of 60 lb and block B has a
weight of 10 lb. Determine the speed of block A after it
moves 5 ft down the plane, starting from rest. Neglect
friction and the mass of the cord and pulleys
21‘ + l, = I
2A + A: 0 9‘ T
5.4 l = I
s, to“
2v‘ 4» v, = 0 ‘3 z
1; 4» EU“ = 7': \NA mu. 3 _ 1 .60; z 13 a
0+ 60(3)(5)  l0(0) — 3(322)“ + 292.2)“ VA) v‘ = 7.18 {US All! 193 1429. Roller coasters are designed so that riders will
not experience more than 3.5 times their weight as a
normal force against the seat of the car. Determine the
smallest radius of curvature p of the track at its lowest
point if the car has a speed of 5 ft/s at the crest of the
drops Neglect friction. Principle of Work and Energy : Here, the rider is being displaced vertically
(downward) by 5 = 120— 10 = 110 ft and does positive work. Applying Eq. 147
we have THEM—2:72 gistsvwmmsisw v2 27109 {12/52 Equation of Motion .' It is required thatN = 3.5W, Applying Eq, 13 7. we 11— r 35w w— W 1—109
"M" ' _lm)(p) p = 88.3 ft Ans 1430. The catapulting mechanism is used to propel the
10kg slider A to the right along the smooth track. The
propelling action is obtained by drawing the pulley
attached to rod BC rapidly to the left by means ofa piston
,lf the piston applies a constant force F = 20 RN to rod
.BC such that it moves it 0.2 m, determine the speed
tained by the slider if it was originally at rest. Neglect
e mass of the pulleys, cable, piston, and rod BC. 255+5A=l cm
2Asc + A14 = 0 C‘_'L:_]——> io(lo=)N
,,,,,,,,,,,,,,,,,,,,,,, ,, ’ T
2(01) = — As‘ "9
SA °‘ = “4 Etaw T1 + l:Uiz = I;
1
o + (10000)(o.4) = 5(10)(v‘)1 v‘ = 28.3 mls Ans 201 1431. The collar has a mass of 20 kg and slides along the
smooth rod. Two springs are attached to it and the ends of
the rod as shown. If each spring has an uncompressed length
of l m and the collar has a speed of 2 m/s when 5 = 0,
determine the maximum compression of each spring due to
the back—and—forth (oscillating) motion of the collar. 1]+£U._2=1; .1 2 1 2 1
2(21))(2)  5(50)(:) «5(100)(.i)2 = o S = 0.730 In An 0.25 m *1432. The cyclist travels to point A. pedaling until he ,l + ,i = 2
reaches a speed 12A = 8 m/s. He then coasts freely up the
curved surface. Determine the normal force he exerts on  x" +
the surface when he reaches point B. The total mass of
the bike and man is 75 kg. Neglect friction, the mass of d), _,l l
2 the wheels, and the size of the bicycle. E " F
Fury = I.
y 2.:1 = 2 x .1,y=1(mmn) dy a; = yléx'l) — t'l(%)()"l)(;)
1;: = éylx’l + éé)
For x = y = 1
12 = 1 5:! 3 1
dx ' nix2
.. “H‘nzlm 2.828111 Ti + zUII = 7;
1 75 3’) — 75(9 81m) = 103M)
5( )(  2
v: = 44.38
° 44.33
A u; = "10,: N, — 9310990545 = 759—21828) N, = now An: 1445. An automobile having a mass of 2 Mg travels
up :1 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. + )3 a. = max». F ~ 2(103)(9.81) sin 7° = 2(10‘)(0) F = 2391.08'N l l 1
Power: Here,th speed ofth car is v = X s) = 27.78 m/s. The power output can be obtained using Eq. 1410. I p = F . v = 239l.08(27.78) = 66.418003) W = 66.418 kW ; Using Eq. 14l l. the required power input from the engine to provide
the above power output is power output
5 V power input 2 = 102 kW Ans = 3.75 fl/s2 3
.r = max; F — 325 : (1600)) (3.75) 32.2 F = 2l88.35 lb _ 21883500) "I" r— =
nu. 119 hp Ans 209 — 1451. To dramatize the loss of energy in an automobile,
consider a car having a weight of 5 000 lb that is traveling
at 35 mi/h. lf the car is brought to a stop, determine how
long a lOO—W light bulb must burn to expend the same amount of energy. (1 mi = 5280 ft.) Energy : Here. the speed ol'thecar is v = (35 mi)x(5280 h)x( 1h )
h 1 mi 36005
= 51.33 ills. Thus. the kinetic energy of Ihecar is 1 1 5000 ‘
U = Emu: = ﬂay 51.33:) = 204.59( 10’) ftlb Thepowerol' the bulb is 3,... = 100 w,{ “‘P )){550 ftvlb/s
746 W 1 bl,
= 73.73 rub/s. Thus, U 204.5900!) P5,”, = 2774.98 5 = 46.2 min Ans *1452. The motor M is used to hoist the SOOkg
elevator upward with a constant velocity 05 = 8 m/s. If
the motor draws 60 kW of electrical power, determine
the motor‘s efficiency. Neglect the mass of the pulleys and cab .
le +Tu; = o; 31— 500(9.81) = o
T T T
T = 1635 N a:
2:; + (:2 up) = 1 some)“
3 vI = v,
v, a 24 ml:
5‘. L .
a, = 1635(24) = 39.24 kW ‘L 5'
l—__.
n = 60 kW
1:, 39.24
£=n3.7=0.654 Ans 1453. The 500kg elevator starts from rest and travels upward with a constant acceleration at. = 2 m/sz. Determine the power output of the motor M when T T
I = 3 s. Neglect the mass of the pulleys and cable. +Tzr; 2 ma,; 37— 50()(9.81) = 500(2)
7' = 1968.33 N
soon8:)“ 1
3:, ~ 5, = l y .
5:
3V: = Vr
When I = 3 s, (+Dv = v0 + 4,:
v, = 0+2(3) = 6111/:
v, = 3(6) = 18 m/s
1:, = 19683308) I}, = 35.4 kW Ans *1456. The 50—kg crate is hoisted up the 30° incline by
the pulley system and motor M. If the crate starts from
rest and by constant acceleration attains a speed of 4 m/s
after traveling 8 m along the plane. determine the power
that must be supplied to the motor at this instant.
Neglect friction along the plane. The motor has an efﬁciency of 5 = 0.74. f:\é+2ag(:—sol (4)2 = 0+2a, (8—0) l / 1 T
a: = m s T 50 a.‘
4/25 =ma,; 2T—50(9.81)sm30° = (50)“) T: 147,6N Lsc+3p =! ZVC = —Vp (2)(—4)=‘VP vP = 8 m/s pa :1. VP: 147.6(8) = HS! w P llSl
p.=_°=_=159v9w=1. w
074 5 60k Ana 2 _1457. The sports car has a mass of 2.3 Mg, and while it
IS traveling at 28 m/s the driver causes it to accelerate at
.im/s'. If the drag resistance on the car due to the wind
IS FD (0.3v3) N, where v is the velocity in m/s
determine the power supplied to the engine at this instant: F = 03v: + 11500!)
The engine has a running efficiency of e = 0.68. . Ln: = may; r — 0.3V: = 2.3005(5) Atv = 28 m/s
F = 11735.2 N
pa = (11735.2)(28) = 328.59 kw Po 328.59
p. = ._ = __ =
, e 0.68 483 kW Ans 1458. The sports car has a mass of 2.3 Mg and
accelerates at 6m/sz, starting from rest. If the drag
resistance on the car due to the wind is FD = (IO/u) N,
where v is the velocity in m/s, determine the power
supplied to the engine when r = 5 s. The engine has a
running efficiency of e = 0.68. #135 = my: F 10v = 2.3(io’)(o) F = 13.8(10’) + 10v
(;)v = v0 + at]
v = O+6(5) = 30m/s Pa = 1“" =[138(10’) + 10(30)l(30) = 423.0 kW P 423.0
p = .9 = __ — w
l e (168 “ 622‘“ Ans 214 1474. The collar has a weight of 8 lb. If it is released
from rest at a height of h = 2ft from the top of the
uncompressed Spring, determine the speed of the collar
after it falls and compresses the spring 0.3 ft. 7; + V] = T: + V1 h
'1 0+0 1 8 2 l
g i = 5(a)“ — 3(23) + 5(30)(0.3)2 V; = 11.7 ftls Am "I 1475. The 2kg collar is attached to a spring that has
an unstretched length of 3 m. If the collar is drawn to
point B and released from rest, determine its speed when it arrives at point A. Potential Energy : The and ﬁnal elastic potential energy are
1 1 1
5(3)( J31 +41—3) = 6.00] and i(3)(3—3)1 = 0. respectively. The gravitational potential energy remain: the same since the elevation of collar
does not change when it moves from Em A. Conservation of Energy : E+W=B+K
1
0+ 6.00: E(2)0: +0
0‘ = 2.45 mls A... *1476. The 5lb collar is released from rest at A and
travels along the smooth guide. Determine the speed of
the collar just before it strikes the stop at B. The spring has an unstretched length of 12 in. 73 + "1 = T: + V: 22 1 10 1 5
0 __ _ _ 1 
+ 5(12) + 2(24)(]2) _ #37”: + 0 + 0 v, = 15.0 m A... The 5lb collar is released from rest at A and
g along the smooth guide. Determine its speed when
intgr reaches point C and the normal force it exerts T + V ___ T + V
rod at this point. The spring has an unstretched A A C c b.0112 in ' 
, . and pornt C is located just before the end 1 10
. o o _ _ 2 12 l 5
curved portion of the rod. + +2[2(12)](12) + 5(a) = §<—322)v’ + Eltznznq/(ll—j): + (5’): _ 12 1
 _ 12 12 v = 12.556 ﬂ/s = 12.6 11.15 Ans
‘ {3
4):}; = mg; NC “51111150194411 = L(L2‘55_6)1
32.2 1 ) r.
F:
5 = ks: E = zuzw '3 2 1° ‘2
(12) +(E)1  E] = 7.24101!) “My
Thus.
NC = 13.911: Ans 221 ven an initial velocity of 20 ft/s
g has an unstretched length .of
/ft, determine the velocny 1478. The 2lb block is gi
when it is at A. If the sprin
2 ft and a stiffness of k = 100 lb of the block when 5 = 1 ft. Patentia! Energy .' Datum is set along AB. The collar is l fibelow the datum
when ii ’5 iii C. Thus. its gravitational potential energy at this point is — 2( l) l
: ~2.00 ft lb. The initial and final elastic potential energy are i(lOO)(22)2 i ,_ ._ 7
= (J and ii mm ( J27 +11 2) = 2.786 ft lb. respectively. Conservation of Energy :
TA + Kl = Tc + VC
(i)(202) + o: 1(~2—)u2 +2786+(—2 00)
32.2 2 32.2 C ‘ '
uc = 19.4 Ms W.
1479. The roller—coaster car has a mass of 800 kg,
including its passenger, and starts from the top of the hill
A with a speed 1),, = 3m/s. Determine the minimum
height h of the hill crest so that the car travels around
both inside loops without leaving the track. Neglect 2 friction, the mass of the wheels, and the size of the car. “I ,g Ema“W
. . + . _ What IS the normal reaction on the car when the car is at ' = "a" 80mm) ' Wit?) “1:0
3 and when it is at C? Elf] Thus, v, = 9.90 m/s roomiw
I! = 24.5 in Am
M3 N3 — 0 Ans (Forli to be minimum) 1
2 vc = 14.69 Inls Swim)“
14.691
g = _7—
14 691 5‘
+JEI'; = 1nd,: NC + 8000.81): soq'T) [:3
3000130” mmmmmmmmmm V *1480. The roller—coaster car has a mass of 800 kg.
including its passenger. If it is released from rest at the
top of the hill A, determine the minimum height h of the
hill crest so that the car travels around both inside loops
without leaving the track. Neglect friction, the mass of the
wheels,and the size of the car.What is the normal reaction
on the car when the car is at B and when it is at C? Ans its
WW NR
NW5” k : ioo lb/l't “a” Skuﬁim‘onisnegleued, memwmhvdmmdme7mlooppmvidedilﬁrstuavekamlm theiOmlmp.
mun)» q: 9"”) 5+u a+u i i
$300)”): + 0 = 50300va — 800(9.81)(n—20) 1
(some)z + o = 5(800)(vc)2  800(9.81)(24.514) mm” NC = mam Ans the IOmloop. n+VA:Tl+VB o + o = garcova — sooiasmLzo) ~o
v1 "us
+L2F. = 71:11.: 8000.8!) = soot136) {23'
Thus, v, = 9.90 mls
h = 25.0111 Ans
AIR: N. = 0 Ans (For It to be minimum) a + V‘ = 1;: + Vc
0+ 0 = %(800)(vc)z — 800(93i)(25—14) vc = 14.69 rnls “4.69)2
7 ht”; = my, NC + 30091“) = 800( ) NC = mam Ans 222 1481. Tarzan has a mass of 100 kg and from rest swings
from the cliff by rigidly holding on to the tree vine, which
is 10 m measured from the supporting limb A to his center
of mass. Determine his speed just after the vine strikes
the lower limb at B. Also, with what force must he hold
on to the vine just before and just after the vine contacts the limb at B? Datum at C T. + V1 = T: + ‘6
0+0: illumch — 100(9.8l)(10)(l cos45°) Vf=7_581=7.58m/s Ans Just before smking B. 9 =10 m + TEE. =ma..; T931=100((7'581)2
lO
T=l.56kN Ans
Iusiafiersmking B, p=3m
+TEF. =ma,; T981= 10061538”)
T=2.90kN Am 1482. The spring has a stiffness k = 3lb/ft and an
unstretched length of 2 ft. If it is attached to the 5—H)
smooth collar and the collar is released from rest at A,
determine the speed of the collar just before it strikes the
end of the rod at B. Neglect the size of the collar. 223 Datum T
E
t
981 H 'm N
100%: g
Datum at B. I roll = (l)2 + (4)1 + (6)1 = 7.28 a
l rml = «(1)2 + (3)2 + (2)2 = 3.7411 n+n=n+n
0 +(5)(6—2) + 51(3)(7.28—2)2 = V, = 27.2 n]: A.“ 1 5 3 1 2
_ __ _ I 4..
2(322)” + 2”“3 7 2) 1486. When the 6kg box reaches point A it has a speed
of 11A = 2 m/s. Determine the angle 9 at which it leaves
the smooth circular ramp and the distance 5 to where it
falls into the cart. Neglect friction. “9.80M
¢ N930 /\Af
All AlpOmIB;
‘;
_  =6_!_ 1
yZF.—ma,.. 6(9.8l)cos¢ (L2) () Datum at bottom of curve : n WA = Tu Hi;
—;(¢5)(2)2 +6(9.81)(l.2 c0520°) = éusxmz +6(9.81)(l.2 cos.) 13.062 = 0.5%, +11.772 cost? (2)
Substitute Eq. (I) imo Eq. (2). and solving for v,.
v. = 2.951 m/s  x_.,_._(2'951)1)=4229°
Thus. dzcos (1.2381) . 0 = ¢— 20° = 22.3" Ans (1» T) .r=s.;lv.)t+‘fa¢t2 — 1.2 cos42.29° = 0 2.951(sin42.29°)!+ %(—9.81)r‘
I. Z c4512.: 9" 4.9059 +1.9857!—0.8877 = o P— Solving {or the positive {00! ‘. r: O. 2687 s S=Jo+vol .r = 0+(2.951 cos42.29°)(0.2687) 5 =0.587 in An: 226 ...
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 Fall '08
 Hudyma
 Dynamics, Energy, Force, Friction, Potential Energy, m/s

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