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Unformatted text preview: 153. A 5lb block is given an initial velocity of 10 ft/s
up a 45° smooth slope. Determine the time it will take to travel up the slope before it stops (4) m(u,), +Zﬂlﬁ4‘p: ”(11:02 5 I
ﬁ(0)+(‘5sin45°)1=0 u / 1:0.4395 Ans 159. When the Skg block is 6 m from the wall, it is
sliding at v, = 14 m/s. If the coefﬁcient of kinetic friction
between the block and the horizontal plane is m, = 0.3.
determine the impulse of the wall on the block necessary
to stop the block. Neglect the friction impulse acting on
the block during the collision. Equal"! of Motion : 'I'heweelention ollhe block mustbeobnined ﬁrst
beforeonemdetmnhetheveloeityofmebloekbefomium Ihewall. ”‘6‘“"6: N—5(9.s)=5(o) N=49.05N
lingmy 0.3(49.05)=5¢ annual/5’ Kinematic: : Applying rheumation u2 = v: + 24030) yields
(1’) o‘ = 14‘ + 2(—2.943) (6— 0) v = 12.68 ml: Principle of Linur Impulse and Momentum : Applying Eq.lS‘. we
have m(u,), +2;I"it;d:=m(u,)z (3+) 5(12.68)l=5(0)
l=63.4N~s Am 1511. The particle P is acted upon by its weight of
3 lb and forces F. and F2, where t is in seconds. If
the particle originally has a velocity of v. = {3i + l j +
6k} ft/s, determine its speed after 2 s. 2
”WI +2] Fdr =mv2
n Resolving into scalar components. lr.=(5i+2¢j+mm 3 3 3 _ y
— ’ 1 _.. ._ , F = 12! lb
322(3)+/0— (5+1 )dt  322g”) 2 l l P / X 3 (l)+ 22tdr— 3
«32708) 3 i
2 2 u u, = 138496 his 11,. =43.933 fl/s vl = —36.933 fl/s ~ ? 1515. The 4lb cabinet is subjected to the force F =
120‘ + 1)2 lb where t is in seconds. If the cabinet is initially moving up the plane with a velocity of 10 his,
determine how long it will take before the cabinet comes to a stop. F always acts parallel to the plane. Neglect the
size of the rollers. Principle of Linear Impulse and Momentum: Applying Eq. [5—4,
we have m(v.). +2] Frdt =m(u,.;2
It 4 , 12 4
— —4 ° = __
(32.2)(10H1; (r+t)2d' 5'"20' (32.2)” t = 8.78 5 Anti 15:33. The car A has a weight of 4500 lb and is trav—
eling to the right at 3 ft/s. Meanwhile a 3000lb car 8 is
traveling at 6 ft/s to the left. If the cars crash headon and
become entangled, determine their common velocity just
after the collision. Assume that the brakes are not applied
during collision. 4.
(9) "’A(UA)I +mgtup). = (”M +7113)”: 4500 3000 7500 __ 3 __ __ =
32.2( ) 32.2 (6’ 32.2 ”2 v3 = “0.600 ftls = 0.600 ft/s <— Ans 1545. The lO—lb projectile is ﬁred from ground level
with an initial velocity of 12,. = 80 ft/s in the direction
shown. When it reaches its highest point B it explodes
into two Slb fragments. If one fragment travels vertically
upward at 12 Ms, determine the distance between the
fragments after they strike the ground. Neglect the size of the gun. (—3) (+7) (—2) (#1) (+ J) (3») v, = (Va). v, a: wooeﬁo‘ = 40m
v; a (vow: + 2ac(:,—(:.),) o : (soimm'al Zt—BLZXh—O) h . 74.53 ft Dav, : Dav, 10 5
m‘wm‘“ ' 352W: (9,), = 801V: —9 Sam = Dnv,
S 5
0 = 3‘23"», + 3—2—10” (Vr), = —12m =12n1; t
l 2
t, = (3o), 4 0.),” 5““ 74.53 = o + m + 502.2)? 121.81: R = 800.81) = 145 it An 15—57. Disk A has a mass of 2 kg and is sliding forward
on the smooth surface with a velocity (11A), = 5 m/s“ when
it strikes the Airkg disk 8, which is sliding towards A at
(125)] = 2 m/s, with direct central impact. It the coefﬁcient
of restitution between the disks is c = 0.4, compute the
velocities of A and B just after collision. Conservation of Momentum : mA(UA)I+mB(ul) =’"A(UA);+M,(U5’2 (—p) 2(5)+4l—2)=2(u‘)2+4(u.)2 {ll
Coefficient of Restitution : I: (an: *(vﬂ; (”4). ‘19.). * (umw.)2
0.4: 2
(a) 54—2) [ I Solving Eqs.[l] and [21 yields (1),.)2 =~l.53 mls= l.53 m/s «— (12,.)3 = l.27 m/s .., A,“ 1587. Two smooth disks A and B have the initial
velocities shown just before they collide at 0.1fthey have
masses MA = 8kg and m; = 6kg. determtne their
speeds just after impact. The coefﬁcient of restitution is
e = 0.5. +/Dltv. = Yaw, (V) —6(3cose7.38°) + 8(1eu367.38') = 5“,), + mm. (V; )2 ‘ (VA )2 ‘ ” (vi). — (m. (V )x’ _ (“)1' as ‘ 1coI6738'+3eou67.38' Solving. (v,),.  114 ml:
(VA )l I: 0.220 ml: (v,),.  3:11:6738“ I 2.169111]: 0")" =1sln67.38' #6462 m/s V.  #2401 + (1.769)’ = 3.50 ml:
v. = ((0.220)2 + (6462):  6.47 ml: Ans ...
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 Spring '10
 Alan

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