# hw5 - 15-3 A 5-lb block is given an initial velocity of 10...

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Unformatted text preview: 15-3. A 5-lb 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 15-9. When the S-kg 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 ling-my -0.3(49.05)=-5¢ annual/5’ Kinematic: : Applying rheumation u2 = v: + 240-30) 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 15-11. 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 32-2(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 ~ -? 15-15. The 4-lb 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 3000-lb car 8 is traveling at 6 ft/s to the left. If the cars crash head-on 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 15-45. 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 S-lb 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 : (so-imm'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 Air-kg 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.). * (um-w.)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,“ 15-87. 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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