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Ch15_Sec1-3_Ex-6-15-23-35-47

Ch15_Sec1-3_Ex-6-15-23-35-47 - u-u ‘ ‘ “I w"15...

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Unformatted text preview: u-u ‘ - ‘ - “I. w "15 Uulluuav- u: viable-I‘ll] uu. aw.”- i , i ii Dete i _its speed when! = 4 5 if the horizontal traction F var s 1th time as shown in the graph. 2 o' .1: (w mlu.).+2J F,d1=m(u. )2 l ,, :0+j:(4-0.01:1)(10’)dt= 28(103)u i { ;v=0.564mls An, 1(5) . l s g 1 l 15-15. I The 20-lb cabinet is subjected to the force F = (3 + 2: lb, where t is in seconds if the cabinet is initially movin down the plane with a velocity of 6 ft/s, determine how long it will take the force to bring the cabinet to rest. F always acts parallel to the plane. Neglect the size of the rollers. i .i ' ? Principle of Linear Impulse and Momentum : Applying Eq.15-4, we have '2 m(u,.),+zj E-dt=m(u,.)z 'i ' ; 2 («33%)(4) +J; (3+ 2t)dt— zosin 20°:= (3%)(0) != 4.64 s Ans 15-23. Detemiine the velocities of blocks A and B 2 s _ after they are released from 'rest. Neglect the mass of the ; pulleys and cables. r 7" t ‘3 2‘] + 255 3 l a 1 3' VA = 'VB ‘ 5 A l l (+l) moltajmxwwyh 3 Solving. 1 2 o—r(2)+2(2) = (3.272)" r= 2.67 lb BlockB: v, =21.sn/sL An: (44,) md'y); {.sz dtgmwyh VA =_—2L5ftls=21.5ft/ST AM 1 0 +40) — 7(2) = (3%», 15-35.» ‘lLe two blocks A and B each h 5 and are su‘spended from parallel cords :fPZT-as: “,5 kg ‘ stiffness of k = 60 N/m, is attached to B and is can avlng a g 0.3 m agai st A and B as shown. Determine the m Pressed ' 808165 '9 a d ¢ of the cords when the blocks are 3:22:31! from rest nd the spring becomes unstretched [—4) th‘l 3-me 3' ‘5 '3 ‘. ‘5 i 3‘ O+O=-SVA+5VB : t'Ai = vs = v ForA orB° Just b fore eblocks ‘ ' - ‘ Pl begin to nse . Dammmowest point Tl + K = 75 + K T: + W = 7}1 + l5 1 5(5"°'7343)’ +0 = 0+5(9.8l)(2)(l —c059) 1 (0+0) -60 2-1 1 +2“ no.3) -5(5)<v)’+§(5>(v)'+o 9=¢=9.52° Ans v = 0.7348 lm/s is held at rest on the smooth block at A. If the 10-g bullet is t becomes embedded in the he distan 10-kg bl, ki determine t ce the block will slide up along the plane before momentarily stopping. 15-47. ei 10—kg block inclined lane by the stop travelin at 300 m/s 'when i : If we consider the block and the bullet impact is Ltive force F caused by the ut. Also. the weight of the bullet momentum is Linear Momentum the impu ill cancel 0 As the reSult. linear Conservation of as a syStem. then from the FBD. internal to the system. Therefore. it w and the block are nonimpulst've forces. conserved along the x' axis. )’ =(mb+ml)vx ’"b ( ”I! 1 001000005 30°) = (0.01 + l0) 1) (+ l 1 v = 0.2595 m/s oclt' 5 initial position. When y are it _abave the l)h = 98.l98lh. Energy : The datum is set at the bl is at their highest point the is (10+0.0l) (9.8 Cans erv atilm of the block and the embedded bullet datum. Them gravitational potential energy Applying 311.14 - 21. we have n+w=n+w l 0+i(10+°'°1)(°‘25952) =0+98.1981h h = 0.003433 m = 3.43 mm d = 3.43 / sin 30° = 6.87 mm Ans ...
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