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111-20051-MT2 - QJ A box oFM = 40-kg mass is pulled for a...

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Unformatted text preview: QJ. A box oFM = 40-kg mass. is pulled for a distance ofs = 5.0 tn alonga horizontal floor with a constant horizontal force of F = 140 N. The kinetic coefficient of friction between the box and the floor is pt = 0.30. 3) Calculate the work done by the applied force. I 3' W = Fri Cat-50° = {lhON)(5m)l1}=—-?00 patriot m l? b) Calculate the work done by the friction force. wq :: \fd an—alfi’tfi :: (ltrk mild fiebla‘d Luf=/3.:x1,axlbnj{5m)f-t) = -£00 ,Tou/e». i‘ c) When the force of]: = I40 N is applied initially. the box was moving at v.= 2.0 mr's in the same ' / direction with the force. Calculate the final speed ol‘the box. NM} 2 Kj - Kt. == fmcfil- i-MU‘EL no - 600 = 10 U531 -— 20 extra ea vim/s dl Calculate the instantaneous power delivered by the force F. at the moment when it is applied initially. -_‘5__)! F? —__- F.U\v=_-{/AON){prs)(gson = j.- l e) Calculate the average power delivered by the applied force F. for the distance of s = 5.0 H1. IL. :U‘. +Cl‘l' when: a=fl=ifllfl"l2° :J.ws:. \5 t m 11 O 1 f 50. 3V5=lmfl+il =2; +=zsmm ileum) the axle-tag Fewer 35 EN: 350 LU p,1U = W :- 10-33; :2 zsowotls ” 413+ 25 Q.2. A msssless and frictionless pulley is attached to a spring of force constant k, hanging from the ceiling. A light string. fixed to the floor at one end, paSSes over the pulley and is tied to a block of mass M at the other end. The block sits on a table hinged at the wall as shown in the figure. In this position the height ofthe block from the ground is yo and the spring is in equilibrium length, that is neither streched nor compressed. Then the table is released down suddenly and the system starts to oscillate. a) When the block is at a height y from the floor, how much is the spring stt'eched (Give it in terms of ya and y)? 33) Write down the total mechanical energy ofthe system when the block moves at a speed v at the height y From the. floor{ Give E..1 in terms of M. k. y. yo and g }. view “ii” Pl v IF— ' c} Calculate the minimum height. y“... reached by the block { Give 3:...“ in terms of M. k. yuand g )_ ‘.' ,3 llvtd 'lU(fl:V_'-»\ Eclfll— Gill L‘J‘Jir'l-I ‘J'TU fw‘lt‘ I. 1 _ _‘ a COM fiery; Ham ;~; dqus-l‘vI E L ‘- il \i-l _:1 s _ t, .. \Jytfi’o 0”“ R V55 ' . . \‘d I )J [\I’O " 7’~""-‘:. lvlj \IO :: Lh- rw-Zn '1' 2 ‘ f. 1| 1. l- . llh'ld [‘19'7,“'Y\} — a: R_ 1’4} "nyuiiu J '3. “TM a \ ——"‘="U k :2 Yo _ Iii-ll ‘ d) Calculate the maximum speed. vm, , ofthe block ( Give v...“ in terms ol‘M, k. and g }. .. ‘ 2. eff: l’or amt. M HE: E‘ “5 Mj‘fo EMUMI‘: Mfi' L (“I int-14M u -+ i—ls“fo“"lsz:‘\-lc ’thflfl 9-. ” ' jl 3 l‘ l 'L 1R 1 $431 1513‘ " 5— MV = "ii—"5:7: a luau-Hf) 1. :- .6 (ll ark—Eh "9 Hus m.“ ‘ VIM: Li?“ 1' 1. £ ' "‘19' =0 WW8 H Ix 0.3. Two blocks owaod with masses M|= 500 g and M3: 490 g'are d = 2.0 m apart on a frictionless table. MI M1. A bullet of'rn = 10 g mass is fired toward the blocks as HVL I l I I Shown in the figure. It passes all the way through the first block. then embeds itself in the second block.The speed l‘" ‘ -—3'l j“ ° of the second block immediately after the bullet stops in it, is V2: 2.0 mls. Then the two blocks again make a completely inelastic collision and afterwards move together at V3: 4.0 mls. 6 :1) Calculate the speed ofthe first block immediately after the bullet leaves it. ._:: ._* i e; z, z 'efl l ‘M‘Vt 4' (“‘1 +‘“ Q V2 : (m. +m1+mbl \lg 05W + 0.3 $2.0 : L0 ego «we 5 h) Calculate the speed ofthe bullet immediately after it leaves the first block . 4 U. II E O >3 z 4.9%“; fin/5 5 1;) Calculate the speed ol'the bullet immediately before I1i[S.[lIe first block . DONUL : 0.0 “JOB —t- (Lg-18.0 \Jl: = 4'00 M/{ 2 5 d) How much kinetic energy is lost in these collisions? ! __ I L 2 k: “3: K1 — il*m.b\'lb : ti WHO-+003 A l at: 2. =. K: T'— ‘300 J Q.4. A system consists ofthree particles ofmasscs M1= 0.10 kg, M3: M3: 0.20 kg. Thcir positions are given as r.= (2.0 t i - 3.0 r‘ j) m. r; = 2.0 t i m and r5 = 4.0 t2 j m, respectively where t is in seconds. @ 3) Find the position ofthe center of mass ofthis system as a function oftimc. in unit vector notation. —b -—a —-9 I '9' 1T. . IT. _ ,- t' .—= Mm. +MLC+M3K EM : Walton—3:01: AJW-mE-C‘LWOIE‘QQ rum 2 M‘ +M1+M3 {c.xu+o.1c+o.a.c "' _ (cacti-0,3e-L‘§)+olnnt7+c.3et‘a' fl 0.6DtLi-C15ct1'.l rm“ _ 0.50 _ 5.50 ‘4 -_" 3‘- T: _ I. ' '0 't M "'-—I- ? ch — l l l' ‘rl a ( ) YE“ :lrl‘t l 456—? m @b) Find the velocity ofthc center ofmass of this system as :1 function oftime. in unit vector notation. glam = 3:“ gm :3? “it? +639): i121; .4. Lot} i2.“ :l. Liz's-1.01”: “[3 @ c] Find the acceleration oFtlic center ol'mass of this system. in unit vector notation. _._') r _ 7-9- ”. ___~"_"_ “1 =__ ‘1 1.2..t’+1.ot d‘”_ ell: 0'" El‘T( 3) act" = m/SL --\-'b 7’ . @d} Find the net external force acting. on this system. in unit vector notation. = KM . +M L+ M3>3Lm : (o,'.01—0.Lc +0.‘1&)2..D~5 mlg‘llfii s; r (r = 0.3:: 712110;; =- [ft-3'03) Eur. = to? UK") 0.5. A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at [2 revfs. After 80 more complete revolutions. its angular speed is 28 rew’s. r . 3) Calculate the angular acceleration. Express it in radfsxlmit. I'm. 3. . . [032‘ : LU!- +1xe (5mm: K :3 cowatl'ani') :5 A _ uu‘L 4U? _ (mint—(:2)Z reulgL 2'9 H g" r” = 71' rd 5 % x = L, TIN/5?“: A x11 Tad/S: = XI Foal/5?- flp, . b) Calculate the tirne required to complete the 80 revolutions mentioned. ’ r-r LU3 = LU; -r rC'l' , ' w i— tau/5 w +55)! Lut_ 23r/s :22 _#3 M l. ran/5 {r = l, 5 ('3'! _ c) Calculate the time required to attain the l2 rew's angular speed. -III- . j (d I I +__, “a: =I1WS= 35 x it YrU/sz +’ = 3 5 A. d) Consider a point on the disk at 10 cm from the center. Calculate the cenripetal (radial) acceleration “'1 of'tltis point when the disk rotates at 12 rcvls. I 2— L1" ”' f l" LU ) 2 a, = —— # L‘s—~— r no l‘ l‘ , z 1' m 1.. ar : (01M) ('IZXZJT NJ,ziz) = 5F.§7t “XL ‘3 558 ’4 air 2-] EEWHW 5: a e) Calculate the tangential linear acceleration ofthe above mentioned point. (14 = m :— mmwm “1%) = 0.3% nit/52:2,?” a =03” w/ 1 ...
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