exercise4_sol

Exercise4_sol - 251 College Physics I Exercise 4 Suggested solutions 1(a From the free-body Lliagral'u{Fig.3 1 for the load being lifted write

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Unformatted text preview: 251 College Physics I Exercise 4 Suggested solutions 1. (a) From the free-body Lliagral'u {Fig.3. 1) for the load being lifted. write Newton's 211d law for the Vertical Llireetion. with up being positive. Fr Fr — mg : ma : 0.1601339 =I> FT 2 1.1[img : l.16(285kg}(9.8l11n/1~'2] = 3.24 X 1031\- (its) The net work done on the load is found from the net force. Him 2 Furieoso'; : (UJBUmgH = 0.1609833kg)(9.80111_.f52)('22.0111) = 9.33 >< 103 J (t!) The work rlone by the cable on the load is: ll’Lflbsg = FTU’ eosU' = (1.1[img'1d = 1.10(285 kg‘}(9.80 mfsglfllfi m) = 7.13 x ll)"l . (d) The work done by gravity on the load is: “7.3 : mgdeosllo = —m.grf : —('285 kg)[9.8[}11'1fs2}[22.In} = —l3.l-’l x 104.] {:3} Use the work-energy theor)r to find the final speech with an initial speed of U. ll'mac = ICE-2 — 1031 = grueg — émi'? LEI-17m . 2(9.83 x 103.1] 2 _ ‘-- : I : — : 5 :> 12 m + t. (285 kg) + U 31 inf 2. Fig. 2 shows the motion of the three particles just after 3'! being given a velocity 1" and the two end particles are about to eollide. v V 1 M m M m v v m m M M Fig. 2 (a) Conservation of momentum gives: MV = (AI + 2m)v _ M V _ M + 2771 =>v conservation Of mechanical CRETE" gives: 1 ‘2 1 2 1 2 2 _) / — _A I u _ ‘ l 2 [L +2(2"l)(v +U) When the two end particles are about to collide, the speed of each m is equal to m : ally/'2 “‘ 1"[1'2 = — = V + r") r 2m 2171. A1 + 2172 3. (a) Fig. 3 shows the motion of the wedge and the particle when the particle has moved over a distance .5. ucos60° Fig. 3 Conservation of momentum in the horizontal direction gives: 0 = 2mV - m(ucos 60° - V) =(hflV—:+V 2 11 V. : _ =’ 6 (b) By the conservation of mechanical energy. 1 1 1 mgs sin 60° 2 5771(1er08600 — V)2 + 5mm sin 60°)2 + 5(2111)lr"2 2 \/§ 1 u u 2 l ufi 1 u 2 ’ 957-567) +5 2 +5‘2’(§) u2 3u2 (l2 11 3 =._ __ _.=._.2 => \/_ gs 9 + 4 + 18 12¢: 12¢3fi 2""=:V u ‘1. Use x-mlsvrvutiun uf 011011;}: The: Im-Vt-I of [11:3 bull n11 $110 llllt'UIllpI'!’SSWI wring taken as the Zr'l'n 1: unflinn fur but]: gl’m‘itatinlml Hi (1/ : (I) null r-lnstin' PE LI' 2 U). 'I‘akv up Tu 1w pnsiI’iVel fur hull], Fig. ~l (a) Subsrript 1 rcprt 50111.»; the hall at fhv hmm'h point. and subscript ‘2 I‘I"1')I‘O>‘(‘I]l.5 the Hill at Hm lut'ntium Wht'l't' i? jlls? bank [111‘ wring. 1-1? [110 1111w-1111I1‘INSHI It'llfJIh. E1213: 1 7;, 1k 3 1 2 1]_ 2 Emil +1111)!“ + j .11 _ 3/1142 + Him/2 + T; .12 u 1 2 1 2 0+0+§kx1 =—mv2 +mgy2+0 (950N/m) (0.150m)2 — 2(0.30kg) (9.8m/52)(0.150m) v = 2 8.26m s 2 m / [13} fin'osn'rrzipt :3 represents 711913311 m in: liigllrfE-i point. £1 = +33 1 ._. I .1 I u l 2 :.> Emmi + mflyl + Eh'l : Em H3 + mi'lilm + Ekr'a 1 2 0+0+§kx1 =0+mgy3+0 _ kxlz _ (950N/m)(0.150m)2 3'3 _ 2mg _ 2(0.30kg)(9.8m/52) 2 3.64m . n. (H) Um" ("1’J11~1'-1'\'z1!iu:11 of llH‘L‘lldniVfll nut-I'm”. :Muming Ihvl‘v are no 1nm-t‘ullSt‘rvmivu 1} 1111's. Subscript 1 represents the water at the top of the dam, and subscript 2 represents the water as it strikes the turbine blades. The level of the turbine blades is the zero location for PE (y = 0). we have v1 = U. y1 = 80 m. and y2 = 0. Solve for 19. E1 2 E2 1 ‘2 1 2 => 511115 + mgyl 2 5172-102 + mng 1 2 => 17293]; = inn-1’2 :5 1.9 : ‘l29y1 : \/2(9.81n/52)(81m) = 39.8 In (b) The energy of the water at the level of the turbine blades is all kinetic energy. and so is given by émug. 58% of that. energy gets transferred to the turbine blades. The rate of energy transfer to the turbine blades is the power developed by the water: 1m?) (O.:38)(650kg/s)(39.8kg/s)2 2 2 P = 0.58 (— t = —— = 2.99 x 105W" 6. Consider the free-body diagram for the coaster at the bottom of the loop shown in Fig. 5. The net force must be an upward centripetal force. N bottom "lg Fig. 5 ' mvgouom 2 F bottom = JrVbomoln _‘ mg : —R—- 2 , mv => Mam = $5 + my Now consider the force diagram at the top of the loop shown in Fig. 6. Again, the net force must be centripetal. and so must be downward. mg N”? Fig. 6 _ 2 m stop za=tWW=R _ 2 m ‘L‘t DD R =;> Nmp = — mg Assume that the speed at the top is large enough that Afton ) U" and so own 5 Now apply the conservation of mechanical energy. Subscript 1 represents the. coast-er at the bottom of the loop1 and subscript 2 represents the coaster at the top of the loop. we choose the level of the bottom of the loop to be the zero location for PE (3; : 0). So we have yl : U and ya : 2R. Therefore. E1 = Es 1 2 1 2 2 5111-111 + neg-y; = Eel-v2 + mng :> lfimm : U30” + 491? The difference in apparent weights is the difference in the normal forces. .' 2 371112 Nbonom — Nu], = (w + my) — ( 1;") — mg) = 6mg Notice that the result does not depend on either 1! or R. U! ...
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This note was uploaded on 02/22/2012 for the course CHEM yscn0027 taught by Professor Drtong during the Fall '10 term at HKU.

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