06prob_11

06prob_11 - Assigment #6, 5/6/11 Page 1 of 3 UC Davis...

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Unformatted text preview: Assigment #6, 5/6/11 Page 1 of 3 UC Davis Physics 9A-C Assignment #6 Cole [7.9]a)Thenormalforceisperpendiculartothedisplacement,soitdoesnowork.Thework donebygravityis mgh = mgR =0.98J.b)Whennonconservativeforcesdowork,theychange theenergyofthesystem: W f =E-E W f = 1 2 mv 2-mgh v = 2 gh + W f m = v = 2 9.8m/s 2 ( ) 0.5 m ( ) + 0.22 J 0.2 kg =2.8m/s. [7.30]Themagnitudeofthefrictionforceonthebookis f= k mg =(0.25)(1.5kg)(9.80/s 2 )=3.68N.a)Theworkdoneduringeachpartofthemotion isthesame,andthetotalworkdoneis W =-2(3.68N)(8.0m)=-59J(roundtotwoplaces). b)Themagnitudeofthedisplacementis r = 2 (8m)sotheworkdonebyfrictionis. W f = 2 8.0 m ( ) 3.68 N ( ) =-42Jc)Theworkisthesamebothcomingandgoing,andthetotal workdoneisthesameasinpart(a),-59J.d)Theworkrequiredtogofromonepointtoanother ispathdependent,andtheworkrequiredforaroundtripisnotzero,sofrictionisnota conservativeforce. [7.32]Theforceis F x = dU dx = C 6 d dx 1 x 6 = 6 C 6 x 7 .Onehydrogenatomparticleisattheorigin andtheotherat x. Thisforceisontheparticleat x andbecauseitisnegative,itactstowardthe atomattheorigin,soitisanattractiveforce. [7.37]a) F r = U dr = 12 a r 13 6 b r 7 b)Setting F r =0andsolvingfor r gives r min =(2 a/b ) 1/6 .Thisis theminimumofpotentialenergy,sotheequilibriumisstable. c) U ( r min ) = a r min 12 b r min 6 = a ((2 a / b ) 1/6 ) 12 b ((2 a / b ) 1/6 ) 6 = ab 2 4 a 2 b 2 2 a = b 2 4 a . Toseparatetheparticlesmeanstogiveenoughenergytogetthemtozeroenergy,andrequires E binding =0- E o =b 2 /4a . d)Theexpressionsfor E and r min intermsof a and b are E = b 2 4 a r min 6 = 2 a b Multiplyingthefirstbythesecondandsolvingfor b gives b = 2 E r min 6 ,andsubstitutingthisinto thefirstandsolvingfor a gives a = E r min 12 .Usingthegivennumbers,....
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06prob_11 - Assigment #6, 5/6/11 Page 1 of 3 UC Davis...

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