X2.17a 2050PHYS-205(17) (Kaldon-20619) Name _________S O L U T I O N_______WMU-Summer I 2006 Exam 2A - 100,000 points + 20,000 ¶ points Book Title __________________________________________06/04/2006•Rev.4State Any Assumptions You Need To Make – Show All Work – Circle Any Final AnswersUse Your Time Wisely – Work on What You Can – Be Sure to Write Down EquationsFeel Free to Ask Any Questions¶2a ¶2b ¶2c ¶2eEXAM 2 [FORM - A]PHYS-2050 (KALDON-17)SUMMER I 2006WMUIf F=ma,then does F²=mama?Physics 2050 / Exam 2 [Form-A] Summer 2006 Page 2Problem One – “The One That Goes Up and Down” (35,000 points)1.) (a) A cargo elevator at an open mine shaft is being pulled on by means ofa rope. Draw the Free Body Diagram of the loaded elevator car at rest.Assume all the cable tensions are the same.(b) The weight of the loaded elevator at rest is 10,500 N. With what force must the guy pull on the rope?FTwTwTwNNyå=-==== =303310 50033500111,.(c) If the elevator is to be lowered at 1.11 m/s², with what force must the guy pull on the rope?wmgmwgNmskgFTwmaTwmaTwmaNkgmsNyyyy=== ==-==+=+=+-=å10 500981107033310 500 1070 1113310421112,./.,../bgch(d) If the elevator is to be lowered at 1.11 m/s, with what force must the guy pull on the rope?FTwTwTwNNyå=-==== =303310 50033500111,. Same answer as (b)!Physics 2050 / Exam 2 [Form-A] Summer 2006 Page 3(e) If there was no air resistance, then the elevator would be in free fall the whole way. Use conservationof energy to find the speed of the elevator after it has fallen 933 meters.NOTE: Missing part of the sentence here! If the cable breaks and there was no air resistance…mgh mv mgh mvmgh mvgh vvghvghms mms11212212221122211222221212222 9 81 9331353+=+======././chbgNOTE: OK if student starts with v1 = -1.11 m/s from part (d) rather than from rest…mgh mv mgh mv hmgh mv mvgh v vvghvvghvms m msms11212212222112121222112121222221122112220222 9 81 933 111135 3+=+ =+=+==+=+=+-=;./ ././chbgb gPhysics 2050 / Exam 2 [Form-A] Summer 2006 Page 4Problem Two – “The One With The Star Problems” (30,000 points)2.) ¶(a) A small red ball (mass m = 0.200 kg) is moving to the right with a speed 10.0 m/s at time zero.It is attached to an elastic cord which applies a force to the left with rFCxielastic=-$. Find the distanced it takes the ball to come to a stop, where x0 = 0 and the coefficient of the cord is C = 5.00 N/m , usingwork and energy.Kmv kg ms JWFdsFdx Cxdx C xdxCxCdWKKK JCdJCddJCJNmdJNmmxdd ddfi== ==×==-=-=- =-==-=- =-=====zzz z12212200 02022220200 100 10002210 00210 0022100020 0050020 005002000../....../../.bgbgbgbgrrDAn object of mass m = 12.0 kg begins its motion with the following equation. All other constants arezero.¶(b) Find the force, F, acting on this object at time t = 1.00 sec.adxdtms tFma kg mstkg m s tFt kgm s N=====×== × =226464646412 012 0 12 0144 0100 144 0 100 144 0(. /).(./)(. /)( . sec) ( . / ) . sec .bgbgdxdtms t226412 0= (. / )Physics 2050 / Exam 2 [Form-A] Summer 2006 Page 5¶(c) A block of mass m = 1.00 kg starts out at rest at a position x0 = 4.00 meters, and moves with a forceFt txNs() .= 20 022di. Find the position x of the block after 5.00 seconds. Assume any other constantsare zero.Ft tFmaaFmtkgms tdxdtvdxdtdxdtdt m s t dtms t C Cxdxdtdt m s t dtms t C C x mms txNsxxxxNsx() ..../././ ( )././ ( .)./==== = === ==+====+===+zzzz×20 020 010020 0020 0020 00 020 0020 00 4 001667 42222422222421343134313444044didichchchchchch.(.sec). / .sec .00500 1667 500 400104644mxt m s mm== +=chbg(d) Jerry is putting grocery carts away at Meijer. He takes one of thecarts (m = 10.2 kg) and gives it a shove so it is traveling at 4.75 m/s.It crashes into and sticks with a group of five grocery carts that werejust sitting there. Find the speed V of the new group of grocery carts.ppmv mv mv mv mv mv m m m m m m m Vmv mVVmvmvmsmsbefore after=+++++=+++++++==== =11 22 33 44 55 66 1 2 2 3 4 5 60666475607917bg././Physics 2050 / Exam 2 [Form-A] Summer 2006 Page 6¶(e) An object of mass 1.00 kg has a motion that follows the following equations. Find the magnitude Kof the kinetic energy of the object at time t = 0 and t = 10.0 seconds. Then find how much Work wasdone to move the object – It Won’t Take Calculus for this last bit.xt m m s m s() . . / . /=+ -700 600 50023tt23xt m m s m svdxdtddtmms msms msms msvvt m s m smsKKmvif() . . / . /../ ./(. / ) (5. / )(. /) (. /)()( . sec) ( . / ) . sec) ( . / ) . sec)./=+ -== + -=+ -=-=== -=-===700 600 500700 600 50002600 30012 00 15 000010 0 12 00 10 0 15 00 10 0138002323232323122tttttttt((2323222ch122100 1380 952 200952 200 0 952 200../,,,kg m s JWKKK J Jfibgb g-===-= -=DPhysics 2050 / Exam 2 [Form-A] Summer 2006 Page 7Problem Three – “The One With All The Angles” (35,000 points)3.) Frida has to get a refrigerator ( m = 135 kg ) loaded onto a truck, using aramp that is angled at 22°. If the refrigerator glides on frictionless rollers,find (a) the force needed to push the refrigerator up the ramp at a constantspeed.FFwFw mgkg m sNxxx¢¢¢å=- ====°=1120135 9 81 224961sin./sin.qbgch (b) The ramp is 1.50 m tall, while the angled part is 4.00 m long. Show that the work to raise therefrigerator 1.50 m is the same as the work in pushing the refrigerator up the slope without friction.Wmgh kg ms mJWFd N m J===== =135 9 81 15019874961 4 00 19842bgchbgbgbg./ ...(c) If the elevator did not have the rollers, that it could sit on the ramp and be held in place by staticfriction without being held by Frida. Find the minimum coefficient of static friction, ms .FFwFF wFFwwmg mgmgmgyNyxfs xfs s N s y xss¢¢¢¢¢¢åå=- ==-=== =====°=0022 0 4040,,max,,maxcos sinsincostan tan .mmmq qmqqqPhysics 2050 / Exam 2 [Form-A] Summer 2006 Page 8(d) The actual coefficients of friction between the refrigerator and the ramp are 0.857 and 0.657. What isthe maximum force Frida can push the stopped refrigerator without having it move up the ramp?FFw FwFFF wFF wwwmg mgmgkg m sNyNy Nyxfsxfs xsy xss¢¢ ¢¢¢¢¢¢åå=- = ==- - ==+=+=+=+=°+°=00135 9 81 0857 22 221548112;cos sincos sin./ . cos sin,,max,,maxmmq qmq qbgbgchbgch(e) What is the force that Frida needs to apply to the refrigerator to get it up the ramp if is sliding at aconstant speed with friction?FFw FwFFF wFF wwwmg mgmgkg m sNyNy Nyxfkxfk xky xkk¢¢ ¢¢¢¢¢¢åå=- = ==- - ==+=+=+=+=°+°=00135 9 81 0 657 22 221303112;cos sincos sin./ . cos sin,,mmq qmq