Solutions Practice Test PHYS 211 Exam 2

Transcription

1 Solution Pactice Tet PHYS 11 Exam 1A We can plit thi poblem up into two pat, each one dealing with a epaate axi. Fo both the x- and y- axe, we have two foce (one given, one unknown) and we get the following equation: F x =.5 + x = ma x F x =.5 + x = (1.7)(3.1) x = 7.77N F y =1.4 + y = ma y F y =1.4 + y = (1.7)(.5) y = 5.65N Since we now have the x and y component of the foce, we ue Pythagoean equation to get F = 9.61N

2 A A omewhat confuing poblem without the coect fee-body diagam, hown below: A alway, it bet to put all you foce vecto into thei x- and y-component (if they aen t aleady) o we hould beak apat the applied foce, F, into it y-component, Finθ, and it x-component, Fcoθ. Now, and hee the impotant pat, ince the book in t liding down (o moving anywhee eally) thee no net foce acting on it in eithe axi/dimenion. Thi baically mean that the um of all the foce pointing up mut equal the um of all thoe pointing down. And alo that the um of all the foce pointing left equal the um of thoe pointing ight. Fom that we get the following two equation: (1) f + F inq = mg () n = F coq The fit equation can be e-witten to how how fiction can be boken down into it contituent pat, and afte we ubtitute n fom the econd equation, we can then olve fo F: f + F inq = mg µ n + F inq = mg µ F coq + F inq = mg F( µ coq + inq ) = mg F = mg = 5.3N µ coq + inq

3 3D The 100N foce will be applied to the ytem (which ha total ma 10kg) cauing it to acceleate with a = F/m = 100/10 = 10 m/. Thu we know that the 6kg ma will be acceleating at 10 m/, and o the foce acting on it mut be given by F = ma = (6)(10) = 60N 4E All of the tatement could be tue: i. If the elevato wee acceleating in eithe diection the nomal foce would not be equal to the man weight, howeve ince they don t mention acceleation we mut aume the acceleation i zeo, and the upwad velocity i contant. ii. iii. The acceleation would have to be in the upwad diection fo nomal foce to be geate, which would mean that although the elevato i moving downwad it i lowing down due to an upwad acceleation. Thi i alway tue when acceleating downwad. 5B Since the cate i at et, we would need to ovecome the foce of tatic fiction in ode to get it moving. The maximum foce of tatic fiction i given by f = µn = (0.5)(40) = 0N Thu, it i clea that the 1N foce applied will not be geat enough to ovecome tatic fiction and the block will not move. In fact, the foce of tatic fiction will imply match the applied foce of 1N, enuing that the object emain till.

4 6C Looking at the pulley ytem, we can geneate the equation fo the entie ytem that ay the net foce acting on the pulley ytem i: F Net = m 3 g = (m 3 + m 9 )a a =.45 m / Now we have acceleation we can et up the equation fo net foce on eithe object to olve fo tenion. Fo the 3kg ma the equation become: F 3 = m 3 g T = ma T =.05N 7A Thi i actually a athe ticky quetion. Given the infomation povided we cannot actually calculate the aveage fictional foce, howeve we do know that ince the bea i able to pick up peed duing it decent, the weight mut be geate than the foce of fiction. The weight of the bea i mg = 45N, o the aveage fictional foce mut be le than 45N, and only one anwe obey that ule. 8A Since the ball motion i vetical, the weight component will not alway be in the ame diection elative to the tenion (which will alway point towad the cente). Fo example, at the top of the cicle both the tenion and weight will point down towad the cente, while at the bottom the tenion will point up and the weight vecto will point down. Ou goal i to figue out what i the laget contant peed that we can ue at all point along the cicle without the ting beaking. Expeience tell u that the bottom of the cicle i whee the tenion will mot be put to the tet ince it will have to contend with the weight component acting againt it, o that i likely whee we will have to calculate ou peed. At the bottom we get the following equation: F C = T mg = mv With the ma, tenion, and adiu aleady given, we can olve fo velocity and get v = 11.1 m/.

5 9E It impotant fo quetion like thi not to get too bogged down on the detail of the poblem. Thi quetion i imila to that of the man in the elevato acceleating down, except thi time we ve eplaced nomal foce with tenion. The ball peed i inceaing in the downwad diection, o it mut be acceleating downwad. We can calculate the acceleation eaily uing: a = DV/Dt = [(-4) (-)]/ = -1 m/ A alway we can et the net foce equal to ma x acceleation and then olve fo tenion. Since the weight vecto point down and the tenion point upwad, the weight mut be lage a we ae acceleating down, and we get: F = mg + T = ma T = 4400N 10E A dicued at the eview, we want to otate the fee-body diagam fo cenaio like thi to minimize the amount of algeba equied. In doing o, the nomal foce now point along the +y-axi, while fiction goe along the x-axi (up the incline) and only the weight vecto i now not aligned with eithe axi. A uch, the weight vecto (mg) i boken into it x- and y- component, with mgco40 diectly oppoing the nomal foce, and mgin40 diectly oppoing fiction (down the incline). Since thee i no net foce along the y-axi we olve fo nomal foce, n = mgco40, and plug it to the equation below. The eulting net foce along the incline (which i now the x-axi) become: F Net = mginθ f k = ma F Net = mginθ µ k n = ma F Net = mginθ µ k mgcoθ = ma a =.9 m / With acceleation now given, we can ue the kinematic equation to olve fo final velocity: V = V 0 + ad V = 6.03 m /

6 11B The fee-body diagam how the net foce along each axi would be a follow: F Net,y = n mg = 0 Thu : n = mg F Net,x = ma F f k = ma F µ k n = ma F µ k mg = ma F = ma + µ k mg Since we aleady have the ma, and the coefficient of fiction, all we need to olve fo the applied foce, F, i the acceleation, a. Auming the cate tat fom et, it initial kinetic enegy would be zeo, thu the final kinetic enegy mut be equal to 600J. Thi allow u to olve fo the final velocity, vf, uing the following equation: K final = 1 mv f v f = K final m = m / Uing the kinematic equation we can now olve fo acceleation: v i = 0 v f = m / a =? t = Δx = 75 m v f = v i + aδx a = 8 m/ Latly, we plug in the acceleation into the above equation to olve fo applied foce, F: F = ma + µ k mg = 9.96N

7 1A The only foce acting on the ca ae it weight (pointing down), and the nomal foce (pependicula to the incline). Thee no need to otate the diagam hee, jut plit the nomal foce into x- and y- component, with ncoq oppoite weight along the y-axi, and ninq pointing along the x-axi diectly towad the cente of the cicula tack, thu making it the centipetal foce. Since thee no net foce along the y-axi, we can et ncoq = mg, and then olve fo nomal foce, n = mg/coq, which i plugged into the equation below. We can now et up the equation fo centipetal foce: F C = ninθ = mv F C = mg mv inθ = coθ which implifie to: gtanθ = v v = g tanθ 13B Hee we e looking fo centipetal acceleation, which only equie velocity and adiu. We get velocity by conveting evolution pe econd to mete pe econd: (p ) (p 0.5) m v = = = p m t 1 And o we get acceleation: v ac = t (p ) = 0.5 = 8p m

8 14E Befoe we can olve fo the dag foce at 30 m/ we need to ecognize that the dag foce at teminal velocity i equal to the weight of the object, mg. Since dag foce i popotional to the quae of the velocity, and change the velocity mut be quaed befoe we can pedict the change to the dag foce. Let D epeent the dag foce at teminal velocity (v = 80 m/), and let D epeent the dag foce at 30 m/. Since D µ v, " D' $ 30 # 80 v % ' & Thu : " = 30 % $ ' v = 0.141v # 80 & D' = 0.141D = 0.141mg = 7.6N 15C Fo a quetion like thi we need to take all the foce acting on the object and put them into the x- and y- component, that way we can et up equation fo net foce along each axi. In thi paticula quetion, we ae only eally inteeted in the y-axi, ince that i whee we ll find the nomal foce: The foce pointing up will be the nomal foce, n, and the foce pointing down will be the weight, mg, and the vetical component of the 00N foce, 00in0 = 68.4N Since the y-axi ha no net foce, the um of foce pointing up equal the um of foce pointing down, thu we et up the equation fo the y-axi: F Net,y = 0 : n = mg + 00in0 0 n = 313 N

9 16A It i impotant to undetand that vetical cale, uch a thoe hown in the poblem, will have a tenion foce ditibuted evenly thoughout the length of the ping. The eult i imila to what we ee when any ma i hung fom a ting o ope, thee i a continuou tenion foce pulling it in both diection. The fact that thee ae two cale make no diffeence compaed to if thee wa only one cale. In both cae, thee i continuou tenion foce equal to the weight of the fih ditibuted along the ping. Thu each cale will ead 16 N. Note that even if thee wee 0 cale all in a ow, auming they wee all male (like in thi quetion) then they would all ead the ame value of 16 N. 17C Remembe that on flat uface the only foce eponible fo tuning a vehicle uch that it doen t lip i tatic fiction, which point along the adial axi and i equal to the centipetal foce. Along the vetical axi the weight vecto, mg, and the nomal foce, n, oppoe one anothe and cancel out, uch that we can wite n = mg. So, when etting up the equation fo centipetal foce we get: F C = f = µ n = µ mg = mv v = µ g =10 m/ 18B Accoding to dag foce, the equation fo dag foce i popotional to uface aea. Given the volume of each cube we can detemine the ide length of Cube A and Cube B ae 3 cm and 4 cm, epectively, meaning thei co-ectional aea ae 9 cm and 16 cm, epectively. Since all the othe vaiable ae contant fo both cube, the atio of Dag Foce fo Cube B to Dag Foce fo Cube A i DB / DA = 16 / 9.