kean (msk955) – Homework 03 – yao – (56725) 1This print-out should have 24 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsTwo identi cal massless springs are hung froma horizontal supp ort. A block of mass 3.8 kg issuspended from the pair of springs, as shown.The acceleration of gravity is 9.8 m/s2.k k3.8 kgWhen the block is in equilibrium, eachspring is stret ched an additional 0.39 m.The force constant of each spring is mostnearly1. k ≈ 502. k ≈ 163. k ≈ 364. k ≈ 585. k ≈ 48 N/m correct6. k ≈ 147. k ≈ 448. k ≈ 779. k ≈ 18010. k ≈ 70Explanation:Let : m = 3.8 kg ,x = 0.39 m , andg = 9.8 m/s2.k x k xm gmDue to equilibrium of the forces,2 k x = m gk =m g2 x=(3.8 kg ) ( 9.8 m/s2)2 (0.39 m)= 47.7436 N/m .002 (part 1 of 2) 10.0 pointsConsider the 613 N weight held by two cablesshown below. The left-hand cable had ten-sion T2and makes an angle of 41◦with t heceiling. The right-hand cable had tension T1and makes an angle of 51◦with the ceiling.613 NT2T151◦41◦a) What is the t ension in the cable labeledT1slanted at an angle of 51◦?Correct answer: 462.919 N.Explanation:Observe the free-body diagram below.kean (msk955) – Homework 03 – yao – (56725) 2F2F1θ1θ2WgNote: The sum of the x- andy-components o f F1, F2, andWgare equal to zero.Given : Wg= 613 N ,θ1= 51◦, andθ2= 41◦.Basic Concept: Vertically and Horizontally,we haveFxnet= Fx1−Fx2= 0= F1cos θ1−F2cos θ2= 0 (1)Fynet= Fy1+ Fy2− Wg= 0= F1sin θ1+ F2sin θ2−Wg= 0 (2)Solution: Using Eq. 1, we haveF2= F1cos θ1cos θ2.Substituting F2from Eq. 2 into Eq. 1, we haveF1sin θ1+ F2sin θ2= WgF1sin θ1+ F1cos θ1cos θ2sin θ2= WgF1sin θ1+ F1cos θ1tan θ2= WgF1=Wgsin θ1+ cos θ1tan θ2F1=613 Nsin 51◦+ cos 51◦tan 41◦= 462.919 N003 (part 2 of 2) 10.0 pointsa) What is the tension in the cable l abeled T2slanted at an angle of 41◦?Correct answer: 386.009 N.Explanation:Using Eq. 1, we haveF2= F1cos θ1cos θ2= (462.919 N)cos 51◦cos 41◦= 386.009 N .004 10.0 pointsGiven: The surface is frictionless,Two blocks in contact with each other moveto the right across a horizontal surface by thetwo forces shown.4 kg 6 kg80 N 20 NWhat is t he magnitude of the force exertedon the block with mass 4 kg by the block withmass 6 kg?1. 62 N2. 59 N3. 53 N4. 65 N5. 56 N correctExplanation:Given : F1= 80 N ,F2= 20 N ,M1= 4 kg , andM2= 6 kgM1M2F1F2akean (msk955) – Homework 03 – yao – (56725) 3The common acceleration of the system i sF1− F2= (M1+ M2) aa =F1−F2M1+ M2. (1)This result is obtai ned by treating the twomasses as a single object.M1+ M2F2F1(M1+ M2) gNTo determine the force exerted on M1byM2, we need t o analyze M1separately.M1F1F12M1gNFrom the free body diagram of M1we seethat the only two horizontal forces acting onM1are the external force F1and the force F12exerted by M2, resulting in the a ccelerat ionof the massF1−F12= M1a , soF12= F1− M1aF12= F1− M1F1−F2M1+ M2F12= (80 N) − ( 4 kg)×(80 N) − (20 N)(4 kg) + (6 kg)F12= (80 N) − ( 24 N)= 56 N .005 10.0 pointsTwo blocks m1(9 kg) and m2(18 kg) are con-nected by strings, as shown in the diagram.String 2 breaks when T2= 4.7787 N. Find T1when string 2 breaks.m1m2T2T1Correct answer: 14.3361 N.Explanation:Find T1when T2breaks at 4.7787 N.XFxon m1= T2= m1a⇒ a =T2m1=4.7787 N9 kg= 0.530967 m/s2XFxon the whole system= T1= (m1+ m2) a= (m1+ m2)T2m1= (m1+ m2m1) T2= (1 +m2m1) T2= (1 + 2) 4.7787 N= (3) (4.7787 N) = 14.3361 N006 10.0 pointsA 5.4 kg object hangs at one end of a rope thatis attached to a support on a railroad boxcar.When the car accelerates to the right, therope makes an angle of 32◦with the verticalThe acceleration of gravity is 9.8 m/s2.a5.4 kg32◦Find the acceleration of the car. (Hint:~aobject= ~acar)Correct answer: 6.12372 m/s2.Explanation:Given : m = 5.4 kg ,θ = 32◦, andg = 9.8 m/s2.kean (msk955) – Homework 03 – yao – (56725) 4TT sin θT cos θθm gVerticallyXFy= T cos θ −m g = 0T cos θ = m g . (1)Horizontally,XFx= T sin θ = m a . (2)Dividing Eqs 1 and 2, we haveT sin θT cos θ=agtan θ =aga = g tan θ=9.8 m/s2tan 32◦=6.12372 m/s2.007 (part 1 of 2) 10.0 pointsA block of mass 4.63 kg lies on a frictionlesshorizontal surface. The block is connected bya cord passing over a pulley to another blockof mass 2.46 kg which hangs in the air, asshown. Assume the cord to be light (masslessand weightless) and unstretchable and thepulley to have no friction and no rotationalinertia.4.63 kg2.46 kgCalculate the acceleration of the first block.The acceleration of gravity is 9 .8 m/ s2.Correct answer: 3.40028 m/s2.Explanation:Let : m1= 4.63 kg andm2= 2.46 kg .m1m2aTNm1gaTm2gSince the cord is unstretchable, the firstblock accelerates to the right at exactly t hesame rate a as the second (hanging) block ac-celerates downward. Also, the cord’s tensionpulls t he first block to the right with exactl ythe same tension T as it pulls the second blockupward.The only horizontal force acting on the firstblock is the cord’s tension T ,so by Newton’sSecond Lawm1a = Fnet→1= T .The second block feels two vertical forces:The cord’s tension T (upward) and the block’sow n weight W2= m2g ( downward). Conse-quently,m2a = Fnet↓2= m2g − T .Adding,(m1+ m2) a = m2ga =m2m1+ m2g=2.46 kg4.63 k g + 2.46 kg(9.8 m/s2)=3.40028 m/s2.008 (part 2 of 2) 10.0 pointsCalculate the tension in the cord.Correct answer: 15.7433 N.Explanation:kean (msk955) – Homework 03 – yao – (56725) 5T = m1a = (4.63 kg) (3.40028 m/s2)=15.7433 N .009 (part 1 of 3) 10.0 pointsThree masses are connected by light stringsas shown in the figure.m2m3m1The string connecting the m1and the m2passes over a light frictionless pulley.Given m1= 4.63 kg, m2= 1.49 kg, m3=4.49 kg, and g = 9.8 m/ s2. The accelerationof gravity i s 9.8 m/s2.Find the downward acceleration of m2mass.Correct answer: 1.24694 m/s2.Explanation:Consider the free body diagrams:m2m3m1T1a ↓T2a ↓T1↑ aApplying Newton’ s second law to each ofthese ma sses we getm1a = T1−m1g (1)m2a = T2+ m2g − T1(2)m3a = m3g − T2(3)Adding these equatio ns yields(m1+ m2+ m3) a = (−m1+ m2+ m3) g ,soa =−m1+ m2+ m3m1+ m2+ m3g=−4.63 kg + 1.49 kg + 4.49 kg4.63 kg + 1.49 kg + 4.49 k g× (9.8 m/s2)=1.24694 m/s2.010 (part 2 of 3) 10.0 pointsFind the tension in the string connecting them1and the m2masses.Correct answer: 51.1473 N.Explanation:From equation (1),T1= m1(a + g)= (4.63 kg)1.24694