Practice Exam 1.1 Solutions

Practice Exam 1.1 Solutions - Name SOHJ‘EQ‘D—é _...

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Unformatted text preview: Name SOHJ‘EQ‘D—é _ Problem 1. (20 points) A crate is at rest at point A when it is nudged down the incline. The distance from point A to point B is 7 m. The. distance t‘mm point B to point (.2 is also 7 m. If the coefficient of kinetic friction (#3:) between the crate and the incline is 0.30 from A to B and 0.22 from B to C, detem'tine its Speeds at points B and C. meflwool 4 f1 FBI) N Rh “X Fictiti- Zp‘jzma?’ iF-x’MCKx N‘macoie =0 M38119 —F]a,.u:m@x / N =mjcose Mg game Vial-:MQCOSG-‘W’iax Fffici- Z/MKM WeMchcz-E e amneH/(«ikg C0§8 :ax A41.- 3 ax: “$1.81 (9m $10”-O.Sccs 510‘): c.5899 Lja v51: uoz flax (his) : 3035899060 = 3. Bic C 0w = (1,8ICS-in10°*0.99 (‘03 :05) = 'O.42!9’ 2 “In 536: => 1333;);8? L3- SDt/t Vca —. V32 + am (x~ x0) -= ease: +a(-o. 4;: mm =7 V5,“: 1.53 3% “(18+th 9 »work energy A +0 B 2 u : AT:T‘_ITV ‘3 (mafia 30“)(=})—(o.3m3cog :10“)(;t):_;13muE % U3 :3=8:}' u=AT:Tc-/n/° ' Mac "‘3 (#sin :51“; ?-.s‘m10“)~(o¢3mgcca MN?) - 4- -\C. _ 2 Answer: SpeedatB 1;; “(O‘QQmC‘jCO‘a IC' —é~mL/C A m :. .. 22],. 11’] Answer: Speed atC _] .83 ‘5 3 UL 6 3’ *1: Name 80 {ASH 10$ Prublem 2. (30 points) A 3-kg block at point A is released from rest in the 60" position Shown and subsequently strikes the l-kg cart at point B. (a) If the coefficient of restitution for the collision is c = 0.7. determine the velocities of the block and cart just after the collision. (b) Determine the maximum displacement s of the can beyond point C. Neglect friction. (0*) _,-n -,- V30 fifflr ’j‘j-x m9“ \‘5 N block A Stunt): down um" ‘ . 9'15(i'3_1-8CGS cc>°)=E—/T,Z¥%yitg (61,31)(i.3)(1—ms c0”) : gm: uqz‘: L:- "“i‘ Congo!" c cch' on 0(3 meme niu m u 3-"? ma VA, + mg':J : rYlAVflz +maVs2 {3)(4.aoa) :('3)VAZ+(1)V32 E z 09 l} : UBZ‘VA-L : VBZ'VAL VAE'L/r'1_ O 4.,309 d— _ .. c - 2 (0.71 M4409): fizz-VA; 3.15431—355m 3c. : 3(5ofiel Sch/e $ch VAZ 4— v n Solve L2,) 5,:- aas m ‘ $01K S m Va; 2 .2. 4; ‘5 —» V32“: 5.3% LE —.» . m . “(g m Answers: Velomty of block 0? - 4 I») S Velomty of cart f) " 5 Answer: Maximum displacement 9' 3 m Name SCiL‘x-Tiofié Problem 3. (35 points) A 2-kg block of wood with an initial velocity of 15 me’s slides along a surface for 10 at before reaching a cliff. The coefficient of kinetic friction between the surface and the block is to; = 0.35. (a) Find the velocity v] of the block at the edge of the cliff. (5 points) (b) Calculate the distance d from the edge n’r'the cliff to where the block hits the ground. (10 points) (c) Calculate the velocity v; of the block when it hits the ground. (10 points) (d) A platform supported by a spring constrained to move only in the vertical direction is placed at the distance d calculated in part (b) even with the ground. The block lands on the platform at the speed calculated in part (c) and strikes and maintains contact with the platform. The acceleration of the body after impact is a = g—cy, where c is a positive constant and y is the displacement measured from the original platform position. If the maximum compression of the spring is observed to he ym, determine an expression for the constant c in terms of ym and v0. Assume v0 is the vertical component of velocity v; that you calculated in part (c). Assume that the mass of the platform is negligible. (10 points) “" __ W‘IMC’J . . Y . (at kBD . -- L es"- Pas-ca Tatuo-_1= ti épy:l’h \f «nus-:5 KNflO :* U1 —m3+;\;=0 Misti—Z at}; frtiw" Ngmrd = 90%”: W02 N -. - to - . . ‘ I Ffiucr = 03506.:22): (5.36:?UN! vi - ‘3-503 De 9 h—-—.+ . d (b = (+6016 X— X1 ' ' ) Y O) abbzcoszes rgfiagfrlhflj Ufl m Vt: :3": :13- 5—53 9% Yaw “Byte-Eeit- 6‘0. «a0 f; Raffle-ta )7- calculated in part (c) forvn X;~IOO o::oo+o+€5)t-fl)éz _ 41/0 7 Z {1:4,er sec ._ -too:-‘l£i d ) VXZL’OX+%?f*Cc) g l a. 12.933 V‘xTUay : la.S'C'ES M —» oi: bot 4S4 m T 5’ VHF“? HM (f—éc) V?! 2 0‘3 (4.5lf—Ci) Vy: Mia-Mag. 4r : ~443¢H 9% at: bfi-‘(RSCE)(4.S‘:S‘): sets m _._‘3 VA; MATC- i‘-— 4.40;;93‘ m 6 Name I I . a Problem 3. Continued (workspace) if 2%" M I T - ow Y 9:0 \f m 5:1 mm: n. v:c V0 0 (Chanel-n: diféCf‘fonS) w “(=0 " - _I_ ~ _ _ I 2 7m J J aV ) _ 9y 30,! Va 0 "—l ' :. - .. I 1 3‘0 3?“ 3C)?!” [ '2. :iCYm L jymi-évo—‘L 0" Ym" . "~ m Answer: Part(a) velocity Vloftheblock IQ-56‘3’ ‘3 Answer: Part (b) distanced r‘ 4 m Answer: Part(c)velociryv;0ftheblock IQ: ?—44.9~43 ID \ 3 Sky.“ +4," 2 2 Answer: Pan ((1) expression for oomlzmt c ( 'J o J m Name SO‘U I 11%. Problem 4. ( 15 points total) (a) At an instant, given the position, velocity and acceleration vectors shown below. accurately sketch and label the normal and tangential components of the acceleration. (2 points) 3 I 1/ a l \ (b) At the instant shown above, is the speed of the particle increasing, decreasing, or staying the same? (2 points) Answer: .l I"! C (8013"! ('1 C! (c) The car is rounding a curve with a radius of curvature p at a constant speed. The resultantforce acting on the car is in the direction of which arrow shown? (write A, B, C, D, E, or F). (2 points) AHSWer: (d) A force P is applied in two separate impulses during a 2—second period as recorded by the instrumentation, which measures the force as shown on the graph. Calculate the value of the total linear impulse. (2 points) genteel -— ézosxs) : 5. 95 n. . .- 31(— CLr-eot C- filitzknéjld .1. 44mm 2 Answer: 5-: 3L5 ' 3 Name 80 ILL“ ot’l ~$ (e) Suppose you are dcsigning a roller-coaster track that will take cars through a vertical ioop of 40—ft radius. If you decide that for safety the dovmward force exerted on a passenger by his Seat at the top of the loop should be at least one-half the passenger’s weight, what is the minimum safe velocity of the cars at the top of the loop? (4 points) (oar +013 .243 if 00am: leap) N W19 éFfiZW‘Qn N _.‘: =sz?“ I _. ,1 +m9 mom 1 E fi/zgm/g—M'Qt LuanfflUzim [513(- :L2 ‘9 g vr/ZI=5X3;.;)(4C)1 Answer: lg: 44 3.: _ h a V: 43.0511.- 342. C (D At the instant depicted, aSSume that the particle, which moves on a curved path, has the velocity v and acceleration :1 indicated. Accurater sketch and label the polar components of the velocity (3 points). ...
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This note was uploaded on 04/08/2010 for the course MECHENG 240 taught by Professor Perkins during the Fall '09 term at University of Michigan.

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Practice Exam 1.1 Solutions - Name SOHJ‘EQ‘D—é _...

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