2007f_2 - Name: E Y Instructor: Louisiana State University...

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Unformatted text preview: Name: E Y Instructor: Louisiana State University Physics 2101, Exam 2, September 25, 2007. ° Please be sure to write your name and class instructor above. - The test consists of 3 questions (multiple choice, no partial credit), and 3 problems. - For the problems: Show your reascming and your work. Note that in many of the problems, you can do parts (b), (c) and (d) even if you get stuck on (a) or (b). - You may use scientific or graphing calculators, but you must derive and explain your answer fully on paper so we can grade your work. - Feel free to detach, use, and keep the formula sheet pages. No other reference material is allowed during the exam. Good Luck! Qua/m a (a) pays 3 (m We) 3 E TQZjL ML Question 1 (11 points) A stationary block attached to a spring of spring constant K is displaced Ax by an applied force as shown in the figure. Circle the correct dimer below. (b) (5 points) The spring force, when the block is displaced A? from the relaxed positionéad—is—statioaefiI-anhe—fiaal—posifiot) a Kw) I; e k A? 3m? lJJ mam : 3 Em) [Elm e em) (b) (6 points) Work done by the applied force to displace the block from the relaxed position to A)? , where the block is stationary at the final position, is; 3- mgAx “it-{AK t: EW:'—Wzflfié b. —mgAx W :: W3 i“ WT: F g 69 em)” W; rat‘s“ Problem 1 (22 points) In the following figure mass of the block is 10kg and it slides 4m down the ramp, starting from rest. 1:: mzio f ,8: m / R 03:0 /. " u? 40° (a)(11 points) What is the coefficienth friction, if the kinetic energy of the block at the. bottomofthe4mramgis SOJ? M;AK1—bu+ =AK H #1. AK: Kgy/l = K19 r ‘96} ( , (b) (11 points) In addition, if an external force is pushing the block down and the kinetic energy of the block at the bottom of the ramp is 100.1, find the work doue by the external force. (Note: Friction is not zero.) MrzSK W ¥fiwk+wf 2: JOB} ;@ M, 43$“; 5‘0 Question 2 (12 points) The following Figure shows a plot of potential energy U versus x of a 0.200 kg particle that can travel only along an x axis under the influence of a conservative force, where UA=5J, UB=7.5J, Uc=15J, and UD=20J. The particle is released at the point x=2.51n where U= U3, with kinetic energy 2.5 J. 1? EM: u 5 5+] 9' Us awn- “ : l0} 0 - 4 . 2 i 5 i L {(111) ‘ ' (a) (4 pts) The particle stops momentarily at, apgro i mately i x22.5m and x=7.67m; i =2.0m and x=7.5m; m Jar-5W 45¢ +utrmlm? US (iii) x=1.0m and x=8.0m; (iv) the particle never stops. (b) (4 pts) The particle has maximum kinetic energy at: x=6.5m E 2': Ki" 71. -- = . W “0" K: mm were a: M (iii) x=2.5m; (iv) the particle has constant kinetic energy, so it has the same energy at all points. (d) (4 pts) The particle is never found at the following point}: (i) x=2.5m (ii) x=5m (iii) x=7m x=8m -—;. myom “inn/rim? from ‘ “5 Krvtellic «my? uremia! bf tie-3M) 4L6?) Midi is two; b L6 Problem 2 (22 points) A breadbox of mass M on a frictionless incline surface of angle 9 is connected, by a cord that runs over a pulley, to a light spring of spring constant K, as shown in the figure. The box is released from rest when the spring is unstrctched. Assume that the pulley is massless and frictionless. 4)..” 0 a un— (a) (6 pts) Assume the breadbox has moved a distance “d” down the incline (when the spring is not maximally stretched). Write an expression for the speed of the x at that point, in terms of 9,M,g,d,andK. £3be :59 ,2— .‘ 137/1131! +5145 am?” 4- ._. Wdflugzfi “(SM (1)) (6 pts) Write an expression, in terms of 6, M, g and K, for the distance L down the incline 4); 5 O A K: 0 that the mass M moves from its point of release to the point where the box stops momentarily. 4}“ :2 D £9), ts) Express the magnitude of the net force on the block M when it has '_, < stopped in term of 9, M, g, L and K. ' " Question 3 (I 1 points) A 2.0 kg object is acted upon by a force in the x direction in a manner described by the graph shown. The object is initially at rest. Circle the correct answer below. F (N) +2, timesm (a) (3 pts) The impulse of the force on the object is M O iii: ‘1': 8 > (i) 128N-s. (ii) 96N-s l” llzdL’SJ—é’ + 2XI£ + 25%): SZ+SZHE (iii) 64N-s 7i 2 Z N . { #80 s SON-s (b) (4 pts) The momentum acquired by the object in the first four seconds (t=Os to F45) is o (c) (4 pts) What is the object doing between t=63 and t=83? \®‘ t is moving in positive x direction and speeding up. 3F Vi; 56+er =? cunfdl‘o “— ii Hat is moving in positive x direction and slowing down. 0; (iii) It is moving in negative x direction and speeding up. M6 (iv) It is moving in negative x direction and slowing down. ' Moi—DOLL firm lamb“ Problem 3 (22 points) A spring, with spring constant of 120 Mm, is fixed on a horizontal part of a track that is 2.00 to higher than the rest. A block with mass m; = 0.30 kg is placed against the spring, compressing it by x = 0.40 m. The block is not attached to the spring, so after the spring is released the block slides down the track and hits a box of mass m; = 0.50 kg that is at rest on the lower portion of the track. The track is frictionless. Just after the collisiomthe box In; moves forward with a speed of 2.0 mfs. Hint: It is not known :fthe collision is elastic! l 2. (a) (8 pts) galculate the speed by which the block in; leaves the spring. .1. 145me 2. k" ’3 2— % Rodouermfi‘on o’fl magemu‘agflww o (b) (7 pts) Calculate the speed of the block In; just before the -_11ision. '° “3 246:.ng 'ng): " Haw/s is (c) (7 pts) Calculate the spec of the leer the collision. 279.; #279,; l)- : 2sz 54 “1/973, 5* Wfliflwflizr muffle-Isak m ‘1} 4r : “(Nola 2- E? ’ ML” : IOJéMQt—gffz):+6o€3‘z :4 r 9‘ —— W L' 0.3 M ’3 W ‘l i r o | A42}: 6.9.3 1% GOflJEVT/wfim of firm M : Formula Sheet for LSU Physics 2101, Test 2, September 25, Fall ’07. -—b:l: Vb! — 40c Quadratic formula: for 03:2 + bx: + c = 0, 931,2 = 2a Dot Product: ('1' - 3 = (tab: + (ruby + (1:53;!z = |6| cos(q5) («p is smaller angle between (1' and 5) Cross Product: ii X 5 = (my!)z — azbyfi + (nzbz -— ambzbi + (azby -— aybzflE, l6 x 5‘ = It'il sin(¢) Equations of Constant Acceleration: linear equation missing 1: = on + at 93 — $0 1 m—moz'vot-i-Eatz v 1:2 = 1:: + 2a(a: — mo) 1: 1 m—mo=§(vo+v)t a. l m—mo=vt—§at3 9., Vector Equations of Motion for Constant Acceleration: 1?" = 1"}, + fiat + gift}, 13’ = 1'1}, + at Projectile Motion: . m=vo¢t y=voyt—§gt2 R=flgfi 11¢ = on = constant '09 = you — gt Force of Friction: Static: f, S jaw = psFN, Kinetic: fk = kaN Elastic (Spring) Force: Hooke‘s Law F = -—km (E: = spring (force) constant) 1 1 Kinetic Energy (nonrelativistic): Translational: K = 5-va Rotational: K = 51:»: Work: W t E - J(const. force), W = f mi Work - Kinetic Energy Theorem: W = AK = Kr — K‘- ! F(:r)da: (variable 1D force)1 W = f 1! E0") - d1?" (variable 3D force) 1,. Work done by weight (gravity close to the Earth surface): W = m 9‘ - (I m! 93:; at? Work done by spring force F = - k x: W = —k/ a: dz: = —k ~— 2‘4 Power: W - _. dW _. _. Average: Pang = P = F - vans (const. force) Instantaneous: P = —, P = F - 1; (const. force) 3’ Potential Energy Change: AU = —W (conservative force) . . til-((3) Potential-Force Relation: = — do: Gravitational (near Earth) Potential Energy: U(y) = mgy (at the height 3;) 1 Elastic (Spring) Potential Energy: U = 516332 (relative to the relaxed spring) Conservation of Energy: W = AK + AU + AEth + AEint, where W is the external work done on the system, and AB“, = fkd. N 1 N 1 N 1 N Center of Mass: M = 2 mi, 93mm = — Z mime, ycom = ——~ 2 mm, 2mm = — Z mgz; ‘- M i=1 M 12—1 M 1 g 1 1i 1 g: 1 it, _. Fem-n = — mifii ficom = — mifii acom = — midi = — F12 M i=1 M i=1 M i=1 M i=1 _. N Definition of Linear Momentum: one particle: 33’ = mfi‘, system of particles: P = 35‘; = M 6mm i=1 d . -o _, dP' Newton’s 2“ Law for a System of Particles: Fact = M acom = 3 Conservation of Linear Momentum of an Isolated System: 215;- = 215} Impulse - Linear Momentum Theorem (one dimension): A131 = J12 = F12 (t)dt = Favg,1gAt £1 m — m 2m 2m m -- m Elastic Collision (1 Dim): 111,: = 1—21)“ + —202,-, 132; = —1—'v1,; + —2-——1‘Ugi m1+m2 m1+m2 m1+m2 m1+mz . . 41:... ...
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2007f_2 - Name: E Y Instructor: Louisiana State University...

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