Solutions for Exam 2 - Spring 2007

Solutions for Exam 2 - Spring 2007 - W 21 February 2007...

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Unformatted text preview: W 21 February 2007 CODE # 1 1:00 pm. — 1:50 pm. PHYSICS 114 Second Hour Exam INSTRUCTIONS: Answer each question. All questions carry equal weight. You should calculate your answers to three (3) significant figures. Your answer may not agree exactly with one of the possible answers, but your answer, if correct, should agree with one of the possible answers to at least two 12! significant figures. Be sure to enter the code number, found in the upper right comer of this page, in the appropriate box in your scantron answer sheet. All information supplied by you on your scantron sheet must be left iustified, that is it must be entered beginning at the left hand end of the entry box. 1. A body rests on the bed of a truck which travels at constant speed. If the body is found suddenly to slide towards the front of the true the truck has ‘ s , . l 5 (a) increased speed ‘ (b) tumedtotheright (M «(M M %(c) decreased speed W) D, W, (d) turned to the lefi \ M W M M (e) unchanged speed 5 [074‘O I "z W7 ‘ , , i M9142 9mm 2. A constant net force, Fnet acts on a body of ass, m, and produces an acceleratlon of :3 5 W a . If another net force acts on this body and produces a resulting acceleration of 1/3 a the second net force is, “ \ r (a) it, We NW8 WW: FM=M€ WWMWB (b) 15F... Fmsmiflt mmmm (c) a. :7 m: Mufti/re (d) 313m :} (e) ‘AFM £11! 3. A vertical force, F? = 44.7 N, is applied, as shown, to a body of mass, 9.52 kg. Determine the normal force exerted by the surface on the body. 3 Wm amawlww —— 1.2.; 22:22 WW szfi=mdq=0 a +2, 3 £32222 MW. (e) 0 :} if“ +11?) ‘3 70) =7 Fu‘fil‘trjffl 01;; F :m — : MISML— ' F: é N: 613.2%) — Ira/ZN; Liz-WM = 48-6“ W/ a mi Phst 14-spO7—Exam 2_el, Page 1 of 3 W 21 February 2007 CODE # 1 1:00 pm. — 1:50 pm. LL FT 4. A body of mass, m = 7.85 kg, is suspended by a cord from the ceiling of an elevator. The tension in th cord is measured to be 64.7 Which of e follo ’ g statements ell/1571‘] ZWMM” F - is correct? WW) AA 5W [14,2 N , -W =r FT—Fazmg gag: FL—mqr;6%-1U—J-3f News”; any (a) The elevator is at rest m 7.8: ('1 :79 (b) The elevator is accelerating downwards at a rate of 1.56 rn/s2 : £9le — 4W: «( 21.2.31} (c) The elevator is accelerating upwards at a rate of 1.56 rn/sZ ;.gy 4.3314 (d) The elevator is moving downwards with constant speed :) as l . egg/0 A l. : __ ,.g M LW ‘ mm (14' (e) The elevator is either at rest or moving upwards or downwards at constant speed 5. The system shown is pulled to the right by an applied force, FT FT = 25.7 N. The masses have values, m1 = 11.7 kg and m m; = 17.6 kg. The cord connecting the two bodies is light an , inextensible. No frictional forces act on this sy tern. The (/5sz F1“ 3? tension in the connecting cord is: W W 3 M W N 2 a t a 9.1- , 3 m I If”? 71:1]:le =1F“~F(+F’/ & FAIMt # (a) 15'4N 9 ml»: ffie #51:. MW: FT 6‘ @1+LML)¢“\ L2, #411” Fl (b) 25'7N ") (A: r—lF mew/14% H‘s/mat r (c) 53.5 N : - thflm— M mme d 11.9N kit. 1 ’ FT = x 25410 : {mm—N =m Ee; 12.9N ‘ [H mm 24') = MW 94) Fm 6. A force, Fp, is applied to a body as shown, where F}: = 25.7 N, 0 = 36.90 and m = 3.74 kg. The coefficient of kinetic friction 6 between the body and the surface is 0.2%,}: eration 1 a ofthebodyis: Wm M S WWWS 2“ _‘___ i = FN~Fpsln0— ‘1 _ 3 O . a . —F‘_r—‘ 9L (a) 2.52m/s2tothengh :2) =F whet-F :T‘p 3w9+m P U L (b) 2.57 m/sztotherightfl Fx 32:4” “$364+ HWMJUMIS“ e“ g! :9 (c) 2.47 m/sztotheright :3 g, = ,ngoanr 35.65)”: EZOSZSN (d) 2.72 m/s:totheright[r ‘ awe _F/V = 5mg VLF” : Ml. JFK (e) 2464 m/s tothenght 2, (w. agewxmbé'Lo-LWSMW: M 31qu M “Mg/v 7. A body of mass, m, rests on top of a 4.00 kg body, w $c moves on a level 7” 1:52. “mud 3 fi'ictionless surface. A horizontal force, F, acts on the combination, which acceT rates 2'4 at 2.00 m/sz. When the body of mass, m, is removed and the same force is applied, Fm the 4.00 kg mass accelerates at 5.00 m/s2. What was the value of the mass, m, of the bod tha ori 'nall re edonth 4.00k d Maia any: t M mgfifaflUXI/W “0 65°“ W: £11: F=Qm+ut)a (b) 2.001: (c) 2.50 k: :3 F ._ QAHWOWX Loom/IL ’51 ’ c A» 38311:: firm MLW: 241:1. : may fl FWD Hoot/r =ZO-0N FM EWW Maw 40713:) :ng fimg "H' X L = ZOOM 2) : ’lez-sw‘fi lel_c1,Page20f3 Ex 74: LouMtho/oh : 600M $41“ 10024 m W 21 February 2007 CODE # 1 1:00 pm. — 1:50 pm. 8. Two bodies of masses, 2m and 3m, collide head on with each other on a frictionless surface. At the height of the impact the body of mass, 3m, exerts a force, F , on the body of mass, 2m. At this time the force exerted by the body of mass, 2m, on the body of mass, Sin, is - g E8 WWWWMZTM, - mam. 3W, , (c) 2£3F ‘ Wm M pm Wm #91) “F fl 2% \QW We MIMI ZS EE (e) 1’” we WM. M. =a’ #2 ~ F’ 9. A body rests on a plane. The coefficient of kinetic friction between the body and the plane is 0.676. The plane is slowly tilted until the body slides down the plane with a if constant veloci .At this time the lane make anan leto e orizo of: Cit/yaw mm m c mgA/Wi M1130, (a) 335" WM 4 so 6Lon M 60 yufi/ 33:32: WW gingwgmww d37:4°“=\h\)‘ W “m “0 \ _ >0 #8 m < Fiawhmame—m “2:... = {e =1 M3W9‘I‘LFW/‘vm “’39 2%“.gflgr =38: “(yum/flow 10..Abody lS attachedtoacord of Ian 1.50mandlswh1rl macrrcle, one end of 03%;“, cord being held fixed at the center of the circle. The period of this motion is 2.17 s. How many revolutions does this body perform in 15.0 seconds? :30 (5% - .1“ 2 4—, a» are W l. WW (b) “7 Wow/1w Wm (M W03 = MW» 6 83%? = era—rm: :64tmm7641W W) (e) 5.03 “4 S V 3 11. The tires on a car begin to slip when the car travels at a constant speed, v = 17.5 m/s, F 3 around an unbanked curve of radius 73.5 m. The coefficient of static friction between A] )C tiresuwhefimhes amftbe rofiils: mil/[w a1 :0 I :54“ (a) 0.477217; Fvvfi=mavco :5 Fbttwfl/Wr 4 333 3333 {fie/Wu = W 1 W? a W = "U? Vv =%(d) 0:425 e/b 7”: = 35:)" =§leL = 0mm 3.01%}; W) fit /~ (c) 0.462 M r (Mm X 416% K 0 \U, 12. A mass is attached to a cord of length 1.58 m and moves in a vertical circular loop. a}? The minimum spzjd at the top of the l p or cord not go slafk ‘s \ \— ( ) 4Wmm J %)%9 m/ 9%:wa 0A,] 7 C 1% a . s . S : z s (e) 4.55 m/s (d) 7.74 m/s L h 4—11 My]; \, L e \ ‘ '3) [~—+ :WLL AS‘F—Taolmfim? :fi) W L Ubifi- 7—Exam2 01,1) e30f3 1d- :a (r: J g1— ‘: QIXOMZSLM'WM mlwvlbstgb: fimtt‘Hfl/X 451W 34% ...
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This note was uploaded on 04/09/2008 for the course PHSX 114 taught by Professor Davis during the Spring '08 term at Kansas.

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Solutions for Exam 2 - Spring 2007 - W 21 February 2007...

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