Coefficient of Kinetic Friction - General Physics - Solved Exam, Exams of Physics

Download Coefficient of Kinetic Friction - General Physics - Solved Exam and more Exams Physics in PDF only on Docsity! Blue Exan PHYSICS 220 EXAM 1 February 23 2005 This exam consists of 13 problems. Please check that you have all of them. One (both sides) 8 1/2” x 11” crib sheet is allowed. It must be of your own creation. Useful formulas and constants: Also uséat ¥-Yo = £LVox tr) t z.3 32 _41'r” GMe “1 fy O= tan e) g=98m/s° Constant acceleration xox ty lar vay tat v? avy +2a(x—Xx)) Fes mii Figg = HN F pion = HN Gmm - Fas = * 2 G =6.67x10" Nin? / kg? a,=V/r Epping = HE o=2nf f=l/T Constant angular acceleration 1 O=8, tay sat? O= 0, +at @ =a, +2a(O-) O=sir @=vir Please sign the opscan sheet and print your name on it. Use a #2 pencil to fill in your full name, your student identification number (old one), and finally the answers for problems 1~13. Please be prepared to show your Purdue ID when you hand in your opscan sheet. Page 1 Oz=25 ° 1. A box is sliding down an incline at constant acceleration. The angle of the incline is 25 degrees and the coefficient of kinetic friction is 0.2. The acceleration ¥ the box is, KrO2 hae: d SR =m ax ify = Mk 2 _ - =O soem ero | C 13 m/s’ i <p Vangel “ey | Ams! yy WoinP=mgsub E) 2.2 m/s’ Wy = = WlOP = mq CoO aelygeod +m $b = = dx Ay = MngtosO 9s 42,37 Wz 2. A locust jumps with a velocity of 3 m/s at an angle of 55 degrees above the horizontal. The locust lands 0.8 m from where it jumped. What is the time it takes the locust to reach its pe Bs | : at ma height: Or 5S? ve tR= 04m A) 0.051 s wae Y lye ts § Gjoass OE bs nas Ug? Vyy = V,C0502 1 72"Is ty ; E) 0.50s en ]vog Vy 2th 38 = 2.46 %s re Aste es cab, Voy it? want 0,15 3. Two masses are suspended by cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, mi, has a mass of 5.0 kg (i the other mass, mp, has a mass of 3.0 kg. The acceleration of my is: 7 (mg Hmyayng = nd el mania ghd A(t tM) = Hg Ma§ A) 3.25 m/s? upward é m4 m4 a: ing -m4 a B) 3.25 mis? d d fl tM; 's* downwar fa 025-08 f m, ti C) 2.45 m/s? upward z yeas Tengeme ea (RE 2SEE D)°2.45 m/s? downward : _ “Wid 2 daonwuad | E) 1,05 m/s? upward m,=4 k4 @ fy 2M, a yt? > | m, moves dana m2 73 \g (ra) , wil Page 217g on ‘f 50 Mh, will Mov downtwafd. T= tagtM& 9. A 2000 kg car is traveling on a banked curved icy road. The velocity of the car is 25 m/s and the road has a radius of curvature of 500 m. If the car is to travel on the icy road without sliding, then the angle of the banked road is, uM v= 25s tn (ee A) 25.7 degrees O=tan ( 4 r= 50m B) 21.0 degrees g= 43 G5* C) 12.7 degrees b= 7,269 ° D) 10.5 degrees (®))7.27 degrees 10. Ifthe length of the Achilles tendon increases 0.50 cm when the force exerted on it by the muscle increases from 3200 N to 4800 N, what is the "spring constant" of the tendon? Fe kx ())3200 Niem « B) 6400 N/em A St tynalorce —~tddrtronel lengihenng C) 9600 N/cem ~32tv v D) 16,000Nom £ = waapn Sv) (3000 “em E) 8,000 N/cm 11. During normal operation, a computer's hard disk spins at 7,200 rpm. If it takes the hard disk 4 s to reach this angular velocity starting from rest, what is the average angular acceleration of the hard disk in rad/s*? 224d | a §,=0 We 7200 $i * a * see A) 1800 rad/s? B® w= 753.98 ls B) 900 rad/s? pO C) 380 rad/s? = 75345 Ys ) 190 rad/s’ a> want E) 6.82 rad/s’ £: 4s 12. An airplane is climbing at an angle of 5 degrees above the horizontal at a constant a=O velocity. The weight of the airplane is 40,000 N. The wings (which are also making an angle of 5 degrees above the horizontal) produce a lift force that is perpendicular to the wings and a drag force that is parallel] to the wings. The drag force is 1,600 N. The forward thrust force the engine produces is: A) 3,200 N \o0o.N 5,100 N oe Fay =| ) 8,300 N 9 D) 10,500 N ~— y=wosit E) 12,000N R <t B04 wx= 348ON Fyn “8 _ Uy = sated x i mike 13. Grant Hill jumps 1.3 m straight up into the air to slam-dunk a basketball into the net. With what speed did he leave the floor? _ ; Fhrosd ~ Fey Wy = 0 In A)3.6m/s y= 13m Frost > Farag +x Bem. a want [Bhs = = 5086N , D)12.7ms y, = 0 (yt peak) E) 9.8 m/s Vy (op dy = -48 g%s2 bt: 7? ee > b yi 2) Voy * “Lay Vay 2 Vayy 21-49% (13m) Page 6