Geneal Physics I Exam 2 - Chs. 4,5,6 - Foces, Cicula Motion, Enegy Oct. 10, 2012 Name Rec. Inst. Rec. Time Fo full cedit, make you wok clea to the gade. Show fomulas used, essential steps, and esults with coect units and significant figues. Patial cedit is available if you wok is clea. Points shown in paenthesis. Fo TF and MC, choose the best answe. 1. (3) A block of mass 1.5 kg is acted on by two foces: one of magnitude 3.0 N diected east and anothe of magnitude 3.0 N diected noth. What can you say about the block s instantaneous motion? a) It is moving to the east. b) It is moving to the noth. c) It is moving towads notheast. d) It must be at est. e) It could be moving in any diection. 2. (3) A 1.5 kg block is moving at constant speed towads due noth. What can you conclude? a) Thee is just one foce towads the noth acting on it. b) Thee is no net foce acting on the block. c) The net foce on the block is diected towads the noth. d) Thee ae no foces acting on the block. 3. (3) A 1.5 kg block is acted on by two foces: one of magnitude 3.0 N diected east and anothe of magnitude 3.0 N diected noth. What can you say about the block s instantaneous acceleation? a) The acceleation has a magnitude of 2.0 m/s 2. b) The acceleation has a magnitude of 2.8 m/s 2. c) The acceleation has a magntiude of 4.0 m/s 2. d) The acceleation is zeo. 4. (3) A block slides to the ight on a level suface. One foce that acts on it is the nomal foce N. The 3d law pai foce associated with N is the a) foce opposite to N that acts on the level suface. b) fictional foce on the block. c) foce of gavity on the block due to the Eath. d) foce of gavity on the Eath due to the block. 5. (3) The two foces in a 3d law action-eaction pai a) can act on diffeent objects o on the same object. b) always act on the same object. c) always act on diffeent objects. 6. (2) T F On a level oad, the nomal foce of the oad on the ties can cause you ca to acceleate. 7. (2) T F When pesent, static fiction always pevents an object fom acceleating. 8. (2) T F A body in fee fall has no net foce acting on it. 9. (2) T F The net foce on a body that is not acceleating is zeo. 10. (2) T F While sitting in you chai, you mass exets a gavitational foce on the Eath. 11. (10) An elevato togethe with its passenges weighs 4900 N. At a cetain instant, the tension in its suppoting cable is 5900 N. Detemine the magnitude and diection of its instantaneous acceleation. 1
12. (16) A 1400-kg ca initially has a velocity of 33.3 m/s due south. It bakes to a stop ove a 180 m distance. a) (6) What is the magnitude of the ca s acceleation, in m/s 2? b) (4) What aveage net foce magnitude was necessay to stop the ca? c) (6) Assuming the ties do not skid, what coefficient of static fiction between ties and pavement is needed? 13. (12) A 32-kg filled gocey cat is to be pushed up a fictionless amp by a pushing foce P acting along the amp, tilted at an angle of 6.0 above hoizontal. a) (6) What magnitude foce P is needed to push the cat at constant speed? b) (6) What magnitude P is needed to push the cat up the amp with an acceleation of 2.0 m/s 2? 2
Name Rec. Inst. Rec. Time 14. (18) A centifuge is used to swing some mass (usually a fluid) aound in a hoizontal cicle at high speed, causing a lage acceleation. Suppose the adius of the cicle is 12.0 cm and the mass is 25.0 gams. a) (6) At what speed v should the mass move so that its centipetal acceleation is 2.00 10 3 g? (whee g = 9.80 m/s 2 is the acceleation due to gavity at Eath s suface) b) (6) What is the ate of otation, measued in evolutions pe minute? c) (6) Calculate the magnitude of the centipetal foce acting on the 2.50 gam mass. 15. (12) Jupite s adius is 11.2 times Eath s adius and its mass is 318 times Eath s mass. a) (6) Using the fomula g = GM/ 2, which gives 9.80 m/s 2 at Eath s suface, how lage is g in m/s 2 at the suface of Jupite? b) (6) A 12.0-kg satellite is to be placed into a cicula obit aound Jupite at an altitude above the suface equal to Jupite s adius. Detemine the centipetal acceleation of the satellite. 3
16. (12) A 1600-kg ca acceleates fom est to a final speed of 40.0 m/s in a time of 6.00 s. a) (6) What is the minimum wok output of the moto, in kj, duing the 6.00 s acceleation? b) (6) What minimum aveage powe output of the moto is equied, in kw? 17. (2) T F When a mass is moved to a highe altitude, its gavitational PE inceases. 18. (2) T F Fiction is an example of a non-consevative foce. 19. (2) T F Fiction acting on a mass always does negative wok. 20. (2) T F When you dive a ca down a hill, gavity does no wok on the ca. 21. (2) T F A foce does the most wok on a mass when paallel to the displacement of the mass. 22. (18) A 70.0-kg skie slides down a unifom 30.0 slope of snow, stating fom est at the top of the hill. The distance taveled along the slope is 250 m. a) (6) Assuming no fiction, how lage is he kinetic enegy, in joules, at the bottom of the hill? b) (6) How fast will the skie be going when she eaches the bottom of the hill? c) (6) Suppose instead that the skie is slowed down by a kinetic fiction foce equal to f k = 60.0 N up the slope. What would be he kinetic enegy, in joules, at the bottom of the hill? 4 Scoe = /133.
Pefixes a=10 18, f=10 15, p=10 12, n=10 9, µ = 10 6, m=10 3, c=10 2, k=10 3, M=10 6, G=10 9, T=10 12, P=10 15 Physical Constants g = 9.80 m/s 2 (gavitational acceleation) M E = 5.98 10 24 kg (mass of Eath) m e = 9.11 10 31 kg (electon mass) c = 299792458 m/s (speed of light) G = 6.67 10 11 N m 2 /kg 2 (Gavitational constant) R E = 6380 km (mean adius of Eath) m p = 1.67 10 27 kg (poton mass) Units and Convesions 1 inch = 1 in = 2.54 cm (exactly) 1 foot = 1 ft = 12 in = 30.48 cm (exactly) 1 mile = 5280 ft 1 mile = 1609.344 m = 1.609344 km 1 m/s = 3.6 km/hou 1 ft/s = 0.6818 mile/hou 1 ace = 43560 ft 2 = (1 mile) 2 /640 1 hectae = 10 4 m 2 Tig summay sin θ = (opp) (hyp), (adj) cos θ = (hyp), (opp) tan θ = (adj), (opp)2 + (adj) 2 = (hyp) 2. sin θ = sin(180 θ), cos θ = cos( θ), tan θ = tan(180 + θ), sin 2 θ + cos 2 θ = 1. Chapte 1 Equations Pecent eo: If a measuement = value ± eo, the pecent eo = eo value 100 %. Chapte 2 Equations Motion: v = x t, x = x x 0, slope of x(t) cuve = v(t). ā = v t, v = v v 0, slope of v(t) cuve = a(t). Fo constant acceleation in one-dimension: v = 1 2 (v 0 + v), v = v 0 + at, x = x 0 + v 0 t + 1 2 at2, v 2 = v 2 0 + 2a(x x 0 ). Fo fee fall on Eath, using an upwad y-axis, with g = 9.80 m/s 2 downwad: v y = 1 2 (v y0 + v y ), v y = v y0 gt, y = y 0 + v y0 t 1 2 gt2, v 2 y = v 2 y0 2g y. Chapte 3 Equations Vectos Witten V o V, descibed by magnitude=v, diection=θ o by components (V x, V y ). V x = V cos θ, V y = V sin θ, V = Vx 2 + V 2 tan θ = Vy θ is the angle fom V to x-axis. y, V x. Addition: A + B, head to tail. Subtaction: A B is A + ( B), B is B evesed. Pojectiles a x = 0, v x = v x0, x = x 0 + v x0 t. Fo a hoizontal x-axis. a y = g, v y = v y0 gt, y = y 0 + v y0 t 1 2 gt2. Fo an upwad y-axis. R = v2 0 g sin 2θ 0, Relative Motion V BS = V BW + V WS, (Fo level gound only.) B=Boat, S=Shoe, W=Wate. BS means boat elative to shoe, etc. Must be applied as a vecto equation! Eq.-1
Chapte 4 Equations Newton s Second Law: F net = m a, means ΣF x = ma x and ΣF y = ma y. Fnet = Fi, sum ove all foces on a mass. Fiction (magnitude): f s µ s N o F f µ s F N (static fiction). f k = µ k N o F f = µ k F N. (kinetic o sliding fiction) Gavitational foce nea Eath: F G = mg, downwad. Chapte 5 Equations Centipetal Acceleation: a R = v2, towads the cente of the cicle. Cicula motion: speed v = 2π T = 2πf, fequency f = 1 T, whee T is the peiod of one evolution. Gavitation: F = G m1m2 ; 2 g = GM, 2 whee G = 6.67 10 11 Nm 2 /kg 2 ; Obits: v 2 = g = GM 2 ; v = Chapte 6 Equations GM. centipetal acceleation = fee fall acceleation. Wok & Kinetic & Potential Enegies: W = F d cos θ, KE = 1 2 mv2, PE gavity = mgy, PE sping = 1 2 kx2. θ = angle btwn F and d. Consevation o Tansfomation of Enegy: Wok-KE theoem: KE = W net = wok of all foces. Powe: P ave = W t, o use P ave = enegy time. Geneal enegy-consevation law: KE + PE = W NC = wok of non-consevative foces. Eq.-2