1 atm = Pa = 14.7 psi 9. I disk = (1/2) mr 2 I rod (center) = (1/12) ml 2 I rod (end) = (1/3) ml 2

Transcription

1 Fall 009 RED barcode here Physics 05, sections and Exam Colton Please write your CID here No time limit. No notes. No books. esting Center calculators only. Constants: g = 9.8 m/s but you may use 0 m/s in nearly all cases G = Nm /kg k B = J/K N A = R = k B N A = 8.34 J/mol K = W/m K 4 Mass of Sun = kg Mass of Earth = kg Radius of Earth = m Radius of Earth s orbit = m Density of water: 000 kg/m 3 Density of air:.9 kg/m 3 Young s modulus of steel: N/m Linear exp. coeff. of copper: /C Linear exp. coeff. of steel: 0-6 /C Specific heat of water: 486 J/kgC Specific heat of ice: 090 J/kgC Specific heat of steam: 00 J/kgC Specific heat of aluminum: 900 J/kgC Latent heat of melting (water): J/kg Latent heat of boiling (water): J/kg hermal conduct. of aluminum: 38 J/smC sin(30) = 0.5 cos(30) tan(30) Conversion factors inch =.54 cm m 3 = 000 L atm = Pa = 4.7 psi 9 F C 3 5 K 73.5 C Other equations b b 4ac x a Surface area of sphere = 4 r 3 Volume of sphere = 43 r vi vf vave v vo at x xo vot at vf vo ax w =, PE g = y F = -kx, PE s = ½ kx f k N (or f sn ) P F// v Fvcos Ft p Elastic: (v - v ) bef = (v - v ) after arc length: s r v r atan r a c = v /r GMm GMm Fg, PE g r r I pt mass = mr I sphere = (/5) mr Ihoop = mr I disk = (/) mr I rod (center) = (/) ml I rod (end) = (/3) ml L rp rp rpsin stress = F/A; strain = LL Y = stress/strain P P 0 gh VFR = Av = Av P v gy P v gy L L0 V V0 ; 3 3 KEave mv ave k B Qmc; Q ml Q ka L 4 P e A W on gas = area under P-V curve = P V (constant pressure) 3 3 U NkB nr (monatomic ideal gas) Q h = W net + Q c Wnet Qc e Qadded Qh c emax h k, m g, L v, ml Y v, v B m k L g v air = 343 m/s at 0 C I 0 log I 0 = 0 - W/m I 0 v v0 f ' f v v S v f n n n,,3,... L v f n n n,3,5,... 4L Did you write down your CID at the top of the page? If not, you may not get this test booklet back.

2 Instructions: Record your answers for problems -8 on the bubble sheet. You may write on this exam booklet, and are strongly encouraged to do so. I strongly encourage you to mark your answers in this test booklet. You will get this test booklet back (but only if you write your CID at the top of the first page). In all problems, ignore friction, air resistance, and the mass of all springs, pulleys, ropes, cables, strings etc., unless specifically stated otherwise. Use g = 9.8 m/s only if there are 9.8 numbers in the answer choices; otherwise use g = 0 m/s. Problems -8 will be scaled to be worth 9 total points; problems 9 and 30 are worth 4 points each. Problem. he figure shows a velocity vs time graph of a car moving along a road, and positive means to the right. According to the graph, what is the car doing from 0 to 5 seconds? a. moving to the left, slowing down, and stopping b. moving to the left and speeding up c. moving to the right, slowing down, and stopping d. moving to the right and speeding up e. first moving right, then moving left f. first moving left, then moving right velocity time. Negative velocity = moving to left ; positive velocity = moving to right. Choice F Problem. A man holding a rifle at height y = 0 tries to hit the side of a large barn 90 m away. he rifle is shot at 4.7º above the horizontal. he initial velocity of the bullet is 00 m/s. What is the y-position in meters of the hole the bullet makes in the side of the barn? Use sin(4.7) = 0.4, cos(4.7) = 0.9. Hint: first find the time it will take. a m b c d e f g m. First divide v 0 into components: v 0x = v 0 cos = 00*0.9 = 80 m/s. v 0y = v 0 sin = 00*0.4 = 80 m/s. hen, x-direction gives time: x = v0xt 90 = 80 t t = 0.5 seconds Finally, y-direction gives final height: y = v 0y t ½ gt y = 80*0.5 ½*0*(0.5) y = 40 5*0.5 = 0.5 = m. Choice F Problem 3. From the perspective of Newton s Second Law, what force on a car causes it to accelerate forward when the gas pedal is pushed? a. he car pushing backward on the road b. he car pushing forward on the road c. he road pushing backward on the car d. he road pushing forward on the car 3. his must be a force on the car, which is pushing forward. Choice D. Problem 4. You slide a block of ice down a 45ramp. here is no friction between the ice and the ramp. How will the acceleration of the ice block down the ramp compare to the standard acceleration of gravity, 9.8 m/s? a. It will be equal to 9.8 m/s. b. It will be greater than 9.8 m/s. c. It will be less than 9.8 m/s. 4. As discussed in class, it must be between 0 and 9.8 m/s (the values when = 0 and = 90). Choice C.

3 Problem 5. A boat moves through the water with two forces acting on it. One is a 00 N forward push by the water on the propeller, and the other is a 500 N resistive force due to the water around the bow. What is the acceleration of the 000 kg boat? a. Less than 0.53 m/s b c d e f. More than 0.93 m/s F = ma = 000a a = 700/000 = 0.7 m/s. Choice C. Problem 6. You need to carry a suitcase up a flight of stairs. In which case will you do the most work on it against gravity? a. You carry the suitcase up quickly. b. You carry the suitcase up slowly. c. Both cases involve the same amount of work. 6. he work against gravity will be equal to F // x, which will also equal the change in potential energy: h. Doesn t matter has fast you carry the suitcase. Choice C. Problem 7. A woman with a parachute is falling at constant speed. How does the magnitude of the force of gravity pulling down on the woman compare to the magnitude of the force of the parachute pulling up on her? a. Gravity is grater b. Parachute is greater c. hey are the same 7. his is just like the monkey sliding down a pole problem. If the speed is constant, there is no acceleration and the forces must add up to zero. herefore the forces from gravity and the parachute are equal (and opposite direction, obviously). Choice C. Problem 8. On a frozen pond, an 8 kg sled has an initial speed of v 0 = 0 m/s. he coefficient of kinetic friction between sled and ice is k = 0.. Find the distance the sled moves before coming to rest. a. Less than 9 m b. 9 5 c. 5 d. 7 e f g. More than 39 m 8. E bef + W net = E aft ½ mv fx = ½ mv f Note that f = N, and from the y-direction, N =. herefore, ½*8*0 0.*8*0*x = 0 Divide by 8, and you get ½*00 x = 0 x = 50/ = 5 m. Choice D 3

4 Problem 9. he amount of potential energy possessed by an object that has been lifted up is equal to: a. the distance the object is lifted b. the force used to lift the object c. the object s acceleration due to gravity d. the weight of the object e. the work done in lifting the object 9. Potential energy accounts for the work done by conservative forces. Choice E Problem 0. Jen throws three balls from the top of a building towards a large flat parking lot: one straight up, one straight down and one horizontally, all with the same initial speed. Which ball will have the highest speed just before it hits the ground? (As usual, ignore air resistance.) a. he one thrown horizontally b. he one thrown straight down c. he one thrown straight up d. All three will all have the same speed 0. hey all have the same initial kinetic and potential energy. hey all have the same final potential energy. herefore if there was no work done by non-conservative forces (i.e., air resistance), the final kinetic energy must be the same. Choice D. Problem. A box slides down a ramp. he work done on the box by the normal force is: a. positive b. negative c. zero. W = F // x. Since the normal force is perpendicular to x, there is no parallel component and the work done is zero. Problem. he parachute on a race car of mass 600 kg opens when the car is traveling at 50 m/s. What average force (magnitude) must be supplied by the parachute to stop the car in a time of 0 seconds? a. Less than 550 N b c d e f g. More than 050 m. First, get acceleration through kinematics: v f = v 0 + at 0 = 50 + a*0 a = -5/ m/s Now use N to get force: F = ma F parachute = ma F parachute = 600*(-5/) = -500 N, 500 N magnitude 4

5 Problem 3. A skier of mass 7 kg is pulled up a 5 frictionless slope by a motor-driven cable at a constant speed of 3 m/s. How many horsepower must a motor have to perform this task? ( hp = W) 3 73 a. hp e sin sin b sin sin5 3 c d. 79.8sin f. g sin sin hp 3. Start with: P = Fv, since both force and velocity are constant. From FBD, F tension - sin = 0, F = sin So P = (sin)v = 79.8sin53 watts Convert to horsepower by multiplying by ( hp)/(745.7 W) 79.8 sin5 3 P =. Choice D Problem 4. You are a passenger in a car and not wearing your seat belt. Without increasing or decreasing its speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door. Which is the correct analysis of the situation according to Newton s laws? a. Before and after the collision with the door, there is a rightward force pushing you into the door. b. Starting at the time of collision with the door, the door exerts a leftward force on you. c. Both of the above d. Neither of the above 4. Ignoring the fictitious forces which seem to be present in the accelerating reference frame of inside the car, it s clear that your own inertia wants to keep going straight, which is why you collide with the door when the car goes left. After you collide, though, the door exerts a force on you (otherwise you would fly out of the car). Choice B. Problem 5. In the racing balls demo, two balls were raced on different tracks. he closer track had a section that was lower for part of the race. Which ball won the race, and why? a. he closer ball, because it was going a greater speed during the lower section. b. he closer ball, because it had a greater change of potential energy from start to finish. c. he farther away ball, because it was going a greater speed during the lower section. d. he farther away ball, because it had a greater change of potential energy from start to finish. e. he balls tied, because they both ended with the same kinetic energy. 5. Choice A describes the situation exactly. (hey both end up with the same v f, though.) Problem 6. Newton s 3 rd Law says that forces come in pairs. A book lying on a table feels a force of gravity and a normal force. What is the Newton s 3 rd Law partner force with the force of gravity on the book? a. the gravitational force of the book pulling up on the earth b. the normal force pushing up on the book c. the pressure of the book pushing down on the table d. none of the above 6. he partner force of F - is F -. hus, the partner of F earth-book is F book-earth. hat is, the force of the book pulling up on the earth. Choice A 5

6 Problem 7. wo cables support a cat burglar of weight W. One cable is at an angle,, as shown in the figure. Find the tension in the cable connected to the left wall, in terms of W and. a. w cos b. w sin c. w tan d. w cos e. w sin f. w tan sin 7. FBD of junction: cos N in y-direction: F y = ma y sin - w = 0 = w/sin N in x-direction: F x = ma x cos = 0 (w/sin)*cos = w/tan. Choice C. Problem 8. A hanging mass, m, is attached via a pulley to another mass, m, which is resting on a horizontal frictionless table as shown in the figure. Find the acceleration of the masses in terms of m, m, and g. a. g b. m g c. m g d. m m e. m m f. m m g. m m g m 8. N for group, in the bent x-direction: F group = m group a m g = (m +m )a a = m g/(m +m ). Choice D w m m 6

7 Problem 9. Repeat, but with coefficient of friction between the table and m. You may take it as given that the friction force is small enough that m still does fall. Note: A free body diagram and an equation will be required for this situation, in problem 9. a. m m FBD: b. N m m m m c. m m d. e. m m m m f. g. m 9. N for group, in the bent x-direction: F group = m grou pa m g f = (m +m)a mg - N = (m +m)a Get normal force from y-direction: F my = 0 N = m g m g - m g = (m +m )a a = (m g- m g)/(m +m). Choice A Problem 0. Same situation as the last problem. his time, however, the friction force is strong enough to prevent m from falling. Specifically, m = kg, m = 7 kg, = 0.5, and both masses are at rest. What is the tension in the line connecting the two masses? a. Less than 8 N b. 8 6 c. 6 4 d. 4 3 e f g. More than 48 N 0. N for m : F m = m a. Note a = 0, since masses are not moving - m g = 0 = m g = *0 = 0N. Choice B m m g m m g f 7

8 Problem. You want to measure the spring constant of a hanging spring. You add a 0.5 kg mass to the initially unmoving spring and quickly release the mass so that it starts oscillating up and down with no additional forces on it. At the lowest point of the first oscillation, the mass has stretched the spring out to a distance of 0 cm below where it started. Wha t i s th e spring constant k? a. Less than 30 N/m b c d e f. More than 50 N/m. Before: After y=0 h E bef + W net = E aft h + 0 = ½ kh k = /h = *0.5*0/0. = 50 N/m. Choice B Problem. A 3 gram ping-pong ball starts at rest and is pushed 0.5 meters along an evacuated tube by a constant force of 30 N from air entering the tube behind it. How long does it take the ball to exit the tube? a. 0.0 s b. 0.0 s c s d s e s f s g s F = ma 30 = 0.003a a = 0000 m/s x = x 0 + v 0x t + ½at 0.5 = / *0000t t=sqrt(/0000) = /00 s. choice A 8

9 Problem 3. A block (m = 0.4 kg) accelerates down a 35 ramp (measured from the horizontal) at a rate of 0.5 m/s. here is friction on the ramp. What is k? 0.5 a. 49.8cos37 b. 9.8sin c. d. e. f cos cos sin cos37 9.8tan free body diagram f N 3. F y = 0 N = cos f = N = cos cos sin F x = ma sin f = ma sin cos = ma Cancel the m s: gsin gcos = a = (gsin a)/(gcos). Choice E Problem 4. wo identical cars, coasting from rest down two hills having the same vertical height but one hill steeper than the other will arrive at the bottom of the hills with the same speed. Yet the acceleration of the car down the steep hill is greater than the car going down the other hill. What is the best explanation for this seeming contradiction? a. he acceleration when we neglect friction and air resistance is always 9.8 m/s b. he car going down the less-steep hill accelerates for a longer distance. c. he car going down the steeper hill experiences more nonconservative forces. d. here is no contradiction: the cars will arrive at the bottom of the hills at the same time. 4. Similar to the racing balls question, although they both end up with the same final speed, the time is different. his is possible because the one experiencing less acceleration, accelerates for a longer distance (and a longer time). Choice B 9

10 Problem 5. arzan (80 kg) swings on a vine (length 0 m) initially inclined at an angle of 36.9 from the vertical. He pushes off with an initial speed of 4 m/s. What is his speed at th e botto m of the swing? he picture may help; arzan is the ball on the end of the vine. Note: sin(36.9) = 0.60, cos(36.9) = 0.80, tan(36.9) = L a. 88 m/s b. 9 m/s c. 96 m/s d. 00 m/s e. 04 m/s his distance is Lcos f. 08 m/s his distance is h = L - Lcos = 0-0* Ebef + Wnet + Eaft = 0-6 = 4 m ½mv + h = ½mv f Cancel the m s, multiple both sides by, plug in the numbers: 4 + *0 *4 = v f = v f v f = sqrt(96) m/s. Choice C y y=0 Problem 6. A crate (mass m) is being p ushed across a level floor at a constant speed by a force F at an angle of below the horizontal as shown in the figure. In terms of F,, m, and g, find the coefficient of kinetic friction between the crate and the floor. Note: A free body diagram and some equations will be required for this situation, in Problem 30. a. tan F b. c. d. e. f. F F sin F F tan F cos F cos F sin free body diagram 6. F y = 0 N = + Fsin Constant speed: F x = 0 Fcos = N + Fsin = Fcos = Fcos/ + Fsin. Choice F f=n N Fsin Fcos 0

11 Problem 7. A block of mass m =.7 kg is held without moving on a frictionless incline of angle of 9 by the horizontal force F, as shown in the figure. Determine the value of the force F, in Newtons. a Newtons b. cos c. sin d. tan9 free body diagram e cos9 Fcos f sin9 N g tan9 Newtons F 7. F y = 0 Fcos sin = 0 F = sin/cos = tan. Choice G. sin Fcos cos Problem 8. Vern, the famous pulley ski-jumper is 00 kg, and is pulled via a pulley by a 300 kg hanging block up a 50 m incline that has a vertical height change of 0 m; see the figure. Friction provides a constant backwards force on Vern of 00 N. How fast will Vern be going when he takes off from the incline, assuming he misses the giant pulley? a. Less than 60 m/s b. Between 60 and 640 c. Between 640 and 660 d. Between 660 and 680 e. Between 680 and 700 f. More than 700 m/s 50 m 50 m Final positions 0 m 8. E bef + W net = E aft PE big mass F friction *x = KE both masses + PE skier Note that while the skier is raised 0 m, the big mass falls 50 m. See figure. Also, friction acts over 50 m. 300kg*0m/s *50m 00N*50m = ½ (00kg+300kg)v f + 00kg*0m/s *0m = 00v f v f = 5000/00 v f = sqrt(65). Choice B. Note if you use g = 9.8 instead of 0 for this one, you get sqrt( 6) as the answer. Although I said specifically to use g = 0 in the exam instructions at the top, I have decided to accept answer choice A as well as choice B. I had the esting Center automatically regrade everyone s exams, so you should automatically get credit for answer A if that s what you answered.

12 Write your CID here: Problem 9 (4 pts). (a) Draw a free-body diagram for each mass in Problem 9. Be sure to label all forces. FBD: N m m f m g m g (b) Write down the N blueprint equation for the two masses taken as a group, and fill in the blueprint with the given forces. Don t solve the equation, just take it one or two steps past the blueprint. [If you don t know how to use N for a group, do two N equations for the two masses sep arately.] Blueprint: F group = m group a Filled in: m g f = (m +m )a m g - N = (m +m )a ( step further, not needed) or F m = m a and F m = m a or m g = m a and - f = m a or m g = m a and - N = m a Problem 30 (4 pts). (a) Draw a free-body diagram of the crate from Problem 6. Be sure to label all forces. free body diagram N Fcos f=n Fsin (b) Based on your FBD, write down and fill in the N blueprint equations for the x- and the y-directions. Don t solve the equations, just take them one or two steps past the blueprint. Be sure to fill in the acceleration(s) if known. x-direction Blueprint: F x = ma x y-direction F y = ma y Filled in: Fcosf - - Fsin = 0 FcosN ( step further, not neeed)