Final Exam April 26, 2016

Transcription

1 PHYS 050 Spring 016 Name: Final Exam April 6, 016 INSTRUCTIONS: a) No books or notes are permitted. b) You may use a calculator. c) You must solve all problems beginning with the equations on the Information Sheet provided with this test. Please write down the equation you are using, or the principle you are applying to a given situation. d) You must show your work to receive credit!!!! If a problem requires only a simple answer, you must explain your answer. Please show all of your work on the sheets provided (use the back if necessary). d) Please draw a box or a circle around each of your answers. 1

2 1. It is desired to construct a simple pendulum with a mass of 3 kg that has a period of seconds. (a) (10 points) How long must the pendulum be? (b) (5 points) If the amplitude of the displacement from equilibrium is doubled, what will the period of the pendulum be? (c) (10 points) If the pendulum is now placed on a elevator that accelerates upward at 4.0 m/s, what is the period of the pendulum? (d) (5 points) If the pendulum is placed on an earth satellite that is circling the earth, what is the period of the pendulum?

3 . A 100-kg satellite is put up that is always above the same point on the earth as it orbits. (M earth = 6.0 x 10 4 kg; R earth = 6.4 x 10 6 m; G = 6.67 x N m /kg ) (a) (10 points) At what altitude above the earth is the satellite? (Hint: Consider the time it takes the satellite to orbit.) (b) (10 points) What is the value of g for the satellite in orbit? (c) (10 points) How much energy would have to be put into the satellite to have it just escape from the earth? 3

4 3. A 30-kg bag of mail is placed at the top of a 10-m long chute inclined at 30 o above the horizontal. The coefficient of static friction is 0.0 and the coefficient of kinetic friction is The bag is initially held at rest at the top of the chute. (a) (5 points) What is the maximum force of static friction on the bag? (b) (10 points) What additional force, directed along the chute, is needed to hold the bag at rest? (c) (5 points) If the bag is released from rest and allowed to slide down the chute, how much work is done by friction as the bag slides the length of the chute? (d) (5 points) How much work is done by gravity as the bag slides the length of the chute? (e) (10 points) What is the speed of the bag when it reaches the bottom of the chute? 4

5 4. A child of mass 30 kg runs at a speed of 3.0 m/s tangent to the edge of a playground merry-go-round, which is initially at rest. The child jumps onto the edge of the merry-go-round and then remains at rest relative to it. The merry-go-round has a mass of 150 kg and a radius of 1.5 m, and it may be considered to be a uniform solid disk (I = ½mr ). (a) (10 points) Find the angular velocity of the merry-go-round plus child just after the child jumps on. (b) (10 points) What is the magnitude and direction of the force required to keep the child on the merry-go-round immediately after jumping on? (c) (10 points) If the combination of merry-go-round plus child comes to a stop in 10 complete revolutions, how long does it take to come to rest assuming constant deceleration? 5

6 5. (10 points) A 10-kg fish swimming at 0.5 m/s swallows a -kg fish swimming directly toward it at a velocity that brings both fish to a halt immediately. Find the velocity of the smaller fish before being eaten. Neglect any resistance with the water. 6

7 6. A 0-kg block slides 1 m along a frictionless ramp inclined at 30 0 to the horizontal. At the bottom, after reaching a horizontal surface, the block makes a perfectly inelastic collision with a 30-kg block attached to a spring. The horizontal surface is also frictionless, and the spring constant is 5000 N/m. (a) (5 points) What is the speed of the 0-kg block when it reaches the bottom of the ramp? (b) (5 points) What is the speed of the pair of blocks just after they collide? (c) (5 points) How much is the spring compressed? 7

8 EXTRA CREDIT MULTIPLE CHOICE (There are 5 questions. Each one counts as points.) 1. A ball falls freely for 1 second starting from rest. Neglecting air resistance, after seconds (starting from rest) (a) Its acceleration will be two times greater. (b) It will fall twice as far. (c) Its speed will be four times greater. (d) It will fall four times as far.. An ice skater gliding on smooth ice eventually slows down and stops because (a) Gravity does this. (b) Her inertia is responsible for slowing down. (c) Friction between her skates and the ice causes this. (d) All objects naturally slow down and stop after a while. 3. You lift a large box from the floor and place it on a table. The amount of work you do (a) Depends equally on the weight of the box and the height of the table. (b) Depends only on the height of the table. (c) Depends primarily on the weight of the box. (d) Depends on fast you raise the box. 4. You are water skiing and you travel in a straight line at constant speed with the boat exerting the force to pull you forward. In skiing this way (a) The water exerts a very small force on you. (b) There is no net force acting on you. (c) The net force on you is the pull of the boat. (d) You do not pull on the boat because there is no way for you to do that. 5. While driving your car at constant speed on an interstate highway you round a long curve. (a) There is no acceleration because you travel at a constant speed. (b) The faster your speed, the smaller your acceleration. (c) The tighter the curve, the greater your acceleration. (d) The centrifugal force acting on your car keeps it on the highway. 8

9 sin = opp/hyp cos = adj/hyp tan = opp/adj INFORMATION SHEET v v 0 at r r f ri f s s N 1 x x0 (v v0) t 1 c = a + b x x0 v0t at r v f k = k N r v lim 0 W = (F-comp.) x (d) = F d x = x f - x i v v v 0 ax a W net = KE = ½mv - ½mv 0 KE = ½ mv x x f x v i v A cos t f t a lim x 0 t i U gravity = PE gravity = mgy F spring = -kx x v lim A y Asin vac vab v bc 0 U spring = PE spring = ½kx v v a t v a lim 0 f f v t i i A F net F ma tan A x A y A A y x E = KE + PE v W dw a c P F v r dt p = mv = s/r t F c = ma c = mv t /r 0 = r F p dp F 1 ( ) t dt t 0 0 = rfsin = rfsin = I ; I = m i r i Impulse F 1 0 0t KE = ½ I Impulse = p = mv f - mv i v t = r Gm 0 1m F ; r F net = 0 P initial = P final a t = r a c = v t /r = r m1m U(r) G r v 1i v i = -(v 1f v f ) L = r p; L = I 1 E KEmax PEmax ka dl ; net = 0 L init = L fin x Acos( ) dt m L g = 9.8 m/s =3 ft/s k f ; T 1/ f ; T k ; T g m 9