# y(t) = y 0 t! 1 2 gt 2. With y(t final ) = 0, we can solve this for v 0 : v 0 A ĵ. With A! ĵ =!2 and A! = (2) 2 + (!

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1 1. The angle between the vector! A = 3î! 2 ĵ! 5 ˆk and the positive y axis, in degrees, is closest to: A) 19 B) 71 C) 90 D) 109 E) 161 The dot product between the vector! A = 3î! 2 ĵ! 5 ˆk and the unit vector along the positive y-axis ĵ is given by A! ĵ = A! ĵ cos!, where!, is the angle between the! A ĵ two vectors and thus cos! =! A ĵ. With A! ĵ =!2 and A! = (2) 2 + (!5) 2 = 38! A ĵ and ĵ =1 we have cos! =! A ĵ = (!2) =!0.32 and thus! = arccos("0.32) = 109! A rock is thrown directly upward from the edge of the roof of a building that is 34.9 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what speed was the rock thrown? A) 6.4 m/s B) 12.1 m/s C) 3.6 m/s D) 8.7 m/s E) 10.9 m/s The vertical position of the rock is given by the equation for free fall: y(t) = y 0 + v 0 t! 1 2 gt 2. With y(t final ) = 0, we can solve this for v 0 : v 0 = 1 2 gt 2! y 0 t final Putting in y 0 = 34.9 m, t final = 4.00 s, g = 9.81 m/s, we get v 0 = 10.9 m/s.

2 3. The figure on the right shows the graph of the position x as a function of time for an object moving in a straight line (along the x-axis). Which of the following graphs best describes the velocity along the x- axis as a function of time for this object? The object starts with a large positive velocity given by the slope of the x(t) curve. At large t the slope goes to zero. Therefore, the correct graph is B).

3 4. A motorist traveling at a constant speed of 41.7 m/s in a 50 km/h speed zone passes a police car. Immediately after the car passes, the police car starts off in pursuit. The police officer accelerates at 2 m/s 2 up to a speed of 70 m/s, and then continues at this speed until she overtakes the speeding motorist. How long from the time she started does it take the police car to overtake the motorist? The motorist continues at a constant speed during this process. A) 99 s B) 43 s C) 38 s D) 35 s E) 56 s Position of motorist as a function of time t after the chase started x M (t) = v M t Position of police officer while she is accelerating: x O (t) = at 2 2 The time it takes her to reach v=70 m/s is v / a = (70m/s)/(2m/s 2 ) = 35 s At t = 35 s, x O = 1225 m and x M = (41.7m/s)(35s) = 1460 m Distance between police and motorist: 235 m Extra time required for officer to catch motorist = (235m)/(70m/s-41.7m/s) = 8 s Total time = (35 + 8) s = 43 s 5. Three people are pushing a stalled car up a hill at constant velocity. The net force on the car is A) down the hill and greater than the weight of the car. B) up the hill and greater than the weight of the car. C) up the hill and equal to the weight of the car. D) down the hill and equal to the weight of the car. E) zero. Newton s first law states that if no net force acts on an object it will move with constant velocity.

4 6. A girl throws a stone from a bridge. Consider the following ways she might throw the stone. Case A: Thrown straight up. Case B: Thrown straight down. Case C: Thrown at an angle of 45 o above horizontal. Case D: Thrown horizontally. The speed of the stone as it leaves her hand is the same in each case. In which case will the speed of the stone be greatest when it hits the ground below. A) Case A B) Case B C) Case C D) Case D E) The speed will be the same in all cases. Assuming the absence of air resistance, the easiest way to analyze this question is with the work-energy theorem. The only force acting is the weight, which is a constant force downwards. So, the net work done is force times the vertical displacement in the downward direction, which is the height of the bridge in all cases. Therefore the same work is done in all cases, leading to the same final kinetic energy and the same landing speed. (However, the direction of the final velocity might be different in different cases.) If air resistance is taken into account the stone with the shortest trajectory will be first hitting the ground. This would be case B.

5 7. For general projectile motion, when the projectile is at the highest point of its trajectory A) its acceleration is zero. B) its velocity and acceleration are both zero. C) the horizontal and vertical components of its velocity are zero. D) the horizontal component of its velocity is zero. E) its velocity is perpendicular to the acceleration. At the highest point of the trajectory for projectile motion the projectile s vertical velocity v y goes from being positive to being negative and is zero. The vertical acceleration is the entire time a y = 9.81 m/s 2 and pointing down. Since there s no acceleration in the horizontal direction, the horizontal velocity v x is equal to its initial value v 0x. Thus, at the highest point the velocity! v has only a horizontal component and is perpendicular to! a which has only a vertical component. 8. A projectile is fired from point 0 at the edge of a high cliff, with initial velocity components of v 0x = 60.0 m/s and v 0y = 175 m/s, as shown in the figure. The projectile rises and then falls into the sea at point P. The time of flight of the projectile is 40.0 s, and it experiences no appreciable air resistance in flight. The height of the cliff is closest to. A) 120 m B) 180 m C) 230 m D) 410 m E) 850 m : v 0 y = 175 m/s t = 40.0 s a y =!9.81 m/s 2 "y = v 0 y t a y t 2 =!848 m

6 9. Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 5.5 times that of particle B. The period of particle B is 2.7 times the period of particle A. The ratio r A / r B of the radius of the motion of particle A to that of particle B is closest to A) 2.0 B) 15 C) 4.1 D) 0.75 E) 0.49 : For uniform circular motion the radial acceleration is given by a r =! 2 r with! = 2" T, where r is the radius, T the period and ω the angular speed of the motion. Thus, r A r B = a A a B! # "! B! A 2 \$ & % = a A a B! # " T A T B 2 \$ & %! = 5.5a \$! B T # & A \$ # & " a B %" 2.7T A % 2 = = An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 30.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? A) 75.5 west of north B) 14.5 east of north C) 14.0 west of north D) 14.0 east of north E) 14.5 west of north We have v! plane,air = 120 m/s and v! air,ground = 30 m/s. From the sketch on the right, we see sin! =! v air,ground /! v plane,air = 30m/s and thus! = arcsin(0.250) = 14.5!. ( ) / ( 120m/s) = 0.250

7 11. A man pushes against a rigid, immovable wall. Which of the following is the most accurate statement concerning this situation? A) If the man pushes on the wall with a force of 200N, we can be sure that the wall is pushing back with a force of exactly 200N on him. B) The friction force on the man s feet is directed to the left. C) Since the wall cannot move, it cannot exert any force on the man. D) The man cannot be in equilibrium since he is exerting a net force on the wall. E) The man can never exert a force on the wall that exceeds his weight. Newton s 3 rd law if the man pushes on the wall with a force 200N (action), then the wall must be pushing back on him with an equal and opposite force (reaction). 12. A cylinder slides down a frictionless iceberg as shown in the figure. Which statement describes the magnitude of the cylinder s velocity! v and the cylinder s acceleration! a best while it is sliding down the hill. A) both v! and a! remain constant. B) both v! and a! increase. C) v! decreases and a! increases. D) v! increases and a! decreases. E) both v! and a! decrease. : The slope of the hill is negative and there will be a downward component of the weight at any place on the slope. Therefore, the acceleration down the hill, will always be positive and! v will increase all the time. However, since the slope flattens at the bottom of the hill, the downward force and the magnitude of the tangent acceleration decreases. However, the total acceleration, the vector sum of tangent and radial acceleration, can increase. For example, if the iceberg slope was circular, the radial acceleration at the bottom would be larger than the tangent acceleration at the top.

8 13. In the figure a block of mass M hangs in equilibrium. The rope, which is fastened to the wall is horizontal and has a tension of 27N. The rope, which is fastened to the ceiling has a tension of 83N, and makes an angle θ with the ceiling. The angle θ is A) 55 o B) 19 o C) 71 o D) 45 o E) 18 o Consider balance of forces in the horizontal direction: (83 N) cos θ=27 N θ=71 o 14. Dick and Jane stand on a platform of negligible weight, as shown in the figure. Dick weighs 500N and Jane weighs 400N. Jane is supporting some of her weight by holding a rope. Assume that all ropes and pulleys are ideal. What is the downward force Jane is exerting on the platform? A) 0 B) 300 N C) 100 N D) 240 N E) 50 N Draw the free body diagram for the whole system: the tension in the rope pulls on the system three times (once on Jane s hand, and twice on the lower pulley) Therefore, 3T = 900N T=300 N. Consider free body diagram for Jane: three forces are acting on her vertically, weight 400 N downward, tension in rope 300 N upward and the force from the platform the platform exerts an upward force of 100 N on her. By Newton s 3 rd law, she is exerting an equal and opposite force on the platform.

9 15. A worker is dragging a packing crate of mass 100kg across a rough floor where the coefficient of kinetic friction is He exerts a force sufficient to give the crate a positive acceleration. At what angle above the horizontal should his pulling force be directed in order to achieve the maximum acceleration? A) 34.5 o B) 27.7 o C) 45.0 o D) 21.8 o E) 30.0 o F cos! " F f = ma F N + F sin! " mg = 0 # F N = mg " F sin! ( ) F f = µ k F N # ma = F cos! " µ k mg " F sin! Taking derivative d(ma)/d! and setting it to zero, we get "F sin! + µ k F cos! = 0 Thus, the acceleration is maximum when tan! = µ k #! = tan "1 µ k = 21.8 The next two questions pertain to the following scenario: The Tornado is a carnival ride that consists of a big vertical cylinder that rotates rapidly about its vertical axis with constant angular speed. As The Tornado rotates, the riders are pressed against the inside wall of the cylinder by the rotation. During the ride, the floor of the cylinder drops away. 16. The force preventing the riders from falling downward is A) the centripetal force. B) the normal force. C) the centrifugal force. D) the gravitational force. E) the friction force. The gravitational force pulls downward on the riders. The force preventing them to fall down is the friction force between the riders and the wall of The Tornado.

10 17. While The Tornado rotates, the net force acting on a rider A) points radially inward. B) points down. C) is zero. D) points in the direction of the riders velocity. E) points radially outward. There a are several forces acting on a rider. Gravity is pulling down while the friction force between a rider and the wall of the The Tornado pulls up. These two are equal in magnitude since the rider neither moves up nor down. The net force is equal to the centripetal force causing the rider to move in a circular trajectory and provided by the normal force of the wall of the The Tornado. It points radially inward to the center of The Tornado.

11 18. A 5.00-kg box slides 3.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s? A) B) C) D) E) We use the work-kinetic energy theorem to solve this problem. The work done by the friction force is W =! f k "x =!µ k mg"x and the change in kinetic energy is!k = 1 2 mv 2 " 0. Setting these two equal and solving for the coefficient of kinetic 0 friction gives W =!K " #µ k mg!x = # 1 2 mv 2 " µ 0 k = v 2 0 2g!x = ( 3.00m/s) 2 2 \$ 9.81m/s 2 \$ 3.00m = A ball of mass 5.0 kg is suspended by two wires from a horizontal arm that is attached to a vertical shaft, as shown in the figure. The shaft is in uniform rotation about its axis. The rate of rotation is adjusted until the tensions in the two wires are equal. At that speed, the radial acceleration of the ball is closest to A) 6.9 m/s 2 B) 7.9 m/s 2 C) 5.9 m/s 2 D) 9.9 m/s 2 E) 4.9 m/s 2 From the y-coordinate, we get T 1 + T 2, y! mg = 0 and from the x-coordinate we get T 2, y! ma rad = 0. With T 1 = T 2 and T 2,y = 0.6T 2 and T 2,x = 0.8T 2, we get T T 2 = mg and 0.8T 2 = ma rad. Division of these two equations gives a rad = g/2 = 4.9 m/s 2.

12 20. In order to do work on an object, A) the object must move. B) the force doing the work must be directed perpendicular to the motion of the object. C) it is necessary that friction not be present. D) the applied force must be greater than the reaction force of the object. E) it is necessary that friction be present. Work done W =! F! "! r "! r = 0 # W = A system comprising blocks, a massless frictionless pulley, a frictionless incline, and connecting massless ropes is shown. The 9- kg block accelerates downward when the system is released from rest. The tension in the rope connecting the 6-kg block and the 4-kg block is closest to A) 39 N B) 30 N C) 36 N D) 42 N E) 33 N To solve problem, we begin by solving for the acceleration of the system m 1 = 6 kg, m 2 = 4 kg, m 3 = 9 kg Write down equations of motion for m 3 and m 1 + m 2 : m 3 g! T 2 = m 3 a T 2! (m 1 + m 2 )g sin" = (m 1 + m 2 )a Adding the 2 eqns together: m 3 g! (m 1 + m 2 )g sin" = (m 1 + m 2 + m 3 )a # a = m g! (m + m )g sin" = m/s 2 (m 1 + m 2 + m 3 ) Now consider free body diagram for m 1 T 1! m 1 g sin" = m 1 a # T 1 = m 1 a + m 1 g sin" = 42 N

13 22. Which of the free-body diagrams below represents a block sliding down a frictionless inclined surface? Answer D represents a block sliding down a frictionless inclined surface. Weight is directed down, and the normal force is directed perpendicular to the inclined surface. These are the only two forces present. 23. Two blocks with masses m 1 = 3.0 kg and m 2 = 2.0 kg rest on a horizontal, frictionless table. A horizontal force F is applied on block 1 from the left, as shown. The magnitude of the force exerted by block 1 on block 2 is 30 N. What is the magnitude of F? A) 30 N B) 45 N C) 50 N D) 65 N E) 75 N F m 1 m2 For block 2, Newton s second law gives F 2,1 = m 2 a, where F 2,1 is the normal force on block 2 by block 1. The acceleration is all the same for block 2, block 1 and the system consisting of blocks 1 and 2. Thus, for the total force F that accelerates the 2-block system, we have F = (m 1 + m 2 )a = m 1 + m 2 m 2 F 2,1 = 3.0kg+2.0kg (30N)=75N 2.0kg

14 24. A conical pendulum made with a bead of mass 4.0 kg and a 1.00 m long massless rope completes 1.2 turns every second. Determine the angle θ between the string and the vertical. A) 40 o B) 50 o C) 60 o D) 70 o E) 80 o x :T sin! = ma rad = m" 2 R = m" 2 Lsin! y :T cos! # mg = 0 \$ T = mg cos! \$ mg sin! cos! = m" 2 Lsin! \$ cos! = g " 2 L = 0.17 where we have used! = 1.2 rev/s=1.2 2" rad rev rev s =7.5 rad s 25. The work performed by an engine as a function of time for a process is given by W = at 3, where a = 2.4 J/s 3. What is the instantaneous power output of the engine at t = 1.4 s? A) 7 W B) 10 W C) 14 W D) 20 W E) 30 W The instantaneous power of the engine is given by P = dw dt = d ( at ) 3 = 3at 2. dt At t = 1.4 s, we have an instantaneous power of P = 3( 2.4J/s ) 3 ( 1.4s) 2 =14W.

15 26. When a parachutist jumps from an airplane, she eventually reaches a constant speed, called the terminal speed. Which statement describes the jump correctly? A) Once the parachutist has reached terminal speed the force of air drag on her is zero. B) Once the parachutist has reached terminal speed the force of air drag on her is equal to her weight. C) Once the parachutist has reached terminal speed the net force on her is equal to her weight. D) When the parachutist jumps out of the plane her acceleration vector points upward. E) Airplanes cannot fly high enough for a parachutist to reach terminal speed. When the parachutist has reached (constant) terminal speed she does not accelerate anymore and thus the net force on her is zero. Her weight is pulling her towards earth, while the force of air drag opposes the motion of the fall. Since the net force on the parachutist is zero once she has reached terminal speed, the force of air drag on her is equal to her weight. 27. A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and between the lower box and the table is 0.600, while the coefficient of static friction between these same surfaces is A horizontal pull to the right is exerted on the lower box, as shown in the figure, and the boxes move together. What is the friction force on the upper box? A) 19.3 N to the left B) 31.4 N to the left C) 19.3 N to the right D) 31.4 N to the right E) 23.5 N to the right Newton s second law for the system of two boxes in the x-direction is given by F! f k = (m + M )a F! µ k (m + M )g = (m + M )a " a = F m + M! µ k g = 150N 4kg+10kg! 0.6(9.8m/s2 ) = 4.83m/s 2 For the top box, we get f s = ma = 4kg! 4.83m/s 2 =19.3N

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