Version PREVIEW Semester 1 Review Slade (22222) 1

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1 Version PREVIEW Semester 1 Review Slade () 1 This print-out should have 48 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Holt SF 0Rev 10A 001 (part 1 of ) 10.0 points Consider the position-time graph for a squirrel running along a clothesline. position (m) time (s) What is the squirrel s displacement at 1.5 s? Your answer must be within ± 5.0% Correct answer: 3 m. The final position is Let : x i 0 m. x f x 1 + x 3.0 m the displacement is m + (4 m) x x f x i 3.0 m (0 m) 3.0 m. 00 (part of ) 10.0 points What is the squirrel s average velocity during the time interval between 0.0 s 1.5 s? Your answer must be within ± 5.0% Correct answer: m/s. Let : t 1.5 s. v avg x t 3.0 m 1.5 s m/s. Holt SF 0A points If Joe rides south on his bicycle in a straight line for 5 min with an average speed of 1.6 km/h, how far has he ridden? Correct answer: 5.5 km. Let : x v avg t t 5 min v avg 1.6 km/h. (1.6 km/h)(5 min) 5.5 km toward the south. 1 h 60 min Velocity Relationships points Consider three position curves between time points t A t B. s s A A 3 s 1 B B 0 t A t B t Choose the correct relationship among quantities v 1, v, v 3. v v A + v B when a is a constant. 1. v 1 v v 3 correct. v 1 > v > v 3 3. v 1 < v < v 3

2 Version PREVIEW Semester 1 Review Slade () The average velocity of an object is v displacement time s B s A t B t A. All three curves have exactly the same change in position s s B s A in exactly the same time interval t t B t A, so all three average velocities are equal: v 1 v v 3. Serway CP (part 1 of 4) 10.0 points The position versus time for a certain object moving along the x-axis is shown. The object s initial position is 4 m. position (m) time (s) Find the instantaneous velocity at 0.5 s. Correct answer: 13 m/s. The instantaneous velocity is the slope of the tangent line at that point. v 9 m ( 4 m) 1 s 0 s 13 m/s. 006 (part of 4) 10.0 points Find the instantaneous velocity at s. Correct answer:.5 m/s. 007 (part 3 of 4) 10.0 points Find the instantaneous velocity at 4.5 s. Correct answer: 0 m/s. v 4 m (4 m) 6 s 3 s 0 m/s. 008 (part 4 of 4) 10.0 points Find the instantaneous velocity at 8 s. Correct answer: m/s. v 0 m ( 4 m) 9 s 7 s m/s. Holt SF 0Rev points A tennis ball with a velocity of 11.3 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of 8.50 m/s to the left. If the ball is in contact with the wall for s, what is the average acceleration of the ball while it is in contact with the wall? Correct answer: 1800 m/s. Let : v i m/s, v f 8.50 m/s, t s. ā v f v i t 1800 m/s. 8.5 m/s 11.3 m/s s v 4 m (9 m) 3 s 1 s.5 m/s. Holt SF 0B 03

3 Version PREVIEW Semester 1 Review Slade () points With an average acceleration of 1.5 m/s, how long will it take a cyclist to bring a bicycle with an initial speed of 14.5 m/s to a complete stop? Correct answer: s. Let : v m/s g 9.8 m/s. Basic Concept: For constant acceleration, we have v v 0 + a t. (1) Let : a avg 1.5 m/s, v i 14.5 m/s, v f 0 m/s. Solution: The velocity at the top is zero. Since we know velocities acceleration, Eq. 1 containing v, a, t. Choose the positive direction to be up; then a g a v t v f v i t t v i a v i t 14.5 m/s 1.5 m/s s. Ball N points A ball is thrown upward. Its initial vertical speed is 10.1 m/s, acceleration of gravity is 9.8 m/s, maximum height h max are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9.8 m/s m/s hmax 9.8 m/s What is its time interval, t up, between the release of the ball the time it reaches its maximum height? Correct answer: s. or 0 v 0 + ( g) t up t up v 0 g (10.1 m/s) (9.8 m/s ) s. Ball Thrown Up points A ball is thrown straight up passes point B (at a height of 48.8 m above its starting point O) in 5. s. The acceleration of gravity is 9.8 m/s. h t A A is the time y A to reach its maximum B height h A 48.8 m v 0 O 5. s t A Figure is not drawn to scale. What was its initial speed v 0? Correct answer: m/s. Basic Concept: Motion under uniform acceleration y v 0 t + 1 a t. t

4 Version PREVIEW Semester 1 Review Slade () 4 Solution: The initial velocity is directed upward gravity acts downward, so y v 0 t 1 g t, v 0 t y + 1 g t, v 0 y + 1 g t t 48.8 m + 1 (9.8 m/s ) (5. s) 5. s m/s. Drop a Rock points If you drop a rock from a height of 1 m, it accelerates at g strikes the ground s later. The acceleration of gravity is 9.8 m/s. If you drop the same rock from half that height, its acceleration will be 1. more unable to determine. 4. about half. 5. the same. correct The acceleration of an object due to gravity is virtually constant at the earth s surface is not affected by height. Holt SF 03Rev points The fastest recorded pitch in Major League Baseball was thrown by Nolan Ryan in If this pitch were thrown horizontally, the ball would fall m (.65 ft) by the time it reached home plate, 18.3 m (60 ft) away. The acceleration of gravity is 9.81 m/s. How fast was Ryan s pitch? Correct answer: m/s. Basic Concepts: Horizontally, since a x 0 m/s. Vertically, since v i,y 0 m/s. Given: x v x t y 1 g ( t) y m x 18.3 m g 9.81 m/s Solution: From the horizontal motion, so that t x v x ) y 1 ( x g v x g( x) v x y (9.81 m/s )(18.3 m) ( m) m/s. Accelerating Elevator points An elevator starts from rest with a constant upward acceleration moves 1 m in the first 1.9 s. A passenger in the elevator is holding a 4.8 kg bundle at the end of a vertical cord. The acceleration of gravity is 9.8 m/s. What is the tension in the cord as the elevator accelerates? Correct answer: N.

5 Version PREVIEW Semester 1 Review Slade () 5 g T mg a elevator Let h be the distance traveled a the acceleration of the elevator. With the initial velocity being zero, we simplify the following expression solve for acceleration of the elevator: h v 0 t + 1 a t 1 a t a h t. The equation describing the forces acting on the bundle is F net ma T mg T m(g + a) ( m g + h ) t [ ] (4.8 kg) 9.8 m/s (1 m) + (1.9 s) N. Tipler PSE points A bullet of mass kg moving at 550 m/s impacts a large fixed block of wood travels 6.8 cm before coming to rest. Assuming that the acceleration of the bullet is constant, find the force exerted by the wood on the bullet. Correct answer: kn. Let : m kg, v o 550 m/s, x 6.8 cm m. The deceleration of the bullet is constant, so v v o + a x 0 a v v o x v o x. Apply F m a to express the force exerted on the bullet by the wood: F wood ma mv o x (0.004 kg)(550 m/s) (0.068 m) kn. 1 kn 1000 N Jumping From the Fire points A person of mass 73.4 kg escapes from a burning building by jumping from a window situated 39.4 m above a catching net. The acceleration of gravity is 9.8 m/s. If air resistance exerts a force of 106. N on him as he falls, determine his speed just before he hits the net. Correct answer: m/s. The forces acting on him are the gravitational force mg acting downward, the air resistance, F a, acting upward. The acceleration is downward the net force is F net ma mg F a a mg F a m The person is in free fall, so his final speed is defined by v f a h (mg Fa ) h m [(73.4 kg)(9.8 m/s ) 106. N] (39.4 m) 73.4 kg m/s.

6 Version PREVIEW Semester 1 Review Slade () 6 Hewitt CP9 04 E points In the orbiting space shuttle you are hed four identical boxes. The first one is filled with s. The second one is filled with iron. The third one is filled with water. The last one is filled with feathers. Shake the boxes. Which one offers the greatest resistance which one offers the smallest resistance? 1. iron, water. s, water 3. iron, feathers correct 4. feathers, iron 5. All are wrong. Among these three materials, iron has the largest density feathers have the smallest density. So the box filled with iron has largest mass offers greatest resistance while the box filled with feathers has the smallest mass offers the smallest resistance. Conceptual 04 Q points Two 30 N forces a 60 N force act on a hanging box as shown. 30 N 30 N 3. No; it is balanced. 4. Yes; upward. The horizontal components of the two 30 N forces cancel, leaving an upward force that is less than 60 N. Thus, the net force on the box is down, causing it to accelerate downward. Conceptual 05 Q points The Earth exerts an 760 N gravitational force on a man. What force does the man exert on the Earth? Correct answer: 760 N. By Newton s third law, the man exerts an equal but opposite force on the Earth. Hewitt CP9 05 E points Consider a stone at rest on the ground. There are two interactions that involve the stone. One is between the stone the Earth; Earth pulls down on the stone the stone pulls up on the Earth. What is the other interaction? 1. between the ground air. between the stone the ground correct 3. between the ground the Earth 4. All are wrong. 60 N Will the box experience acceleration? 1. Unable to determine without the angle.. Yes; downward. correct 5. between the Earth air If the action is the stone pushing down on the ground surface, the reaction is the ground pushing up on the stone. This upward force on the stone is called the normal force. Force Motion 15 0 (part 1 of ) 10.0 points

7 Version PREVIEW Semester 1 Review Slade () 7 The following questions refer to the collisions between a car a truck whose weight is much heavier than the car (M m). For each description of a collision below, choose the one answer from the possibilities that best describes the size (or magnitude) of the forces between the car the truck. Assume: Friction is so small that it can be ignored. M v They are both moving at the same speed when they collide. 1. Neither exerts a force on the other; the car gets smashed simply because it is in the way of the truck.. The truck exerts a greater amount of force on the car than the car exerts on the truck. 3. Not enough information is given to pick one of these answers. 4. The truck exerts the same amount of force on the car as the car exerts on the truck. correct 5. None of these answers describes the situation correctly. 6. The car exerts a greater amount of force on the truck than the truck exerts on the car. By Newton s third law, action reaction are of the same magnitude in the opposite direction. 03 (part of ) 10.0 points The car is moving much faster than the heavier truck when they collide. v m on the car than the car exerts on the truck.. The truck exerts the same amount of force on the car as the car exerts on the truck. correct 3. None of these answers describes the situation correctly. 4. The car exerts a greater amount of force on the truck than the truck exerts on the car. 5. Neither exerts a force on the other; the car gets smashed simply because it is in the way of the truck. 6. Not enough information is given to pick one of these answers. The same reason as Part 1. Holt SF 04Rev (part 1 of 5) 10.0 points Three blocks are in contact with each other on a frictionless horizontal surface. A 335 N horizontal force is applied to the block with mass of.9 kg as shown in the figure below. The acceleration of gravity is 9.8 m/s. F.9 kg 5.7 kg 7.5 kg a) What is the net force on the block with mass.9 kg? Correct answer: N. F m 1 m m 3 1. The truck exerts a greater amount of force

8 Version PREVIEW Semester 1 Review Slade () 8 Given : F 335 N, m 1.9 kg, m 5.7 kg, m kg, Basic Concepts: Solution: F g 9.8 m/s. F net ma m total m 1 + m + m 3 N m 1 + m + m 3 g (m 1 + m + m 3 ) g F (m 1 + m + m 3 ) a. The acceleration of the system is to the right. F a m 1 + m + m N.9 kg kg kg m/s F 1,net m 1 a (.9 kg) ( m/s ) N 05 (part of 5) 10.0 points b) What is the resultant force on the block with mass 5.7 kg? Correct answer: N. Solution: F,net m a (5.7 kg) ( m/s ) N 06 (part 3 of 5) 10.0 points c) What is the resultant force on the block with mass 7.5 kg? Correct answer: N. Solution: F 3,net m 3 a (7.5 kg) ( m/s ) N 07 (part 4 of 5) 10.0 points d) What is the magnitude of the force between the block with mass 5.7 kg 7.5 kg? Correct answer: N. F,3 m 3 g a m 3 Basic Concept: Solution: F 3,net m 3 a F,3 F,3 m 3 a (7.5 kg) ( m/s ) N 08 (part 5 of 5) 10.0 points e) What is the magnitude of the force between the block with mass.9 kg 5.7 kg? Correct answer: N. a F 1, F,3 m m g

9 Version PREVIEW Semester 1 Review Slade () 9 Basic Concept: F,net m a F 1, F,3 Solution: Since F, N is provided in the previous part, we have F 1, m a + F,3 (5.7 kg) ( m/s ) N N Down a Smooth Incline 09 (part 1 of 3) 10.0 points A 3.98 kg block slides down a smooth, frictionless plane having an inclination of 4. The acceleration of gravity is 9.8 m/s..88 m 3.98 kg 4 Find the acceleration of the block. Correct answer: m/s. Given : m 3.98 kg θ 4. Consider the free body diagram for the block mg sin θ Basic Concepts: N N mg cos θ W mg F x,net F cos θ W 0 W mg sin θ ma Motion has a constant acceleration. Recall the kinematics of motion with constant acceleration. Solution: Because the block slides down along the plane of the ramp, it seems logical to choose the x-axis in this direction. Then the y-axis must emerge perpendicular to the ramp, as shown. Let us now examine the forces in the x- direction. Only the weight has a component along that axis. So, by Newton s second law, Fx ma mg sin θ Thus a g sin θ With our particular value of θ, a (9.8 m/s ) sin m/s 030 (part of 3) 10.0 points What is the block s speed when, starting from rest, it has traveled a distance of.88 m along the incline. Correct answer: m/s. Since v 0 0, a m/s L.88 m, v f v 0 + a (x x 0 ) ( m/s ) (.88 m) v f m/s. It is very important to note at this point that neither of these values depended on the mass of the block. This may seem odd at first, but recall what Galileo discovered 300 years ago objects of differing mass fall at the same rate. 031 (part 3 of 3) 10.0 points What is the magnitude of the perpendicular force that the block exerts on the surface of the plane at a distance of.88 m down the incline? Correct answer: N.

10 Version PREVIEW Semester 1 Review Slade () 10 By examining the free body diagram again, we see that the force in the y direction is given by F mg cos θ FN. T T a 16 kg 7.6 kg a F N mg cos θ 3.98 kg (9.8 m/s ) cos N. Atwood Machine points A light, inextensible cord passes over a light, frictionless pulley with a radius of 9.5 cm. It has a(n) 16 kg mass on the left a(n) 7.6 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 1. m apart. The acceleration of gravity is 9.8 m/s. m g m1 g Since the larger mass will move down the smaller mass up, we can take motion downward as positive for m motion upward as positive for m 1. Apply Newton s second law to m 1 m respectively then combine the results: For mass 1: F1 : T m 1 g m 1 a (1) For mass : F : m g T m a () 9.5 cm ω We can add Eqs. (1) () above obtain: m g m 1 g m 1 a + m a 16 kg 1. m 7.6 kg At what rate are the two masses accelerating when they pass each other? Correct answer: m/s. Let : R 9.5 cm, m kg, m 16 kg, h 1. m, v ω R. Consider the free body diagrams a m m 1 g m 1 + m 16 kg 7.6 kg 16 kg kg (9.8 m/s ) m/s. T m 1 (g + a) (7.6 kg) (9.8 m/s m/s ) N. Centripetal Acceleration points A car rounds a curve while maintaining a constant speed. Is there a net force on the car as it rounds the curve? 1. It depends on the sharpness of the curve speed of the car.

11 Version PREVIEW Semester 1 Review Slade () 11. No its speed is constant. 3. Yes. correct Acceleration is a change in the speed /or direction of an object. Thus, because its direction has changed, the car has accelerated a force must have been exerted on it. Horizontal Circle points A 3.08 kg mass attached to a light string rotates on a horizontal, frictionless table. The radius of the circle is m, the string can support a mass of 30 kg before breaking. The acceleration of gravity is 9.8 m/s. What maximum speed can the mass have before the string breaks? Suppose an automobile has a kinetic energy of 500 J. When it moves with five times the speed, what will be its kinetic energy? Correct answer: 6500 J. Let E old 500 J n 5. The kinetic energy K v, so for v new n v old, K new K old v new v old (n v old) v old n Correct answer: m/s. The string will break when the tension exceeds the weight corresponding to 5 kg, so T max M g (30 kg)(9.8 m/s ). K new n K old (5) (500 J) 6500 J. When the smaller mass rotates in a horizontal circle, the string tension provides the centripetal force, so T mv r r T v m. Thus the maximum velocity before the string breaks is r Tmax v max m (0.691 m)(30 kg)(9.8 m/s ) 3.08 kg m/s. Kinetic Energy Comparison points Dragging a Block (part 1 of 5) 10.0 points An 18.1 kg block is dragged over a rough, horizontal surface by a constant force of 178 N acting at an angle of angle 8.1 above the horizontal. The block is displaced 96. m, the coefficient of kinetic friction is kg µ N 8.1 Find the work done by the 178 N force. The acceleration of gravity is 9.8 m/s. Correct answer: J. Consider the force diagram

12 Version PREVIEW Semester 1 Review Slade () 1 n F What is the sign of the work done by the frictional force? θ 1. zero f k. positive mg Work is W F s, where s is the distance traveled. In this problem s 5 ˆx is only in the ˆx direction. W F F x s x F cos θ s x (178 N) cos8.1 (96. m) J. 037 (part of 5) 10.0 points Find the magnitude of the work done by the force of friction. Correct answer: J. To find the frictional force, F friction µ N, we need to find N from vertical force balance. N is in the same direction as the y component of F opposite the force of gravity, so F sin θ + N mg N mg F sin θ. Thus the friction force is F friction µ N ˆx µ (mg F sin θ ˆx the work done by friction is W µ F friction s F f s µ (mg F sin θ) s x (0.11)[(18.1 kg)(9.8 m/s ) (178 N) sin8.1 ](96. m) J W µ J. 038 (part 3 of 5) 10.0 points 3. negative correct 039 (part 4 of 5) 10.0 points Find the work done by the normal force. Correct answer: 0 J. Since the normal force points in the ŷ direction s is in the ˆx direction W N N s N ŷ ˆx (part 5 of 5) 10.0 points What is the net work done on the block? Correct answer: J. The net work done on the body is the algebraic sum of the work done by the external force F the work done by the frictional force W net W F + W µ J J J. Holt SF 06Rev (part 1 of ) 10.0 points A constant force of 3.0 N to the right acts on a 1.4 kg mass for 0.48 s. a) Find the final velocity of the mass if it is initially at rest. Correct answer: m/s. Let to the right be positive. Let : F 3.0 N, m 1.4 kg, t 0.48 s.

13 Version PREVIEW Semester 1 Review Slade () 13 Since v i 0 m/s, F t m v m v f m v i m v f to the right. v f F t m (3 N) (0.48 s) 1.4 kg m/s 04 (part of ) 10.0 points b) Find the final velocity of the mass if it is initially moving along the x-axis with a velocity of 3.9 m/s to the left. Correct answer: m/s. Let : v i 3.9 m/s F t m v m v f m v i v f F t + mv i m F t m + v i (3 N) (0.48 s) 1.4 kg m/s, which is m/s to the left. + ( 3.9 m/s) Serway CP points A football punter accelerates a 0.57 kg football from rest to a speed of 9 m/s in 0. s. What constant force does the punter exert on the ball? Correct answer: N. Applying impulse, since v i 0 m/s. F t m v f m v i m v f F mv f t (0.57 kg) (9 m/s) 0. s N. Spring Between Blocks points Two blocks of masses M 3 M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, the blocks are pushed together with the spring between them. v M M Before (a) After (b) 3M 3M A cord holding them together is burned, after which the block of mass 3 M moves to the right with a speed of 49 m/s. What is the speed of the block of mass M? Correct answer: 147 m/s. Given : m 0.57 kg, v f 9 m/s, t 0. s. Let : m 1 M, m 3 M, v 49 m/s.

14 Version PREVIEW Semester 1 Review Slade () 14 From conservation of momentum p 0, so 0 m 1 v 1 + m v v 1 m v m 1 (3 M) (49 m/s) M 147 m/s. Pitching Machine Recoil points A baseball player uses a pitching machine to help him improve his batting average. He places the 66.1 kg machine on a frozen pond. The machine fires a kg baseball horizontally at a speed of 31.9 m/s. What is the magnitude of the recoil velocity of the machine? Correct answer: m/s. Let : m kg, m 66.1 kg, v 1f 31.9 m/s. We take the system to consist of the baseball the pitching machine. Because of the force of gravity the normal force, the system is not really isolated. However, both of these forces are directed perpendicularly to the motion of the system. Therefore, momentum is constant in the x direction because there are no external forces in this direction (assuming the surface is frictionless). The total momentum of the system before firing is zero. Therefore, the total momentum after firing must be zero, that is, or, in components, m 1 v 1f + m v f 0, m 1 v 1f m v f 0. v f m 1 m v 1f kg 66.1 kg (31.9 m/s ) m/s. Blocks Compress a Spring points Two blocks of masses 4 kg 15 kg are placed on a horizontal, frictionless surface. A light spring is attached to one of them, the blocks are pushed together with the spring between them. A cord holding them together is burned, after which the block of mass 15 kg moves to the right with a speed of 7 m/s. What is the velocity of the other mass in m/s? Correct answer: m/s. By momentum conservation m 1 v 1 + m v 0 so the velocity of m 1 is v 1 m v m 1 Serway CP (part 1 of ) 10.0 points Consider a 60 N cat burglar supported by a cable as in the figure Find the tension in the inclined cable. Correct answer: N.

15 Version PREVIEW Semester 1 Review Slade () 15 Let : W 60 N θ y T 1 T W θ x Since the burglar is held in equilibrium, Fy 0 T sin θ W 0 T W sin θ 60 N sin N. 048 (part of ) 10.0 points Find the tension in the horizontal cable. Correct answer: N. Since the burglar is held in equilibrium, Fx 0 T 1 T cos θ 0 T 1 T cos θ (105.8 N) cos N.