Chapter 4 Quick Check Problems

Chapter 4 Quick Check Problems

Quick Check 4.1 The net force on an object points to the left. Two of three forces are shown. Which is the missing third force? Answer: C. Reason: The net force of the two forces points up and to the right. Using vector addition by the head-to-tail method, it will become clear that once you add the third vector given in answer C, the vertical components will cancel out and there will be a resulting net force that points to the left. A. B. C. D. © 2015 Pearson Education, Inc.

Quick Check 4.2 A ball rolls down an incline and off a horizontal ramp. Ignoring air resistance, what force or forces act on the ball as it moves through the air just after leaving the horizontal ramp? The weight of the ball acting vertically down. A horizontal force that maintains the motion. A force whose direction changes as the direction of motion changes. The weight of the ball and a horizontal force. The weight of the ball and a force in the direction of motion. Answer: A Reason: Once the ball leaves the ramp, when ignoring air resistance, the only force acting on the ball is that of gravity (weight). Much like in projectile motion, we always stated that the only acceleration was that due to gravity. This force of weight and the acceleration due to gravity both point exclusively downward. © 2015 Pearson Education, Inc.

Quick Check 4.3 A steel beam hangs from a cable as a crane lifts the beam. What forces act on the beam? Gravity Gravity and tension in the cable Gravity and a force of motion Gravity and tension and a force of motion Answer: B Reason: First identify the contact forces. In this case, the only contact force is that of tension in the rope. Then, identify the long range forces. In this case, the only one is weight (or the force due to gravity). © 2015 Pearson Education, Inc.

Quick Check 4.4 A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows to a halt. What forces act on the sled just after she’s jumped in? Gravity and kinetic friction Gravity and a normal force Gravity and the force of the push Gravity, a normal force, and kinetic friction Gravity, a normal force, kinetic friction, and the force of the push Answer: D Reason: There is the long range force of gravity (weight) of the bobsledder. There are also two contact forces. The first is the normal force since the person is on a surface (the surface exerts a normal force on the person). The second is kinetic friction since the person is sliding across the surface. This force opposes the person’s motion. © 2015 Pearson Education, Inc.

Quick Check 4.5 A cart is pulled to the right with a constant, steady force. How will its acceleration graph look? Answer: C Reason: Newton’s second law has shown that an object that is pulled with a constant force will experience a constant acceleration. In this case, the force is constant and so the acceleration is constant. Graph C shows such a situation where the acceleration is, indeed, not changing. A. B. C. © 2015 Pearson Education, Inc.

Quick Check 4.6 A constant force causes an object to accelerate at 4 m/s2. What is the acceleration of an object with twice the mass that experiences the same force? 1 m/s2 2 m/s2 4 m/s2 8 m/s2 16 m/s2 Answer: B Reason: Acceleration is inversely proportional to mass, meaning that as one increases, the other decreases. In this case, we double the mass, so the acceleration is halved. One half of 2 is 1 and the correct answer is then B. © 2015 Pearson Education, Inc.

Quick Check 4.7 An object, when pushed with a net force F, has an acceleration of 2 m/s2. Now twice the force is applied to an object that has four times the mass. Its acceleration will be ½ m/s2 1 m/s2 2 m/s2 4 m/s2 Answer: B Reason: ai = Fi/mi = 2  af = 2Fi/4mi = ½ Fi/mi = ½ ai = 1 Thus, the correct answer is B. © 2015 Pearson Education, Inc.

Quick Check 4.8 A 40-car train travels along a straight track at 40 mph. A skier speeds up as she skis downhill. On which is the net force greater? The train The skier The net force is the same on both. There’s not enough information to tell. Answer: B Reason: F = ma. If there is no acceleration, there is no force. In this example, the train is moving at a constant speed so it had no acceleration so its force is zero. The skier, however, is accelerating and, even though we don’t know the exact values, we know this implies there must be some force. Thus, the answer is B. © 2015 Pearson Education, Inc.

Quick Check 4.9 An object on a rope is lowered at constant speed. Which is true? The rope tension is greater than the object’s weight. The rope tension equals the object’s weight. The rope tension is less than the object’s weight. The rope tension can’t be compared to the object’s weight. Answer: B Reason: First identify the forces. In this case we have the weight pointing down and the tension from the rope pointing up. Now, determine whether or not there is a net force so you know how large the arrows must be (the magnitude). The problem states that it is moving at a constant speed. Constant speed implies no acceleration. No acceleration implies no net force. As a result, the two forces must be equal in magnitude. Therefore, the answer is B. © 2015 Pearson Education, Inc.

Quick Check 4.10 An object on a rope is lowered at a steadily decreasing speed. Which is true? The rope tension is greater than the object’s weight. The rope tension equals the object’s weight. The rope tension is less than the object’s weight. The rope tension can’t be compared to the object’s weight. Answer: A Reason: Same as last problem, but now the object is decreasing in speed. It is moving down and slowing down. Since it is slowing down, the acceleration points in the opposite direction of the motion (upward). Since there is an acceleration, there is a force and the force points in the same direction as the acceleration. Thus, there must be a net force pointing upward. The only way this happens is if the tension in the rope is greater than the object’s weight. The correct answer is A. © 2015 Pearson Education, Inc.

Quick Check 4.11 An elevator, lifted by a cable, is moving upward and slowing. Which is the correct free-body diagram? Answer: C Reason: Identify the forces. Here, we have the weight of the elevator pointing downward and the tension in the cable pointing upward. The elevator is moving up, but slowing down. This implies that the acceleration points downward. This further implies that there is a net force pointing downward, which will occur when the weight is greater than the tension. The correct answer is C. A. B. C. D. E. © 2015 Pearson Education, Inc.

Quick Check 4.12 A ball has been tossed straight up. Which is the correct free-body diagram just after the ball has left the hand? Ignore air resistance. Answer: D Reason: The ball is tossed into the air. Once it leaves the person’s hand, the only force acting upon the ball is that due to gravity; the weight of the ball. This points straight down and the answer is, thus, D. © 2015 Pearson Education, Inc. A. B. C. D.

Quick Check 4.13 A ball, hanging from the ceiling by a string, is pulled back and released. Which is the correct free-body diagram just after its release? Answer: B Reason: There are two forces acting on the ball. The contact force is that of tension, which points along the rope (up and to the right). The long range force is that of gravity which points straight down. The correct answer is B. We do not tilt axes as it is not an inclined plane problem. A. B. C. D. E. © 2015 Pearson Education, Inc.

Quick Check 4.14 A car is parked on a hill. Which is the correct free-body diagram? Answer: C Reason: There are three forces at play in this example. Two contact forces and one long range force. The long range force is that of gravity (the weight). The two contact forces are the normal force (of the surface on the car, perpendicular to the ramp), and the static friction force, which points in the direction necessary to oppose the motion (so backward, up the ramp to the left). The correct answer is C. © 2015 Pearson Education, Inc.

Quick Check 4.15 A car is towed to the right at constant speed. Which is the correct free-body diagram? Answer: D Reason: There are four forces at play in this example. Three contact forces and one long range force. The long range force is the weight of the car. The contact forces are the normal force upward, the kinetic friction to the left, and the tension in the rope to the right. It is moving at a constant speed so there is no acceleration and no net force, so the magnitudes are the same and the answer is D. © 2015 Pearson Education, Inc.

Quick Check 4.17 A mosquito runs head-on into a truck. Splat! Which is true during the collision? The mosquito exerts more force on the truck than the truck exerts on the mosquito. The truck exerts more force on the mosquito than the mosquito exerts on the truck. The mosquito exerts the same force on the truck as the truck exerts on the mosquito. The truck exerts a force on the mosquito but the mosquito does not exert a force on the truck. The mosquito exerts a force on the truck but the truck does not exert a force on the mosquito. Answer: C Reason: Newton’s third law summarizes as “for every action, there is an equal but opposite reaction. The bug will exert the same force on the truck as the truck exerted on the bug. The answer is C. © 2015 Pearson Education, Inc. 18

Quick Check 5.1 A ring, seen from above, is pulled on by three forces. The ring is not moving. How big is the force F? 20 N 10 cos N 10 sin N 20 cos N 20 sin N Answer: E Reason: To find F, sum the forces in the y-direction. There are two forces acting in the positive y-direction; the y-component of the left force and the y-component of the right force. ∑F_y = 10sinθ + 10sinθ = 20sinθ. The answer is E. © 2015 Pearson Education, Inc.

Quick Check 5.4 What are the components of in the coordinate system shown? Answer: D Reason: The answer can be determined nearly instantly by realizing that the weight in this case would point in both the negative x- and y-directions. Furthermore, using trig., it is clear to see that the angle the weight makes with the y-axis is also θ and so we use sin along x and cos along y. Both ways confirm the answer D. © 2015 Pearson Education, Inc.

Quick Check 5.5 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. As the elevator accelerates upward, the scale reads > 490 N 490 N < 490 N but not 0 N 0 N Answer: A Reason: If the acceleration is upward, so is the net force. Since the normal force is that which points up in this case, the normal force must be larger than the weight. We know, by definition, that apparent weight is equal to the contact forces supporting an object/person. In this case, the apparent weight equals the normal force (a common occurrence). Thus, w_app = n > w. The scale reads more than the actual weight of 490 N. The answer is A. © 2015 Pearson Education, Inc. 22

Quick Check 5.6 A 50-kg student (mg = 490 N) gets in a 1000-kg elevator at rest and stands on a metric bathroom scale. Sadly, the elevator cable breaks. What is the reading on the scale during the few seconds it takes the student to plunge to his doom? > 490 N 490 N < 490 N but not 0 N 0 N Answer: D Reason: As soon as the cable breaks, the elevator, person, and scale are in free-fall. In freefall, the apparent weight is zero since there are no longer any supporting contact forces. © 2015 Pearson Education, Inc. 23

Quick Check 5.2 The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is n > mg n = mg n < mg n = 0 Not enough information to tell Answer: A Reason: If the acceleration is upward, so is the net force. Since n points up and w points down, and the net force points up, we know n must be larger than w =mg. Thus, the answer is A. © 2015 Pearson Education, Inc. 24

Quick Check 5.3 A box is being pulled to the right at steady speed by a rope that angles upward. In this situation: n > mg n = mg n < mg n = 0 Not enough information to judge the size of the normal force Answer: C Reason: There are now two forces that act in the positive y-direction. As a result, the normal force must be smaller so that together with the tension they add up to the weight. Proof: ∑F_y = n + Tsinθ – w = 0  w = n + Tsinθ. This shows that the normal force PLUS something else equals the weight, so n < w = mg. The answer is C. © 2015 Pearson Education, Inc. 25

Quick Check 5.7 A box on a rough surface is pulled by a horizontal rope with tension T. The box is not moving. In this situation: fs > T fs = T fs < T fs = smg fs = 0 Answer: B Reason: A force (tension) is applied to the right but the box is not moving. This, static friction must be opposing the motion. Set up N2L and solve: ∑ F_x = T – f_s = 0  T = f_s. © 2015 Pearson Education, Inc. 26

Quick Check 5.8 A box with a weight of 100 N is at rest. It is then pulled by a 30 N horizontal force. Does the box move? Yes No Not enough information to say Answer: B Reason: In order for the box to move it must overcome static friction. Thus, we need to find the force of friction and compare it to that of the pulling force. By definition, we have: f_s = mu_s * n = 0.4 * 100 = 40 > 30. As a result, the box will not move since the friction force is greater than that applied to the box. © 2015 Pearson Education, Inc. 27

Quick Check 5.9 A box is being pulled to the right over a rough surface. T > fk, so the box is speeding up. Suddenly the rope breaks. What happens? The box Stops immediately. Continues with the speed it had when the rope broke. Continues speeding up for a short while, then slows and stops. Keeps its speed for a short while, then slows and stops. Slows steadily until it stops. Answer: E Reason: As soon as the rope breaks, the box will start to slow down due to friction (but it won’t immediately come to a complete stop). The friction doesn’t wait to kick in at some later time, so all other answers are incorrect. © 2015 Pearson Education, Inc. 28

Quick Check 5.11 Consider the situation in the figure. Which pair of forces is an action/reaction pair? The tension of the string and the friction force acting on A The normal force on A due to B and the weight of A The normal force on A due to B and the weight of B The friction force acting on A and the friction force acting on B Answer: B Reason: An action/reaction pair is when two objects in contact exert “equal and opposite forces” on one another. Answer A: Incorrect. ∑F_x = f_k – T = ma_x. The object will accelerate backward relative to block B and, thus, T does not equal f_k. Answer C: Incorrect. The normal force of B on A would equal the magnitude of A’s weight, not B’s. The weight of B is actually the weight of A + B in this case. Answer D: There is no friction force acting on B since the surface is frictionless. If they mean the friction force on top of B, it’s the same friction force as that at the bottom of A and so, while they are equal in magnitude, they are also equal in direction…not an action/reaction pair. This leaves us with answer B, where we have a clear action/reaction pair. © 2015 Pearson Education, Inc.

Quick Check 5.12 Boxes A and B are being pulled to the right on a frictionless surface; the boxes are speeding up. Box A has a larger mass than Box B. How do the two tension forces compare? T1 > T2 T1 = T2 T1 < T2 Not enough information Answer: C Reason: Easy solution: F=ma. Rope 2 is pulling both masses while rope 1 is just pulling the mass of A. More mass, more force. Proper solution: ∑ F_x_A = T_1 = ma_x. ∑ F_x_B = T_2 – T_1 = ma_x  T_2 = T_1 + ma_x. Thus, T_2 equals T_1 plus some additional force, so it must be larger than T_1. The answer is C. © 2015 Pearson Education, Inc. 30

Quick Check 5.13 Boxes A and B are sliding to the right on a frictionless surface. Hand H is slowing them. Box A has a larger mass than Box B. Considering only the horizontal forces: FB on H = FH on B = FA on B = FB on A FB on H = FH on B > FA on B = FB on A FB on H = FH on B < FA on B = FB on A FH on B = FH on A > FA on B Answer: B Reason: Answer D can be ruled out immediately since H on A is not an action/reaction pair. All that needs to be done is to determine whether the pair between the hand and B is equal to, less than, or greater than the pair between A and B. In this case, noting that F=ma, the forces between the hand and B are greater since both mass A and B are pushing against the hand, whereas only the mass of A is pushing against B. The answer is B. © 2015 Pearson Education, Inc. 31

Quick Check 5.10 All three 50-kg blocks are at rest. The tension in rope 2 is Greater than the tension in rope 1. Equal to the tension in rope 1. Less than the tension in rope 1. Answer: B Reason: Draw a FBD of the masses. The blocks are in static equilibrium if at rest. Thus T=w for each of the masses. Since the weights are identical, so is the tension. The answer is B. © 2015 Pearson Education, Inc. 32

Quick Check 5.15 The top block is accelerated across a frictionless table by the falling mass m. The string is massless, and the pulley is both massless and frictionless. The tension in the string is T < mg T = mg T > mg Answer: A Reason: The top block accelerates along x and so the mass m must also be accelerating (downward in this case). As a result, the net force will also point downward. Since tension acts upward and the weight downward, the weight must be greater than the tension so that the net force does indeed point down. The answer is A. © 2015 Pearson Education, Inc. 33

Quick Check 3.19 A car is traveling around a curve at a steady 45 mph. Is the car accelerating? Yes No Answer: A Reason: The care is moving with a constant speed. However, the car is constantly changing direction (in this case, as it follows a circular path). By definition, we know that changing direction results in an acceleration. © 2015 Pearson Education, Inc.

Quick Check 3.20 A car is traveling around a curve at a steady 45 mph. Which vector shows the direction of the car’s acceleration? Answer: B Reason: Acceleration in circular motion always points toward the center of the circle. E. The acceleration is zero. © 2015 Pearson Education, Inc.

Quick Check 3.21 A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a result, the centripetal acceleration changes by a factor of 1/4 1/2 No change since the radius doesn’t change. 2 4 Answer: E Reason: Recall the expression for acceleration in circular motion (a = v^2 / r). If the speed v is doubled, then you quadruple the acceleration since 2^2 = 4. As a further example, what happens if you instead double the radius? In this case the acceleration would be ½ its original value since you are multiplying by two in the denominatior. © 2015 Pearson Education, Inc. 37

Quick Check 6.6 An ice hockey puck is tied by a string to a stake in the ice. The puck is then swung in a circle. What force is producing the centripetal acceleration of the puck? Gravity Air resistance Friction Normal force Tension in the string Answer: E Reason: Net force always points inward toward the center of the circle (centripetal force). In this example, the force pointing inward is the tension in the string (recall that tension points along the string). © 2015 Pearson Education, Inc. 38

Quick Check 6.7 A coin is rotating on a turntable; it moves without sliding. At the instant shown in the figure, which arrow gives the direction of the coin’s velocity? Answer: A Reason: Velocity is always tangent to the path of the object. The path here is that of a circle moving counterclockwise. At the very right where the coin is in this image, the tangent line would point straight up. The answer is A. © 2015 Pearson Education, Inc.

Quick Check 6.8 A coin is rotating on a turntable; it moves without sliding. At the instant shown in the figure, which arrow gives the direction of the frictional force on the coin? Answer: D Reason: There is only one force that acts along the surface of the turntable; friction. If the net force must point inward, then the friction must be pointing inward toward the center. The answer is D. © 2015 Pearson Education, Inc.

Quick Check 6.9 A coin is rotating on a turntable; it moves without sliding. At the instant shown, suppose the frictional force disappeared. In what direction would the coin move? Answer: A Reason: If the force of friction suddenly disappeared, the coin would continue to move in a straight line path until acted upon by another force (Newton’s First Law). If velocity points up, then it will move up in the direction of A. © 2015 Pearson Education, Inc.

Conceptual Example 6.4 Engineers design curves on roads to be segments of circles. They also design dips and peaks in roads to be segments of circles with a radius that depends on expected speeds and other factors. A car is moving at a constant speed and goes into a dip in the road. At the very bottom of the dip, which of the following is true? n > w n = w n < w Not enough information Answer: A Reason: At the bottom of the dip, the car is at the bottom of a circular path. Here, there is a normal force pointing upward and the weight pointing downward. Net force points toward the center of the circle. At the bottom of the circle, the net force would then point straight upward. Since net force is upward and the normal force is upward, it must be that n > w. © 2015 Pearson Education, Inc.

Quick Check 6.10 A physics textbook swings back and forth as a pendulum. Which is the correct free-body diagram when the book is at the bottom and moving to the right? Answer: C Reason: Same reason as last example. Note: it is not D or E as there are no horizontal forces acting upon the book. © 2015 Pearson Education, Inc. 43

Quick Check 6.11 A car that’s out of gas coasts over the top of a hill at a steady 20 m/s. Assume air resistance is negligible. Which free-body diagram describes the car at this instant? Answer: A Reason: Opposite situation as last problem. Now at the top of a circular path, the net force points down along with the weight. Thus, w >n. © 2015 Pearson Education, Inc. 44

Quick Check 6.12 A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected. Answer: E Reason: Tricky one here. The cart is on the INSIDE of the circular path. As a result of the fact that the normal force is always perpendicular to the surface, the normal force will point DOWN along with the weight. This is only depicted in answer E. © 2015 Pearson Education, Inc. 45

Quick Check 6.14 A coin sits on a turntable as the table steadily rotates counterclockwise. The free-body diagrams below show the coin from behind, moving away from you. Which is the correct diagram? Answer: C Reason: This is an example of horizontal circular motion so the FBD is drawn edge-on with the x-axis pointing toward the center (to the left). That alone rules out A or B since it shows it pointing rightward. In the vertical direction, there is the normal force and the weight. In the horizontal direction, there is only the friction which has to act inward. The answer is then C. © 2015 Pearson Education, Inc. 46

Quick Check 6.16 The force of Planet Y on Planet X is ___ the magnitude of One quarter One half The same as Twice Four times 2M M Planet X Planet Y Answer: C Reason: Newton’s law of gravitation defines an action/reaction pair among two objects. In this example, the two objects are forming an action reaction pair, regardless of mass, and so the magnitude of their forces are equal. © 2015 Pearson Education, Inc. 47

Quick Check 6.17 The gravitational force between two asteroids is 1,000,000 N. What will the force be if the distance between the asteroids is doubled? 250,000 N 500,000 N 1,000,000 N 2,000,000 N 4,000,000 N Answer: A Reason: Gravitational force is given by F = GMm/r2 If we increase the distance, based upon the equation, the force will decrease. Hence, if we doubled it, the force would be divided by 22 = 4. Thus, the force is ¼ its original value. The answer is A. © 2015 Pearson Education, Inc. 48

Quick Check 6.18 Planet X has free-fall acceleration 8 m/s2 at the surface. Planet Y has twice the mass and twice the radius of planet X. On Planet Y g = 2 m/s2 g = 4 m/s2 g = 8 m/s2 g = 16 m/s2 g = 32 m/s2 Answer: B Reason: The free-fall acceleration for other planets is given by g = GM/r2. Doubling the mass will double the acceleration. However, doubling the radius will divide the acceleration by 4. Thus we multiply by two and divide by 4. 2/4 = ½. The answer is B. © 2015 Pearson Education, Inc. 49

Quick Check 6.22 A 60-kg person stands on each of the following planets. On which planet is his or her weight the greatest? Answer: A Reason: Weight = mass*acceleration due to gravity. Same concept as last problem. g = GM/r2. 1/1 B) 2/4 = 1/2 C) 3/9 = 1/3 . Thus, Each case decreases the acceleration more and more. The answer is A. © 2015 Pearson Education, Inc.

Quick Check 6.20 Two satellites have circular orbits with the same radius. Which has a higher speed? The one with more mass. The one with less mass. They have the same speed. Answer: C Reason: Orbital speed is best given by v = sqrt(GM/R) where M is the mass of the Planet. Note that the mass of the orbiting object does not appear in the equation. Thus, the small mass has no effect on the speed. The answer is C. © 2015 Pearson Education, Inc. 51

Quick Check 6.21 Two identical satellites have different circular orbits. Which has a higher speed? The one in the larger orbit The one in the smaller orbit They have the same speed Answer: B Reason: Orbital speed is best given by v = sqrt(GM/R). The answer is B since as you decrease radius, you increase the speed (see equation above). © 2015 Pearson Education, Inc. 52

Quick Check 6.23 A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher orbit. Which of these quantities increases? Speed Period Centripetal acceleration Gravitational force of the earth Answer: C Reason: v = sqrt(GM/R) T2 = (4*pi2 / GM)*r3 a = v2 / r F = GMm/r2 The one equation out of the four that doesn’t have the radius in the denominator is that for period. The speed, acceleration, and force are all divided by r (thus, increasing r will decrease them). The answer is then B. © 2015 Pearson Education, Inc.

Chapter 4 Quick Check Problems

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