Ch 05 HW

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 3/53 For the same special case (the block of mass m 1 not present), what is the acceleration of the block of mass m 2 ? Express your answer in terms of g , and remember that an upward acceleration should be positive. ANSWER: Correct Part D Next, consider the special case where only the block of mass m 1 is present. Find the magnitude, T , of the tension in the rope. ANSWER: Correct Part E For the same special case (the block of mass m 2 not present) what is the acceleration of the end of the rope where the block of mass m 2 would have been attached? Express your answer in terms of g , and remember that an upward acceleration should be positive. ANSWER: Correct Part F Next, consider the special case m 1 = m 2 = m . What is the magnitude of the tension in the rope connecting the two blocks? Use the variable m in your answer instead of m 1 or m 2 . ANSWER: Correct Part G For the same special case ( m 1 = m 2 = m ), what is the acceleration of the block of mass m 2 ? ANSWER: Correct Part H Finally, suppose m 1 → ∞ , while m 2 remains finite. What value does the the magnitude of the tension approach? a 2 = -9.80 T = 0 a 2 = 9.80 T = mg a 2 = 0

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 6/53 Hint 2. Apply Newton's 2nd law to block 2 in the direction parallel to the incline What is Newton's 2nd law for block 2 in the direction parallel to the incline? (Assume the positive direction is going up the incline.) Express your answer in terms of m 2 m_2 , T T , f f , and θ theta , as well as the magnitude of the free-fall acceleration g g . ANSWER: Hint 3. Find an expression for the friction force What is the magnitude f f of the friction force acting on block 2? ANSWER: Hint 4. Find the normal force By applying Newton's 2nd law to block 2 in the direction perpendicular to the incline determine the magnitude of the normal force n n . Express your answer in terms of m 2 m_2 and θ theta , as well as the magnitude of the free-fall acceleration g g . ANSWER: Hint 5. Apply Newton's 2nd law to block 1 in the vertical direction Write an expression for Newton's 2nd law in the vertical direction for block 1. Take the positive direction to point downward. Express your answer in terms of the variables m 1 m_1 and/or T T , as well as the magnitude of the free-fall acceleration g g . ANSWER: Hint 6. Solve for the unknown tension T T Using your result from the previous hint, express T T in terms of g g , a a , and m 1 m_1 . ANSWER: Hint 7. Putting it all together By applying Newton's law to both block 1 and block 2, as you did in Hints 2 and 4, you found two equations where the masses m 1 m_1 , m 2 m_2 , the tension T T , and the acceleration a a , all appear, along with g g and θ theta . (Note that the friction force can be expressed in terms of the normal force, which, in turn, can be written as m 2 g cos( θ ) , as you found in Hint 3.) Choose one of the two equation and solve for T T ; substitute this result into the other equation. You will then have an equation with factors of m 1 m_1 and m 2 m_2 . You can then deduce the ratio. ANSWER: m 2 a = T − f − m 2 g sin( θ ) f = n / μ f = μ / n f = μn → f = μ → n n n = m 2 g cos( θ ) m 1 a = m 1 g − T T T = m 1 ( g − a ) m 1 / m 2 = a + ( μ cos θ + sin θ ) g g − a

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 9/53 Correct The key point is that contact forces must be determined from Newton's equation. In the problem described above, there is not enough information given to determine the normal force (e.g., the acceleration is unknown). Each of the answer options is valid under some conditions ( θ = 0 , the car is sliding down an icy incline, or the car is going around a banked turn), but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. Do not memorize values for the normal force valid in different problems--you must determine → n n_vec from ∑ → F = m → a . Centripetal Acceleration Explained Learning Goal: To understand that centripetal acceleration is the acceleration that causes motion in a circle. Acceleration is the time derivative of velocity. Because velocity is a vector, it can change in two ways: the length (magnitude) can change and/or the direction can change. The latter type of change has a special name, the centripetal acceleration . In this problem we consider a mass moving in a circle of radius R with angular velocity ω , → r ( t ) = R cos( ωt ) ˆ i + sin( ωt ) ˆ j = R cos( ωt ) ˆ i + R sin( ωt ) ˆ j . The main point of the problem is to compute the acceleration using geometric arguments. Part A What is the velocity of the mass at a time t ? You can work this out geometrically with the help of the hints, or by differentiating the expression for → r ( t ) given in the introduction. Express this velocity in terms of R R , ω omega , t t , and the unit vectors ˆ i i_unit and ˆ j j_unit . n = Mg n = Mg cos( θ ) n = Mg cos ( θ ) is found using ∑ → F = M → a [ ]

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 12/53 Correct That's right; the true statements are therefore: a. The centripetal acceleration might better be expressed as − ω 2 → r ( t ) because it is a vector. c. The magnitude of the centripetal acceleration is v 2 tangential / R . d. A particle that is going along a path with local radius of curvature R at speed s experiences an acceleration − s 2 / R . There is so much confusion about centrifugal force that you should probably ban this term from your vocabulary and thought processes. If you are in a car turning left, your centripetal acceleration is to the left (i.e., inward) and some real force must be applied to you to give you this acceleration--typically this would be provided by friction with the seat. The force you "feel" pushing you to the right is not a real force but rather a "fictitious force" that is present if you are in an accelerating coordinate system (in this case the car). It is best to stick to inertial (i.e., nonaccelerating) coordinate systems when doing kinematics and dynamics (i.e., → F = m → a calculations). Video Tutor: Ball Leaves Circular Track First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A Consider the video demonstration that you just watched. Which of the following changes would make it more likely for the ball to hit both the white can and the green can? Hint 1. How to approach the problem To answer this question, you first have to decide whether changing the ball’s mass or its speed can change the path it follows after it leaves the track. Newton’s second law says that a net force acting on the ball will change the ball’s motion—that is, its speed and/or direction. Newton’s first law says that, in the absence of a net force, the ball’s motion won’t change. After the ball leaves the track, does a net force act on it? Draw a free-body diagram for the ball if you’re not sure. To hit the green can, the ball must continue following a curved path. What would be needed to make that happen? a only b only c only d only e only b and e c and e d and e

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 15/53 ANSWER: ANSWER: ANSWER: Correct Part B Now, suppose that the curve is level ( θ = 0 ) and that the ice has melted, so that there is a coefficient of static friction μ mu between the road and the car's tires as shown in . What is μ min mu_min , the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 45.0 km/hour and that the radius of the curve is 43.8 m . Express your answer numerically. Hint 1. How to approach the problem You need to apply Newton's 2nd law to the car. Because you do not want the car to slip as it goes around the curve, the car needs to have a net acceleration of magnitude v 2 / r pointing radially inward (toward the center of the curve). Hint 2. Identify the correct free-body diagram ∑ F x = n cos θ ∑ F x = n sin θ ∑ F x = n cos θ + Mv 2 r ∑ F x = n cos θ − Mv 2 r a = 3.57 m/s 2 r = 43.8 m

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 18/53 ANSWER: Hint 2. Find the sum of forces in the y direction Now type the corresponding equation relating the y components of the forces acting on the chandelier, again using the coordinate system shown. Express your answer in terms of some or all of the variables m m , T 1 T_1 , T 2 T_2 , θ 1 theta_1 , and θ 2 theta_2 , as well as the magnitude of the acceleration due to gravity g g . You must use parentheses around θ 1 theta_1 and θ 2 theta_2 , when they are used as arguments to any trigonometric functions in your answer. ANSWER: Hint 3. Putting it all together There are two unknowns in this problem, T 1 T_1 and T 2 T_2 . Each of the previous two hints leads you to an equation involving these two unknowns. Eliminate T 2 T_2 from this pair of equations and solve for T 1 T_1 . ANSWER: Correct Exercise 5.18 - Enhanced - with Feedback A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is 700 kg , and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to 2500 N . The tension in the towrope between the transport plane and the first glider is not to exceed 12000 N . Part A If a speed of 40 m/s is required for takeoff, what minimum length of runway is needed? Express your answer in meters to two significant figures. ANSWER: ∑ F x = 0 = T 2 cos θ 2 − T 1 cos θ 1 ( ) ( ) ∑ F y = 0 = T 1 sin θ 1 + T 2 sin θ 2 − mg ( ) ( ) T 1 = mg sin θ 1 + cos θ 1 tan θ 2 ( ) ( ) ( )

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 21/53 Correct EXECUTE the solution as follows Part C Find both the x and y components of the acceleration of the box. Enter the x and y components of the acceleration in meters per second squared, separated by commas. Hint 1. Find the x component of the pulling force The pulling force → F p F_p_vec points in a direction 30 above the positive x axis. Use this to determine the value of F px F_px , the x component of the pulling force. Express your answer in newtons. ANSWER: Hint 2. Find the y component of the pulling force The pulling force → F p F_p_vec points in a direction 30 above the positive x axis. Use this to determine the value of F py F_py , the y component of the pulling force. Express your answer in newtons. ANSWER: No elements selected F px F_px = 208 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 22/53 Hint 3. Find the magnitude of the weight You know that the mass of the box is 30.0 kg and that g g = 9.80 m/s 2 . Use this information to calculate the magnitude w w of the weight of the box. Express your answer in newtons. ANSWER: Hint 4. Set up Newton's second law for the x direction Newton's second law states that Σ F x = ma x . Write an expression for Σ F x in terms of the forces acting on the box. Use f f , n n , and w w for the magnitudes of the friction force, the normal force, and the weight, respectively. Use F px F_px and F py F_py for the magnitudes of the x and the y components of the pulling force, respectively. Express your answer in terms of some or all of the variables F px F_px , F py F_py , f f , n n , and w w . ANSWER: Hint 5. Newton's second law in the y direction You are told that the box moves horizontally, which means that the acceleration in the y direction must be zero. Therefore, from Newton's second law in the y direction, you see that the net force in the y direction must also be zero. As long as the upward component F py F_py of the pulling force has a smaller magnitude than the weight (which is true in this case), the normal force will take a value that leads to a net force of zero in the y direction. The preceding paragraphs explain why both sides of Σ F y = ma y are zero. ANSWER: Correct EVALUATE your answer Part D A normal walking speed is around 2.0 m/s . How much time t t does it take the box to reach this speed if it has the acceleration 5.54 m/s 2 that you calculated above? Express your answer in seconds to two significant figures. ANSWER: Correct This time should match roughly to your experience with pulling boxes or furniture around. If you found a time longer than 10 s , you should be skeptical of your work, since it would generally not take you so long to get a box up to a normal walking pace. Note that once you get up to a normal walking pace, you generally reduce the force you exert so that it is just enough to balance the force of friction. In this way, the box will continue to move with you at a comfortable pace. Board Pulled Out from under a Box F py F_py = 120 N w w = 294 N Σ F x = F px − f = ma x a x a_x , a y a_y = 5.54,0 m/s 2 , m/s 2 t t = 0.36 s

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 23/53 A small box of mass m 1 m_1 is sitting on a board of mass m 2 m_2 and length L L . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is μ s mu_s . The coefficient of kinetic friction between the board and the box is, as usual, less than μ s mu_s . Throughout the problem, use g g for the magnitude of the free-fall acceleration. In the hints, use f for the magnitude of the friction force between the board and the box. Part A Find F min F_min , the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board). Express your answer in terms of some or all of the variables μ s mu_s , m 1 m_1 , m 2 m_2 , g g , and L L . Do not include f in your answer. Hint 1. Condition for the board sliding out from under the box The board will slide out from under the box if the magnitude of the board's acceleration exceeds the magnitude of the maximum acceleration that friction can give to the box. Hint 2. Find the acceleration of the box in terms of f Assume that the coefficient of static friction between the board and the box is not known at this point. What is the magnitude of the acceleration of the box in terms of the friction force f ? Express your answer in terms of f and m 1 m_1 . ANSWER: Hint 3. Find the largest acceleration of the box Now take the coefficient of static friction between the board and the box to be μ s mu_s . What is the largest possible magnitude of the acceleration of the box? Express your answer in terms of some or all of the variables μ s mu_s , g g , and m 1 m_1 . Hint 1. Maximum force on the box Friction is the only horizontal force on the box. What is the largest possible value for f f ? ANSWER: Hint 4. Find the sum of horizontal forces on the board Write down the sum of all the horizontal forces acting on the board. Take the positive x direction to be to the right. Give your answer in terms of F F , f , and any constants necessary. Hint 1. Friction and Newton's 3rd law Remember, by Newton's 3rd law, if there is a force of magnitude f f acting on the box due to the board, there is a force of equal magnitude and opposite direction acting on the board due to the box. a box a_box = f m 1 a box a_box = μ s g

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 24/53 ANSWER: Hint 5. Find the acceleration of the board for large f In Hint 4 you found the net horizontal force on the board. Now, find the acceleration of the board when the force of static friction reaches its maximum possible value. Express your answer in terms of F F , μ s mu_s , m 1 m_1 , m 2 m_2 , and g g . ANSWER: Hint 6. Putting it all together Reread Hint 1. In Hint 3, you found the largest possible acceleration of the box, a box a_box . In Hint 5, you found the acceleration of the board, a board a_board . What is the minimum value of the constant force, F min F_min , so that a board > a box ? ANSWER: Correct Exercise 5.34 - Enhanced - with Feedback Consider the system shown in the figure . Block A weighs 44.3 N and block B weighs 30.2 N . Once block B is set into downward motion, it descends at a constant speed. Part A Calculate the coefficient of kinetic friction between block A and the tabletop. ANSWER: Correct ∑ F x = F − f a board a_board = F − μ s m 1 g m 2 | | | | F min F_min = m 1 + m 2 μ s g ( ) μ = 0.682

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 25/53 Part B A cat, also of weight 44.3 N , falls asleep on top of block A . If block B is now set into downward motion, what is its acceleration magnitude? Express your answer in meters per second squared. ANSWER: Correct Part C A cat, also of weight 44.3 N , falls asleep on top of block A . If block B is now set into downward motion, what is its acceleration direction? ANSWER: Correct Exercise 5.54 - Enhanced - with Feedback The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m . Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s ). Part A Find the speed of the passengers when the Ferris wheel is rotating at this rate. Express your answer in meters per second. ANSWER: Correct Part B A passenger weighs 902 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel? Express your answer in newtons. ANSWER: Correct Part C What is his apparent weight at the highest point on the Ferris wheel? Express your answer in newtons. ANSWER: a = 2.49 m/s 2 upwards downwards v = 5.2 m/s w lowest = 950 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 26/53 Correct Part D What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? Express your answer in seconds. ANSWER: Correct Part E What then would be the passenger's apparent weight at the lowest point? Express your answer in newtons. ANSWER: Correct Normal Force and Centripetal Force Ranking Task A roller-coaster track has six semicircular "dips" with different radii of curvature. The same roller-coaster cart rides through each dip at a different speed. Part A For the different values given for the radius of curvature R R and speed v v , rank the magnitude of the force of the roller-coaster track on the cart at the bottom of each dip. Rank from largest to smallest. To rank items as equivalent, overlap them. Hint 1. Newton’s 2nd law Newton’s 2nd law is valid for any motion, whether the motion is along a straight line or a circular path. The net force acting on the cart must be equal to the product of its mass and acceleration, w highest = 850 N t = 14 s w lowest = 1800 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 27/53 ∑ → F = m → a , and since the cart is moving along a circular path, its acceleration (toward the center of the circular path) is given by a = v 2 R . Hint 2. Determine the net force on the cart Defining up as the positive direction, which of the following expressions describes the net force acting on the cart in the vertical direction while it's at the bottom of the dip? The variables F gravity F_gravity and F track F_track are the magnitudes of the forces. ANSWER: ANSWER: Correct Problem 5.84 - Enhanced - with Feedback Part A If the coefficient of static friction between a table and a uniform massive rope is μ s , what fraction of the rope can hang over the edge of the table without the rope sliding? Express your answer in terms of μ s mu_s . ∑ F = F gravity + F track ∑ F = F gravity − F track ∑ F = − F track − F gravity ∑ F = F track − F gravity Reset Help The correct ranking cannot be determined. R = 30 m v = 16 m/s R = 15 m v = 8 m/s R = 15 m v = 12 m/s R = 30 m v = 4 m/s R = 60 m v = 16 m/s R = 45 m v = 4 m/s smallest largest

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 28/53 ANSWER: Correct Problem 5.110 - Enhanced - with Feedback A road heading due east passes over a small hill. You drive a car of mass m at constant speed v over the top of the hill, where the shape of the roadway is well approximated as an arc of a circle with radius R . Sensors have been placed on the road surface there to measure the downward force that cars exert on the surface at various speeds. The table gives values of this force versus speed for your car is shown in the table below. Treat the car as a particle. Speed ( m/s ) 6.00 8.00 10.0 12.0 14.0 16.0 Force ( N ) 8100 7690 7050 6100 5200 4200 Part A Select a way to represent the data as a straight line. You might need to raise the speed, the force, or both to some power. ANSWER: Correct Part B Select the correct graph with the equation for the straight line that is the best fit to the data points described in the previous part. ANSWER: r = μ s 1 + μ s n vs. v 2 n 2 vs. v n vs. v n 2 vs. v 2

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 29/53 . . .

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 30/53 Correct Part C Use the graph to calculate m . Express your answer with the appropriate units. ANSWER: Correct Part D Use the graph to calculate R . Express your answer with the appropriate units. ANSWER: Correct Part E What maximum speed can the car have at the top of the hill and still not lose contact with the road? Express your answer with the appropriate units. ANSWER: Correct Key Example Variation Problem 5.5 Be sure to review Example 5.5 (Section 5.1) before attempting these problems. In all problems, ignore air resistance. VP 5.5.1 Part A In a modified version of the cart and bucket in , the angle of the slope is α alpha_1 = 36.9 and the bucket weighs 295 N . The cart moves up the incline and the bucket moves downward, both at constant speed. The cable has negligible mass, and there is no friction. What is the weight of the cart? Express your answer with the appropriate units. . m = 897 kg R = 49.6 m v = 22.0 m s

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 31/53 ANSWER: Correct Part B What is the tension in the cable? Express your answer with the appropriate units. ANSWER: Correct VP 5.5.2 Part C You change the angle of the slope in a to 26.0 and use a different cart and a different bucket. You observe that the cart and bucket remain at rest when released and that the tension in the cable of negligible mass is 155 N . There is no friction.What is the weight of the cart? Express your answer with the appropriate units. ANSWER: Correct Part D What is the combined weight of the cart and bucket? w cart 1 = 491 N T 1 = 295 N w cart 2 = 354 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 32/53 Express your answer with the appropriate units. ANSWER: Correct VP 5.5.3 Part E You construct a version of the cart and bucket in , but with a slope whose angle can be adjusted. You use a cart of mass 185 kg and a bucket of mass 59.0 kg . The cable has negligible mass, and there is no friction. What must be the angle of the slope so that the cart moves downhill at a constant speed and the bucket moves upward at the same constant speed? Express your answer in degrees. ANSWER: Correct Part F With this choice of angle, what will be the tension in the cable? Express your answer with the appropriate units. ANSWER: Correct VP 5.5.4 Part G In the situation shown in , let α be the angle of the slope and suppose there is friction between the cart and the track. You find that if the cart and bucket each have the same weight w , they remain at rest when released. In this case, what is the magnitude of the friction force on the cart? Express your answer in terms of α alpha and w w . w combined = 509 N θ 3 = 18.6 T 3 = 578 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 33/53 ANSWER: Correct Part H Is the friction force less than, greater than, or equal to w ? ANSWER: Correct Key Example Variation Problem 5.15 Be sure to review Examples 5.13, 5.14, and 5.15 (Section5.3) before attempting these problems. VP 5.15.1 Part A You pull on a crate using a rope as in , except the rope is at an angle of 20.0 above the horizontal. The weight of the crate is 325 N , and the coefficient of kinetic friction between the crate and the floor is 0.250. What must be the tension in the rope to make the crate move at a constant velocity? Express your answer with the appropriate units. ANSWER: f = w (1 − sin α ) f = w f > w f < w

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 34/53 Correct Part B What is the normal force that the floor exerts on the crate? Express your answer with the appropriate units. ANSWER: Correct VP5.15.2 Part C You pull on a large box using a rope as in , except the rope is at an angle of 15.0 below the horizontal. The weight of the box is 325 N , and the coefficient of kinetic friction between the box and the floor is 0.250. What must be the tension in the rope to make the box move at a constant velocity? Express your answer with the appropriate units. ANSWER: Correct Part D What is the normal force that the floor exerts on the box? Express your answer with the appropriate units. ANSWER: Correct VP 5.15.3 Part E You are using a lightweight rope to pull a sled along level ground. The sled weighs 435 N , the coefficient of kinetic friction between the sled and the ground is 0.200, the rope is at an angle of 12 above the horizontal, and you pull on the rope with a force of 125 N . Find the normal force that the ground exerts on the sled. T 1 = 79.3 N n 1 = 298 N T 2 = 90.2 N n 2 = 348 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 35/53 Express your answer with the appropriate units. ANSWER: Correct Part F Find the acceleration of the sled. Is the sled speeding up or slowing down? Express your answer with the appropriate units. Enter positive value if the sled is speeding up and negative value if the sled is slowing down. ANSWER: Correct VP 5.15.4 Part G A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above the horizontal, and pull. You find that the minimum tension you can apply to the rope in order to make the box start moving is T min . Find the coefficient of static friction between the floor and the box. Express your answer in terms of the variables m m , θ theta , T min T_min , and acceleration due to gravity g g . ANSWER: Correct Key Example Variation Problem 5.22 Be sure to review Examples 5.20, 5.21, and 5.22 (Section 5.4) before attempting these problems. VP 5.22.1 Part A You make a conical pendulum using a string of length 0.900 m and a bob of mass 0.450 kg . When the bob is moving in a circle at a constant speed, the string is at an angle of 30.0 from the vertical. What is the radius of the circle around which the bob moves? Express your answer with the appropriate units. n = 409 N a = 0.91 m s 2 μ s = T min cos θ mg − T min sin θ

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 36/53 ANSWER: Correct Part B How much time does it take the bob to complete one circle? Express your answer with the appropriate units. ANSWER: Correct Part C What is the tension in the string? Express your answer with the appropriate units. ANSWER: Correct VP 5.22.2 Part D A competition cyclist rides at a constant 12.5 m/s around a curve that is banked at 38.0 . The cyclist and her bicycle have a combined mass of 69.0 kg . What must be the radius of her turn if there is to be no friction force pushing her either up or down the banked curve? Express your answer with the appropriate units. ANSWER: Correct Part E R = 0.450 m t = 1.77 s T = 5.09 N R = 20.4 m

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 37/53 What is the magnitude of her acceleration? Express your answer with the appropriate units. ANSWER: Correct Part F What is the magnitude of the normal force that the surface of the banked curve exerts on the bicycle? Express your answer with the appropriate units. ANSWER: Correct VP 5.22.3 Part G An aerobatic airplane flying at a constant 60.0 m/s makes a horizontal turn of radius 275 m . The pilot has mass 80.0 kg . What is the bank angle of the airplane? Express your answer in degrees. ANSWER: Correct Part H What is the pilot's apparent weight during the turn? Express your answer with the appropriate units. ANSWER: Correct Part I How many times greater than his actual weight is this? ANSWER: Correct VP 5.22.4 a rad = 7.66 m s 2 n = 858 N β b = 53.2 n = 1310 N n / w = 1.67

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 38/53 Part J A sports car moves around a banked curve at just the right constant speed v so that no friction is needed to make the turn. During the turn, the driver (mass m ) feels as though she weighs x times her actual weight. Find the magnitude of the net force on the driver during the turn in terms of m , g , and x . Express your answer in terms of the variables m m , x x , and acceleration due to gravity g g . ANSWER: Correct Part K Find the radius of the turn. Express your answer in terms of the variables v v , x x , and acceleration due to gravity g g . ANSWER: Correct Exercise 5.15 - Enhanced - with Solution A load of bricks with mass m 1 m_1 = 15.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m 2 m_2 = 31.0 kg is suspended from the other end of the rope, as shown in . The system is released from rest. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Two bodies with the same magnitude of acceleration . Part A Draw a free-body diagram for the load of bricks. Draw the vectors starting at the black dot. The location, orientation, and relative length of the vectors will be graded. The exact length of the vectors will not be graded. ANSWER: F net = mg √ x 2 − 1 r = v 2 g √ x 2 − 1

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 39/53 Correct Part B Draw a free-body diagram for the counterweight. Draw the vectors starting at the black dot. The location, orientation, and relative length of the vectors will be graded. The exact length of the vectors will not be graded. ANSWER: No elements selected

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 40/53 Correct Part C What is the magnitude of the upward acceleration of the load of bricks? Express your answer with the appropriate units. ANSWER: Correct Part D What is the tension in the rope while the load is moving? Express your answer with the appropriate units. ANSWER: Correct Part E No elements selected a = 3.41 m s 2 T = 198 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 41/53 How does the tension compare to the weight of the load of bricks? To the weight of the counterweight? Drag the terms on the left to the appropriate blanks on the right to complete the sentence. ANSWER: Correct Exercise 5.46 - Enhanced - with Feedback A small car with mass 0.640 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 m . Part A If the normal force exerted by the track on the car when it is at the top of the track (point B ) is 6.00 N , what is the normal force on the car when it is at the bottom of the track (point A )? Express your answer with the appropriate units. ANSWER: Correct Exercise 5.39 - Enhanced - with Solution As shown in , block A (mass 2.22 kg ) rests on a tabletop. It is connected by a horizontal cord passing over a light, frictionless pulley to a hanging block B (mass 1.28 kg ). The coefficient of kinetic friction between block A and the tabletop is 0.448. The blocks are released then from rest. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Friction in horizontal motion . Reset Help smaller than greater than The tension in the rope is greater than the weight of the load of bricks and smaller than the weight of the counterweigh. F = 18.5 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 42/53 Part A Draw the free-body diagram for block A . Draw the force vectors with their tails at the center of the block A. The location and orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: No elements selected

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 43/53 Correct IDENTIFY: Apply Σ → F = m → a to each block. The target variables are the tension T in the cord and the acceleration a of the blocks. Then a can be used in a constant acceleration equation to find the speed of each block. The magnitude of the acceleration is the same for both blocks. SET UP: The system is sketched in the figure below. For each block take a positive coordinate direction to be the direction of the block’s acceleration. block on the table: The free-body is sketched in the figure below EXECUTE: Σ F y = ma y n − m A g = 0 n = m A g f k = μ k n = μ k m A g Σ F x = ma x T − f k = m A a T − μ k m A g = m A a Part B Draw the free-body diagram for block B . Draw the force vectors with their tails at the center of the block B. The location and orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER:

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 44/53 Correct SET UP: hanging block: The free-body is sketched in the figure below. EXECUTE: Σ F y = ma y m B g − T = m B a T = m B g − m B a Use the second equation in the first m B g − m B a − μ k m A g = m A a ( m A + m B ) a = ( m B − μ k m A ) g a = ( m B − μ k m A ) g m A + m B = ( 1.28 kg − ( 0.448 ) ( 2.22 kg ) ) ( 9.80 m / s 2 ) 2.22 kg + 1.28 kg = 0.7992 m/s 2 Part C No elements selected

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 45/53 Find the speed of each block after moving 2.30 cm . Express your answer in meters per second. ANSWER: Correct SET UP: Now use the constant acceleration equations to find the final speed. Note that the blocks have the same speeds. x − x 0 = 0.0230 m , a x = 0.7992 m/s 2 , v 0 x = 0 , v x = ? v 2 x = v 2 0 x + 2 a x ( x − x 0 ) EXECUTE: v x = 2 a x ( x − x 0 ) = √ 2(0.7992 m/s 2 )(0.0230 m) = 0.192 m/s = 19.2 cm/s . Part D Find the tension in the cord. Express your answer in newtons. ANSWER: Correct T = m B g − m B a = m B ( g − a ) = 1.28 kg(9.80 m/s 2 − 0.7992 m/s 2 ) = 11.5 N Or, to check, T − μ k m A g = m A a . T = m A ( a + μ k g ) = 2.22 kg(0.7992 m/s 2 + (0.448)(9.80 m/s 2 )) = 11.5 N , which checks. EVALUATE: The force T exerted by the cord has the same value for each block. T < m B g since the hanging block accelerates downward. Also, f k = μ k m A g = 9.75 N . T > f k and the block on the table accelerates in the direction of T . Exercise 5.56 A 51.0 kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. Part A If the plane's speed at the lowest point of the circle is 95.0 m/s , what is the minimum radius of the circle so that the acceleration at this point will not exceed 4.00 g ? Express your answer with the appropriate units. ANSWER: Correct Part B What is the apparent weight of the pilot at the lowest point of the pullout? Express your answer with the appropriate inits. ANSWER: v = 0.19 m/s √ T = 12 N R = 230 m

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 46/53 Correct Problem 5.60 - Enhanced - with Hints An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope . The rope will break if the tension in it exceeds 2.60×10 4 N , and our hero's mass is 85.4 kg . You may want to review (Page) . For help with math skills, you may want to review: Vector Components For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Two dimensional equilibrium . Part A If the angle between the rope and the horizontal is θ theta = 10.5 , find the tension in the rope. Express your answer in newtons. Hint 1. How to approach the problem Start by drawing a diagram with the forces acting on the person, as he rests at the middle of the rope. Create an xy coordinate system that best matches the forces and angles on the diagram. Consider that the suspended person is in equilibrium in both the x and y directions and that a tension is exerted by each half of the rope on either side of the person. Use this information, and apply Newton's 1st law in both the x and y directions separately. Then, solve for the tension in the rope. ANSWER: Correct Part B What is the smallest value the angle θ can have if the rope is not to break? Express your answer in degrees. Hint 1. How to approach the problem From Part A, you obtained an expression for the tension in the rope for a given angle θ . Use this expression in conjunction with the value given for the maximum tension the rope can bear to solve for the angle at this limit. ANSWER: n = 2500 N 2300 N

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 47/53 Correct Problem 5.73 An 3.00 kg box sits on a ramp that is inclined at 33.0 above the horizontal. The coefficient of kinetic friction between the box and the surface of the ramp is μ k = 0.300. A constant horizontal force F F = 26.0 N is applied to the box , and the box moves down the ramp. Part A If the box is initially at rest, what is its speed 2.00 s after the force is applied? Express your answer with the appropriate units. ANSWER: Correct Bridging Problem: In a Rotating Cone A small block with mass m is placed inside an inverted cone that is rotating about a vertical axis such that the time for one revolution of the cone is T . The walls of the cone make an angle β with the horizontal. The coefficient of static friction between the block and the cone is μ s . If the block is to remain at a constant height h above the apex of the cone, what are (a) the maximum value of T and (b) the minimum value of T ? (That is, find expressions for T max and T min in terms of β and h .) 0.922 v = 23.1 m s

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 48/53 IDENTIFY and SET UP Part A Although we want the block not to slide up or down on the inside of the cone, this is not an equilibrium problem. The block rotates with the cone and is in uniform circular motion, so it has an acceleration directed toward the center of its circular path. Identify the forces on the block. Select all that apply. ANSWER: Correct Part B What is the direction of the friction force when the cone is rotating as slowly as possible, so T has its maximum value T max ? ANSWER: Correct Part C What is the direction of the friction force when the cone is rotating as rapidly as possible, so T has its minimum value T min ? ANSWER: Correct Part D the normal force the weight of the block the tension force the friction force the weight of the cone down along the direction of cone's inner surface clockwise counterclockwise up along the direction of cone's inner surface horizontally towards the rotational axis at the cone's center down along the direction of cone's inner surface counterclockwise up along the direction of cone's inner surface horizontally towards the rotational axis at the cone's center clockwise

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 49/53 When T = T max and T = T min does the static friction force have its maximum magnitude? Why or why not? ANSWER: Correct The normal force is in direct proportion to the static friction force. Thus, at the minimum period T of rotation the normal force has its maximum and so does the static friction force. Part E Draw a free-body diagram for the block when the cone is rotating with T = T max . Draw the force vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. Hint 1. The normal force Recall that the normal force is always perpendicular to the surface. ANSWER: When T = T min , the radial acceleration is maximal, so the normal force exerted on the block, and hence, the static friction force has its maximum magnitude. When T = T max , the static friction force has its minimum. The static friction force has its maximum magnitude in both cases, when T = T max and T = T min . The only difference is in the direction of the friction force. The static friction force has its minimum magnitude in both cases, when T = T max and T = T min . The only difference is in the direction of the friction force. When T = T max , the radial acceleration is maximal, so the normal force exerted on the block, and hence, the static friction force has its maximum magnitude. When T = T min , the static friction force has its minimum.

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 50/53 Correct Part F Draw a free-body diagram for the block when the cone is rotating with T = T min . Draw the force vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded. ANSWER: No elements selected

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 51/53 Correct Part G What is the radius of the circular path that the block follows? Express your answer in terms of variables β beta and h h . ANSWER: Correct Part H Choose the unknown quantities, and decide which of these are the target variables. ANSWER: No elements selected R = h tan β

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 52/53 Correct EXECUTE Part I In case of the rotation with a period T = T max and with a period T = T min , write the acceleration of the block. Express your answer in terms of the variables T T , β beta , and h h . ANSWER: Correct Part J Find T max and T min . Express your answer in terms of the variables μ s mu_s , β beta , h h , and constants g g , π pi separated by a comma. Hint 1. Newton's second law Choose coordinate axes, and remember that it’s usually easiest to choose one of the axes to be in the direction of the acceleration. Write Newton's second law in component form. Hint 2. Static friction Recall that the relationship for when maximum static friction acts is f s = μ s n . Hint 3. Simplification It will be enough to solve T max or T min -- the other one can be obtained with the change of a sign before μ s . ANSWER: Correct EVALUATE Part K What do your expressions for T max and T min become if μ s = 0? Express your answer in terms of the variables β beta , h h , and constants g g , π pi separated by a comma. f s , n , β , R , T ; the target variables are T and n . f s , n , R , T ; the target variable is T . f s , n , w , R , T ; the target variable is T . f s , n , w , β , R , T ; the target variable are T and R . a = 4 π 2 h T 2 tan ( β ) T max , T min = 4 π 2 h cos ( β ) + μ s sin ( β ) g tan ( β ) sin ( β ) − μ s cos ( β ) , 4 π 2 h cos ( β ) − μ s sin ( β ) g tan ( β ) sin ( β ) + μ s cos ( β ) √ ( ) ( ) √ ( ) ( )

3/24/23, 2:08 PM Ch 05 HW https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10460269 53/53 ANSWER: Correct Substituting the earlier relationships v = 2 πR T and h = R tan β , this result can be rewritten as tan β = v 2 gR , which is the same result as Example 5.22 for a car rounding a banked curve with no friction. For the steps and strategies involved in solving this problem, you may view the Video Tutor Solution . Score Summary: Your score on this assignment is 30.4%. You received 7.91 out of a possible total of 26 points. T max , T min = 4 π 2 h g tan ( β ) 2 , 4 π 2 h g tan ( β ) 2 √ √