2/14/12MasteringPh sics: Assignments Assignment 04 Due: 11:59pm on Tuesday, February 7, 2012 Note: To understand how points are aw arded, read your instructor's Grading Policy. [Sw itch to Standard Assignment View ] Problem 1 Learning Goal: To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by , that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed (although they can be smaller than this or even zero). Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by and (or , or something similar) respectively. These are the components of the overall contact force: contact. perpendicular to and parallel to the plane of Kinetic friction when surfaces slide When one surface is sliding past the other, experiments show three things about the friction force (denoted ): The frictional force opposes the relative motion at the point of contact, is proportional to the normal force, and the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds. The constant of proportionality is called the coefficient of kinetic friction, often designated long as the sliding continues, the frictional force is then (valid when the surfaces slide by each other). . As Static friction when surfaces don't slide When there is no relative motion of the surfaces, the frictional force can assume an value from zero up to a maximum , where is the coefficient of static friction. Invariably, session.masteringph sics.com/m ct/assignments 1/18 2/14/12 MasteringPh sics: Assignments is larger than . it often accelerates. then the object will slide subject to the resistance of kinetic friction. .com/m ct/assignments 2/18 . The frictional force is determined by other forces on the objects so it can be either equal to or less than . which of the following statements about the force of friction between them. or equal to depends on the magnitude of the other forces (if any) as well as the . Part A When two objects slide by one another. together with any other forces present. in agreement with the observation that when a force is large enough that something breaks loose and starts to slide. Whether the actual magnitude of the friction force is 0. the actual magnitude and direction of the friction force are such that it. will cause the object to have the observed acceleration. Correct For static friction.. Do not automatically assume that unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i. is true? ANSWER: The frictional force is always equal to The frictional force is always less than . The equation is valid onl when the surfaces are on the verge of sliding. acceleration of the object through Part C When a board with a box on it is slowly tilted to larger and larger angle. when determining at what point an object will just begin to slip). is true? ANSWER: The frictional force is always equal to The frictional force is always less than . The magnitude of the force cannot exceed . which of the following statements about the frictional force between them. If the magnitude of static friction needed to keep acceleration equal to zero exceeds . less than . common experience session. . Correct Part B When two objects are in contact with no relative motion. The frictional force for surfaces with no relative motion is therefore (valid when the contacting surfaces have no relative motion).e. The frictional force is determined by other forces on the objects so it can be either equal to or less than . The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed .masteringph sics. . there is not enough information given to determine the normal force. The box begins to slide once the component of gravity acting parallel to the board equals the force of static friction. but remains the same. friction.2/14/12 MasteringPh sics: Assignments shows that the box will at some point "break loose" and start to accelerate down the board. is on a road tilted at an angle ." you know that the component of the box's weight that is parallel to the board is equal to (i. the sliding reduces the normal force. Once the box is moving. can no longer oppose it.) For the box to then accelerate. In the problem described above. which in turn reduces the friction. Correct At the point when the box finally does "break loose. Once the box is moving. there must be a net force on the box along the board.e. the component of the box's weight parallel to the board must be greater than the force of kinetic friction. this component of gravitational force on the box has just reached a magnitude such that the force of static friction. equals the force of static is larger than the force of static friction. Part D Consider a problem in which a car of mass Select the best answer. which has a maximum value of . When the box is stationary. Therefore the force of kinetic friction must be less than the force of static friction which implies . Which of the following is the most general explanation for why the box accelerates down the board? ANSWER: The force of kinetic friction is smaller than that of static friction. as expected. is smaller than the force of static friction but larger than the force of kinetic friction. Each of the answer options is valid under some conditions.masteringph sics. but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. but once the box starts moving. The normal force ANSWER: is found using Correct The key point is that contact forces must be determined from Newton's equation.com/m ct/assignments 3/18 . Thus. Do not memorize session. The board rests on a frictionless horizontal surface.masteringph sics. Hint A. 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). use for the magnitude of the acceleration due to gravity.3.com/m ct/assignments 4/18 . What is the magnitude of the acceleration of the box in terms of the friction force ? Express your answer in terms of ANSWER: = Correct and . as usual.1 Maximum force on the box Hint not displa ed session. from Problem 2 A small box of mass is sitting on a board of mass and length . The coefficient of static friction between the board and the box is .you must determine . less than .2 Find the acceleration of the box in terms of Assume that the coefficient of static friction between the board and the box is not known at this point.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. use for the magnitude of the friction force between the board and the box. Throughout the problem. What is .3 Find the largest acceleration of the box Now take the coefficient of static friction between the board and the box to be the largest possible magnitude of the acceleration of the box? Hint A.2/14/12 MasteringPh sics: Assignments values for the normal force valid in different problems -. Part A Find . In the hints. Hint A. The coefficient of kinetic friction between the board and the box is. Hint A. Hint 5.4 Find the sum of horizontal forces on the board Write down the sum of all the horizontal forces acting on the board. and . In Hint 3. . . Take the positive direction to be to the right. . In . = Correct Problem 3 A sky diver of mass 80.4. = Correct Hint A. constant force. session.1 Friction and Newton's 3rd law Hint not displa ed Give your answer in terms of ANSWER: .2/14/12 MasteringPh sics: Assignments Express your answer in terms of some or all of the variables ANSWER: = . Hint A. find the acceleration of the board when the force of static friction reaches its maximum possible value. . Now. so that ? .0 (including parachute) jumps off a plane and begins her descent. Correct Hint A.6 Putting it all together Reread Hint 1.com/m ct/assignments 5/18 .masteringph sics. and any constants necessary. Express your answer in terms of ANSWER: = Correct . What is the minimum value of the Express your answer in terms of some or all of the variables Do not include ANSWER: in your answer. and .5 Find the acceleration of the board for large In Hint 4 you found the net horizontal force on the board. Hint A. . . . . you found the largest possible acceleration of the box. . and . . you found the acceleration of the board. Yes and her acceleration is directed downward. ANSWER: = 784 Correct Part C For an object falling through air at a high speed can be expressed as . You can see how her acceleration drops to zero over time. Part B At some point during her free fall.80 for the magnitude of the acceleration due to gravity. the numerical value for Using this value for .com/m ct/assignments 6/18 . For a human body.250 . giving constant speed after a long time.1 Dynamic equilibrium Hint not displa ed Express your answer in newtons. the sky diver falls at constant speed.1 Free fall Hint not displa ed ANSWER: No. Part A At the beginning of her fall. the sky diver reaches her terminal speed. where the coefficient depends on the shape and size of the falling object and on the density of is about 0. Correct This applet shows the sky diver (not to scale) with her position.masteringph sics.2/14/12 MasteringPh sics: Assignments Throughout this problem use 9. . the drag force acting on it due to air resistance air.1 Terminal speed Hint not displa ed session. does the sky diver have an acceleration? Hint A. what is the terminal speed of the sky diver? Hint C. What is the magnitude of the drag force due to air resistance that acts on the sky diver when she has reached terminal speed? Hint B. Yes and her acceleration is directed upward. speed. and acceleration graphed as functions of time. does the skydiver have a nonzero acceleration? Hint D. their terminal speed is typically around 45 meters per second (100 miles per hour).2 Find the speed of the sky diver when the parachute is deployed Hint not displa ed Express your answer in newtons.2/14/12 MasteringPh sics: Assignments Express you answer in meters per second.88 105 Correct Part F session.masteringph sics.0 Correct Recreational sky divers can control their terminal speed to some extent by changing their body posture. the sky diver keeps falling at constant speed. the effective drag coefficient of the sky diver plus parachute increases to 60.com/m ct/assignments 7/18 . For maximum drag and stability. When oriented in a headfirst dive. ANSWER: = 1. (Usually the parachute is deployed when the sky diver reaches an altitude of about 900 --3000 . Yes and her acceleration is directed upward. Correct Part E When the parachute is fully open. Part D When the sky diver descends to a certain height from the ground.0 . a sky diver can reach speeds of about 54 meters per second (120 miles per hour)." In this position.1 How to approach the problem Hint not displa ed Hint E. she deploys her parachute to ensure a safe landing. What is the drag force acting on the sky diver immediately after she has opened the parachute? Hint E.1 How to approach the problem Hint not displa ed ANSWER: No. sky divers often will orient themselves "belly-first. ANSWER: = 56. Yes and her acceleration is directed downward.) Immediately after deploying the parachute. that may be because the sky divers you saw were using high-performance parachutes. ANSWER: = 3. without slipping.320 the platform when it is brought up to a rotational speed of 40.140 Part A What is the coefficient of static friction between the button and the platform? ANSWER: 0. Each tug exerts a constant force of 2. provided the button is Part B How far from the axis can the button be placed. Problem 4 A small button placed on a horizontal rotating platform with diameter 0.251 Correct from the axis.82 Part A What is the total work they do on the supertanker? Express your answer using two significant figures. If this seems slow based on video or real-life sky divers you have seen.masteringph sics.2/14/12 MasteringPh sics: Assignments What is the terminal speed of the sky diver when the parachute is opened? Hint F. will revolve with . these offer the sky divers more maneuverability in the air but increase the terminal speed up to 4 meters per second (10 miles per hour). if the platform rotates at 65.3 109 8/18 toward the north. session.1 How to approach the problem Hint not displa ed Express your answer in meters per second.61 Correct A typical "student" parachute for recreational skydiving has a drag coefficient that gives a terminal speed for landing of about 2 meters per second (5 miles per hour). as they pull the tanker a distance 0.0 ? ANSWER: 5.com/m ct/assignments .1 106 .0 a distance no more than 0. one an angle 18 west of north and the other an angle 18 east of north. ANSWER: 3.30 10−2 Correct Problem 5 Two tugboats pull a disabled supertanker. above the bottom of the incline.4 What is the final kinetic energy? Hint not displa ed Express your answer in terms of some or all of the variables ANSWER: = Correct .1 How to approach the problem Hint not displa ed . .com/m ct/assignments 9/18 . The coefficient of kinetic friction between the plane and the block is . so that it reaches a stranded skier who is a vertical . but there is some friction that you must give the box Hint A.masteringph sics. with mass up an incline of constant slope angle distance present. Problem 7 A block of weight sits on an inclined plane as shown. . and .2/14/12 MasteringPh sics: Assignments = Correct Problem 6 You are a member of an alpine rescue team and must project a box of supplies. .2 Find the total work done on the box Hint not displa ed Hint A. Hint A. A force of magnitude is applied to pull the block up the incline at constant speed.3 What is the initial kinetic energy? Hint not displa ed Hint A. with kinetic friction coefficient Part A Use the work-energy theorem to calculate the minimum speed at the bottom of the incline so that it will reach the skier. The incline is slippery. session. . . . . session. .1 How to start Hint not displa ed Hint A. the force pulls the block down the incline at a constant speed. done on the block by the applied force as the block moves a Express your answer in terms of any or all of the variables . and .masteringph sics. ANSWER: = Correct Now the applied force is changed so that instead of pulling the block up the incline. . ANSWER: = Correct Part B What is the total work distance up the incline? . .2 Find the magnitude of the friction force Hint not displa ed Express the work done by friction in terms of any or all of the variables .2/14/12 MasteringPh sics: Assignments Part A What is the total work distance up the incline? done on the block by the force of friction as the block moves a Hint A.com/m ct/assignments 10/18 . and . .1 Find the final kinetic energy Hint not displa ed Express the final speed in terms of session. and . . . ANSWER: = Correct Part D What is the total work done on the box by the appled force in this case? . and . Express your answer in terms of any or all of the variables . the final speed of the particle after it has traveled a distance . Hint A. . ANSWER: = Correct Problem 8 A particle of mass moves along a straight line with initial speed along the direction of its motion. A force of magnitude pushes the particle a distance Part A Find .masteringph sics. . .com/m ct/assignments . . .2/14/12 MasteringPh sics: Assignments Part C What is the total work distance done on the block by the force of friction as the block moves a down the incline? . 11/18 . and . Express your answer in terms of any or all of the variables . . If a quantity stays the same. assume that the particle's mass is increased to parameters in the problem introduction remain the same. and speed ).2/14/12 MasteringPh sics: Assignments ANSWER: = Correct Increase in mass For the next two parts. Hint B. assume that the initial speed of the particle is increased to the particle's mass once again equal to Part D By what factor with mass increase? session.1 Find the work done Hint not displa ed You should enter the two factors separated by a comma. does the initial kinetic energy increase (with respect to the first situation. it would increase by a factor of 2. Part B By what multiplicative factor factor does the initial kinetic energy increase. while all other does the work done by the force increase (with respect to the case when the particle )? had a mass If one of the quantities doubles. = 3. ANSWER: . with . and by what multiplicative . for instance. and by what factor does the work done by the force . then the multiplicative factor would be 1.masteringph sics.1 Correct Part C The particle's change in speed over the distance when it had a mass equal to ANSWER: less than Correct . will be ______ the change in speed Increase in initial velocit For the final two parts.com/m ct/assignments 12/18 . 1 Correct Part E The particle's change in speed over the distance when it had an initial velocity equal to Hint E.com/m ct/assignments 13/18 . ANSWER: . or .2/14/12 MasteringPh sics: Assignments Again. The suitcase travels a distance 3. The coefficient of kinetic friction between the ramp and the incline is 0. An elegant way to see this is to take the derivative of kinetic energy with respect to velocity: .9 above the horizontal by a force of magnitude 150 up a ramp inclined at an angle 24. This says that when the velocity of an object increases by an amount increases by . Part A Calculate the work done on the suitcase by the force ANSWER: = 585 Correct .254. so if you have a very fast moving object (large . enter the two factors.1 Some math help Hint not displa ed will be ______ the change in speed . separated by a comma. = 9. Part B session. it takes a large amount of work or energy to increase its velocity even just a little bit! Problem 9 A luggage handler pulls a suitcase of mass 16.masteringph sics.90 along the ramp.0 that acts parallel to the ramp. ANSWER: less than Correct Make sure you understand this result: The amount of energy needed to increase an object's velocity by a certain fixed amount increases the faster the object is already moving. its kinetic energy ). what is its speed after it has traveled 3. ANSWER: = -263 Correct Part C Calculate the work done on the suitcase by the normal force. The 20. session.90 along the ramp? ANSWER: = 4.com/m ct/assignments 14/18 . ANSWER: = -150 Correct Part E Calculate the total work done on the suitcase.block moves 75.masteringph sics. ANSWER: = 0 Correct Part D Calculate the work done on the suitcase by the friction force.block moves 75.2/14/12 MasteringPh sics: Assignments Calculate the work done on the suitcase by the gravitational force.0 to the right and the 12. ANSWER: = 172 Correct Part F If the speed of the suitcase is zero at the bottom of the ramp.0 downward.52 Correct Problem 10 Two blocks are connected by a very light string passing over a massless and frictionless pulley (Figure ).0.0. 500 and =0.2/14/12 MasteringPh sics: Assignments Part A Find the total work done on 20.58 Correct block if =0. = 5.62 Correct block if there is no friction between the table and the ANSWER: Part B Find the total work done on 12.0ANSWER: block.0and the 20.38 Correct block if there is no friction between the table and the ANSWER: Part C Find the total work done on 20.325 between the table and Part D Find the total work done on 12.0the 20. = 1.020. = 3.0block.masteringph sics.0ANSWER: block. = 2.54 Correct block if and between the table session.com/m ct/assignments 15/18 .0block.020. but instead had the constant value of 0.50 on a horizontal surface when. at point P.050 length? Express your answer using two significant figures.304 Correct Part C How far would the box have slid if the friction coefficient didn't increase. Part A Use the work-energy theorem to find how far this box slides before stopping. the coefficient of friction is not constant.2/14/12 MasteringPh sics: Assignments Problem 11 A box is sliding with a speed of 4. it encounters a rough section. On the rough section. Part A How much work must be done to stretch this spring by 0.masteringph sics.11 Correct from its unstretched must be applied to the free end. One end of the spring is fixed. reaching a value of 0. Note that . To keep the spring stretched or compressed an amount .050 length? session.600 at 12.3 Correct past point P Problem 12 Consider a spring that does not obey Hooke's law very faithfully. ANSWER: = 5.100 at P and increases linearly with distance past P. but starts at 0. and when it is compressed. ANSWER: = 0.com/m ct/assignments 16/18 from its unstretched . a force along the -axis with -component .11 Correct past point P Part B What is the coefficient of friction at the stopping point? ANSWER: = 0.5 past point P.100? ANSWER: = 10. when the spring is stretched and Part B How much work must be done to compress this spring by 0. Here . ANSWER: 10. Part A What should be the force constant of the spring? Assume that the spring has negligible mass. ANSWER: = 8.0 ? Give your answer in kilowatts .masteringph sics.0 (37.3 17/18 session.2/14/12 MasteringPh sics: Assignments Express your answer using two significant figures.com/m ct/assignments . a constant velocity of magnitude 60. If the force of rolling friction is independent of speed.66 is to compress the spring no more than 8. and the force of air resistance is proportional to the square of the speed. Express your answer using two significant figures.0 Part A What is the resisting force acting on the truck? ANSWER: = 1680 Correct Part B Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance.5 ) to the driving wheels when the truck is traveling at ( 37. ANSWER: = 0.17 Correct Part C Is it easier to stretch or compress this spring? ANSWER: to stretch to compress Correct Problem 13 You are asked to design spring bumpers for the walls of a parking garage.3 ) on a level road.1 10−2 before stopping. A freely rolling 1300 car moving at 0.6 104 Correct Problem 14 A truck engine transmits 28. what power will drive the truck at 30. and the force of air resistance is proportional to the square of the speed.2/14/12 MasteringPh sics: Assignments = Correct Part C Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. and the force of air resistance is proportional to the square of the speed. session. If the force of rolling friction is independent of speed.masteringph sics. ANSWER: = 115 Correct Part E Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance.08 out of a possible total of 140 points. ANSWER: = 13. You received 130.0 ? Give your answer in kilowatts.9%.com/m ct/assignments 18/18 .8 Correct Part D Assume that 65% of the resisting force is due to rolling friction and the remainder is due to air resistance. what power will drive the truck at 30. If the force of rolling friction is independent of speed. and the force of air resistance is proportional to the square of the speed.0 ? Give your answer in horsepower. what power will drive the truck at 120. If the force of rolling friction is independent of speed.0 ? Give your answer in horsepower. what power will drive the truck at 120. ANSWER: = 154 Correct Score Summary: Your score on this assignment is 92.