In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

Arterioles transporting blood to external capillaries beneath the surface of … Read More...
The 20-kg homogeneous smooth sphere rests on the two inclined smooth surfaces as shown. Draw relevant free-body diagram(s), and determine the support forces on the sphere at A and B.

The 20-kg homogeneous smooth sphere rests on the two inclined smooth surfaces as shown. Draw relevant free-body diagram(s), and determine the support forces on the sphere at A and B.

The 20-kg homogeneous smooth sphere rests on the two inclined … Read More...
1 | P a g e Lecture #2: Abortion (Warren) While studying this topic, we will ask whether it is morally permissible to intentionally terminate a pregnancy and, if so, whether certain restrictions should be placed upon such practices. Even though we will most often be speaking of terminating a fetus, biologists make further classifications: the zygote is the single cell resulting from the fusion of the egg and the sperm; the morula is the cluster of cells that travels through the fallopian tubes; the blastocyte exists once an outer shell of cells has formed around an inner group of cells; the embryo exists once the cells begin to take on specific functions (around the 15th day); the fetus comes into existence in the 8th week when the embryo gains a basic structural resemblance to the adult. Given these distinctions, there are certain kinds of non-fetal abortion—such as usage of RU-486 (the morning-after “abortion pill”)—though most of the writers we will study refer to fetal abortions. So now let us consider the “Classical Argument against Abortion”, which has been very influential: P1) It is wrong to kill innocent persons. P2) A fetus is an innocent person. C) It is wrong to kill a fetus. (Note that this argument has received various formulations, including those from Warren and Thomson which differ from the above. For this course, we will refer to the above formulation as the “Classical Argument”.) Before evaluating this argument, we should talk about terminology: A person is a member of the moral community; i.e., someone who has rights and/or duties. ‘Persons’ is the plural of ‘person’. ‘Person’ can be contrasted with ‘human being’; a human being is anyone who is genetically human (i.e., a member of Homo sapiens). ‘People’ (or ‘human beings’) is the plural of ‘human being’. Why does this matter? First, not all persons are human beings. For example, consider an alien from another planet who mentally resembled us. If he were to visit Earth, it would be morally reprehensible to kick him or to set him on fire because of the pain and suffering that these acts would cause. And, similarly, the alien would be morally condemnable if he were to propagate such acts on us; he has a moral duty not to act in those ways (again, assuming a certain mental resemblance to us). So, even though this alien is not a human being, he is nevertheless a person with the associative rights and/or duties. 2 | P a g e And, more controversially, maybe not all human beings are persons. For example, anencephalic infants—i.e., ones born without cerebral cortexes and therefore with severely limited cognitive abilities—certainly do not have duties since they are not capable of rational thought and autonomous action. Some philosophers have even argued that they do not have rights. Now let us return to the Classical Argument. It is valid insofar as, if the premises are true, then the conclusion has to be true. But maybe it commits equivocation, which is to say that it uses the same word in multiple senses; equivocation is an informal fallacy (i.e., attaches to arguments that are formally valid but otherwise fallacious). Consider the following: P1) I put my money in the bank. P2) The bank borders the river. C) I put my money somewhere that borders the river. This argument equivocates since ‘bank’ is being used in two different senses: in P1 it is used to represent a financial institution and, in P2, it is used to represent a geological feature. Returning to the classical argument, it could be argued that ‘person’ is being used in two different senses: in P1 it is used in its appropriate moral sense and, in P2, it is inappropriately used instead of ‘human being’. The critic might suggest that a more accurate way to represent the argument would be as follows: P1) It is wrong to kill innocent persons. P2) A fetus is a human being. C) It is wrong to kill a fetus. This argument is obviously invalid. So one way to criticize the Classical Argument is to say that it conflates two different concepts—viz., ‘person’ and ‘human being’—and therefore commits equivocation. However, the more straightforward way to attack the Classical Argument is just to deny its second premise and thus contend that the argument is unsound. This is the approach that Mary Anne Warren takes in “On the Moral and Legal Status of Abortion”. Why does Warren think that the second premise is false? Remember that we defined a person as “a member of the moral community.” And we said that an alien, for example, could be afforded moral status even though it is not a human being. Why do we think that this alien should not be tortured or set on fire? Warren thinks that, intuitively, we think that membership in the moral community is based upon possession of the following traits: 3 | P a g e 1. Consciousness of objects and events external and/or internal to the being and especially the capacity to feel pain; 2. Reasoning or rationality (i.e., the developed capacity to solve new and relatively complex problems); 3. Self-motivated activity (i.e., activity which is relatively independent of either genetic or direct external control); 4. Capacity to communicate (not necessarily verbal or linguistic); and 5. Possession of self-concepts and self-awareness. Warren then admits that, though all of the items on this list look promising, we need not require that a person have all of the items on this list. (4) is perhaps the most expendable: imagine someone who is fully paralyzed as well as deaf, these incapacities, which preclude communication, are not sufficient to justify torture. Similarly, we might be able to imagine certain psychological afflictions that negate (5) without compromising personhood. Warren suspects that (1) and (2) are might be sufficient to confer personhood, and thinks that (1)-(3) “quite probably” are sufficient. Note that, if she is right, we would not be able to torture chimps, let us say, but we could set plants on fire (and most likely ants as well). However, given Warren’s aims, she does not need to specify which of these traits are necessary or sufficient for personhood; all that she wants to observe is that the fetus has none of them! Therefore, regardless of which traits we want to require, Warren thinks that the fetus is not a person. Therefore she thinks that the Classical Argument is unsound and should be rejected. Even if we accept Warren’s refutation of the second premise, we might be inclined to say that, while the fetus is not (now) a person, it is a potential person: the fetus will hopefully mature into a being that possesses all five of the traits on Warren’s list. We might then propose the following adjustment to the Classical Argument: P1) It is wrong to kill all innocent persons. P2) A fetus is a potential person. C) It is wrong to kill a fetus. However, this argument is invalid. Warren grants that potentiality might serve as a prima facie reason (i.e., a reason that has some moral weight but which might be outweighed by other considerations) not to abort a fetus, but potentiality alone is insufficient to grant the fetus a moral right against being terminated. By analogy, consider the following argument: 4 | P a g e P1) The President has the right to declare war. P2) Mary is a potential President. C) Mary has the right to declare war. This argument is invalid since the premises are both true and the conclusion is false. By parity, the following argument is also invalid: P1) A person has a right to life. P2) A fetus is a potential person. C) A fetus has a right to life. Thus Warren thinks that considerations of potentiality are insufficient to undermine her argument that fetuses—which are potential persons but, she thinks, not persons—do not have a right to life.

1 | P a g e Lecture #2: Abortion (Warren) While studying this topic, we will ask whether it is morally permissible to intentionally terminate a pregnancy and, if so, whether certain restrictions should be placed upon such practices. Even though we will most often be speaking of terminating a fetus, biologists make further classifications: the zygote is the single cell resulting from the fusion of the egg and the sperm; the morula is the cluster of cells that travels through the fallopian tubes; the blastocyte exists once an outer shell of cells has formed around an inner group of cells; the embryo exists once the cells begin to take on specific functions (around the 15th day); the fetus comes into existence in the 8th week when the embryo gains a basic structural resemblance to the adult. Given these distinctions, there are certain kinds of non-fetal abortion—such as usage of RU-486 (the morning-after “abortion pill”)—though most of the writers we will study refer to fetal abortions. So now let us consider the “Classical Argument against Abortion”, which has been very influential: P1) It is wrong to kill innocent persons. P2) A fetus is an innocent person. C) It is wrong to kill a fetus. (Note that this argument has received various formulations, including those from Warren and Thomson which differ from the above. For this course, we will refer to the above formulation as the “Classical Argument”.) Before evaluating this argument, we should talk about terminology: A person is a member of the moral community; i.e., someone who has rights and/or duties. ‘Persons’ is the plural of ‘person’. ‘Person’ can be contrasted with ‘human being’; a human being is anyone who is genetically human (i.e., a member of Homo sapiens). ‘People’ (or ‘human beings’) is the plural of ‘human being’. Why does this matter? First, not all persons are human beings. For example, consider an alien from another planet who mentally resembled us. If he were to visit Earth, it would be morally reprehensible to kick him or to set him on fire because of the pain and suffering that these acts would cause. And, similarly, the alien would be morally condemnable if he were to propagate such acts on us; he has a moral duty not to act in those ways (again, assuming a certain mental resemblance to us). So, even though this alien is not a human being, he is nevertheless a person with the associative rights and/or duties. 2 | P a g e And, more controversially, maybe not all human beings are persons. For example, anencephalic infants—i.e., ones born without cerebral cortexes and therefore with severely limited cognitive abilities—certainly do not have duties since they are not capable of rational thought and autonomous action. Some philosophers have even argued that they do not have rights. Now let us return to the Classical Argument. It is valid insofar as, if the premises are true, then the conclusion has to be true. But maybe it commits equivocation, which is to say that it uses the same word in multiple senses; equivocation is an informal fallacy (i.e., attaches to arguments that are formally valid but otherwise fallacious). Consider the following: P1) I put my money in the bank. P2) The bank borders the river. C) I put my money somewhere that borders the river. This argument equivocates since ‘bank’ is being used in two different senses: in P1 it is used to represent a financial institution and, in P2, it is used to represent a geological feature. Returning to the classical argument, it could be argued that ‘person’ is being used in two different senses: in P1 it is used in its appropriate moral sense and, in P2, it is inappropriately used instead of ‘human being’. The critic might suggest that a more accurate way to represent the argument would be as follows: P1) It is wrong to kill innocent persons. P2) A fetus is a human being. C) It is wrong to kill a fetus. This argument is obviously invalid. So one way to criticize the Classical Argument is to say that it conflates two different concepts—viz., ‘person’ and ‘human being’—and therefore commits equivocation. However, the more straightforward way to attack the Classical Argument is just to deny its second premise and thus contend that the argument is unsound. This is the approach that Mary Anne Warren takes in “On the Moral and Legal Status of Abortion”. Why does Warren think that the second premise is false? Remember that we defined a person as “a member of the moral community.” And we said that an alien, for example, could be afforded moral status even though it is not a human being. Why do we think that this alien should not be tortured or set on fire? Warren thinks that, intuitively, we think that membership in the moral community is based upon possession of the following traits: 3 | P a g e 1. Consciousness of objects and events external and/or internal to the being and especially the capacity to feel pain; 2. Reasoning or rationality (i.e., the developed capacity to solve new and relatively complex problems); 3. Self-motivated activity (i.e., activity which is relatively independent of either genetic or direct external control); 4. Capacity to communicate (not necessarily verbal or linguistic); and 5. Possession of self-concepts and self-awareness. Warren then admits that, though all of the items on this list look promising, we need not require that a person have all of the items on this list. (4) is perhaps the most expendable: imagine someone who is fully paralyzed as well as deaf, these incapacities, which preclude communication, are not sufficient to justify torture. Similarly, we might be able to imagine certain psychological afflictions that negate (5) without compromising personhood. Warren suspects that (1) and (2) are might be sufficient to confer personhood, and thinks that (1)-(3) “quite probably” are sufficient. Note that, if she is right, we would not be able to torture chimps, let us say, but we could set plants on fire (and most likely ants as well). However, given Warren’s aims, she does not need to specify which of these traits are necessary or sufficient for personhood; all that she wants to observe is that the fetus has none of them! Therefore, regardless of which traits we want to require, Warren thinks that the fetus is not a person. Therefore she thinks that the Classical Argument is unsound and should be rejected. Even if we accept Warren’s refutation of the second premise, we might be inclined to say that, while the fetus is not (now) a person, it is a potential person: the fetus will hopefully mature into a being that possesses all five of the traits on Warren’s list. We might then propose the following adjustment to the Classical Argument: P1) It is wrong to kill all innocent persons. P2) A fetus is a potential person. C) It is wrong to kill a fetus. However, this argument is invalid. Warren grants that potentiality might serve as a prima facie reason (i.e., a reason that has some moral weight but which might be outweighed by other considerations) not to abort a fetus, but potentiality alone is insufficient to grant the fetus a moral right against being terminated. By analogy, consider the following argument: 4 | P a g e P1) The President has the right to declare war. P2) Mary is a potential President. C) Mary has the right to declare war. This argument is invalid since the premises are both true and the conclusion is false. By parity, the following argument is also invalid: P1) A person has a right to life. P2) A fetus is a potential person. C) A fetus has a right to life. Thus Warren thinks that considerations of potentiality are insufficient to undermine her argument that fetuses—which are potential persons but, she thinks, not persons—do not have a right to life.

MAE 214 – Fall 2015 Homework 3 Due: October 1, 2015 – Thursday by 1:00 p.m. Total Problems: 4 (including Extra Credit), Total Points: 105 1. Make a solid works part model from the given figure below. All dimensions are in millimeters. All sketches must be fully defined. Also create a drawing sheet and dimension it as shown. You can use a hole call out option under annotation to dimension a counter bore hole. (30 points) Save your part files as follows: My Documents/Homework 3 Folder/Prob1_LastName.SLDPRT My Documents/ Homework 3 Folder/Prob1_LastName.SLDDRW 2. Make a solid works part of the given figure below and also make a drawing sheet – front, top and right side views using 3rd angle projection method. Dimension the views with appropriate dimension technique. All dimensions are in mm. (30 points) Save your part file and drawing sheet as follows: Documents/Homework 3 folder/Problem 2_Last Name.SLDPRT Documents/Homework 3 folder/Problem 2_Last Name.SLDDRW 3. Make a solid works part file for the given figure below. All sketches must be fully defined. Your design tree menu must have advanced features i.e. plane, mirror, and fillet. The spot facing (SF) must be defined in a problem. The inclined cut must be created with an offset sketch and extrude cut or a suitable sketch that uses “up to surface” option. (40 points) Your part model must stick to the isometric view as it is shown here. Save your part file into: My documents/Homework 3 Folder/Problem3_Last Name.SLDPRT Given: A = 76 B = 127 Unit: MMGS ALL ROUNDS (FILLET) EQUAL 6 MM 4. (Extra Credit) Make a solid works part from the given figure below. All sketches must be fully defined. Save your part file to Documents/Homework3 Folder/Prob#4_Last Name.SLDPRT All dimensions are in millimeters. (5 points)

MAE 214 – Fall 2015 Homework 3 Due: October 1, 2015 – Thursday by 1:00 p.m. Total Problems: 4 (including Extra Credit), Total Points: 105 1. Make a solid works part model from the given figure below. All dimensions are in millimeters. All sketches must be fully defined. Also create a drawing sheet and dimension it as shown. You can use a hole call out option under annotation to dimension a counter bore hole. (30 points) Save your part files as follows: My Documents/Homework 3 Folder/Prob1_LastName.SLDPRT My Documents/ Homework 3 Folder/Prob1_LastName.SLDDRW 2. Make a solid works part of the given figure below and also make a drawing sheet – front, top and right side views using 3rd angle projection method. Dimension the views with appropriate dimension technique. All dimensions are in mm. (30 points) Save your part file and drawing sheet as follows: Documents/Homework 3 folder/Problem 2_Last Name.SLDPRT Documents/Homework 3 folder/Problem 2_Last Name.SLDDRW 3. Make a solid works part file for the given figure below. All sketches must be fully defined. Your design tree menu must have advanced features i.e. plane, mirror, and fillet. The spot facing (SF) must be defined in a problem. The inclined cut must be created with an offset sketch and extrude cut or a suitable sketch that uses “up to surface” option. (40 points) Your part model must stick to the isometric view as it is shown here. Save your part file into: My documents/Homework 3 Folder/Problem3_Last Name.SLDPRT Given: A = 76 B = 127 Unit: MMGS ALL ROUNDS (FILLET) EQUAL 6 MM 4. (Extra Credit) Make a solid works part from the given figure below. All sketches must be fully defined. Save your part file to Documents/Homework3 Folder/Prob#4_Last Name.SLDPRT All dimensions are in millimeters. (5 points)

Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F  F push F push Draw the vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Tactics Box 5.2 Identifying Forces Learning Goal: To practice Tactics Box 5.2 Identifying Forces. The first basic step in solving force and motion problems generally involves identifying all of the forces acting on an object. This tactics box provides a step-by-step method for identifying each force in a problem. TACTICS BOX 5.2 Identifying forces Identify the object of interest. This is the object whose motion 1. you wish to study. 2. Draw a picture of the situation. Show the object of interest and all other objects—such as ropes, springs, or surfaces—that touch it. 3. Draw a closed curve around the object. Only the object of interest is inside the curve; everything else is outside. 4. Locate every point on the boundary of this curve where other objects touch the object of interest. These are the points where contact forces are exerted on the object. Name and label each contact force acting on the object. There is at least one force at each point of contact; there may be more than one. When necessary, use subscripts to distinguish forces of the same type. 5. 6. Name and label each long-range force acting on the object. For now, the only long-range force is the gravitational force. Apply these steps to the following problem: A crate is pulled up a rough inclined wood board by a tow rope. Identify the forces on the crate. Part A Which of the following objects are of interest? Check all that apply. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Conceptual Questions on Newton’s 1st and 2nd Laws Learning Goal: To understand the meaning and the basic applications of Newton’s 1st and 2nd laws. In this problem, you are given a diagram representing the motion of an object–a motion diagram. The dots represent the object’s position at moments separated by equal intervals of time. The dots are connected by arrows representing the object’s average velocity during the corresponding time interval. Your goal is to use this motion diagram to determine the direction of the net force acting on the object. You will then determine which force diagrams and which situations may correspond to such a motion. crate earth rope wood board Part A What is the direction of the net force acting on the object at position A? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D upward downward to the left to the right The net force is zero. This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Understanding Newton’s Laws Part A An object cannot remain at rest unless which of the following holds? You did not open hints for this part. ANSWER: Part B If a block is moving to the left at a constant velocity, what can one conclude? You did not open hints for this part. ANSWER: The net force acting on it is zero. The net force acting on it is constant and nonzero. There are no forces at all acting on it. There is only one force acting on it. Part C A block of mass is acted upon by two forces: (directed to the left) and (directed to the right). What can you say about the block’s motion? You did not open hints for this part. ANSWER: Part D A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be __________. You did not open hints for this part. ANSWER: There is exactly one force applied to the block. The net force applied to the block is directed to the left. The net force applied to the block is zero. There must be no forces at all applied to the block. 2 kg 3 N 4 N It must be moving to the left. It must be moving to the right. It must be at rest. It could be moving to the left, moving to the right, or be instantaneously at rest. Part E Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown. The net force acting on the object __________. Check all that apply. You did not open hints for this part. ANSWER: Tactics Box 5.3 Drawing a Free-Body Diagram Learning Goal: To practice Tactics Box 5.3 Drawing a Free-Body Diagram. A free-body diagram is a diagram that represents the object as a particle and shows all of the forces acting on the object. Learning how to draw such a diagram is a very important skill in solving physics problems. This tactics box explains the essential steps to construct a correct free-body diagram. TACTICS BOX 5.3 Drawing a free-body diagram Identify all forces acting on the object. This step was described 1. in Tactics Box 5.2. continuously changing direction moving at constant velocity moving with a constant nonzero acceleration moving with continuously increasing acceleration 4 N 10 N cannot have a magnitude equal to cannot have a magnitude equal to cannot have the same direction as the force with magnitude must have a magnitude greater than 5 N 10 N 10 N 10 N Draw a coordinate system. Use the axes defined in your pictorial representation. If those axes are tilted, for motion along an incline, then the axes of the free-body diagram should be similarly tilted. 2. Represent the object as a dot at the origin of the coordinate axes. This is 3. the particle model. 4. Draw vectors representing each of the identified forces. This was described in Tactics Box 5.1. Be sure to label each force vector. Draw and label the net force vector . Draw this vector beside the diagram, not on the particle. Or, if appropriate, write . Then, check that points in the same direction as the acceleration vector on your motion diagram. 5. Apply these steps to the following problem: Your physics book is sliding on the carpet. Draw a free-body diagram. Part A Which forces are acting on the book? Check all that apply. You did not open hints for this part. ANSWER: F  net F =  net 0 F  net a Part B Draw the most appropriate set of coordinate axes for this problem. The orientation of your vectors will be graded. ANSWER: gravity normal force drag static friction tension kinetic friction spring force Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points.

Chapter 5 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 5.1 Drawing Force Vectors Learning Goal: To practice Tactics Box 5.1 Drawing Force Vectors. To visualize how forces are exerted on objects, we can use simple diagrams such as vectors. This Tactics Box illustrates the process of drawing a force vector by using the particle model, in which objects are treated as points. TACTICS BOX 5.1 Drawing force vectors Represent the object 1. as a particle. 2. Place the tail of the force vector on the particle. 3. Draw the force vector as an arrow pointing in the proper direction and with a length proportional to the size of the force. 4. Give the vector an appropriate label. The resulting diagram for a force exerted on an object is shown in the drawing. Note that the object is represented as a black dot. Part A A book lies on a table. A pushing force parallel to the table top and directed to the right is exerted on the book. Follow the steps above to draw the force vector . Use the black dot as the particle representing the book. F  F push F push Draw the vector starting at the black dot. The location and orientation of the vector will be graded. The length of the vector will not be graded. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Tactics Box 5.2 Identifying Forces Learning Goal: To practice Tactics Box 5.2 Identifying Forces. The first basic step in solving force and motion problems generally involves identifying all of the forces acting on an object. This tactics box provides a step-by-step method for identifying each force in a problem. TACTICS BOX 5.2 Identifying forces Identify the object of interest. This is the object whose motion 1. you wish to study. 2. Draw a picture of the situation. Show the object of interest and all other objects—such as ropes, springs, or surfaces—that touch it. 3. Draw a closed curve around the object. Only the object of interest is inside the curve; everything else is outside. 4. Locate every point on the boundary of this curve where other objects touch the object of interest. These are the points where contact forces are exerted on the object. Name and label each contact force acting on the object. There is at least one force at each point of contact; there may be more than one. When necessary, use subscripts to distinguish forces of the same type. 5. 6. Name and label each long-range force acting on the object. For now, the only long-range force is the gravitational force. Apply these steps to the following problem: A crate is pulled up a rough inclined wood board by a tow rope. Identify the forces on the crate. Part A Which of the following objects are of interest? Check all that apply. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Conceptual Questions on Newton’s 1st and 2nd Laws Learning Goal: To understand the meaning and the basic applications of Newton’s 1st and 2nd laws. In this problem, you are given a diagram representing the motion of an object–a motion diagram. The dots represent the object’s position at moments separated by equal intervals of time. The dots are connected by arrows representing the object’s average velocity during the corresponding time interval. Your goal is to use this motion diagram to determine the direction of the net force acting on the object. You will then determine which force diagrams and which situations may correspond to such a motion. crate earth rope wood board Part A What is the direction of the net force acting on the object at position A? You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D upward downward to the left to the right The net force is zero. This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Understanding Newton’s Laws Part A An object cannot remain at rest unless which of the following holds? You did not open hints for this part. ANSWER: Part B If a block is moving to the left at a constant velocity, what can one conclude? You did not open hints for this part. ANSWER: The net force acting on it is zero. The net force acting on it is constant and nonzero. There are no forces at all acting on it. There is only one force acting on it. Part C A block of mass is acted upon by two forces: (directed to the left) and (directed to the right). What can you say about the block’s motion? You did not open hints for this part. ANSWER: Part D A massive block is being pulled along a horizontal frictionless surface by a constant horizontal force. The block must be __________. You did not open hints for this part. ANSWER: There is exactly one force applied to the block. The net force applied to the block is directed to the left. The net force applied to the block is zero. There must be no forces at all applied to the block. 2 kg 3 N 4 N It must be moving to the left. It must be moving to the right. It must be at rest. It could be moving to the left, moving to the right, or be instantaneously at rest. Part E Two forces, of magnitude and , are applied to an object. The relative direction of the forces is unknown. The net force acting on the object __________. Check all that apply. You did not open hints for this part. ANSWER: Tactics Box 5.3 Drawing a Free-Body Diagram Learning Goal: To practice Tactics Box 5.3 Drawing a Free-Body Diagram. A free-body diagram is a diagram that represents the object as a particle and shows all of the forces acting on the object. Learning how to draw such a diagram is a very important skill in solving physics problems. This tactics box explains the essential steps to construct a correct free-body diagram. TACTICS BOX 5.3 Drawing a free-body diagram Identify all forces acting on the object. This step was described 1. in Tactics Box 5.2. continuously changing direction moving at constant velocity moving with a constant nonzero acceleration moving with continuously increasing acceleration 4 N 10 N cannot have a magnitude equal to cannot have a magnitude equal to cannot have the same direction as the force with magnitude must have a magnitude greater than 5 N 10 N 10 N 10 N Draw a coordinate system. Use the axes defined in your pictorial representation. If those axes are tilted, for motion along an incline, then the axes of the free-body diagram should be similarly tilted. 2. Represent the object as a dot at the origin of the coordinate axes. This is 3. the particle model. 4. Draw vectors representing each of the identified forces. This was described in Tactics Box 5.1. Be sure to label each force vector. Draw and label the net force vector . Draw this vector beside the diagram, not on the particle. Or, if appropriate, write . Then, check that points in the same direction as the acceleration vector on your motion diagram. 5. Apply these steps to the following problem: Your physics book is sliding on the carpet. Draw a free-body diagram. Part A Which forces are acting on the book? Check all that apply. You did not open hints for this part. ANSWER: F  net F =  net 0 F  net a Part B Draw the most appropriate set of coordinate axes for this problem. The orientation of your vectors will be graded. ANSWER: gravity normal force drag static friction tension kinetic friction spring force Part C This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points.

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Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

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Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks 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 −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

Chapter 3 Practice Problems (Practice – no credit) Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Tactics Box 3.1 Determining the Components of a Vector Learning Goal: To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and . TACTICS BOX 3.1 Determining the components of a vector The absolute value of the x component is the magnitude of the 1. component vector . 2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction. 3. The y component is determined similarly. Part A What is the magnitude of the component vector shown in the figure? Express your answer in meters to one significant figure. A A x A y A Ax Ay |Ax| Ax A x Ax A x A x Ay A x ANSWER: Answer Requested Part B What is the sign of the y component of vector shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure. |Ax| = 5 m Ay A positive negative Bx By B Express your answers, separated by a comma, in meters to one significant figure. ANSWER: Correct Vector Components–Review Learning Goal: To introduce you to vectors and the use of sine and cosine for a triangle when resolving components. Vectors are an important part of the language of science, mathematics, and engineering. They are used to discuss multivariable calculus, electrical circuits with oscillating currents, stress and strain in structures and materials, and flows of atmospheres and fluids, and they have many other applications. Resolving a vector into components is a precursor to computing things with or about a vector quantity. Because position, velocity, acceleration, force, momentum, and angular momentum are all vector quantities, resolving vectors into components is the most important skill required in a mechanics course. The figure shows the components of , and , along the x and y axes of the coordinate system, respectively. The components of a vector depend on the coordinate system’s orientation, the key being the angle between the vector and the coordinate axes, often designated . Bx, By = -2,-5 m, m F  Fx Fy  Part A The figure shows the standard way of measuring the angle. is measured to the vector from the x axis, and counterclockwise is positive. Express and in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER:  Fx Fy F  Fx, Fy = Fcos, Fsin Correct In principle, you can determine the components of any vector with these expressions. If lies in one of the other quadrants of the plane, will be an angle larger than 90 degrees (or in radians) and and will have the appropriate signs and values. Unfortunately this way of representing , though mathematically correct, leads to equations that must be simplified using trig identities such as and . These must be used to reduce all trig functions present in your equations to either or . Unless you perform this followup step flawlessly, you will fail to recoginze that , and your equations will not simplify so that you can progress further toward a solution. Therefore, it is best to express all components in terms of either or , with between 0 and 90 degrees (or 0 and in radians), and determine the signs of the trig functions by knowing in which quadrant the vector lies. Part B When you resolve a vector into components, the components must have the form or . The signs depend on which quadrant the vector lies in, and there will be one component with and the other with . In real problems the optimal coordinate system is often rotated so that the x axis is not horizontal. Furthermore, most vectors will not lie in the first quadrant. To assign the sine and cosine correctly for vectors at arbitrary angles, you must figure out which angle is and then properly reorient the definitional triangle. As an example, consider the vector shown in the diagram labeled “tilted axes,” where you know the angle between and the y axis. Which of the various ways of orienting the definitional triangle must be used to resolve into components in the tilted coordinate system shown? (In the figures, the hypotenuse is orange, the side adjacent to is red, and the side opposite is yellow.) F  /2 cos() sin() F  sin(180 + ) = −sin() cos(90 + ) = −sin() sin() cos() sin(180 + ) + cos(270 − ) = 0 sin() cos()  /2 F  |F| cos() |F| sin() sin() cos()  N  N N  Indicate the number of the figure with the correct orientation. Hint 1. Recommended procedure for resolving a vector into components First figure out the sines and cosines of , then figure out the signs from the quadrant the vector is in and write in the signs. Hint 2. Finding the trigonometric functions Sine and cosine are defined according to the following convention, with the key lengths shown in green: The hypotenuse has unit length, the side adjacent to has length , and the   cos() side opposite has length . The colors are chosen to remind you that the vector sum of the two orthogonal sides is the vector whose magnitude is the hypotenuse; red + yellow = orange. ANSWER: Correct Part C Choose the correct procedure for determining the components of a vector in a given coordinate system from this list: ANSWER: sin() 1 2 3 4 Correct Part D The space around a coordinate system is conventionally divided into four numbered quadrants depending on the signs of the x and y coordinates . Consider the following conditions: A. , B. , C. , D. , Which of these lettered conditions are true in which the numbered quadrants shown in ? Write the answer in the following way: If A were true in the third quadrant, B in the second, C in the first, and D in the fourth, enter “3, 2, 1, 4” as your response. ANSWER: Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with as the included angle. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with as the included angle. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with as the included angle.     x > 0 y > 0 x > 0 y < 0 x < 0 y > 0 x < 0 y < 0 Correct Part E Now find the components and of in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector and the angle , with the components separated by a comma. ANSWER: Answer Requested ± Resolving Vector Components with Trigonometry Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think of vectors; however, to calculate results with vectors, it is best to select a coordinate system and manipulate the components of the vectors in that coordinate system. Nx Ny N N  Nx, Ny = −Nsin(),Ncos() T  T  Part A Find the components of the vector with length = 1.00 and angle =10.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. What is the x component? Look at the figure shown. points in the positive x direction, so is positive. Also, the magnitude is just the length . ANSWER: Correct Part B Find the components of the vector with length = 1.00 and angle =15.0 with respect to the x axis as shown. Enter the x component followed by the y component, separated by a comma. A a  A x Ax |Ax| OL = OMcos( ) A  = 0.985,0.174 B b   Hint 1. What is the x component? The x component is still of the same form, that is, . ANSWER: Correct The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing. Part C Find the components of the vector with length = 1.00 and angle 35.0 as shown. Enter the x component followed by the y component, separated by a comma. Hint 1. Method 1: Find the angle that makes with the positive x axis Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with respect to the x axis. What is the angle that makes with the positive x axis? Express your answer numerically in degrees. ANSWER: Hint 2. Method 2: Use vector addition Look at the figure shown. Lcos() B = 0.966,0.259 B (Lcos(), Lsin()) B C c  =  C  C 125 1. . 2. . 3. , the x component of is negative, since points in the negative x direction. Use this information to find . Similarly, find . ANSWER: Answer Requested ± Vector Addition and Subtraction In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors , , and . Calculate the following, and express your answers as ordered triplets of values separated by commas. Part A ANSWER: Correct C = C + x C y |C | = length(QR) = c sin() x Cx C C x Cx Cy C  = -0.574,0.819 A = (1, 0,−3) B = (−2, 5, 1) C = (3, 1, 1) A − B  = 3,-5,-4 Part B ANSWER: Correct Part C ANSWER: Correct Part D ANSWER: Correct B − C  = -5,4,0 −A + B − C  = -6,4,3 3A − 2C  = -3,-2,-11 Part E ANSWER: Correct Part F ANSWER: Correct Video Tutor: Balls Take High and Low Tracks 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 −2A + 3B − C  = -11,14,8 2A − 3(B − C) = 17,-12,-6 Consider the video demonstration that you just watched. Which of the following changes could potentially allow the ball on the straight inclined (yellow) track to win? Ignore air resistance. Select all that apply. Hint 1. How to approach the problem Answers A and B involve changing the steepness of part or all of the track. Answers C and D involve changing the mass of the balls. So, first you should decide which of those factors, if either, can change how fast the ball gets to the end of the track. ANSWER: Correct If the yellow track were tilted steeply enough, its ball could win. How might you go about calculating the necessary change in tilt? Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. A. Increase the tilt of the yellow track. B. Make the downhill and uphill inclines on the red track less steep, while keeping the total distance traveled by the ball the same. C. Increase the mass of the ball on the yellow track. D. Decrease the mass of the ball on the red track.

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Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0 ASSESS: Check if your result has the correct units, is reasonable, and answers the question. Model The trunk is moving at a constant velocity. This means that you can model it as a particle in dynamic equilibrium and apply the strategy above. Furthermore, you can ignore the masses of the ropes and the ring because it is reasonable to assume that their combined weight is much less than the weight of the trunk. Visualize Part A The most convenient coordinate system for this problem is one in which the y axis is vertical and the ropes both lie in the xy plane, as shown below. Identify the forces acting on the trunk, and then draw a free-body diagram of the trunk in the diagram below. The black dot represents the trunk as it is lifted by the students. Draw the 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: Part B This question will be shown after you complete previous question(s). Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). A Gymnast on a Rope A gymnast of mass 70.0 hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value for the acceleration of gravity. Part A Calculate the tension in the rope if the gymnast hangs motionless on the rope. Express your answer in newtons. You did not open hints for this part. ANSWER: Part B Calculate the tension in the rope if the gymnast climbs the rope at a constant rate. Express your answer in newtons. You did not open hints for this part. kg 9.81m/s2 T T = N T ANSWER: Part C Calculate the tension in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: Part D Calculate the tension in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: T = N T m/s2 T = N T m/s2 T = N Applying Newton’s 2nd Law Learning Goal: To learn a systematic approach to solving Newton’s 2nd law problems using a simple example. Once you have decided to solve a problem using Newton’s 2nd law, there are steps that will lead you to a solution. One such prescription is the following: Visualize the problem and identify special cases. Isolate each body and draw the forces acting on it. Choose a coordinate system for each body. Apply Newton’s 2nd law to each body. Write equations for the constraints and other given information. Solve the resulting equations symbolically. Check that your answer has the correct dimensions and satisfies special cases. If numbers are given in the problem, plug them in and check that the answer makes sense. Think about generalizations or simplfications of the problem. As an example, we will apply this procedure to find the acceleration of a block of mass that is pulled up a frictionless plane inclined at angle with respect to the horizontal by a perfect string that passes over a perfect pulley to a block of mass that is hanging vertically. Visualize the problem and identify special cases First examine the problem by drawing a picture and visualizing the motion. Apply Newton’s 2nd law, , to each body in your mind. Don’t worry about which quantities are given. Think about the forces on each body: How are these consistent with the direction of the acceleration for that body? Can you think of any special cases that you can solve quickly now and use to test your understanding later? m2  m1 F = ma One special case in this problem is if , in which case block 1 would simply fall freely under the acceleration of gravity: . Part A Consider another special case in which the inclined plane is vertical ( ). In this case, for what value of would the acceleration of the two blocks be equal to zero? Express your answer in terms of some or all of the variables and . ANSWER: Isolate each body and draw the forces acting on it A force diagram should include only real forces that act on the body and satisfy Newton’s 3rd law. One way to check if the forces are real is to detrmine whether they are part of a Newton’s 3rd law pair, that is, whether they result from a physical interaction that also causes an opposite force on some other body, which may not be part of the problem. Do not decompose the forces into components, and do not include resultant forces that are combinations of other real forces like centripetal force or fictitious forces like the “centrifugal” force. Assign each force a symbol, but don’t start to solve the problem at this point. Part B Which of the four drawings is a correct force diagram for this problem? = 0 m2 = −g a 1 j ^  = /2 m1 m2 g m1 = ANSWER: Choose a coordinate system for each body Newton’s 2nd law, , is a vector equation. To add or subtract vectors it is often easiest to decompose each vector into components. Whereas a particular set of vector components is only valid in a particular coordinate system, the vector equality holds in any coordinate system, giving you freedom to pick a coordinate system that most simplifies the equations that result from the component equations. It’s generally best to pick a coordinate system where the acceleration of the system lies directly on one of the coordinate axes. If there is no acceleration, then pick a coordinate system with as many unknowns as possible along the coordinate axes. Vectors that lie along the axes appear in only one of the equations for each component, rather than in two equations with trigonometric prefactors. Note that it is sometimes advantageous to use different coordinate systems for each body in the problem. In this problem, you should use Cartesian coordinates and your axes should be stationary with respect to the inclined plane. Part C Given the criteria just described, what orientation of the coordinate axes would be best to use in this problem? In the answer options, “tilted” means with the x axis oriented parallel to the plane (i.e., at angle to the horizontal), and “level” means with the x axis horizontal. ANSWER: Apply Newton’s 2nd law to each body a b c d F  = ma  tilted for both block 1 and block 2 tilted for block 1 and level for block 2 level for block 1 and tilted for block 2 level for both block 1 and block 2 Part D What is , the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. Express your answer in terms of some or all of the variables tension , , the magnitude of the acceleration of gravity , and . You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Lifting a Bucket A 6- bucket of water is being pulled straight up by a string at a constant speed. F2x T m2 g  m2a2x =F2x = kg Part A What is the tension in the rope? ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Friction Force on a Dancer on a Drawbridge A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer’s foot are and , respectively. represents the normal force exerted on the dancer by the bridge, and represents the gravitational force exerted on the dancer, as shown in the drawing . For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges. about 42 about 60 about 78 0 because the bucket has no acceleration. N N N N μs μk n F  g Part A Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, ? Express your answer in terms of some or all of the variables , , and/or . You did not open hints for this part. ANSWER: Part B The drawbridge then starts to rise and the dancer continues to stand on one leg. The drawbridge stops just at the point where the dancer is on the verge of slipping. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. F  f n μs μk Ff = Ff n μs μk  You did not open hints for this part. ANSWER: Part C Then, because the bridge is old and poorly designed, it falls a little bit and then jerks. This causes the person to start to slide down the bridge at a constant speed. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. ANSWER: Part D The bridge starts to come back down again. The dancer stops sliding. However, again because of the age and design of the bridge it never makes it all the way down; rather it stops half a meter short. This half a meter corresponds to an angle degree (see the diagram, which has the angle exaggerated). What is the force of friction now? Express your answer in terms of some or all of the variables , , and . Ff = Ff n μs μk  Ff =   1 Ff  n Fg You did not open hints for this part. ANSWER: Kinetic Friction Ranking Task Below are eight crates of different mass. The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates. Ff = Part A Rank the ropes on the basis of the force each exerts on the crate immediately to its left. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Pushing a Block Learning Goal: To understand kinetic and static friction. A block of mass lies on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with . Part A m μs μk μk < μs If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction? You did not open hints for this part. ANSWER: Part B Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force must you be pushing the block just before the block begins to move? Express the magnitude of in terms of some or all the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Part C Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force? Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. Ffriction = F F μs μk m g F = μs μk m g ANSWER: Part D Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration of the block after it begins to move. Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Ffriction = a μs μk m g a =

Chapter 6 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, March 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy PSS 6.1 Equilibrium Problems Learning Goal: To practice Problem-Solving Strategy 6.1 for equilibrium problems. A pair of students are lifting a heavy trunk on move-in day. Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity . Each rope makes an angle with respect to the vertical. The gravitational force acting on the trunk has magnitude . Find the tension in each rope. PROBLEM-SOLVING STRATEGY 6.1 Equilibrium problems MODEL: Make simplifying assumptions. VISUALIZE: Establish a coordinate system, define symbols, and identify what the problem is asking you to find. This is the process of translating words into symbols. Identify all forces acting on the object, and show them on a free-body diagram. These elements form the pictorial representation of the problem. SOLVE: The mathematical representation is based on Newton’s first law: . The vector sum of the forces is found directly from the free-body diagram. v  FG T F  = = net i F  i 0 ASSESS: Check if your result has the correct units, is reasonable, and answers the question. Model The trunk is moving at a constant velocity. This means that you can model it as a particle in dynamic equilibrium and apply the strategy above. Furthermore, you can ignore the masses of the ropes and the ring because it is reasonable to assume that their combined weight is much less than the weight of the trunk. Visualize Part A The most convenient coordinate system for this problem is one in which the y axis is vertical and the ropes both lie in the xy plane, as shown below. Identify the forces acting on the trunk, and then draw a free-body diagram of the trunk in the diagram below. The black dot represents the trunk as it is lifted by the students. Draw the 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: Part B This question will be shown after you complete previous question(s). Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). A Gymnast on a Rope A gymnast of mass 70.0 hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value for the acceleration of gravity. Part A Calculate the tension in the rope if the gymnast hangs motionless on the rope. Express your answer in newtons. You did not open hints for this part. ANSWER: Part B Calculate the tension in the rope if the gymnast climbs the rope at a constant rate. Express your answer in newtons. You did not open hints for this part. kg 9.81m/s2 T T = N T ANSWER: Part C Calculate the tension in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: Part D Calculate the tension in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.10 . Express your answer in newtons. You did not open hints for this part. ANSWER: T = N T m/s2 T = N T m/s2 T = N Applying Newton’s 2nd Law Learning Goal: To learn a systematic approach to solving Newton’s 2nd law problems using a simple example. Once you have decided to solve a problem using Newton’s 2nd law, there are steps that will lead you to a solution. One such prescription is the following: Visualize the problem and identify special cases. Isolate each body and draw the forces acting on it. Choose a coordinate system for each body. Apply Newton’s 2nd law to each body. Write equations for the constraints and other given information. Solve the resulting equations symbolically. Check that your answer has the correct dimensions and satisfies special cases. If numbers are given in the problem, plug them in and check that the answer makes sense. Think about generalizations or simplfications of the problem. As an example, we will apply this procedure to find the acceleration of a block of mass that is pulled up a frictionless plane inclined at angle with respect to the horizontal by a perfect string that passes over a perfect pulley to a block of mass that is hanging vertically. Visualize the problem and identify special cases First examine the problem by drawing a picture and visualizing the motion. Apply Newton’s 2nd law, , to each body in your mind. Don’t worry about which quantities are given. Think about the forces on each body: How are these consistent with the direction of the acceleration for that body? Can you think of any special cases that you can solve quickly now and use to test your understanding later? m2  m1 F = ma One special case in this problem is if , in which case block 1 would simply fall freely under the acceleration of gravity: . Part A Consider another special case in which the inclined plane is vertical ( ). In this case, for what value of would the acceleration of the two blocks be equal to zero? Express your answer in terms of some or all of the variables and . ANSWER: Isolate each body and draw the forces acting on it A force diagram should include only real forces that act on the body and satisfy Newton’s 3rd law. One way to check if the forces are real is to detrmine whether they are part of a Newton’s 3rd law pair, that is, whether they result from a physical interaction that also causes an opposite force on some other body, which may not be part of the problem. Do not decompose the forces into components, and do not include resultant forces that are combinations of other real forces like centripetal force or fictitious forces like the “centrifugal” force. Assign each force a symbol, but don’t start to solve the problem at this point. Part B Which of the four drawings is a correct force diagram for this problem? = 0 m2 = −g a 1 j ^  = /2 m1 m2 g m1 = ANSWER: Choose a coordinate system for each body Newton’s 2nd law, , is a vector equation. To add or subtract vectors it is often easiest to decompose each vector into components. Whereas a particular set of vector components is only valid in a particular coordinate system, the vector equality holds in any coordinate system, giving you freedom to pick a coordinate system that most simplifies the equations that result from the component equations. It’s generally best to pick a coordinate system where the acceleration of the system lies directly on one of the coordinate axes. If there is no acceleration, then pick a coordinate system with as many unknowns as possible along the coordinate axes. Vectors that lie along the axes appear in only one of the equations for each component, rather than in two equations with trigonometric prefactors. Note that it is sometimes advantageous to use different coordinate systems for each body in the problem. In this problem, you should use Cartesian coordinates and your axes should be stationary with respect to the inclined plane. Part C Given the criteria just described, what orientation of the coordinate axes would be best to use in this problem? In the answer options, “tilted” means with the x axis oriented parallel to the plane (i.e., at angle to the horizontal), and “level” means with the x axis horizontal. ANSWER: Apply Newton’s 2nd law to each body a b c d F  = ma  tilted for both block 1 and block 2 tilted for block 1 and level for block 2 level for block 1 and tilted for block 2 level for both block 1 and block 2 Part D What is , the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. Express your answer in terms of some or all of the variables tension , , the magnitude of the acceleration of gravity , and . You did not open hints for this part. ANSWER: Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Lifting a Bucket A 6- bucket of water is being pulled straight up by a string at a constant speed. F2x T m2 g  m2a2x =F2x = kg Part A What is the tension in the rope? ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Friction Force on a Dancer on a Drawbridge A dancer is standing on one leg on a drawbridge that is about to open. The coefficients of static and kinetic friction between the drawbridge and the dancer’s foot are and , respectively. represents the normal force exerted on the dancer by the bridge, and represents the gravitational force exerted on the dancer, as shown in the drawing . For all the questions, we can assume that the bridge is a perfectly flat surface and lacks the curvature characteristic of most bridges. about 42 about 60 about 78 0 because the bucket has no acceleration. N N N N μs μk n F  g Part A Before the drawbridge starts to open, it is perfectly level with the ground. The dancer is standing still on one leg. What is the x component of the friction force, ? Express your answer in terms of some or all of the variables , , and/or . You did not open hints for this part. ANSWER: Part B The drawbridge then starts to rise and the dancer continues to stand on one leg. The drawbridge stops just at the point where the dancer is on the verge of slipping. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. F  f n μs μk Ff = Ff n μs μk  You did not open hints for this part. ANSWER: Part C Then, because the bridge is old and poorly designed, it falls a little bit and then jerks. This causes the person to start to slide down the bridge at a constant speed. What is the magnitude of the frictional force now? Express your answer in terms of some or all of the variables , , and/or . The angle should not appear in your answer. ANSWER: Part D The bridge starts to come back down again. The dancer stops sliding. However, again because of the age and design of the bridge it never makes it all the way down; rather it stops half a meter short. This half a meter corresponds to an angle degree (see the diagram, which has the angle exaggerated). What is the force of friction now? Express your answer in terms of some or all of the variables , , and . Ff = Ff n μs μk  Ff =   1 Ff  n Fg You did not open hints for this part. ANSWER: Kinetic Friction Ranking Task Below are eight crates of different mass. The crates are attached to massless ropes, as indicated in the picture, where the ropes are marked by letters. Each crate is being pulled to the right at the same constant speed. The coefficient of kinetic friction between each crate and the surface on which it slides is the same for all eight crates. Ff = Part A Rank the ropes on the basis of the force each exerts on the crate immediately to its left. Rank from largest to smallest. To rank items as equivalent, overlap them. You did not open hints for this part. ANSWER: Pushing a Block Learning Goal: To understand kinetic and static friction. A block of mass lies on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with . Part A m μs μk μk < μs If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction? You did not open hints for this part. ANSWER: Part B Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force must you be pushing the block just before the block begins to move? Express the magnitude of in terms of some or all the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Part C Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force? Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. Ffriction = F F μs μk m g F = μs μk m g ANSWER: Part D Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration of the block after it begins to move. Express your answer in terms of some or all of the variables , , and , as well as the acceleration due to gravity . You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. Ffriction = a μs μk m g a =

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Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I Typesetting math: 91% This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Work-Energy Theorem Reviewed Learning Goal: Review the work-energy theorem and apply it to a simple problem. If you push a particle of mass in the direction in which it is already moving, you expect the particle’s speed to increase. If you push with a constant force , then the particle will accelerate with acceleration (from Newton’s 2nd law). Part A Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied for a fixed interval of time , then the _____ of the particle will increase by an amount . You did not open hints for this part. ANSWER: M F a = F/M t at Typesetting math: 91% Part B Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied over a given distance , along the path of the particle, then the _____ of the particle will increase by . ANSWER: Part C If the initial kinetic energy of the particle is , and its final kinetic energy is , express in terms of and the work done on the particle. ANSWER: Part D In general, the work done by a force is written as . Now, consider whether the following statements are true or false: The dot product assures that the integrand is always nonnegative. The dot product indicates that only the component of the force perpendicular to the path contributes to the integral. The dot product indicates that only the component of the force parallel to the path contributes to the integral. Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t). ANSWER: D FD Ki Kf Kf Ki W Kf = F W =  ( ) d f i F r r Typesetting math: 91% Part E Assume that the particle has initial speed . Find its final kinetic energy in terms of , , , and . You did not open hints for this part. ANSWER: Part F What is the final speed of the particle? Express your answer in terms of and . ANSWER: ± The Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 , one at an angle 10.0 west of north, and the other at an angle 10.0 east of north, as they pull the tanker a distance 0.660 toward the north. Part A What is the total work done by the two tugboats on the supertanker? Express your answer in joules, to three significant figures. vi Kf vi M F D Kf = Kf M vf = N km Typesetting math: 91% You did not open hints for this part. ANSWER: Energy Required to Lift a Heavy Box As you are trying to move a heavy box of mass , you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box. Use for the magnitude of the acceleration due to gravity and neglect friction forces. Part A Once you have pulled hard enough to start the box moving upward, what is the magnitude of the upward force you must apply to the rope to start raising the box with constant velocity? Express the magnitude of the force in terms of , the mass of the box. J m g F m Typesetting math: 91% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Pulling a Block on an Incline with Friction A block of weight sits on an inclined plane as shown. A force of magnitude is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is . Part A F = mg F μ Typesetting math: 91% What is the total work done on the block by the force of friction as the block moves a distance up the incline? Express the work done by friction in terms of any or all of the variables , , , , , and . You did not open hints for this part. ANSWER: Part B What is the total work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed. Wfric L μ m g  L F Wfric = WF F L μ m g  L F WF = Typesetting math: 91% Part C What is the total work done on the block by the force of friction as the block moves a distance down the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Part D What is the total work done on the box by the appled force in this case? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: When Push Comes to Shove Two forces, of magnitudes = 75.0 and = 25.0 , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position = -1.00 . At some later time, the block has moved to the right, and its center is at a new position, = 1.00 . Wfric L μ m g  L F Wfric = WF μ m g  L F WF = F1 N F2 N xi cm xf cm Typesetting math: 91% Part A Find the work done on the block by the force of magnitude = 75.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Part B Find the work done by the force of magnitude = 25.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: W1 F1 N xi cm xf cm W1 = J W2 F2 N xi cm xf cm Typesetting math: 91% Part C What is the net work done on the block by the two forces? Express your answer numerically, in joules. ANSWER: Part D Determine the change in the kinetic energy of the block as it moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Work from a Constant Force Learning Goal: W2 = J Wnet Wnet = J Kf − Ki xi cm xf cm Kf − Ki = J Typesetting math: 91% To understand how to compute the work done by a constant force acting on a particle that moves in a straight line. In this problem, you will calculate the work done by a constant force. A force is considered constant if is independent of . This is the most frequently encountered situation in elementary Newtonian mechanics. Part A Consider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force . The force (think of it as a field, having a magnitude and direction at every position ) is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical only because the force is constant along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making and angle (radians) with the positive x axis). Find , the work that the force performs on the particle as it moves from point B to point A. Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. You did not open hints for this part. ANSWER: Part B Now consider the same force acting on a particle that travels from point A to point B. The displacement vector now points in the opposite direction as it did in Part A. Find the work done by in this case. Express your answer in terms of , , and . F( r) r F r F L L  WBA F L F  WBA = F L WAB F Typesetting math: 91% L F  You did not open hints for this part. ANSWER: ± Vector Dot Product Let vectors , , and . Calculate the following: Part A You did not open hints for this part. ANSWER: WAB = A = (2, 1,−4) B = (−3, 0, 1) C = (−1,−1, 2) Typesetting math: 91% Part B What is the angle between and ? Express your answer using one significant figure. You did not open hints for this part. ANSWER: Part C ANSWER: Part D ANSWER: A B = AB A B AB = radians 2B 3C = Typesetting math: 91% Part E Which of the following can be computed? You did not open hints for this part. ANSWER: and are different vectors with lengths and respectively. Find the following: Part F Express your answer in terms of You did not open hints for this part. ANSWER: 2(B 3C) = A B C A (B C) A (B + C) 3 A V 1 V 2 V1 V2 V1 Typesetting math: 91% Part G If and are perpendicular, You did not open hints for this part. ANSWER: Part H If and are parallel, Express your answer in terms of and . You did not open hints for this part. ANSWER: ± Tactics Box 11.1 Calculating the Work Done by a Constant Force V = 1 V 1 V 1 V 2 V = 1 V 2 V 1 V 2 V1 V2 V = 1 V 2 Typesetting math: 91% Learning Goal: To practice Tactics Box 11.1 Calculating the Work Done by a Constant Force. Recall that the work done by a constant force at an angle to the displacement is . The vector magnitudes and are always positive, so the sign of is determined entirely by the angle between the force and the displacement. W F  d W = Fd cos  F d W  Typesetting math: 91% TACTICS BOX 11.1 Calculating the work done by a constant force Force and displacement Work Sign of Energy transfer Energy is transferred into the system. The particle speeds up. increases. No energy is transferred. Speed and are constant. Energy is transferred out of the system. The particle slows down. decreases. A box has weight of magnitude = 2.00 accelerates down a rough plane that is inclined at an angle = 30.0 above the horizontal, as shown at left. The normal force acting on the box has a magnitude = 1.732 , the coefficient of kinetic friction between the box and the plane is = 0.300, and the displacement of the box is 1.80 down the inclined plane.  W W 0 F(“r) + K < 90 F("r) cos  + 90 0 0 K > 90 F(“r) cos  − K 180 −F(“r) − FG N  n N μk d m Typesetting math: 91% Part A What is the work done on the box by gravity? Express your answers in joules to two significant figures. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Wgrav Wgrav = J Typesetting math: 91% Work and Potential Energy on a Sliding Block with Friction A block of weight sits on a plane inclined at an angle as shown. The coefficient of kinetic friction between the plane and the block is . A force is applied to push the block up the incline at constant speed. Part A What is the work done on the block by the force of friction as the block moves a distance up the incline? Express your answer in terms of some or all of the following: , , , . You did not open hints for this part. ANSWER: w  μ F Wf L μ w  L Wf = Typesetting math: 91% Part B What is the work done by the applied force of magnitude ? Express your answer in terms of some or all of the following: , , , . ANSWER: Part C What is the change in the potential energy of the block, , after it has been pushed a distance up the incline? Express your answer in terms of some or all of the following: , , , . ANSWER: Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). W F μ w  L W = “U L μ w  L “U = Typesetting math: 91% Part F This question will be shown after you complete previous question(s). Where’s the Energy? Learning Goal: To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy and potential energy . For such systems where no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as , where the quantities with subscript “i” refer to the “initial” moment and those with subscript “f” refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass \texttip{m}{m} and velocity \texttip{v}{v} is given by \large{K=\frac{1}{2}mv^2}. Potential energy, instead, has many forms. The two forms that you will be dealing with most often in this chapter are the gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as U_{\rm g}=mgh, where \texttip{m}{m} is the mass of the object, \texttip{g}{g} is the acceleration due to gravity, and \texttip{h}{h} is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be (you guessed it) zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant \texttip{k}{k}, stretched or compressed a distance \texttip{x}{x}, the associated elastic potential energy is \large{U_{\rm e}=\frac{1}{2}kx^2}. When all three types of energy change, the law of conservation of energy for an object of mass \texttip{m}{m} can be written as K U Ki + Ui = Kf + Uf Typesetting math: 91% \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy would change. The law of conservation of energy is then written as \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2+W_{\rm nc}=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}, where \texttip{W_{\rm nc}}{W_nc} represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work \texttip{W_{\rm nc}}{W_nc} is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system. In this problem, we will consider the following situation as depicted in the diagram : A block of mass \texttip{m}{m} slides at a speed \texttip{v}{v} along a horizontal, smooth table. It next slides down a smooth ramp, descending a height \texttip{h}{h}, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor). You will analyze the motion of the block at different moments using the law of conservation of energy. Part A Which word in the statement of this problem allows you to assume that the table is frictionless? ANSWER: Part B straight smooth horizontal Typesetting math: 91% This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H Typesetting math: 91% This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Sliding In Socks Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor. Part A If Zak’s speed is 3.00 \rm m/s when he starts to slide, what distance \texttip{d}{d} will he slide before stopping? Express your answer in meters. ANSWER: Typesetting math: 91% Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm m Typesetting math: 91%

Chapter 11 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Understanding Work and Kinetic Energy Learning Goal: To learn about the Work-Energy Theorem and its basic applications. In this problem, you will learn about the relationship between the work done on an object and the kinetic energy of that object. The kinetic energy of an object of mass moving at a speed is defined as . It seems reasonable to say that the speed of an object–and, therefore, its kinetic energy–can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object. First, let us consider a sled of mass being pulled by a constant, horizontal force of magnitude along a rough, horizontal surface. The sled is speeding up. Part A How many forces are acting on the sled? ANSWER: Part B This question will be shown after you complete previous question(s). Part C K m v K = (1/2)mv2 m F one two three four This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H This question will be shown after you complete previous question(s). Part I Typesetting math: 91% This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Work-Energy Theorem Reviewed Learning Goal: Review the work-energy theorem and apply it to a simple problem. If you push a particle of mass in the direction in which it is already moving, you expect the particle’s speed to increase. If you push with a constant force , then the particle will accelerate with acceleration (from Newton’s 2nd law). Part A Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied for a fixed interval of time , then the _____ of the particle will increase by an amount . You did not open hints for this part. ANSWER: M F a = F/M t at Typesetting math: 91% Part B Enter a one- or two-word answer that correctly completes the following statement. If the constant force is applied over a given distance , along the path of the particle, then the _____ of the particle will increase by . ANSWER: Part C If the initial kinetic energy of the particle is , and its final kinetic energy is , express in terms of and the work done on the particle. ANSWER: Part D In general, the work done by a force is written as . Now, consider whether the following statements are true or false: The dot product assures that the integrand is always nonnegative. The dot product indicates that only the component of the force perpendicular to the path contributes to the integral. The dot product indicates that only the component of the force parallel to the path contributes to the integral. Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t). ANSWER: D FD Ki Kf Kf Ki W Kf = F W =  ( ) d f i F r r Typesetting math: 91% Part E Assume that the particle has initial speed . Find its final kinetic energy in terms of , , , and . You did not open hints for this part. ANSWER: Part F What is the final speed of the particle? Express your answer in terms of and . ANSWER: ± The Work Done in Pulling a Supertanker Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 , one at an angle 10.0 west of north, and the other at an angle 10.0 east of north, as they pull the tanker a distance 0.660 toward the north. Part A What is the total work done by the two tugboats on the supertanker? Express your answer in joules, to three significant figures. vi Kf vi M F D Kf = Kf M vf = N km Typesetting math: 91% You did not open hints for this part. ANSWER: Energy Required to Lift a Heavy Box As you are trying to move a heavy box of mass , you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box. Use for the magnitude of the acceleration due to gravity and neglect friction forces. Part A Once you have pulled hard enough to start the box moving upward, what is the magnitude of the upward force you must apply to the rope to start raising the box with constant velocity? Express the magnitude of the force in terms of , the mass of the box. J m g F m Typesetting math: 91% You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Pulling a Block on an Incline with Friction A block of weight sits on an inclined plane as shown. A force of magnitude is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is . Part A F = mg F μ Typesetting math: 91% What is the total work done on the block by the force of friction as the block moves a distance up the incline? Express the work done by friction in terms of any or all of the variables , , , , , and . You did not open hints for this part. ANSWER: Part B What is the total work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed. Wfric L μ m g  L F Wfric = WF F L μ m g  L F WF = Typesetting math: 91% Part C What is the total work done on the block by the force of friction as the block moves a distance down the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: Part D What is the total work done on the box by the appled force in this case? Express your answer in terms of any or all of the variables , , , , , and . ANSWER: When Push Comes to Shove Two forces, of magnitudes = 75.0 and = 25.0 , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position = -1.00 . At some later time, the block has moved to the right, and its center is at a new position, = 1.00 . Wfric L μ m g  L F Wfric = WF μ m g  L F WF = F1 N F2 N xi cm xf cm Typesetting math: 91% Part A Find the work done on the block by the force of magnitude = 75.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Part B Find the work done by the force of magnitude = 25.0 as the block moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: W1 F1 N xi cm xf cm W1 = J W2 F2 N xi cm xf cm Typesetting math: 91% Part C What is the net work done on the block by the two forces? Express your answer numerically, in joules. ANSWER: Part D Determine the change in the kinetic energy of the block as it moves from = -1.00 to = 1.00 . Express your answer numerically, in joules. You did not open hints for this part. ANSWER: Work from a Constant Force Learning Goal: W2 = J Wnet Wnet = J Kf − Ki xi cm xf cm Kf − Ki = J Typesetting math: 91% To understand how to compute the work done by a constant force acting on a particle that moves in a straight line. In this problem, you will calculate the work done by a constant force. A force is considered constant if is independent of . This is the most frequently encountered situation in elementary Newtonian mechanics. Part A Consider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force . The force (think of it as a field, having a magnitude and direction at every position ) is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical only because the force is constant along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making and angle (radians) with the positive x axis). Find , the work that the force performs on the particle as it moves from point B to point A. Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. You did not open hints for this part. ANSWER: Part B Now consider the same force acting on a particle that travels from point A to point B. The displacement vector now points in the opposite direction as it did in Part A. Find the work done by in this case. Express your answer in terms of , , and . F( r) r F r F L L  WBA F L F  WBA = F L WAB F Typesetting math: 91% L F  You did not open hints for this part. ANSWER: ± Vector Dot Product Let vectors , , and . Calculate the following: Part A You did not open hints for this part. ANSWER: WAB = A = (2, 1,−4) B = (−3, 0, 1) C = (−1,−1, 2) Typesetting math: 91% Part B What is the angle between and ? Express your answer using one significant figure. You did not open hints for this part. ANSWER: Part C ANSWER: Part D ANSWER: A B = AB A B AB = radians 2B 3C = Typesetting math: 91% Part E Which of the following can be computed? You did not open hints for this part. ANSWER: and are different vectors with lengths and respectively. Find the following: Part F Express your answer in terms of You did not open hints for this part. ANSWER: 2(B 3C) = A B C A (B C) A (B + C) 3 A V 1 V 2 V1 V2 V1 Typesetting math: 91% Part G If and are perpendicular, You did not open hints for this part. ANSWER: Part H If and are parallel, Express your answer in terms of and . You did not open hints for this part. ANSWER: ± Tactics Box 11.1 Calculating the Work Done by a Constant Force V = 1 V 1 V 1 V 2 V = 1 V 2 V 1 V 2 V1 V2 V = 1 V 2 Typesetting math: 91% Learning Goal: To practice Tactics Box 11.1 Calculating the Work Done by a Constant Force. Recall that the work done by a constant force at an angle to the displacement is . The vector magnitudes and are always positive, so the sign of is determined entirely by the angle between the force and the displacement. W F  d W = Fd cos  F d W  Typesetting math: 91% TACTICS BOX 11.1 Calculating the work done by a constant force Force and displacement Work Sign of Energy transfer Energy is transferred into the system. The particle speeds up. increases. No energy is transferred. Speed and are constant. Energy is transferred out of the system. The particle slows down. decreases. A box has weight of magnitude = 2.00 accelerates down a rough plane that is inclined at an angle = 30.0 above the horizontal, as shown at left. The normal force acting on the box has a magnitude = 1.732 , the coefficient of kinetic friction between the box and the plane is = 0.300, and the displacement of the box is 1.80 down the inclined plane.  W W 0 F(“r) + K < 90 F("r) cos  + 90 0 0 K > 90 F(“r) cos  − K 180 −F(“r) − FG N  n N μk d m Typesetting math: 91% Part A What is the work done on the box by gravity? Express your answers in joules to two significant figures. You did not open hints for this part. ANSWER: Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Wgrav Wgrav = J Typesetting math: 91% Work and Potential Energy on a Sliding Block with Friction A block of weight sits on a plane inclined at an angle as shown. The coefficient of kinetic friction between the plane and the block is . A force is applied to push the block up the incline at constant speed. Part A What is the work done on the block by the force of friction as the block moves a distance up the incline? Express your answer in terms of some or all of the following: , , , . You did not open hints for this part. ANSWER: w  μ F Wf L μ w  L Wf = Typesetting math: 91% Part B What is the work done by the applied force of magnitude ? Express your answer in terms of some or all of the following: , , , . ANSWER: Part C What is the change in the potential energy of the block, , after it has been pushed a distance up the incline? Express your answer in terms of some or all of the following: , , , . ANSWER: Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). W F μ w  L W = “U L μ w  L “U = Typesetting math: 91% Part F This question will be shown after you complete previous question(s). Where’s the Energy? Learning Goal: To understand how to apply the law of conservation of energy to situations with and without nonconservative forces acting. The law of conservation of energy states the following: In an isolated system the total energy remains constant. If the objects within the system interact through gravitational and elastic forces only, then the total mechanical energy is conserved. The mechanical energy of a system is defined as the sum of kinetic energy and potential energy . For such systems where no forces other than the gravitational and elastic forces do work, the law of conservation of energy can be written as , where the quantities with subscript “i” refer to the “initial” moment and those with subscript “f” refer to the final moment. A wise choice of initial and final moments, which is not always obvious, may significantly simplify the solution. The kinetic energy of an object that has mass \texttip{m}{m} and velocity \texttip{v}{v} is given by \large{K=\frac{1}{2}mv^2}. Potential energy, instead, has many forms. The two forms that you will be dealing with most often in this chapter are the gravitational and elastic potential energy. Gravitational potential energy is the energy possessed by elevated objects. For small heights, it can be found as U_{\rm g}=mgh, where \texttip{m}{m} is the mass of the object, \texttip{g}{g} is the acceleration due to gravity, and \texttip{h}{h} is the elevation of the object above the zero level. The zero level is the elevation at which the gravitational potential energy is assumed to be (you guessed it) zero. The choice of the zero level is dictated by convenience; typically (but not necessarily), it is selected to coincide with the lowest position of the object during the motion explored in the problem. Elastic potential energy is associated with stretched or compressed elastic objects such as springs. For a spring with a force constant \texttip{k}{k}, stretched or compressed a distance \texttip{x}{x}, the associated elastic potential energy is \large{U_{\rm e}=\frac{1}{2}kx^2}. When all three types of energy change, the law of conservation of energy for an object of mass \texttip{m}{m} can be written as K U Ki + Ui = Kf + Uf Typesetting math: 91% \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}. The gravitational force and the elastic force are two examples of conservative forces. What if nonconservative forces, such as friction, also act within the system? In that case, the total mechanical energy would change. The law of conservation of energy is then written as \large{\frac{1}{2}mv_{\rm i}^2+mgh_{\rm i}+\frac{1}{2}kx_{\rm i}^2+W_{\rm nc}=\frac{1}{2}mv_{\rm f \hspace{1 pt}}^2+mgh_{\rm f \hspace{1 pt}}+\frac{1}{2}kx_{\rm f \hspace{1 pt}}^2}, where \texttip{W_{\rm nc}}{W_nc} represents the work done by the nonconservative forces acting on the object between the initial and the final moments. The work \texttip{W_{\rm nc}}{W_nc} is usually negative; that is, the nonconservative forces tend to decrease, or dissipate, the mechanical energy of the system. In this problem, we will consider the following situation as depicted in the diagram : A block of mass \texttip{m}{m} slides at a speed \texttip{v}{v} along a horizontal, smooth table. It next slides down a smooth ramp, descending a height \texttip{h}{h}, and then slides along a horizontal rough floor, stopping eventually. Assume that the block slides slowly enough so that it does not lose contact with the supporting surfaces (table, ramp, or floor). You will analyze the motion of the block at different moments using the law of conservation of energy. Part A Which word in the statement of this problem allows you to assume that the table is frictionless? ANSWER: Part B straight smooth horizontal Typesetting math: 91% This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s). Part D This question will be shown after you complete previous question(s). Part E This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Part G This question will be shown after you complete previous question(s). Part H Typesetting math: 91% This question will be shown after you complete previous question(s). Part I This question will be shown after you complete previous question(s). Part J This question will be shown after you complete previous question(s). Part K This question will be shown after you complete previous question(s). Sliding In Socks Suppose that the coefficient of kinetic friction between Zak’s feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor. Part A If Zak’s speed is 3.00 \rm m/s when he starts to slide, what distance \texttip{d}{d} will he slide before stopping? Express your answer in meters. ANSWER: Typesetting math: 91% Part B This question will be shown after you complete previous question(s). Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. \rm m Typesetting math: 91%

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