MasteringPhysics_ ENERGY View
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MasteringPhysics_ ENERGY View

Course: PHYS 2211, Fall 2008

School: Georgia Perimeter

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MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... [ Print View ] [ Print ] Physics 2211-001, Fall 2008 Energy Due at 11:00pm on Monday, September 29, 2008 View Grading Details Problem 10.25 A 48.0 marble moving at 2.30 strikes a 30.0 marble at rest. What is the speed of each marble immediately after the collision? Part A ANSWER: = 0.531 m/s Part B ANSWER: =...

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MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... [ Print View ] [ Print ] Physics 2211-001, Fall 2008 Energy Due at 11:00pm on Monday, September 29, 2008 View Grading Details Problem 10.25 A 48.0 marble moving at 2.30 strikes a 30.0 marble at rest. What is the speed of each marble immediately after the collision? Part A ANSWER: = 0.531 m/s Part B ANSWER: = 2.83 m/s Hooke's Law Learning Goal: To understand the use of Hooke's law for a spring. Hooke's law states that the restoring force the displacement spring is neither stretched nor compressed. Recall that means that is equal to a constant times . For a spring, the proportionality constant is called the on a spring when it has been stretched or compressed is proportional to of the spring from its equilibrium position. The equilibrium position is the position at which the spring constant and denoted by . The spring constant is a property of the spring and must be measured experimentally. The larger the value of , the stiffer the spring. In equation form, Hooke's law can be written . The minus sign indicates that the force is in the opposite direction to that of the spring's displacement from its equilibrium length and is &quot;trying&quot; to restore the spring to its equilibrium position. The magnitude of the force is given by , where is the magnitude of the displacement. In Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment 1 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs. A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck. Part A A 60driver gets into an empty taptap to start the day's work. The springs compress 0.02 . What is the effective spring constant of the spring system in the taptap? Hint A.1 How to approach the problem Hint not displayed Enter the spring constant numerically in newtons per meter. ANSWER: = 2.94 1 0 4 Part B After driving a portion of the route, the taptap is fully loaded with a total of 25 people with an average mass of 60 per person. In addition, there are three 15- goats, five 3- chickens, and a total of 25 of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed? Hint B.1 How to find the compression of the spring Hint not displayed Enter the compression numerically in meters. ANSWER: = 0.528 Part C Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic. Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic? Select the best choice below. ANSWER: No, typical small pickup truck springs are not large enough to compress about a half meter. Yes, typical small pickup truck springs can easily compress about a half meter. The answer to Part B is not physically realistic because the springs of a typical light truck will compress their maximum amount (typically about 10 ) before the total weight of all the passengers and other cargo given in Part B is added to the truck. When this maximum compression is reached, the springs will bottom out, and the ride will be very rough. 2 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Part D Now imagine that you are a Haitian taptap driver and want a more comfortable ride. You decide to replace the springs with new springs that can handle the typical heavy load on your vehicle. What spring constant do you want your new spring system to have? ANSWER: The new springs should have a spring constant that is substantially larger than the spring constant of the old springs. A spring constant with a large value is a stiff spring. It will take more force to compress (or stretch) a stiff spring. On a taptap, stiffer springs are less likely to bottom out under a heavy load. However, with a lighter load, for most vehicles, very stiff springs will not compress as much for a bump in the road. Hence very stiff springs will give a better ride with a very heavy load, but less-stiff springs (lower spring constant) will give a smoother ride with a light load. This is why larger vehicles need stiffer springs than smaller vehicles. Baby Bounce with a Hooke One of the pioneers of modern science, Sir Robert Hooke (1635-1703), studied the elastic properties of springs and formulated the law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibrium the end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of the spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or . In its scalar form, this equation is simply . The negative sign indicates that the force that the spring exerts and its displacement have opposite directions. The value of depends on the geometry and the material of the spring; it can be easily determined experimentally using this scalar equation. Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well-known application is a baby bouncer,which consists of a harness seat for a toddler, attached to a spring. The entire contraption hooks onto the top of a doorway. The idea is for the baby to hang in the seat with his or her feet just touching the ground so that a good push up will get the baby bouncing, providing potentially hours of entertainment. Part A 3 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... The following chart and accompanying graph depict an experiment to determine the spring constant for a baby bouncer. Displacement from equilibrium, () 0 0.005 0.010 0.015 0.020 What is the spring constant Force exerted on the spring, () 0 2.5 5.0 7.5 10 of the spring being tested for the baby bouncer? Hint A.1 How to approach the problem Hint not displayed Part A.2 Find the spring constant from the graph Part not displayed Express your answer to two significant figures in newtons per meter. ANSWER: = 500 Part B One of the greatest difficulties with setting up the baby bouncer is determining the right height above the floor so that the child can push off and bounce. Knowledge of physics can be really helpful here. If the spring constant , the baby has a mass , and the baby's legs reach a distance from the bouncer, what should be the height of the &quot;empty&quot; bouncer above the floor? Hint B.1 How to approach the problem 4 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Hint not displayed Hint B.2 Which force to use Hint not displayed Part B.3 Find the force exerted by the baby Part not displayed Part B.4 Find the displacement of the spring Part not displayed Express your answer in meters to two significant figures. ANSWER: = 0.37 A displacement of for the spring holding up a baby may not seem very large but you must consider how small babies are. Also, once the baby begins jumping up and down, the extra energy allows the spring to stretch further than 0.22 and a resonant frequency may be achieved. At resonance the bouncing may become too violent, leading to a potentially dangerous situation for the little bouncer. A One-Dimensional Inelastic Collision Block 1, of mass 2, of mass = 2.90 , moves along a frictionless air track with speed = 15.0 . It collides with block = 49.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Hint A.1 How to approach the problem Hint not displayed Express your answer numerically. 5 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... ANSWER: = 43.5 Part B Find , the magnitude of the final velocity of the two-block system. Hint B.1 How to approach the problem Hint not displayed Express your answer numerically. ANSWER: = 0.838 Part C What is the change in the system's kinetic energy due to the collision? Part C.1 Find the <a href="/keyword/initial-kinetic-energy/" >initial kinetic energy</a> Find , the <a href="/keyword/initial-kinetic-energy/" >initial kinetic energy</a> of the system. Express your answer numerically in joules. ANSWER: = 326 Express your answer numerically in joules. ANSWER: = -308 J Not Quite around the Globe A large globe, with a radius of about 5 , was built in Italy between 1982 and 1987. Imagine that such a globe has a radius and a frictionless surface. A small block of mass slides starts from rest at the very top of the globe and slides along the surface of the globe. The block leaves the surface of the globe when it reaches a height ground. The geometry of the situation is shown in the figure for an arbitrary height . above the 6 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Part A Consider what happens at the moment when the block leaves the surface of the globe. Which of the following statements are correct? a. The net acceleration of the block is directed straight down. b. The component of the force of gravity toward the center of the globe is equal to the magnitude of the normal force. c. The force of gravity is the only force acting on the block. Part A.1 How is the normal force changing? Part not displayed Register to View Answeronly b only c only a and b a and c b and c a and b and c Part B Which of the following statements is also true at the moment when the block leaves the surface of the globe? ANSWER: The centripetal acceleration is zero. The normal force is zero. The net acceleration of the block is parallel to its velocity. The kinetic energy of the block equals its potential energy. Part C Using Newton's 2nd law, find the globe. , the speed of the block at the critical moment when the block leaves the surface of . Assume that the height at which the block leaves the surface of the globe is Hint C.1 How to approach this problem Hint not displayed Part C.2 Find the centripetal acceleration Part not displayed Part C.3 Find the radial component of the gravitational force 7 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Part not displayed Part C.4 What is ? Part not displayed Express the speed in terms of Do n o t use ANSWER: in your answer. = , , and , the magnitude of the accleration due to gravity. Part D Use the law of conservation of energy to find in the previous part. . This will give you a difference expression for than you found Hint D.1 How to apply conservation of energy Hint not displayed Express ANSWER: in terms of = , , and . Part E Find , the height from the ground at which the block leaves the surface of the globe. Hint E.1 How to approach this question Hint not displayed Express ANSWER: in terms of = . Energy in a Spring Graphing Question A toy car is held at rest against a compressed spring, as shown in the figure. When released, the car slides across the room. Let be the initial position of the car. Assume that friction is negligible. 8 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Part A Sketch a graph of the total energy of the spring and car system. There is no scale given, so your graph should simply reflect the qualitative shape of the energy vs. time plot. ANSWER: View Part B Sketch a plot of the elastic potential energy of the spring from the point at which the car is released to the equilibrium position of the spring. Make your graph consistent with the given plot of total energy (the gray line given in the graphing window). Part B.1 Determine the sign of the initial elastic potential energy At the instant the car is released, the spring is compressed. Therefore, is the spring's initial elastic potential energy positive, negative, or zero? ANSWER: positive negative zero Part B.2 Determine the sign of the <a href="/keyword/initial-kinetic-energy/" >initial kinetic energy</a> Is the <a href="/keyword/initial-kinetic-energy/" >initial kinetic energy</a> of the cart positive, negative, or zero? ANSWER: positive negative zero Part B.3 Determine the sign of the final elastic potential energy When the car reaches the equilibrium position of the spring, is the elastic potential energy positive, negative, or zero? ANSWER: positive negative zero Hint B.4 The shape of the elastic potential energy graph 9 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... The elastic potential energy of a spring with spring constant given by that is stretched or compressed to position is , where is the equilibrium position of the spring. ANSWER: View Part C Sketch a graph of the car's kinetic energy from the moment it is released until it passes the equilibrium position of the spring. Your graph should be consistent with the given plots of total energy (gray line in graphing window) and potential energy (gray parabola in graphing window). ANSWER: View Trading Momenta in a Collision Two particles move perpendicular to each other until they collide. Particle 1 has mass , and particle 2 has mass and momentum of magnitude . Note: Magnitudes are not drawn to scale in any of the figures. and momentum of magnitude 10 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Part A Suppose that after the collision, the particles &quot;trade&quot; their momenta, as shown in the figure. That is, particle 1 now has magnitude of momentum , and particle 2 has magnitude of momentum ; furthermore, each particle is now moving in the direction in which the other had been moving. How much kinetic energy, collision? Hint A.1 How to approach the problem Hint not displayed Part A.2 Find the relationship between energy and momentum Part not displayed Part A.3 Find the <a href="/keyword/initial-kinetic-energy/" >initial kinetic energy</a> Part not displayed Part A.4 Find the final kinetic energy Part not displayed Express your answer in terms of ANSWER: = and . , is lost in the Part B Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy is lost in this case? Hint B.1 How to approach the problem Hint not displayed Hint B.2 Definition of completely inelastic Hint not displayed Hint B.3 <a href="/keyword/initial-kinetic-energy/" >initial kinetic energy</a> Hint not displayed Part B.4 Find the final kinetic energy Part not displayed Express your answer in terms of ANSWER: = and . 11 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... Circling Ball A ball of mass is attached to a string of length . It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. At the top and bottom of the vertical circle, the ball's speeds are and , and the corresponding tensions in the string are and . and have magnitudes and . Part A Find , the difference between the magnitude of the tension in the string at the bottom relative to that at the top of the circle. Hint A.1 How to approach this problem Hint not displayed Part A.2 Find the sum of forces at the bottom of the circle Part not displayed Part A.3 Find the acceleration at the bottom of the circle Part not displayed Part A.4 Find the tension at the bottom of the circle Part not displayed Part A.5 Find the sum of forces at the top of the circle Part not displayed Part A.6 Find the acceleration at the top of the circle Part not displayed Part A.7 Find the tension at the top of the circle Part not displayed Part A.8 Find the relationship between and Part not displayed Express the difference in tension in terms of and . The quantities and should n o t 12 of 13 10/11/08 11:45 PM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignm... appear in your final answer. ANSWER: = The method outlined in the hints is really the only practical way to do this problem. If done properly, finding the difference between the tensions, , can be accomplished fairly simply and elegantly. Summary 8 of 8 items complete (95.71% avg. score) 20.83 of 22 points 13 of 13 10/11/08 11:45 PM

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UChicago - ECON - 201
Kennesaw - CHEM - 1152
Practice 4 Alkane Conformation 1) (Your answers may look different and still be correct. If yours look different, hand write these answers and consider them carefully.)i) (H3C)2HC CH3 ii) (H3C)2HC CH2CH3 iii) H3CH2C CH2OH iv) (H3C)2HC CH2OHH3C H3
Kennesaw - CHEM - 1152
Practice - Lewis StructuresProvide Lewis structures for the following molecules, be sure to show all non-bonding electrons. Bear in mind that hydrogen and halogens can make only one bond at a time. a) CCl4 b) CH3Br c) H2NNH2 d) PH3 e) H2S f) CH3CH2
Kennesaw - CHEM - 1152
Practice 1 - Lewis StructuresProvide Lewis structures for the following molecules, be sure to show all non-bonding electrons. Cl a) CCl4 b) CH3Br c) H2NNH2 d) PH3 e) H2S f) CH3CH2OH g) HOCH2CH2OH H HP H HS H HH HCCOH HH l) CO2 m) CF4 O n) Cl2CO o)
Kennesaw - CHEM - 1152
-Provide IUPAC names for the following molecules:BrBr
Kennesaw - CHEM - 1152
Practice 2 - Naming AlkanesProvide IUPAC names for the following molecules:3-ethyl-2,5-dimethylheptane4-ethyl-6-methylnonane1,1-dimethylcyclopentane Br3,5-diethyl-2-methyldecane4-ethylnonane1-bromo-3-ethylcyclohexaneBr propylcyclohept
Kennesaw - CHEM - 1152
Practice - Balancing Combustion Equations Provide a balanced equation for the complete combustion of each substance:a)+ 7 O25 CO2 + 4 H2Ob)+ 11 O27 CO2 + 8 H2OOHc)+ 6 O24 CO2 + 5 H2Od)+ 15 O210 CO2 + 10 H2Oe)O+ 6 O24
Kennesaw - CHEM - 1152
Practice - Balancing Combustion Equations Provide a balanced equation for the complete combustion of each substance:a)b)OHc)d)e)Of)g)Oh)OOO OO
Kennesaw - CHEM - 1152
Practice 4 Alkane Conformation1) For the molecules below, provide a bond-line structure. i) (H3C)2HC CH3 ii) (H3C)2HC CH2CH3 2) a) b) c) d) e) iii) H3CH2C CH2OH iv) (H3C)2HC CH2OHDraw chair cyclohexane 20 times. Cross out incorrect ones. Draw cha