Law of Conservation of Energy Assignments

Ups and Downs Description: Several qualitative and conceptual questions involving objects launched upward in the gravitational field of Earth in the absence of non-conservative forces. Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from the earth. The energy transformations that take place involve the object's kinetic energy and its gravitational potential energy . The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation , where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed . Neglect air resistance. Part A As the projectile goes upward, what energy changes take place? ANSWER: Both kinetic and potential energy decrease. Both kinetic and potential energy increase. n m l k j i Kinetic energy decreases; potential energy increases. n m l k j Kinetic energy increases; potential energy decreases. n m l k j n m l k j Part B At the top point of the flight, what can be said about the projectile's kinetic and potential energy? ANSWER: Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. n m l k j Kinetic energy is at a maximum; potential energy is at a minimum. n m l k j i Kinetic energy is at a minimum; potential energy is at a maximum. n m l k j n m l k j Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system "Earth-ball." Although we will often talk about "the gravitational potential energy of an elevated object," it is useful to keep in mind that the energy, in fact, is associated with the interactions between the earth and the elevated object. Part C The potential energy of the object at the moment of launch __________. ANSWER: is negative is positive n m l k j is zero n m l k j i depends on the choice of the "zero level" of potential energy n m l k j n m l k j Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that at the ground level--but this is not, by any means, the only choice! Part D Using conservation of energy, find the maximum height to which the object will rise. Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: = You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer. Part E At what height above the ground does the projectile have a speed of ? Express your answer in terms of and the magnitude of the acceleration of gravity . ANSWER: = Part F What is the speed of the object at the height of Hint F.1 ? How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height ( ), the speed is . All of the energy is kinetic energy, and so, the total energy is . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to , half of the initial kinetic energy must have been converted to potential energy when the projectile is at . Thus, the kinetic energy must be half of its original value (i.e., when ). You need to determine the speed, as a multiple of , that corresponds to such a kinetic energy. Express your answer in terms of and . Use three significant figures in the numeric coefficient. ANSWER: = Let us now consider objects launched at an angle. For such situations, using conservation of energy leads to a quicker solution than can be produced by kinematics. Part G A ball is launched as a projectile with initial speed at an angle above the horizontal. Using conservation of energy, find the maximum height of the ball's flight. Part G.1 Find the final kinetic energy Find the final kinetic energy of the ball. Here, the best choice of "final" moment is the point at which the ball reaches its maximum height, since this is the point we are interested in. Part G.1.a Find the speed at the maximum height The speed of the ball at the maximum height is __________. ANSWER: n m l k j 0 n m l k j n m l k j i n m l k j n m l k j Express your answer in terms of , , and . ANSWER: = Express your answer in terms of , , and . ANSWER: = Part H A ball is launched with initial speed from ground level up a frictionless slope. The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of , , and . You may or may not use all of these quantities. ANSWER: = Interestingly, the answer does not depend on . The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does. Part I A ball is launched with initial speed from the ground level up a frictionless hill. The hill becomes steeper as the ball slides up; however, the ball remains in contact with the hill at all times. Using conservation of energy, find the maximum vertical height to which the ball will climb. Express your answer in terms of and . ANSWER: = The profile of the hill does not matter; the equation would have the same terms regardless of the steepness of the hill. Potential Energy Graphs and Motion Description: Several questions that require the students to predict the characteristics of the particle's motion (direction of force, acceleration, equilibrium positions, etc.) using a potential enrgy diagram U vs. x. Learning Goal: To be able to interpret potential energy diagrams and predict the corresponding motion of a particle. Potential energy diagrams for a particle are useful in predicting the motion of that particle. These diagrams allow one to determine the direction of the force acting on the particle at any point, the points of stable and unstable equilibrium, the particle's kinetic energy, etc. Consider the potential energy diagram shown. The curve represents the value of potential energy particle's coordinate . The horizontal line above the curve represents the constant value of the total energy of the particle . The total energy is the sum of kinetic ( ) and potential ( ) energies of the particle. as a function of the The key idea in interpreting the graph can be expressed in the equation where is the x component of the net force as function of the particle's coordinate . Note the negative sign: It means that the x component of the net force is negative when the derivative is positive and vice versa. For instance, if the particle is moving to the right, and its potential energy is increasing, the net force would be pulling the particle to the left. If you are still having trouble visualizing this, consider the following: If a massive particle is increasing its gravitational potential energy (that is, moving upward), the force of gravity is pulling in the opposite direction (that is, downward). If the x component of the net force is zero, the particle is said to be in equilibrium. There are two kinds of equilibrium: Stable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle back toward the equilibrium point (think of a ball rolling between two hills). Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle further away from the equilibrium point (think of a ball on top of a hill). In answering the following questions, we will assume that there is a single varying force acting on the particle along the x axis. Therefore, we will use the term force instead of the cumbersome x component of the net force. Part A The force acting on the particle at point A is __________. Hint A.1 Sign of the derivative If a function increases (as increases) in a certain region, then the derivative of the function in that region is positive. Hint A.2 Sign of the component If increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive x component points to the right, and vice versa. ANSWER: directed to the right directed to the left n m l k j equal to zero n m l k j n m l k j i Consider the graph in the region of point A. If the particle is moving to the right, it would be "climbing the hill," and the force would "pull it down," that is, pull the particle back to the left. Another, more abstract way of thinking about this is to say that the slope of the graph at point A is positive; therefore, the direction of is negative. Part B The force acting on the particle at point C is __________. Hint B.1 Sign of the derivative If a function increases (as increases) in a certain region, then the derivative of the function in that region is positive, and vice versa. Hint B.2 Sign of the component If increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive x component points to the right, and vice versa. ANSWER: directed to the right directed to the left n m l k j equal to zero n m l k j i n m l k j Part C The force acting on the particle at point B is __________. Hint C.1 Derivative of a function at a local maximum At a local maximum, the derivative of a function is equal to zero. ANSWER: directed to the right directed to the left n m l k j i equal to zero n m l k j n m l k j The slope of the graph is zero; therefore, the derivative , and . Part D The acceleration of the particle at point B is __________. Hint D.1 Relation between acceleration and force The relation between acceleration and force is given by Newton's 2nd law, . ANSWER: directed to the right directed to the left n m l k j i equal to zero n m l k j n m l k j If the net force is zero, so is the acceleration. The particle is said to be in a state of equilibrium. Part E If the particle is located slightly to the left of point B, its acceleration is __________. Hint E.1 The force on such a particle To the left of B, is an increasing function and so its derivative is positive. This implies that the x component of the force on a particle at this location is negative, or that the force is directed to the left, just like at A. What can you say now about the acceleration? ANSWER: directed to the right directed to the left n m l k j equal to zero n m l k j n m l k j i Part F If the particle is located slightly to the right of point B, its acceleration is __________. Hint F.1 The force on such a particle To the right of B, is a decreasing function and so its derivative is negative. This implies that the x component of the force on a particle at this location is positive, or that the force is directed to the right, just like at C. What can you now say about the acceleration? ANSWER: directed to the right directed to the left n m l k j equal to zero n m l k j i n m l k j As you can see, small deviations from equilibrium at point B cause a force that accelerates the particle further away; hence the particle is in unstable equilibrium. Part G Name all labeled points on the graph corresponding to unstable equilibrium. Hint G.1 Definition of unstable equilibrium Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle further away from the equilibrium point (think of a ball on top of a hill). List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. ANSWER: BF Part H Name all labeled points on the graph corresponding to stable equilibrium. Hint H.1 Definition of stable equilibrium Stable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle back toward the equilibrium point. (Think of a ball rolling between two hills.) List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. ANSWER: DH Part I Name all labeled points on the graph where the acceleration of the particle is zero. Hint I.1 Relation between acceleration and force The relation between acceleration and force is given by Newton's 2nd law, . List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. ANSWER: BDFH Your answer, of course, includes the locations of both stable and unstable equilibrium. Part J Name all labeled points such that when a particle is released from rest there, it would accelerate to the left. Part J.1 Determine the sign of the x component of force If the acceleration is to the left, so is the force. This means that the x component of the force is __________. ANSWER: n m l k j n m l k j i Part J.2 positive negative What is the behavior of ? If the x component of the force at a point is negative, then the derivative of the region around the point is __________. ANSWER: n m l k j i n m l k j at that point is positive. This means that in increasing decreasing List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. ANSWER: AE Part K Consider points A, E, and G. Of these three points, which one corresponds to the greatest magnitude of acceleration of the particle? Hint K.1 Acceleration and force The greatest acceleration corresponds to the greatest magnitude of the net force, represented on the graph by the magnitude of the slope. ANSWER: A E n m l k jG n m l k j i n m l k j If the total energy the particle since of the particle is known, one can also use the graph of to draw conclusions about the kinetic energy of . As a reminder, on this graph, the total energy is shown by the horizontal line. Part L What point on the graph corresponds to the maximum kinetic energy of the moving particle? Hint L.1 , , and Since the total energy does not change, the maximum kinetic energy corresponds to the minimum potential energy. ANSWER: D It makes sense that the kinetic energy of the particle is maximum at one of the (force) equilibrium points. For example, think of a pendulum (which has only one force equilibrium point--at the very bottom). Part M At what point on the graph does the particle have the lowest speed? ANSWER: B As you can see, many different conclusions can be made about the particle's motion merely by looking at the graph. It is helpful to understand the character of motion qualitatively before you attempt quantitative problems. This problem should prove useful in improving such an understanding. Problem 7.12 Tarzan and Jane. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of with the vertical. Part A Calculate Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine. ANSWER: m/s Problem 7.39 A man with mass sits on a platform suspended from a movable pulley, as shown in the figure , and raises himself at constant speed by a rope passing over a fixed pulley. The platform and the pulleys have negligible mass. Assume that there are no friction losses. Part A Find the force he must exert. Take the free fall acceleration to be . ANSWER: Part B Find the increase in the energy of the system when he raises himself a distance . (Answer by calculating the increase in potential energy.) Take the free fall acceleration to be . ANSWER: Part C Find the increase in the energy of the system when he raises himself . (Answer by computing the product of the force on the rope and the length of the rope passing through his hands.) Take the free fall acceleration to be . ANSWER: Problem 7.42 A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope . Part A What is the speed of the block as it slides along the horizontal surface after having left the spring? ANSWER: m/s Part B How far does the block travel up the incline before starting to slide back down? ANSWER: m Problem 7.46 Riding a Loop-the-loop. A car in an amusement park ride rolls without friction around the track shown in the figure . It starts from rest at point at a height above the bottom of the loop. Treat the car as a particle. Part A What is the minimum value of (in terms of ) such that the car moves around the loop without falling off at the top (point )? ANSWER: = Part B If the car starts at height and the radius is , compute the speed of the passengers when the car is at point , which is at the end of a horizontal diameter. Take the free fall acceleration to be . ANSWER: = Part C If the car starts at height and the radius is , compute the radial acceleration of the passengers when the car is at point , which is at the end of a horizontal diameter. Take the free fall acceleration to be . ANSWER: = Part D If the car starts at height and the radius is , compute the tangential acceleration of the passengers when the car is at point , which is at the end of a horizontal diameter. Take the free fall acceleration to be . ANSWER: Problem 7.62 A skier of mass starts from rest at the top of a ski slope of height . Part A If frictional forces do of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be . ANSWER: Part B Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is . If the patch is of width and the average force of air resistance on the skier is , how fast is she going after crossing the patch? ANSWER: Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance What is the average force exerted on her by the snowdrift as it stops her? into it before coming to a stop. ANSWER: Problem 7.66 A truck with mass has a brake failure while going down an icy mountain road of constant downward slope angle the truck is moving downhill at speed . After careening downhill a distance with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle . The truck ramp has a soft sand surface for which the coefficient of rolling friction is . . Initially Part A What is the distance that the truck moves up the ramp before coming to a halt? Use energy methods. ANSWER: Problem 7.68 A variable force is maintained tangent to a frictionless, semicircular surface . By slowly varying the force, a block with weight is moved through the angla , and the spring to which it is attached is stretched from position 1 to position 2. The spring has negligible mass and force constant . The end of the spring moves in an arc of radius . Part A Calculate the work done by the force . ANSWER: = Problem 7.74 A 2.00-kg package is released on a incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline . The coefficients of friction between the package and the incline are and . The mass of the spring is negligible. Part A What is the speed of the package just before it reaches the spring? ANSWER: m/s Part B What is the maximum compression of the spring? ANSWER: m Part C The package rebounds back up the incline. How close does it get to its initial position? ANSWER: m Problem 7.80 A hydroelectric dam holds back a lake of surface area that has vertical sides below the water level. The water level in the lake is a height above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted into electrical energy with efficiency. Part A If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is . Take free fall acceleration to be . ANSWER: Part B What volume of water must pass through the dam to produce an amount of electrical energy totalling of electrical energy? ANSWER: Part C What distance does the level of water in the lake fall when this much water passes through the dam? ANSWER: The Impulse-Momentum Theorem Description: Several qualititative and computational questions involving derivation and applications of the impulse-momentum theorem. Learning Goal: To learn about the impulse-momentum theorem and its applications in some common cases. Using the concept of momentum, Newton's second law can be rewritten as , (1) where is the net force acting on the object, and is the rate at which the object's momentum is changing. If the object is observed during an interval of time between times and , then integration of both sides of equation (1) gives . (2) The right side of equation (2) is simply the change in the object's momentum force and is denoted by . Then equation (2) can be rewritten as . The left side is called the impulse of the net . This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force acting along the direction of motion, the impulse-momentum theorem can be written as . (3) Here , , and are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately. The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass moving along the x axis. The net force is acting on the particle along the x axis. is a constant force. Part A The particle starts from rest at . What is the magnitude of the momentum of the particle at time ? Assume that Express your answer in terms of any or all of , , and . ANSWER: Part B = . The particle starts from rest at . What is the magnitude of the velocity of the particle at time ? Assume that . Express your answer in terms of any or all of , , and . ANSWER: = Part C The particle has momentum of magnitude at a certain instant. What is , the magnitude of its momentum Express your answer in terms of any or all of , , , and ANSWER: seconds later? . = Part D The particle has momentum of magnitude at a certain instant. What is , the magnitude of its velocity Express your answer in terms of any or all of , , , and ANSWER: seconds later? . = Let us now consider several two-dimensional situations. A particle of mass is moving in the positive x direction at speed . After a certain constant force is applied to the particle, it moves in the positive y direction at speed . Part E Find the magnitude of the impulse Hint E.1 delivered to the particle. How to approach the problem This is a two-dimensional situation. It is helpful to find the components theorem to find . Part E.2 Find and separately and then use the Pythagorean Find the change in momentum , the magnitude of the change in the x component of the momentum of the particle. Express your answer in terms of ANSWER: = Similarly, you can find . . Once you have obtained both Express your answer in terms of ANSWER: and . and , you can find the magnitude of the delivered impulse and . Use three significant figures in the numerical coefficient. = Part F Which of the vectors below best represents the direction of the impulse vector ? ANSWER: 1 2 n m l k j3 n m l k j4 n m l k j5 n m l k j6 n m l k j7 n m l k j8 n m l k j i n m l k j Part G What is the angle between the positive y axis and the vector ANSWER: as shown in the figure? 26.6 degrees 30 degrees 60 degrees n m l k j 63.4 degrees n m l k j i n m l k j n m l k j Part H If the magnitude of the net force acting on the particle is , how long does it take the particle to acquire its final velocity, the positive y direction? in Express your answer in terms of , , and . If you use a numerical coefficient, use three significant figures. ANSWER: = So far, we have considered only the situation in which the magnitude of the net force acting on the particle was either irrelevant to the solution or was considered constant. Let us now consider an example of a varying force acting on a particle. Part I A particle of mass kilograms is at rest at seconds. A varying force between seconds and seconds. Find the speed of the particle at Hint I.1 Use the impulse-momentum theorem In this case, and . Therefore, . Part I.2 What is the correct antiderivative? Which of the following is an antiderivative ANSWER: ? n m l k j n m l k j n m l k j i n m l k j Express your answer in meters per second to three significant figures. is acting on the particle seconds. ANSWER: Problem 8.8 Force of a Baseball Swing. A baseball has mass 0.145 kg. Part A If the velocity of a pitched ball has a magnitude of 45.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball. ANSWER: Part B Find the magnitude of the impulse applied to it by the bat. ANSWER: Part C If the ball remains in contact with the bat for 2.00 ms, find the magnitude of the average force applied by the bat. ANSWER: N Problem 8.10 An engine of the orbital maneuvering system (OMS) on a space shuttle exerts a force of exhausting a negligible mass of fuel relative to the shuttle's mass of . Part A What is the y component of the impulse of the force for this time interval of ? ANSWER: Part B What is the y component of the shuttle's change in momentum from this impulse? ANSWER: Part C What is the y component of the shuttle's change in velocity from this impulse? ANSWER: Problem 8.17 over a time interval of , Energy Change During a Hip Check. Ice-hockey star Wayne Gretzky is skating at 13.0 m/s toward a defender, who in turn is skating at 5.0 m/s toward Gretzky . Gretzky's weight is 756 N; that of the defender is 900 N. Immediately after the collision, Gretzky is moving at 1.5 m/s in his original direction. You can ignore external horizontal forces applied by the ice to the skaters during the collision. Part A What is the velocity of the defender immediately after the collision? ANSWER: m/s Part B Calculate the change in the combined kinetic energy of the two players. Take the free fall acceleration to be ANSWER: Summary kJ 0 of 15 problems complete (0% avg. score) 0 of 150 points .