PHYS 210 Homework, Spring 2008

Homework: chapter 8 Potential Energy Homework: chapter 8 Potential Energy 1,3,4,5,6,9,10,20,21,22,25,26,31,33,36, 41,45,46,54,57,61,65 8.1 A 1 000–kg roller coaster train is initially at the top of a rise, at point A. It then moves 135 ft, at an angle of 40.0° below the horizontal, to a lower point B. (a) Choose point B to be the zero level for gravitational potential energy. Find the potential energy of the roller coaster–Earth system at points A and B, and the change in potential energy as the coaster moves. (b) Repeat part (a), setting the zero reference level at point A. (a) With our choice for the zero level for potential energy when the car is at point B, B 0 U  When the car is at point A, the potential energy of the car-Earth system is given by UA mgy  where y is the vertical height above zero level. With135 ft=41.1m , this height is found as:   41.1 m sin 40.0 26.4 m y        2 5 A 1 000 kg 9.80 m s 26.4 m 2.59 10 J U    The change in potential energy as the car moves from A to B is 5 5 B A 0 2.59 10 J 2.59 10 J U  U       With our choice of the zero level when the car is at point A, we have 8.1 Continuation The potential energy when the car is at point B is given by where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5 m. Because this distance is now below the zero reference level, it is a negative number A 0 U      2 5 B 1 000 kg 9.80 m s 26.5 m 2.59 10 J U      The change in potential energy when the car moves from A to B is 5 5 B A 2.59 10 J 0 2.59 10 J U  U       8.3 A person with a remote mountain cabin plans to install her own hydroelectric plant. A nearby stream is 3.00 m wide and 0.500 m deep. Water flows at 1.20 m/s over the brink of a waterfall 5.00 m high. The manufacturer promises only 25.0% efficiency in converting the potential energy of the water- Earth system into electric energy. Find the power she can generate. (Large- scale hydroelectric plants, with a much larger drop, are more efficient.) The volume flow rate is the volume of water going over the falls each second    3 3 m 0.5 m 1.2 m s 1.8 m s     3 3 1 000 kg m 1.8 m s 1 800 kg s m V t t    The mass flow rate is If the stream has uniform width and depth, the speed of the water below the falls is the same as the speed above the falls. Then no kinetic energy, but only gravitational energy is available for conversion into internal and electric energy.     2 4 in energy 1 800 kg s 9.8 m s 5 m 8.82 10 J s mgy m gy t t t       P     4 4 useful in efficiency 0.25 8.82 10 W 2.20 10 W      P P Input power Output power The efficiency of electric generation at Hoover Dam is about 85%, with a head of water (vertical drop) of 174 m. Intensive research is underway to improve the efficiency of low head generators. 8.4 At 11:00 AM on September 7, 2001, more than one million British school children jumped up and down for one minute. The curriculum focus of the “Giant Jump” was on earthquakes, but it was integrated with many other topics, such as exercise, geography, cooperation, testing hypotheses, and setting world records. Children built their own seismographs, which registered local effects. (a) Find the mechanical energy released in the experiment. Assume that 1 050 000 children of average mass 36.0 kg jump twelve times each, raising their centers of mass by 25.0 cm each time and briefly resting between one jump and the next. The free-fall acceleration in Britain is 9.81 m/s2. (b) (b) Most of the energy is converted very rapidly into internal energy within the bodies of the children and the floors of the school buildings. Of the energy that propagates into the ground, most produces high frequency “microtremor” vibrations that are rapidly damped and cannot travel far. Assume that 0.01% of the energy is carried away by a long-range seismic wave. The magnitude of an earthquake on the Richter scale is given by M log E  4.8 1.5 One child in one jump converts chemical energy into mechanical energy in the amount that her body has as gravitational energy at the top of her jump: (a)    36 kg 9.81 m s2 0.25 m 88.3 J mgy   For all of the jumps of the children the energy is   6 9 12 1.05 10 88.3 J 1.11 10 J    8.4 continuation 9 5 0.011.11 10 J 1.11 10 J E  100    The seismic energy is modeled as 5 log 4.8 log1.11 10 4.8 5.05 4.8 0.2 1.5 1.5 1.5 E        making the Richter magnitude . 8.5 A bead slides without friction around a loop-the-loop (Fig. P8.5). The bead is released from a height h = 3.50R. (a) What is its speed at point A? (b) (b) How large is the normal force on it if its mass is 5.00 g? Figure P8.5 (a) i i ff U K U K      2 1 0 2 2 mgh mg R mv        2 1 3.50 2 2 g R g R v   3.00 v gR  v2 F m R  v2 n mg m R      2 3 2 3.00 2.00 2.00 5.00 10 kg 9.80 m s 0.098 0 N downward gR v n m g m g mg R R n                      Circular motion 8.6 Dave Johnson, the bronze medalist at the 1992 Olympic decathlon in Barcelona, leaves the ground at the high jump with vertical velocity component 6.00 m/s. How far does his center of mass move up as he makes the jump? 8.9 A simple pendulum, which we will consider in detail in chapter 15, consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.00 meters long and makes an initial angle of 30.0 with the vertical, calculate the speed of the particle (a) at the lowest point in its trajectory and (b) when the angle is 15.0.     2 2 1 6.00 m s 0 0 9.80 m s 2 m m y    i i ff K U K U        2 2 1 6.00 m s 0 0 9.80 m s 2 m m y         2 2 6.00 m s 1.84 m 2 9.80 m s  y   Energy of the object-Earth system is conserved as the object moves between the release point and the lowest point. We choose to measure heights from at the top end of the string.     g g i f K U K U    2 1 0 2 i ff mgy mv mgy              2 2 2 2 1 9.8 m s 2 mcos30 9.8 m s 2 m 2 2 9.8 m s 2 m 1 cos30 2.29 m s f f v v          Choose I initial point at=30 and final point at  =15           2 2 1 0 cos30 cos15 2 2 cos15 cos30 2 9.8 m s 2 m cos15 cos30 1.98 m s f f mg L mv mg L v gL                 8.10 An object of mass m starts from rest and slides a distance d down a frictionless incline of angle  . While sliding, it contacts an unstressed spring of negligible mass as shown in Figure P8.10. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring. Figure P8.10 Choose the zero point of gravitational potential energy of the object-spring-Earth system as the configuration in which the object comes to rest. Then because the incline is frictionless, we have : B A E E  B gB sB A gA sA K U U K U U        2 1 0 sin 0 0 0 2 mg d x kx        2 2 sin kx d x mg    8.20 Review problem. The system shown in Figure P8.20 consists of a light inextensible cord, light frictionless pulleys, and blocks of equal mass. It is initially held at rest so that the blocks are at the same height above the ground. The blocks are then released. Find the speed of block A at the moment when the vertical separation of the blocks is h. Figure P8.20 When block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm. We then choose the final point to be when B has moved up by h/3 and has speed vA/2 . Then A has moved down 2h/3 and has speed vA:     A B A B 2 2 A A 2 A A 2 1 1 0 0 0 2 2 2 3 3 5 3 8 8 15 g g i f K K U K K U mgh mg h v mv m mgh mv gh v                    8.21 A 4.00-kg particle moves from the origin to position C, having coordinates x = 5.00 m and y = 5.00 m. One force on the particle is the gravitational force acting in the negative y direction (Fig. P8.21). Using Equation 7.3, calculate the work done by the gravitational force in going from O to C along (a) OAC, (b) OBC, (c) OC. Your results should all be identical. Why? Figure P8.21    4.00 kg 9.80 m s2 39.2 N gF mg   Work along OAC= work along OA + work along AC       39.2 N 5.00 m cos180 39.2 N 5.00 m cos90.0 196 J       (b) W along OBC W along OB + W along BC 196J 5.0 39.2 0 cos180 90 cos           m OC F OA F g g     Fg OC cos135  (c ) Work along OC    1 39.2 N 5.00 2 m 196 J 2            All the same: sure the force is conservative 8.22 (a) Suppose that a constant force acts on an object. The force does not vary with time, nor with the position or the velocity of the object. Start with the general definition for work done by a force and show that the force is conservative. (b) As a special case, suppose that the force acts on a particle that moves from O to C in Figure P8.21 (see previuos problem). Calculate the work done by F if the particle moves along each one of the three paths OAC, OBC, and OC. (Your three answers should be identical.)    f i d W r F F  3ˆ i  4ˆ j  N  , which depends only on end points, not path. fi W   d    F  r F r r (a)         5.00 m 5.00 m 0 0 ˆ ˆ ˆ ˆ 3 4 3.00 N 4.00 N W d dx dy dx dy             F r i j i j     5.00 m 5.00 m 0 0 3.00 N 4.00 N 15.0 J 20.0 J 35.0 J W x y      The same calculation applies for all paths. (b) 8.25 A single constant force acts on a 4.00-kg particle. (a) Calculate the work done by this force if the particle moves from the origin to the point having the vector position . Does this result depend on the path? Explain. (b) What is the speed of the particle at r if its speed at the origin is 4.00 m/s? (c) What is the change in its potential energy? F  3ˆ i  5ˆ j  N r  2ˆ i  3ˆ j  m   ˆ ˆ 3.00 5.00 N   F i j       4.00 kg ˆ ˆ 2.00 3.00 m 3.00 2.00 5.00 3.00 9.00 J m W         r i j The result does not depend on the path since the force is conservative. W K  2 2 4.00 4.00 9.00 4.00 2 2 v          (b) 32.0 9.00 3.39 m s 2.00 v    9.00 J U W    (c ) Method like in problem 8.22 (a) 8.26 At time ti, the kinetic energy of a particle is 30.0 J and the potential energy of the system to which it belongs is 10.0 J. At some later time tf, the kinetic energy of the particle is 18.0 J. (a) ) If only conservative forces act on the particle, what are the potential energy and the total energy at time tf ? (b) If the potential energy of the system at time tf is 5.00 J, are there any nonconservative forces acting on the particle? Explain. fi fi U K K U    30.0 18.0 10.0 22.0 J U f     (a) 40.0 J E  (b) Yes there was a nonconservative force acting, because the mechanical energy is not conserved 8.31 The coefficient of friction between the 3.00-kg block and the surface in Figure P8.31 is 0.400. The system starts from rest. What is the speed of the 5.00- kg ball when it has fallen 1.50 m? Figure P8.31 f f f i nc U K U K W     2 2 2 1 2 1 1 2 2 m gh fh m v m v    1 f n m g       2 2 1 1 2 1 2 m gh m gh m m v        2 1 2 1 2 2 m m hg v m m          2 9.80 m s2 1.50 m 5.00 kg 0.400 3.00 kg 3.74 m s 8.00 kg v        8.33 A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s (Fig. P8.33). The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal. For this motion determine (a) the change in the block's kinetic energy, (b) the change in the potential energy of the block-Earth system, and (c) the friction force exerted on the block (assumed to be constant). (d) What is the coefficient of kinetic friction? Figure P8.33   2 2 2 1 1 160 J 2 2 fi i K m v v mv         3.00 m sin 30.0 73.5 J U mg     (c )The mechanical energy converted due to friction is 86.5 J (b) a) 86.5 J 28.8 N f 3.00 m  cos 30.0 28.8 N k k f n mg          2 28.8 N 0.679 5.00 kg 9.80 m s cos30.0 k     (d) 8.36 A 50.0-kg block and 100-kg block are connected by a string as in Figure P8.36. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 50-kg block and incline is 0.250. Determine the change in the kinetic energy of the 50-kg block as it moves from A to B, a distance of 20.0 m. Figure P8.36                      mech 3 4 2 2 2 2 cos37.0 0 cos37.0 400 N 0.250 400 N 100 N 100 20.0 50.0 9.80 20.0sin37.0 5.90 10 100 9.80 20.0 1.96 10 1 2 1 2 y A B A B A A fi B B fi A A fi B B fi F n mg n mg f n f x E U U K K U m g h h U m g h h K m v v K m v v                                             2 B A A A m K K m    3.92 kJ KA  8.41 A single conservative force acts on a 5.00-kg particle. The equation Fx = (2x + 4) N describes the force, where x is in meters. As the particle moves along the x axis from x = 1.00 m to x = 5.00 m, calculate (a) the work done by this force, (b) the change in the potential energy of the system, and (c) the kinetic energy of the particle at x = 5.00 m if its speed is 3.00 m/s at x = 1.00 m.   5.00 m 5.00 m 2 1 1 2 2 4 4 25.0 20.0 1.00 4.00 40.0 J 2 x x W F dx x dx x                   0 K U     a) b) 40.0 J U K W       2 1 2 f mv K K    2 1 62.5 J 2 f mv K K   c) Problem 8.45 For the potential energy curve shown in the Figure. (a) determine whether the force Fx is positive, negative, or zero at the five points indicated. (b) Indicate points of stable, unstable, and neutral equilibrium. (c) Sketch the curve for Fx versus x from x = 0 to x = 9.5 m. (a) The force is given by dx dU Fx   Fx is zero at points A, C and E; is positive at point B and negative at point D. ( c ) x(m) Fx qualitatively (b) A and E are unstable, and C is stable. 8.46 A particle moves along a line where the potential energy of its system depends on its position r as graphed in Figure P8.46. In the limit as r increases without bound, U(r) approaches +1 J. (a) Identify each equilibrium position for this particle. Indicate whether each is a point of stable, unstable or neutral equilibrium. (b) The particle will be bound if the total energy of the system is in what range? Now suppose that the system has energy 3 J. Determine (c) the range of positions where the particle can be found, (d) its maximum kinetic energy, (e) the location where it has maximum kinetic energy, and (f) the binding energy of the system—that is, the additional energy that it would have to be given in order for the particle to move out to r . Figure P8.46 There is an equilibrium point wherever the graph of potential energy is horizontal: At r=1.5mm and 3.2 mm, the equilibrium is stable. At r=2.3 mm , the equilibrium is unstable. A particle moving out toward r approaches neutral equilibrium. (a) stable unstable The system energy E cannot be less than –5.6 J. The particle is bound if (b) 5.6 J 1 J E    If the system energy is –3 J, its potential energy must be less than or equal to –3 J. Thus, the particle’s position is limited to . (c ) 0.6 mm 3.6 mm r   K U E   8.46 Continuation (d )   max min 3.0 J 5.6 J 2.6 J K E U       Then: Kinetic energy is a maximum when the potential energy is a minimum, at (e) 1.5 mm r  W  4 J 3 J 1 J W    Hence, the binding energy is (f) This is what you need to overcome the binding of this system, to escape it Try to solve 8.48: it is a nice problem 8.54 A 2.00-kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m (Fig. P8.54). The pulley is frictionless. The block is released from rest when the spring is unstretched. The block moves 20.0 cm down the incline before coming to rest. Find the coefficient of kinetic friction between block and incline. Figure P8.54 The gain in internal energy due to friction represents a loss in mechanical energy that must be equal to the change in the kinetic energy plus the change in the potential energy. Therefore, 2 1 cos sin 2 kmgx K kx mgx        0 i f v v  But:  K=0              100 0.200 2 2.00 9.80 cos37.0 0.200 2.00 9.80 sin 37.0 0.200 2  k     0.115 k Then:   8.61 A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (Fig. P8.61). The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point B, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vB = 12.0 m/s, and the block experiences an average friction force of 7.00 N while sliding up the track. (a) What is x? (b) What speed do you predict for the for the block at the top of the track (c ) Does the block actually reach the top of the track, or does it fall off before reaching the top? Figure P8.61 (a) 2 2 1 1 2 2 kx mv        2 2 1 1 450 N m 0.500 kg 12.0 m s 2 2 0.400 m x x     8.61 continuation                2 2 2 2 2 2 1 1 2 2 1 1 0.500 kg 9.80 m s 2.00 m 0.500 kg 0.500 kg 12.0 m s 2 2 7.00 N 1.00 m 0.250 4.21 4.10 m s T T B B T T T mgh mv mgh mv f R v v v                         (b) ac  g Does block fall off at or before top of track? Block falls if  2 2 2 4.10 16.8 m s 1.00 T c v a  R   ca  g therefore and the block fails to leave the track 8.65 Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with man-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle  with the vertical (Fig. P8.65). Taking D = 50.0 m, F = 110 N, L = 40.0 m, and = 50.0, (a) with what minimum speed must Jane begin her swing in order to just make it to the other side? (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume that Tarzan has a mass of 80.0 kg. Figure P8.65 Jane wants to save Tarzan, wow Solution on next side The geometry reveals : sin sin D L L       50.0 m 40.0 m sin 50 sin     28.9                      wind 2 2 2 2 2 4 3 4 1 cos 1 0 cos 2 1 50 kg 50 kg 9.8 m s 40 mcos50 110 N 50 m 50 kg 9.8 m s 40 mcos28.9 2 1 50 kg 1.26 10 J 5.5 10 J 1.72 10 J 2 2 947 J 6.15 m s 50 kg g g i f i i i i K U W K U mv mg L FD mg L v v v                                             2 2 2 2 2 4 4 1 cos 1 0 cos 2 1 130 kg 130 kg 9.8 m s 40 mcos28.9 110 N 50 m 2 130 kg 9.8 m s 40 mcos50 1 130 kg 4.46 10 J 5 500 J 3.28 10 J 2 2 6 340 J 9.87 m s 130 kg i i i i mv mg L FD mg L v v v                        (a) (b )