Physics 11 HW #6 Solutions

Transcription

1 Physics HW #6 Solutions Chapter 6: Focus On Concepts:,,, Probles: 8, 4, 4, 43, 5, 54, 66, 8, 85 Focus On Concepts 6- (b) Work is positive when the orce has a coponent in the direction o the displaceent. The orce shown has a coponent along the x and along the +y axis. Thereore, displaceents in these two directions involve positive work. Focus On Concepts 6- (c) The gravitational orce is a conservative orce, and a conservative orce does no net work on an object oving around a closed path. Focus On Concepts 6- (b) The principle o conservation o echanical energy applies only i the work done by the net nonconservative orce is zero, W nc J. When only a single nonconservative orce is present, it is the net nonconservative orce, and a orce that is perpendicular to the displaceent does no work. Focus On Concepts 6- (a) A single nonconservative orce is the net nonconservative orce, and the work it does is given by W nc KE + PE. Since the velocity is constant, KE J. Thereore, W nc PE g(h h ). But h h 35, since h is saller than h. Thus, W nc (9. kg)(9.8 /s )(35 ). Proble 6-8 REASONING AND SOLUTION According to Equation 6., W Fs cos θ, the work is a. W (94. N)(35. ) cos J b. W (94. N)(35. ) cos 39 J Proble 6-4 REASONING The initial speed v o the skier can be obtained by applying the workenergy theore: W v v (Equation 6.3). This theore indicates that the initial kinetic energy v o the skier is related to the skier s inal kinetic energy v and the work W done on the skier by the kinetic rictional orce according to

2 Solving or the skier s initial speed gives v v W v W () v The work done by the kinetic rictional orce is given by W ( cos θ ) s (Equation 6.), where k is the agnitude o the kinetic rictional orce and s is the agnitude o the skier s displaceent. Because the kinetic rictional orce points opposite to the displaceent o the skier, θ 8. According to Equation 4.8, the kinetic rictional orce has a agnitude o k µ kfn, where µ k is the coeicient o kinetic riction and F N is the agnitude o the noral orce. Thus, the work can be expressed as W ( cos θ ) s ( µ F cos8 ) s k k N Substituting this expression or W into Equation (), we have that W kfn cos8 s v v v () Since the skier is sliding on level ground, the agnitude o the noral orce is equal to the weight g o the skier (see Exaple in Chapter 4), so F N g. Substituting this relation into Equation () gives k v k v g( cos8 ) s v kg cos8 s SOLUTION Since the skier coes to a halt, v /s. Thereore, the initial speed is v v kg cos8 s /s /s cos8 4.5 /s Proble 6-4 REASONING Since air resistance is being neglected, the only orce that acts on the gol ball is the conservative gravitational orce (its weight). Since the axiu height o the trajectory and the initial speed o the ball are known, the conservation o echanical energy can be used to ind the kinetic energy o the ball at the top o the highest point. The conservation o echanical energy can also be used to ind the speed o the ball when it is 8. below its highest point. SOLUTION a. The conservation o echanical energy, Equation 6.9b, states that

3 v + gh v + gh KE Solving this equation or the inal kinetic energy, KE, yields KE v + g h h.47 kg 5. / s +.47 kg 9.8 / s J b. The conservation o echanical energy, Equation 6.9b, states that v + gh v + gh E E The ass can be eliinated algebraically ro this equation, since it appears as a actor in every ter. Solving or v and noting that the inal height is h , we have that v v + g h h 5. / s / s / s Proble 6-43 REASONING To ind the axiu height H above the end o the track we will analyze the projectile otion o the skateboarder ater she leaves the track. For this analysis we will use the principle o conservation o echanical energy, which applies because riction and air resistance are being ignored. In applying this principle to the projectile otion, however, we will need to know the speed o the skateboarder when she leaves the track. Thereore, we will begin by deterining this speed, also using the conservation principle in the process. Our approach, then, uses the conservation principle twice. SOLUTION Applying the conservation o echanical energy in the or o Equation 6.9b, we have We designate the lat portion o the track as having a height h and note ro the drawing that its end is at a height o h.4 above the ground. Solving or the inal speed at the end o the track gives v v + g h h 5.4 /s /s /s This speed now becoes the initial speed v 4.6 /s or the next application o the conservation principle. At the axiu height o her trajectory she is traveling

4 horizontally with a speed v that equals the horizontal coponent o her launch velocity. Thus, or the next application o the conservation principle v (4.6 /s) cos 48º. Applying the conservation o echanical energy again, we have Recognizing that h.4 and h.4 + H and solving or H give (.4 ) (.4 ) v + g + H v + g H v v 4.6 /s 4.6 /s cos 48 g 9.8 /s.6 Proble 6-5 REASONING a. Since there is no air riction, the only orce that acts on the projectile is the conservative gravitational orce (its weight). The initial and inal speeds o the ball are known, so the conservation o echanical energy can be used to ind the axiu height that the projectile attains. b. When air resistance, a nonconservative orce, is present, it does negative work on the projectile and slows it down. Consequently, the projectile does not rise as high as when there is no air resistance. The work-energy theore, in the or o Equation 6.6, ay be used to ind the work done by air riction. Then, using the deinition o work, Equation 6., the average orce due to air resistance can be ound. SOLUTION a. The conservation o echanical energy, as expressed by Equation 6.9b, states that The ass can be eliinated algebraically ro this equation since it appears as a actor in every ter. Solving or the inal height h gives h ( v v ) + h Setting h and v /s, the inal height, in the absence o air resistance, is g h ( 8. / s) ( /s) vo v g 9.8 / s 6.5 b. The work-energy theore is

5 F R ( ) W v v + gh gh (6.6) nc where W nc is the nonconservative work done by air resistance. According to Equation 6., the work can be written as W nc ( F R cos 8 ) s, where F R is the average orce o air resistance. As the projectile oves upward, the orce o air resistance is directed downward, so the angle between the two vectors is θ 8 and cos θ. The agnitude s o the displaceent is the dierence between the inal and initial heights, s h h.8. With these substitutions, the work-energy theore becoes Solving or F R gives ( ) + ( ) v v g h h s o R o F s v v + g h h + (.8 ).75 kg /s 8. /s.75 kg 9.8 /s.8.9 N Proble 6-54 REASONING We will use the work-energy theore Wnc E E (Equation 6.8) to ind the speed o the student. W nc is the work done by the kinetic rictional orce and is negative because the orce is directed opposite to the displaceent o the student. SOLUTION The work-energy theore states that nc + + W v gh v gh (( ( ( ( ( ((( (6.8) E E Solving or the inal speed gives W v v g h h 3 ( ) nc J + ( /s) ( 9.8 /s )(.8 ) 8.6 / s 83. kg

6 Proble 6-66 REASONING The average power is deined as the work divided by the tie, Equation 6.a, so both the work and tie ust be known. The tie is given. The work can be obtained with the aid o the work-energy theore as orulated in ( ) W v v + gh gh (Equation 6.6). W nc is the work done by the liting nc orce acting on the helicopter. In using this equation, we note that two types o energy are v and the gravitational potential energy (gh). The changing: the kinetic energy kinetic energy is increasing, because the speed o the helicopter is increasing. The gravitational potential energy is increasing, because the height o the helicopter is increasing. SOLUTION The average power is W nc P (6.a) t where W nc is the work done by the nonconservative liting orce and t is the tie. The work is related to the helicopter s kinetic and potential energies by Equation 6.6: Thus, the average power is ( ) W v v + gh gh nc ( ) + ( ) ( ) + ( ) W v v gh gh v v g h h nc P t t t ( 8 kg) ( 7. /s) ( /s) + ( 8 kg)( 9.8 /s )[ 8. ] 4 P.4 W 3.5 s Proble 6-8 REASONING The change in gravitational potential energy or both the adult and the child is PE gh - gh, where we have used Equation 6.5. Thereore, PE g(h ). In this expression h is the vertical height o the second loor above the irst loor, and its value is not given. However, we know that it is the sae or both staircases, a act that will play the central role in our solution. SOLUTION Solving PE g(h ) or h, we obtain ( PE) ( PE) h h and h h g g Since h is the sae or the adult and the child, we have

7 Solving this result or ( PE) gives ( PE) ( PE) ( PE) g g ( PE) (. 3 J)( 8. kg) 8. kg 444 J Proble 6-85 REASONING Friction and air resistance are being ignored. The noral orce ro the slide is perpendicular to the otion, so it does no work. Thus, no net work is done by nonconservative orces, and the principle o conservation o echanical energy applies. SOLUTION Applying the principle o conservation o echanical energy to the swier at the top and the botto o the slide, we have I we let h be the height o the botto o the slide above the water, h h, and h H. Since the swier starts ro rest, v /s, and the above expression becoes Solving or H, we obtain v + gh gh H h + v g Beore we can calculate H, we ust ind v and h. Since the velocity in the horizontal direction is constant, v x t 5.. /s.5 s The vertical displaceent o the swier ater leaving the slide is, ro Equation 3.5b (with down being negative), y a yt ( 9.8 /s )(.5 s).3 Thereore, h.3. Using these values o v and h in the above expression or H, we ind H h + v (. /s).3 + g (9.8 /s ) 6.33