Pearson Physics Level 20 Unit III Circular Motion, Work, and Energy: Unit III Review Solutions

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1 Pearon Phyic Level 0 Unit III Circular Motion, Work, and Energy: Unit III Review Solution Student Book page 6 9 Vocabulary. artificial atellite: a huan-ade object in orbit around a celetial body axi of rotation: an iaginary line that pae through the centre of rotation perpendicular to circular otion axle: a haft on which a wheel rotate centripetal acceleration: acceleration acting toward the centre of a circle for an object oving in a circular path centripetal force: force acting toward the centre of a circle cauing an object to ove in a circular path conervation of energy: within an iolated yte, energy ay be tranferred fro one object to another or tranfored fro one for to another, but it cannot be increaed or decreaed conervative force: a force that act within a yte but doe not change it total echanical energy; include gravity and elatic force cycle: one coplete ocillation eccentricity: a nuber between 0 and that repreent the degree of elongation of a circle efficiency: ratio of the energy output to the energy input of any yte elatic potential energy: energy reulting fro an object being altered fro it tandard hape, without peranent deforation ellipe: an elongated circle that ha two foci energy: ability to do work frequency: the nuber of ocillation (cycle) per econd gravitational potential energy: energy of an object due to it poition above the urface of Earth iolated yte: group of object aued to be iolated fro all other object in the univere Kepler contant: a contant that relate a planet orbital period and radiu 9 K a /AU or.0 0 / for object orbiting the Sun Kepler law: the three law that Kepler deduced fro Tycho Brahe obervation (ee text for full definition) kinetic energy: energy due to the otion of an object; ybol i E k. ean orbital radiu: the average orbital radiu of a atellite, eaured fro the centre of the body being orbited to the centre of the atellite; for object with elliptical orbit, it i equal to the ei-ajor axi echanical energy: u of potential and kinetic energie; ybol i E. non-iolated yte: yte in which there i an energy exchange with the urrounding orbital period: the tie it take for a atellite to ake one coplete orbit Copyright 009 Pearon Education Canada

2 orbital perturbation: irregularitie or diturbance in the predicted orbit of a planet period: the tie required for an object to ake one coplete ocillation (cycle) potential energy: energy that i tored and held in readine; include gravitational and elatic potential energie; ybol i E p power: rate of doing work reference point: arbitrarily choen point fro which ditance are eaured revolution: one coplete cycle for an object oving in a circular path rp: revolution per inute; a unit of frequency for an object oving in a circular path atellite: a celetial body that i in orbit around another celetial body unifor circular otion: otion in a circular path at a contant peed work: eaure of the aount of energy tranferred when a force act over a given diplaceent; calculated a the product of the agnitude of applied force and the diplaceent of the object in the direction of the force work-energy theore: work done on a yte i equal to the u of the change in the potential and kinetic energie of the yte Knowledge Chapter 5. If the centripetal acceleration reain contant, the velocity will decreae a the radiu decreae.. The car kid when the force of friction i not great enough to provide the neceary centripetal force. The car goe alot in a traight-line otion, which take it to the outide of the curve. 4. (a) When the bucket i in the highet poition, two force work in concert to create the centripetal force: the force of gravity and the tenion. (b) The rope i likely to break when the bucket i in the botto of the wing. In thi poition, the tenion on the rope i greatet becaue it ut provide all the centripetal force neceary to keep the bucket oving around the circle, while alo balancing the weight of the bucket. 5. The force you feel on your hand when pinning an object around in a circle i the reaction force the object i exerting on you. Your hand i providing a centripetal force on the object. 6. The agnitude of the centripetal force for the planet orbiting the Sun change becaue the planet have different ae and are different ditance fro the Sun. Copyright 009 Pearon Education Canada

3 . The orbit are elliptical due to the change in centripetal S P Fc FG G r P acceleration with ditance fro the Sun ( r ). 7. (a) The force of gravity i the centripetal force at the top of the loop (poition C). In the top half of the loop, a coponent of the force of gravity act a the centripetal force along with the track. (b) The roller coater exert a force on the track in all poition except at the very top (poition C). (c) Fc Fg v Fc r v g ae cancel r v g r 8. There i no relationhip between the frequency and radiu of a rotating dic. The frequency i the ae for all radii. 9. A table aw ight ue different blade of different diaeter. The peed at the outer edge of different blade will vary, but the rotational frequency will not. 0. The period of the Moon (T Moon ), radiu of the Moon orbit (r Moon ), and G ut be known in order to deterine Earth a.. (a) The planet with the leat elliptical orbit i Venu with an eccentricity of (b) Venu ei-ajor axi i cloet in length to it ei-inor axi becaue it ha the leat eccentricity.. The three coin are identical except that they are placed at different radii. A the frequency of the platter increae, o doe the centripetal force a hown by thi equation: Fc 4 rf Since 4 i contant, the only factor affecting the centripetal force are: Fc rf The coin at the outer radiu will experience a greater centripetal force. If the force of friction (which i the ae for all coin) i not great enough to provide the centripetal force, the outerot coin (coin C) will lide off firt, followed by coin B then coin A.. Neptune wa dicovered when atronoer noticed perturbation in the orbit of Uranu. They noticed that the orbital velocity of Uranu wa lowed. They reaoned Copyright 009 Pearon Education Canada

4 that the force of gravity exerted on Uranu by another planet farther out ut be the caue of the orbital perturbation. Baed on the degree that Uranu orbital velocity changed, they atheatically deterined where the unknown planet hould be. When they looked for it in the poition predicted, they found it and called it Neptune. 4. She i probably right. If Neptune were not a huge planet, the force of gravity that it exerted on Uranu would probably not have been enough to ake a noticeable perturbation in the orbit of Uranu. Neptune would likely have gone undicovered for quite oe tie. 5. An extraolar planet doe not produce light of it own, and the light fro the tar it orbit i o bright that we can t ee the planet. Viually ighting one with current telecope i very unlikely. A planet ay be detected by the perturbation in the tar oveent if the planet i aive. To date, planet detected in thi way are very large and cloe to their un. Thi ake the likelihood of life on thee planet reote. Chapter 6 6. J = kg / 7. The work done by a force depend on the angle between the direction of the force and the diplaceent over which it act. 8. An object in free fall i contantly loing gravitational potential energy. If there were no air reitance, then the lot gravitational potential energy would all be tranfored into kinetic energy. However, ince there i air reitance, the air reitance tranfor kinetic energy into heat in the urrounding air. A the peed of the object increae, the agnitude of the force of air reitance increae until it reache the agnitude of the force of gravity o that the net force on the object gradually becoe zero. At that tie, the acceleration of the object becoe zero and the object continue it fall at a contant peed known a it terinal velocity. While falling at it terinal velocity, the object' gravitational potential energy i tranfored into kinetic energy at the ae rate a it kinetic energy i tranfored into heat. 9. Since an object' kinetic energy varie a the quare of it peed, doubling the peed will quadruple ( ) the kinetic energy. 0. For a given force, the topping ditance varie directly a the initial kinetic energy. Since the two ae have equal kinetic energie, a force of equal agnitude acting on each of the ae will top the in equal ditance.. A the rollercoater roll down the firt hill, gravitational potential energy i tranfored into kinetic energy. In turn, the force of friction (both rolling and air) act on the trolley to tranfor kinetic energy into theral energy (heat). A the trolley roll up the following hill, gravity tranfor kinetic energy into gravitational potential energy. At the ae tie, friction i tranforing kinetic energy into heat. Thu, the height of the econd hill ut be lower than the height of the firt hill becaue the gravitational potential energy at the top of the econd hill ut be le than the gravitational potential energy at the top of the firt hill by a agnitude at leat equal to the aount of energy that wa converted into heat by friction.. In ter of echanical energy, thi i not an iolated yte. The cart' gravitational potential energy i increaing while the kinetic energy i contant. Thu the force pulling the cart up the hill ut be adding echanical energy to the yte. Therefore, the force pulling the cart up the hill cannot be conidered a conervative force. 4 Copyright 009 Pearon Education Canada

5 . If no force other than the force of gravity and the force of the inclined plane act on the cart, then a cart rolling down an inclined plane could be conidered to be an iolated yte with repect to echanical energy. If, however, friction or oe other force i applied to the cart, then thoe force ay reduce or add to the quantity of echanical energy of the yte. Regardle of whether or not other force act on the cart, the force of gravity i till conidered a conervative force. 4. Auing no friction, a the linghot i pulled back, cheical energy fro your ucle i tranfored into elatic potential energy in the ling. When the ling i releaed, it elatic potential energy i tranfored into kinetic energy of the tone. A the tone rie, it kinetic energy i tranfored into gravitational potential energy until the tone reache it axiu height and no longer ha any kinetic energy. If friction exit, energy i continually being tranfored into heat at all tage except for the intant at which the tone i at ret, jut before it i releaed and at the top of the rie. 5. If an object kinetic energy change, there ut be a force that ha a coponent in the direction of the object diplaceent. 6. On a force-diplaceent graph, the work i equivalent to the area under the curve. 7. According to the work-energy theore, no work i done on an iolated yte. 8. Power decribe the rate at which you are able to do work, not how uch work you are able to do. A otor ay be rated at 750 W; however, the actual work it doe will depend on how long and how fat it run. Application 9. Analyi and Solution T f 60 Hz 0.07 The period of the electron i Analyi and Solution f T Hz The cell phone ha a frequency of 6.5 Hz. 5 Copyright 009 Pearon Education Canada

6 . Analyi and Solution 00.0rev in f Hz in Hz T f Hz The frequency of the toy top i Hz and it period i Analyi and Solution T 7. d d r.8440 k k r v T / The peed of the Moon a it orbit Earth i.0 0 /.. Analyi and Solution r v T r T v (4.0) The period of the ride i Analyi and Solution v ac r / He experience an acceleration of.09 /. 6 Copyright 009 Pearon Education Canada

7 5. Given ac 4.00g r = peed of the airplane ( v ) Analyi and Solution Convert the acceleration of 4.00g to appropriate SI unit. Ue the equation for centripetal acceleration to deterine the peed of the plane. 4 tie the acceleration of gravity i: / 9.4/ Now deterine the peed of the plane. v ac r v acr 9.4 (500.0).40 0 / The airplane i travelling at a peed of.40 0 /. 6. Analyi and Solution v Fc r kg N The centripetal force on the cork i N. 7. Given 50 kg Ff 00 N (a) v.0 (b) v v E 0 k 7 Copyright 009 Pearon Education Canada

8 (a) topping ditance for the given peed (b) topping ditance for a peed that i twice a great Analyi and Solution (a) When the car applie it brake and coe to ret, the force of friction act to tranfor it kinetic energy into heat. Calculate the ditance required for the force of friction to do work equal in agnitude to the car initial kinetic energy. The direction of the force of friction i oppoite to that of the initial velocity. Ek v 50 kg J W Ek Ff co d Ek E k Ek E k d Ff co J 00 Nco (b) Since the kinetic energy varie a the peed quared, doubling the peed quadruple the kinetic energy. When the force i contant, the diplaceent varie directly a the work it doe. Hence, if the energy i four tie larger, then the topping ditance ut be four tie larger. 4d (a) The force of friction bring the car to a top in 8. in the direction of the initial velocity. (b) If the peed were doubled, the topping ditance would be four tie a great: that i,. 8. Given Fapp 50 N [up] 5.0 kg g 9.8 [down] d h 9.60 [up] (a) work done by the applied force (b) change in gravitational potential energy for the a (c) explanation of difference in anwer (a) and (b), auing iolated yte 8 Copyright 009 Pearon Education Canada

9 Analyi and Solution (a) The work done by the applied force i found uing the equation for work. W Fco d 50 N co J =.400 J (b) The change in gravitational potential energy i found uing the following equation. The change in height i equal to the diplaceent. E gh p 5.0 kg J.40 J (c) If the applied force i greater in agnitude than the force of gravity, the object will gain kinetic energy a well a gravitational potential energy a it ove up the incline. A the object ove upward, the applied force doe.400 J of work on it. Of that work,.4 0 J i tranfored into gravitational potential energy and the reainder (about 990 J) i tranfored into kinetic energy. If the yte i not frictionle, oe of the kinetic energy will have been tranfored into heat. A the reult of the.400 J done by the applied force, the object gained.4 0 J of gravitational potential energy and about 990 J of kinetic energy. 9 Copyright 009 Pearon Education Canada

10 9. Given.000 kg v 5.0 [0 ] Fapp.00 N [0 ] d 55.0 [0 ] Ff N (a) free body diagra and net force on the car (b) work done by the net force (c) final kinetic energy (d) final peed of the car Analyi and Solution (a) Draw a diagra howing all the force that act on the car. Find the net force by adding all the force together then ue that force to calculate the work it doe. Since the upward force exerted by road urface and the force of gravity are equal in agnitude but oppoite in direction the u of thoe force i zero. F F F F F net app f g road F F F F F net app f g road.00 N N F F.800 N N Fnet.800 N [0 ] (b) W F co d net.80 0 Nco J g g 0 Copyright 009 Pearon Education Canada

11 (c) The work-energy theore tate that the work done by the net force caue a change in kinetic energy. Calculate the final kinetic energy by adding the anwer fro part (b) to the initial kinetic energy. Ek E k Ek W v J.00 0 kg J (d) Ue the final kinetic energy fro part (c) to calculate the final peed of the car. Ek v Ek v 5 (.79 0 J).000 kg 9.5 (a) The free body diagra i a hown. The net force on the car i.800 N [0 ]. (b) The work done by the net force i equal to J. (c) The final kinetic energy of the car i J. (d) The final peed of the car i 9.5 /. 40. Given kg v 0 Fapp 5.00 N d 4.50 v 6.00 g 9.8 change in height through which block oved Analyi and Solution Since the plane i frictionle, all of the work done by the applied force will be tranfored into kinetic and gravitational potential energy. The difference between Copyright 009 Pearon Education Canada

12 the work done by the force and the gain in kinetic energy will give the gravitational potential energy. Ue the gravitational potential energy to calculate the gain in height. W Ek Ep E W E p k Fd E E k k Fd v v (5.00 N)(4.50 ) kg J Ep gh Ep h g 8.0J 0.800kg9.8.0 A the force ove the block a ditance of 4.50, the block gain.0 in height. 4. Given 5.0 kg.0 [0 ] F 5.0 N [0 ] d 0.0 F 45.0 N [0 ] d 0.0 F 0 d 0.0 (a) work done by the force (b) final peed of the a Analyi and Solution (a) The work i equal to the area under the curve. The area conit of three ection (i) a trapezoid, (ii) a rectangle and (iii) a triangle. Ue the equation for the area for each of thee figure to calculate the work in each region then add to get the total work done. W A A A F Fd Fd Fd 5.0 N 45.0 N N N J.00 J (b) Since the force are net force, the work reult in a change in the object kinetic energy. Calculate the initial kinetic energy and add the change in kinetic energy to Copyright 009 Pearon Education Canada

13 give the final kinetic energy. Ue the final kinetic energy to calculate the final peed. Ek E k Ek W v 00 J 5.0 kg J.000 J Ek v Ek v (000 J) 5.0 kg 5.5 (a) The total work done by the force i.0 0 J. (b) The final peed of the a i 5.5 /. 4. Given A 4.50 kg B 5.50 kg g 9.8 hb.50 da.50 v 0 v.00 (a) change in gravitational potential energy for block A (b) change in height through which block A oved Analyi and Solution (a) When the yte i et in otion, block B will fall loing gravitational potential energy. Since there i no friction, by conervation of echanical energy, the net lo in gravitational potential energy of the yte ut equal it net gain in kinetic energy. A block B fall, the lo in gravitational potential energy i tranfored into a gain in kinetic energy for block A and B a well a a gain in gravitational potential energy for A. The gain in block A' gravitational potential energy can be deterined by ubtracting the gain in kinetic energie for block A and B fro the lo in gravitational potential energy for block B. Copyright 009 Pearon Education Canada

14 Ep Ek 0 E E p E E E E p p k A B A k k E E E E pa pb ka kb B gh v v v v B B A A B B 5.50 kg kg kg0 5.50kg k g J 0.5 J 4.75 J 5.9 J 5.9 J (b) Ue the change in block A' gravitational potential energy to calculate it gain in height. AghA EpA EpA ha A g 5.9 J 4.50 kg (a) A block B fall, it loe 80.9 J of gravitational potential energy, which i tranfored into 45.0 J of kinetic energy for block A and B and 5.9 J of gravitational potential energy for block A. (b) The gain in gravitational potential energy for block A i the reult of a gain in height of Given A 4.50 kg B 5.50 kg g 9.8 v 0 v.00 hb.50 da.50 Graph of potential, kinetic, and echanical energie of the yte under condition of (a) no friction, (b) friction but till acceleration 4 Copyright 009 Pearon Education Canada

15 (a) (b) [NOTE: Since A and B are tied together, they undergo the ae change in gravitational potential energy on both graph. On graph (b), however, the lo in energy to friction ut coe at the expene of the kinetic energie. Student ay decreae the kinetic energie by any arbitrarily choen aount but the ratio of the kinetic energie of A to B hould reain at 4.5/5.5 (9/). At the end of the otion, the u of the kinetic energie of A and B plu the gravitational potential energy of A hould be equal to the total echanical energy.] 5 Copyright 009 Pearon Education Canada

16 44. Given (a) kg (b) Fapp 40.0 N g 9.8 v 0 h 0 h 0.50 v.00 (a) work done on the pendulu by your puh (b) diplaceent over which you exerted force Analyi and Solution (a) The work done on the pendulu reult in it gain in gravitational and potential energie. The u of thee energie ut be equal to the work done to produce the. By auing that the initial height i zero, the gain in energy i equal to the u of the final kinetic and gravitational potential energie. E E E p k gh v 0.750kg kg J.60 J W E E.604 J 0.60 J 6 Copyright 009 Pearon Education Canada

17 (b) Ue the equation for work to calculate the agnitude of the diplaceent given the force. W Fd W d F.604 J 40.0 N (a) To give the pendulu enough energy to produce the pecified condition would require.60 J of work. (b) A force of 40.0 N would need to act over a ditance of to do that aount of work. 45. Given A 0.00 kg B 0.00 kg va.00 v B 0 v A.50 peed of ball B after the colliion Analyi and Solution If echanical energy i conerved and there i no change in potential energy, then the final kinetic energy ut equal the initial kinetic energy. Find the total initial kinetic energy of the two ball and the final kinetic energy of ball A. The final kinetic energy of ball B i the difference between thee two value. Ue the final kinetic energy of ball B to find it final peed. 7 Copyright 009 Pearon Education Canada

18 E E E E k ka k B AvA BvB 0.00 kg kg J Ek E E k ka k B AvA E kb k B k A A E E v E k v J 0.00 kg J v B B B B E k B B 0.75 J 0.00 kg. The final kinetic energy of ball B i equal to 0.75 J, which give it a final peed of. /. [NOTE: In quetion 45, the law of conervation of oentu i obeyed. The colliion i elatic o that both kinetic energy and oentu are conerved. In an elatic colliion of thi nature, the two ball ove away at right angle to each other. When tudent have copleted the unit on conervation of oentu in Phyic 0, they can return to thee quetion and verify that linear oentu ut be conerved in both elatic and inelatic colliion.] 46. Given.00 kg (b) v vco 0 g 9.8 [down] (c) h 0 v 80 [0 ] h 450 (a) initial echanical energy for the cannonball (b) highet point the cannonball attain (c) peed of the cannonball when it land 8 Copyright 009 Pearon Education Canada

19 Analyi and Solution (a) Uing the ocean urface a height = 0, calculate the total echanical energy by calculating the u of the kinetic and gravitational potential energie. E E E p k gh v.00 kg kg J 5.0 J (b) Aue that echanical energy i conerved. To calculate gravitational potential energy, firt find the kinetic energy at the highet point and then ubtract that quantity fro the echanical energy. Ue the gravitational potential energy to calculate the height. The peed at the highet point i the horizontal coponent of it initial velocity. Calculate the horizontal coponent of the peed uing vector analyi. v v co 0 80 co E E k E p E E E p k E v J.00 kg J 9 Copyright 009 Pearon Education Canada

20 gh h p E Ep g J.00 kg (c) At the urface of the ocean the gravitational potential energy of the ball i zero. Thu, auing that echanical energy i conerved, the kinetic energy i equal to it echanical energy. Ue the kinetic energy to calculate the peed of the ball when it hit the ocean. E E k E p Ek E E p J J v Ek v E k J.00 kg (a) Initially, the echanical energy of the cannonball i. 0 5 J. (b) At it highet point above the ocean, the cannon ball ha a peed of 6 / and i 97 above the ocean (or 467 above it tarting point). (c) When the cannonball hit the urface of the ocean, it peed i 95 /. 47. Given h kg g 9.8 v 0 v.00 (a) change in the ball' echanical energy a it fall (b) average force that air friction exert on the ball while it fall 0 Copyright 009 Pearon Education Canada

21 Analyi and Solution (a) A the ball fall, it loe gravitational potential energy and gain kinetic energy. Since there i a force of friction, air friction will tranfor oe of the kinetic energy into heat. When the ball land, it final echanical energy will be le than it initial echanical energy by the aount of energy tranfored into heat. E E E ( E E ) ( E E ) k p k p v gh v gh ( ) ( ) 0.00 kg kg kg kg J (b) To calculate the force of air friction, find the force required to caue the change in echanical energy over the ditance the ball fall. That force will be equal to the average force of friction. W E FhE E F h 8.9 J N (a) During it fall, the ball loe 8.9 J of echanical energy. (b) Friction exert an average force of.78 N in the oppoite direction (up) to the direction of the otion (down). Copyright 009 Pearon Education Canada

22 48. Given 50 kg g 9.8 d h0.0 t 0.0 average power output of the engine Analyi and Solution The work i equal to the increae in gravitational potential energy. Ue the equation for power to calculate the output of the engine. W P t Ep t gh t 50 kg W The average power output of the engine i.680 W or.68 kw. 49. Given k v 08 h h 080 h F 540 N power output (P) Analyi and Solution Ue the equation that relate power to force and peed. P Fv 540 N W For a force of 540 N, to aintain a peed of 08 k/h require a power output 4 of.6 0 W or 6. kw. Copyright 009 Pearon Education Canada

23 50. Given 5 P 50 kw.50 0 W engine Ff.50 0 N peed that reult for power output of 50 kw Analyi and Solution To aintain a contant peed, the force the engine exert to ove the plane forward (thrut) ut be equal in agnitude to the force of friction (drag). The peed generated by a given power output can be found uing the equation for power. Pengine Fenginev Pengine v Fengine Pengine Ff W.50 0 N 60.0 The airplane' peed will be 60.0 / (6 k/h). Extenion 5. Given = 90.9 kg FT 08.8 N r =.0 peed of the gynat when he i above the bar ( v ) Analyi and Solution The gynat ar experience a tug of 08.8 N when he i above the bar. Thi i the tenion in hi ar. The force of gravity and tenion are both down. Take down to be negative, and up to be poitive. Deterine the peed of the gynat body fro the centripetal force that he experience. Firt deterine the centripetal force; then the peed. Copyright 009 Pearon Education Canada

24 Fc Fg FT Fc ( g) ( 08.8N) (90.9 kg) N.00 0 N The centripetal force i downward, but for the purpoe of deterining peed, it i a calar quantity, o the negative ign i not needed. v Fc r Fr c v ( N)(.0) 90.9 kg.6 / The peed of the gynat above the bar i.6 /. 5. (a) Given 7 T 6.0 r.000 a of the planet tar ( Star ) Analyi and Solution Deterine the a of the planet tar fro Newton verion of Kepler third law. 4 Copyright 009 Pearon Education Canada

25 Fc Fg 4 planetr GplanetStar T r 4 r Star TG N kg kg 0 The a of the planet tar i kg. (b) Given The orbital period and radiu of the planet oon: 6 T.70 8 r a of the planet ( ) p Analyi and Solution Deterine the a of the planet in the ae anner a it tar. The orbital period and radiu of the planet oon will be ued to deterine it a. Fc Fg 4 Moonr GMoonp T r 4 r p TG N kg kg 5 The a of the planet i kg. (c) Given The orbital period and radiu of the planet oon: 6 T.70 8 r peed of the planet oon ( v ) Analyi and Solution Deterine the peed of the oon directly fro it period and orbital radiu. 5 Copyright 009 Pearon Education Canada

26 r v T / The peed of the planet oon i.8 0 /. 5. Given r =.05 f = 0.70 Hz deterine if the dirt will fall out at the top of the wheel Analyi and Solution Thi proble i iilar to the roller coater exaple. The centripetal force i a net force that i the u of the force exerted by the bucket wall and the force of gravity. The dirt will not fall out if the centripetal force i greater than the force of gravity, becaue the wall will be providing a portion of the required force. If the force of gravity i greater than the centripetal force, the dirt will fall out. You don t know the a of the dirt o you cannot deterine the force, but a i not needed: Fc F Fg ac a g ae cancel ac a g The ae logic applie to the acceleration. Deterine the centripetal acceleration and copare it with the acceleration of gravity. If a c < g, the dirt will fall out; if a c > g, the dirt will tay in the bucket. a 4 rf c Hz 8.78 / [down] g 9.8 / [down] The centripetal acceleration i le than the acceleration of gravity, o the dirt will fall out. 54. Given A 6.00 kg B 4.00 kg C.00 kg Fapp 48.0 N [90 ] Ek 0 (a) d 7.50 [90 ] (b) d.00 [90 ] (a) peed of the block after the force ha acted over 7.50 (b) peed of the block A after the force ha acted over.00 6 Copyright 009 Pearon Education Canada

27 Analyi and Solution (a) Since there i no friction and the block are oving on a horizontal urface, the applied force i a net force. The work-energy theore ay that the work done by a net force produce a change in the kinetic energy. Calculate the final kinetic energy fro the change in kinetic energy and ue that to calculate the final peed of the block a if they were a ingle block. The echanical energy of thi yte i not contant, o it i not conidered an iolated yte. A B C 6.00kg 4.00kg.00 kg =.00 kg Ek Wnet E F co d k net net v F co d v F co d net 48.0 Nco kg 7.75 (b) When the force ha acted over a diplaceent of.00, the tring between block A and B ha becoe tight o that block B i oving. However, the tring between block B and C i not yet taut o that block C i till at ret. Calculate the work done by the force and olve for the kinetic energy in the ae anner a in part (a) except that the a in which the kinetic energy i tored i that of block A and B. A B 6.00 kg 4.00 kg =0.00 kg Ek Wnet E F co d k net net v F co d v Fnet co d 48.0 N co kg 4.8 (a) After the force ha acted over the diplaceent of 7.50, the three block are oving with a peed of 7.75 /. 7 Copyright 009 Pearon Education Canada

28 (b) After the force ha acted over a diplaceent of.00, only block A and B are oving with a peed of 4.8 /. Skill Practice 55. Satellite don t fall to Earth becaue they have a velocity that i tangent to their circular path. If gravity were to diappear, they would fly off tangentially in a traight path. Gravity i pulling atellite toward the centre of Earth, o atellite are oving both toward the centre of Earth and off tangentially. The reult i that a atellite fall around the Earth. 56. Student will find any webite dealing with eteor ipact crater. Site where data can be found are: The lat ite i an Earth Ipact Databae aintained by the Univerity of New Brunwick. If you click on the link (at left) for the Crater Inventory, by Nae, you get a data table of the ore iportant crater lited alphabetically. By location, give you a world ap howing the location of the crater in the databae. Click on any area of the ap and you will get a ap of that region in ore detail. If you click on Canada, you can eaily locate the nae and location of the crater in Canada and near Alberta. Click on the nuber on the ap and the detail of that crater are given. While thi databae lit the ot iportant ipact ite, there are hundred of other aller ipact ite in Alberta. A bit of web urfing will reveal any ite on thi topic. 57. The work done by external force on a cart rolling down an inclined plane i equivalent to the lo in echanical energy of the cart a it decend. Calculate the cart' echanical energie at the top and botto of the rap. At the top of the rap, the cart' echanical energy i equal to it gravitational potential energy relative to the bae of the rap. At the botto of the rap the cart' echanical energy i equal to it kinetic energy. Meaure the peed of the cart when it reache the bae of the rap uing a photo gate tier or other appropriate ethod. Ue that peed to calculate the final kinetic energy. Other variable that need to be deterined are the initial height of the cart, the a of the cart and the acceleration due to gravity at the location of the experient. Self-aeent Student' anwer in thi ection will vary greatly depending on their experience, depth of undertanding, and the depth of the reearch that they do into each of the topic. 8 Copyright 009 Pearon Education Canada

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