Energy of a System ANSWERS TO QUESTIONS. Q7.1 (a) Positive work is done by the chicken on the dirt.

Transcription

1 7 Energy o a System CHAPTER OUTLINE 7. Work Done by a Constant Force 7. The Scalar Product o Two Vectors 7.4 Work Done by a Varying Force 7.5 Kinetic Energy and the Work- Kinetic Energy Theorem 7.6 Potential Energy o a System 7.7 Conservative and Nonconservative Forces 7.8 Relationship Between Conservative Forces and Potential Energy 7.9 Energy Diagrams and the Equilibrium o a System ANSWERS TO QUESTIONS Q7. (a) Positive work is done by the chicken on the dirt. The person does no work on anything in the environment. Perhaps some etra chemical energy goes through being energy transmitted electrically and is converted into internal energy in his brain; but it would be very hard to quantiy etra. (c) Positive work is done on the bucket. (d) Negative work is done on the bucket. (e) Negative work is done on the person s torso. Q7. Force o tension on a ball moving in a circle on the end o a string. Normal orce and gravitational orce on an object at rest or moving across a level loor. Q7. (a) Tension Air resistance (c) The gravitational orce does positive work in increasing speed on the downswing. It does negative work in decreasing speed on the upswing. *Q7.4 Each dot product has magnitude cosθ where θ is the angle between the two actors. Thus or (a) and we have cos =. For and (g), cos 45 =.77. For (c) and (h), cos 8 =. For (d) and (e), cos 9 =. The assembled answer is a = > b = g > d = e > c = h. Q7.5 The scalar product o two vectors is positive i the angle between them is between and 9. The scalar product is negative when 9 < θ < 8. *Q7.6 (i) The orce o block on spring is equal in magnitude and opposite to the orce o spring on block. The answers are (c) and (e). (ii) The spring tension eerts equal-magnitude orces toward the center o the spring on objects at both ends. The answers are (c) and (e). Q7.7 k = k. To stretch the smaller piece one meter, each coil would have to stretch twice as much as one coil in the original long spring, since there would be hal as many coils. Assuming that the spring is ideal, twice the stretch requires twice the orce. Q7.8 No. Kinetic energy is always positive. Mass and squared speed are both positive. A moving object can always do positive work in striking another object and causing it to move along the same direction o motion _7_ch7_p5-74.indd 5 //6 :46:44 PM

2 54 Chapter 7 Q7.9 Work is only done in accelerating the ball rom rest. The work is done over the eective length o the pitcher s arm the distance his hand moves through windup and until release. He etends this distance by taking a step orward. *Q7. answer (e). Kinetic energy is proportional to mass. *Q7. answer (a). Kinetic energy is proportional to squared speed. Doubling the speed makes an object s kinetic energy our times larger. *Q7. It is sometimes true. I the object is a particle initially at rest, the net work done on the object is equal to its inal kinetic energy. I the object is not a particle, the work could go into (or come out o) some other orm o energy. I the object is initially moving, its initial kinetic energy must be added to the total work to ind the inal kinetic energy. *Q7. Yes. The loor o a rising elevator does work on a passenger. A normal orce eerted by a stationary solid surace does no work. *Q7.4 answer (c). I the total work on an object is zero in some process, its kinetic energy and so its speed must be the same at the inal point as it was at the initial point. *Q7.5 The cart s ied kinetic energy means that it can do a ied amount o work in stopping, namely (6 N)(6 cm) =.6 J. The orward orce it eerts and the distance it moves in stopping must have this ied product. answers: (i) c (ii) a (iii) d Q7.6 As you ride an epress subway train, a backpack at your eet has no kinetic energy as measured by you since, according to you, the backpack is not moving. In the rame o reerence o someone on the side o the tracks as the train rolls by, the backpack is moving and has mass, and thus has kinetic energy. *Q7.7 answer (e). 4. J= k (. m) Thereore k = 8 N m and to stretch the spring to. m requires etra work W = ( 8)(. ) 4. J=. J Q7.8 (a) Not necessarily. It does i it makes the object s speed change, but not i it only makes the direction o the velocity change. *Q7.9 (i) (ii) Yes, according to Newton s second law. The gravitational acceleration is quite precisely constant at locations separated by much less than the radius o the planet. Answer: a = b = c = d The mass but not the elevation aects the gravitational orce. Answer: c = d > a = b (iii) Now think about the product o mass times height. Answer: c > b = d > a Q7. There is no violation. Choose the book as the system. You did work and the Earth did work on the book. The average orce you eerted just counterbalanced the weight o the book. The total work on the book is zero, and is equal to its overall change in kinetic energy. Q7. In stirring cake batter and in weightliting, your body returns to the same conormation ater each stroke. During each stroke chemical energy is irreversibly converted into output work (and internal energy). This observation proves that muscular orces are nonconservative. 794_7_ch7_p5-74.indd 54 //6 :46:45 PM

3 Energy o a System 55 Q7. A graph o potential energy versus position is a straight horizontal line or a particle in neutral equilibrium. The graph represents a constant unction. *Q7. (c) The ice cube is in neutral equilibrium. Its zero acceleration is evidence or equilibrium. *Q7.4 The gravitational energy o the key-earth system is lowest when the key is on the loor letterside-down. The average height o particles in the key is lowest in that coniguration. As described by F = dud, a orce pushes the key downhill in potential energy toward the bottom o a graph o potential energy versus orientation angle. Friction removes mechanical energy rom the key- Earth system, tending to leave the key in its minimum-potential energy coniguration. Q7.5 Gaspard de Coriolis irst stated the work-kinetic energy theorem. Jean Victor Poncelet, an engineer who invaded Russia with Napoleon, is most responsible or demonstrating its wide practical applicability, in his 89 book Industrial Mechanics. Their work came remarkably late compared to the elucidation o momentum conservation in collisions by Descartes and to Newton s Mathematical Principles o the Philosophy o Nature, both in the 6 s. SOLUTIONS TO PROBLEMS Section 7. Work Done by a Constant Force P7. (a) W = F rcos θ =( 6. N)(. m) cos 5. =. 9 J, (c) The normal orce and the weight are both at 9 to the displacement in any time interval. Both do work. (d) W =. 9 J+ + =. 9 J P7. (a) W = mgh= (. 5 )( 9. 8) J=. 8 5 Since R= mg, W air resistance = 8. J P7. METHOD ONE Let φ represent the instantaneous angle the rope makes with the vertical as it is swinging up rom φ i = to φ = 6. In an incremental bit o motion rom angle φ to φ+ d φ, the deinition o radian measure implies that r = ( m) dφ. The angle θ between the incremental displacement and the orce o gravity is θ = 9 + φ. Then cosθ= cos( 9 + φ)=sin φ. The work done by the gravitational orce on Batman is W = F cosθdr = mg( sinφ)( m) dφ i φ= 6 6 φ= =mg( m) sin φ dφ = 8 kg 9. 8 m s m ( cosφ) = ( 784 N)( m )( cos 6 + ) = 4. 7 J FIG. P7. METHOD TWO The orce o gravity on Batman is mg = ( 8 kg)( 9. 8 m s ) = 784 N down. Only his vertical displacement contributes to the work gravity does. His original y-coordinate below the tree limb is m. His inal y-coordinate is ( m) cos 6 = 6 m. His change in elevation is 6 m( m) = 6 m. The work done by gravity is W = F rcosθ =( 784 N)( 6 m) cos 8 = 4. 7 kj J 6 794_7_ch7_p5-74.indd 55 //6 :46:45 PM

4 56 Chapter 7 *P7.4 Yes. Object eerts some orward orce on object as they move through the same displacement. By Newton s third law, object eerts an equal-size orce in the opposite direction on object. In W = F r cos θ, the actors F and r are the same, and θ diers by 8, so object does 5. J o work on object. The energy transer is 5 J rom object to object, which can be counted as a change in energy o 5 J or object and a change in energy o +5 J or object. Section 7. P7.5 The Scalar Product o Two Vectors A B= ( A ˆi+ A ˆj+ A kˆ) ( B ˆi+ B ˆ y z yj+ Bzkˆ) A B= AB ˆi ˆi AB ˆi ˆj AB ˆi kˆ + ( ) + ( ) ( ˆji ˆ) + ( ˆjj ˆ) + ( ˆjk ˆ) ( kˆ i) + AB z y( k j) + AB z z( k k) y z + A B A B A B y y y y z + AB z ˆ ˆ ˆ ˆ ˆ A B= AB + AB + AB y y z z P7.6 A = 5. ; B = 9. ; θ = 5. A B= AB cos θ =( 5. )( 9. ) cos 5. = 8. 9 P7.7 (a) W = F r = F+ Fyy=( 6. )(. ) N m+ (. )(. ) N m = 6. J F r θ = = 6 cos cos F r ( 6. ) + (. ) ( + ) = P7.8 We must irst ind the angle between the two vectors. It is: Then F v θ = =. = Fv cos θ =(. 8 N)(. 7 m s) cos. or Fv = N m 5. = 5. s J s P7.9 (a) A=. ˆ i. ˆj B= 4. ˆ i4. ˆ A B j θ = cos = cos AB (. )(. ) =. FIG. P7.8 (c) B=. ˆ i 4. ˆ j+. kˆ A=. ˆ i+ 4. ˆ A B.. j cosθ = = 6 6 θ = 56 AB (. )( 9. ) A= ˆ i. ˆ j+. kˆ B=. ˆ j+ 4. kˆ A B θ = = cos cos 8. AB = 794_7_ch7_p5-74.indd 56 //6 :46:46 PM

5 Energy o a System 57 P7. *P7. A B= (. ˆi+ ˆjkˆ) ( ˆ i+. ˆ j+ 5. kˆ ) A B= 4. ˆiˆ j6. kˆ C ( AB) = (. ˆ j.kˆ ) ( 4. ˆiˆ j6. kˆ ) = + (. ) + ( + 8. ) = 6. Let θ represent the angle between A and B. Turning by 5 makes the dot product larger, so the angle between C and B must be smaller. We call it θ 5. Then we have A 5 cos θ = and A 5 cos (θ 5 ) = 5 Then A cos θ = 6 and A (cos θ cos 5 + sin θ sin 5 ) = 7 Dividing, cos 5 + tan θ sin 5 = 76 tan θ = (76 cos 5 )sin 5 =.66 θ =.6. Then the direction angle o A is 6.6 = 8.4 Substituting back, A cos.6 = 6 so A = 7.5 m at 8.4 Section 7.4 P7. F ( 8 6) N = Work Done by a Varying Force (a) See igure to the right m N m N W net = (. )( 6. ) + (. )( 8. ) =. J P7. W = Fd = area under curve rom i to i (a) i = = 8. m W = area o triangle ABC = AC altitude, W = m. N=. J i = 8. m =. m W = area o CDE = CE altitude, W = 8 (. m) (. N) =. J (c) W = W 8 + W8 = 4. + (. ) =. J 5 m P7.4 W = F dr= 4ˆi+ yˆj N dˆi i 5 m 5 m 4 Nm d 4 Nm 5. J + = = FIG. P7. FIG. P7. 794_7_ch7_p5-74.indd 57 //6 :46:47 PM

6 58 Chapter 7 P7.5 W = Fd and W equals the area under the Force-Displacement curve (a) For the region 5. m, W = (. N )( 5. m ) = For the region 5.., 75. J FIG. P7.5 W = (. N)( 5. m) = 5. J (c) For the region. 5., W = (. N )( 5. m ) = (d) For the region J W = ( ) J=. J P7.6 (a) Spring constant is given by F = k F k = = ( N ) ( 4 ) = 575. m Work = Favg = ( N)( 4 m) = 46 J.. Nm F P7.7 k = Mg y = y = ( 4. )( 98. ) N = 57. Nm 5. m mg (a) For.5 kg mass y = = ( 5. )( 98. ) = k 57. Work = ky. 98 cm Work = (. 57 N m) ( 4. m) =. 5 J *P7.8 In F = k, F reers to the size o the orce that the spring eerts on each end. It pulls down on the doorrame in part (a) in just as real a sense at it pulls on the second person in part. (a) Consider the upward orce eerted by the bottom end o the spring, which undergoes a downward displacement that we count as negative: k = F = (7.5 kg)(9.8 ms )(.45 m +.5 m) = 7.5 N(.65 m) =. knm Consider the end o the spring on the right, which eerts a orce to the let: = Fk = (9 N)( Nm) =.68 m The length o the spring is then.5 m +.68 m =.58 m *P7.9 ΣF = ma : k = ma k ma = = = 77. (4.7 kg).8(9.8 m s ).5 m Nm FIG. P _7_ch7_p5-74.indd 58 //6 :46:48 PM

7 Energy o a System 59 *P7. The spring eerts on each block an outward orce o magnitude Fs = k =(. 85 Nm)(. 8 m) =. 8 N n F s Take the + direction to the right. For the light block on the let, the vertical orces are given by Fg = mg=( 5. kg )( 98. m s ) = 45. N, F y =, n 45. N =, n = 45. N. Similarly or the heavier block n= =( 5. kg)( 98. m s ) = 49. N F g.45 N FIG. P7. *P7. =, = For the heavier block, + = (a) For the block on the let, F ma. 8 N. 5 kg a, a =. m s.. 8 N. 5 kg a, a =. 66 m s. For the block on the let, k = µ kn=. ( 45. N) = 45. N F = ma. 8 ms. 45 N. 5 kga + = a = 5. m s i the orce o static riction is not too large. For the block on the right, k = µ kn=. 49 N. The maimum orce o static riction would be larger, so no motion would begin and the acceleration is zero. (c) Let block: k =. 46(. 45 N) =. N. The maimum static riction orce would be larger, so the spring orce would produce no motion o this block or o the right-hand block, which could eel even more riction orce. For both a =. Compare an initial picture o the rolling car with a inal picture with both springs compressed K + W = K. Work by both springs changes the car s kinetic energy i K + k ( )+ k ( ) = K mv + i ( 6 Nm)(. 5 m) + ( 4 Nm) (. m) = ( 6 kg) v J 68. J = i ( 68 J) v = 99 i =. ms 6 kg i i i P7. (a) W = F dr i. 6 m ( W = 5 N+ N m5 N m ) d cos 5 W = W = 9. kj + 8. kj 8. kj = 9. kj m F ( N) FIG. P7. continued on net page 794_7_ch7_p5-74.indd 59 //6 :46:49 PM

8 6 Chapter 7 Similarly, W = +. kn m. m ( 5. kn m )(. m) 5. kn. m W =. 7 kj,largerby9.6% P7. The same orce makes both light springs stretch. (a) The hanging mass moves down by mg mg = + = + = mg + k k k k = 5. kg 9.8ms m m + 4 m N 8 N =. We deine the eective spring constant as F mg k = = mg k + k k k = + m m = + N 8 N = 7 Nm P7.4 See the solution to problem 7.. (a) = mg + k k it would i either spring were missing. Both springs stretch, so the load moves down by a larger amount than k = + k k The spring constant o the series combination is less than the smaller o the two individual spring constants, to describe a less sti system, that stretches by a larger etension or any particular load. P7.5 (a) The radius to the object makes angle θ with the horizontal, so its weight makes angle θ with the negative side o the -ais, when we take the -ais in the direction o motion tangent to the cylinder. F = ma F mgcosθ = F = mgcosθ FIG. P7.5 mg W = F dr i We use radian measure to epress the net bit o displacement as dr = Rdθ in terms o the net bit o angle moved through: π W = mgcosθrdθ = mgrsinθ W = mgr( ) = mgr π F P7.6 [ k]= = N = kg m s m m = kg s 794_7_ch7_p5-74.indd 6 //6 :46:5 PM

9 Energy o a System 6 *P7.7 We can write u as a unction o v : (8 N [ N])(5 cm 5 cm) = (u [ N])(v 5 cm) (.5 Ncm)(v 5 cm) = u + N u = (.5 Ncm)v 4.5 N also v = ( cmn)u + 9 cm (a) Then b a udv= ( v ) dv= v / 45. v =. 5( 65 5) 4. 5( 5 5) = 5 9 = 6 Ncm =. 6 J Reversing the limits o integration just gives us the negative o the quantity: udv =. 6 J (c) This is an entirely dierent integral. It is larger because all o the area to be counted up is positive (to the right o v = ) instead o partly negative (below u = ). b 8 8 vdu = ( u + 9) du = u / + 9u = 64( ) + 9( 8 + ) = = 5 N cm =. 5 J a b a *P7.8 I the weight o the irst tray stretches all our springs by a distance equal to the thickness o the tray, then the proportionality epressed by Hooke s law guarantees that each additional tray will have the same eect, so that the top surace o the top tray will always have the same elevation above the loor. =. The orce 4 The weight o a tray is 58. kg 98. m s 568. N Fs 4. N stretch one spring by.45 cm, so its spring constant is k = = =.4 5 m We did not need to know the length or width o the tray. ( 568. N) = 4. N should 6 Nm. Section 7.5 Kinetic Energy and the Work-Kinetic Energy Theorem. 6 kg. m s. J P7.9 (a) K A = = K B mv B = KB: v B = m = ( )( 7. 5 ). 6 = 5. m s (c) W = K = KB KA = m( vb v A) = 75. J. J= 6. J P7. (a) K = mv = (. kg)( 5. m s) =. 8 J K = = = (.8) = 5 J P7. v 6. ˆ i. ˆj ms i = = (a) v = v + v = i i iy 4. m s Ki = mv i = (. kg)( 4. m s ) = 6. J v = 8. ˆ i+ 4. ˆj = v v = = 8. m s v K = K Ki = m v vi. = = J 794_7_ch7_p5-74.indd 6 //6 :46:5 PM

10 6 Chapter 7 P7. (a) K = K Ki = mv = W = (area under curve rom = to = 5. m) ( area) ( 7. 5 J) v = = = 94. ms m 4. kg K = K Ki = mv = W = (area under curve rom = to =. m) v ( area) (. 5 J) = = = m 4. kg 5. ms (c) K = K Ki = mv = W = (area under curve rom = to = 5. m) v = ( area) (. J) = = m 4. kg 87. ms P7. Consider the work done on the pile driver rom the time it starts rom rest until it comes to rest at the end o the all. Let d = 5. m represent the distance over which the driver alls reely, and h =. m the distance it moves the piling. so Thus, W = K: Wgravity + Wbeam = mv mvi + mg h d = ( + ) = d mg ( h d) cos + ( F)( d) cos8 = ( kg)( 9. 8 m s )( 5. m) F. m The orce on the pile driver is upward. = N *P7.4 (a) We evaluate the kinetic energy o the cart and the work the cart would have to do to plow all the way through the pile. I the kinetic energy is larger, the cart gets through. K = ()mv = ()(. kg)(.6 m s) =.54 J The work done on the cart in traveling the whole distance is the net area under the graph, W = ( N) (. m) + [( N)](.4 m) =. J.6 J =.4 J The work the cart must do is less than the original kinetic energy, so the cart does get through all the sand. The work the cart does is +.4 J, so its inal kinetic energy is the remaining.54 J.4 J =.4 J. Another way to say it: rom the work-kinetic energy theorem, K i + W = K.54 J.4 J =.4 J = ()(.5 kg)v v = [(.4 kg m s )(. kg)] =.6 m s *P7.5 (a) Ki + W = K = mv + W = ( 5. kg)( 78 m s) = kj W F r 456. J cosθ (.7 m) cos = 64. kn continued on net page 794_7_ch7_p5-74.indd 6 //6 :46:5 PM

11 Energy o a System 6 (c) v vi a = 78 ms = = (. 7 m) 4 km s ( ) = (d) F = ma= 5 kg 4 m s 6. 4 kn (e) The orces are the same. The two theories agree. = P7.6 (a) v =. 96 ms. 88 K 8 7 ms 7 = mv = ( 9. kg) (. 88 m s ) =.78 6 J K + W = K : + F rcosθ = i K F (. 8 m) cos = J F = 5. 4 N F 5. (c) F = ma; a = = m 9. 4 N kg = 48. (d) v = v + at ms= ms t i t = s + 6 ms Check: = i + i + t v v. 8 m = + ( m s)t t = s Section 7.6 Potential Energy o a System P7.7 (a) With our choice or the zero level or potential energy when the car is at point B, U B = When the car is at point A, the potential energy o the car-earth system is given by U A = mgy FIG. P7.7 where y is the vertical height above zero level. With 5 t = 4. m, this height is ound as: y = ( 4. m) sin 4. = 6. 4 m Thus, U A = = kg 98. m s 64. m J The change in potential energy as the car moves rom A to B is 5 5 UB UA =. 59 J=. 59 J continued on net page 794_7_ch7_p5-74.indd 6 //6 :46:5 PM

12 64 Chapter 7 With our choice o the zero level when the car is at point A, we have U A =. The potential energy when the car is at point B is given by UB = mgy where y is the vertical distance o point B below point A. In part (a), we ound the magnitude o this distance to be 6.5 m. Because this distance is now below the zero reerence level, it is a negative number. Thus, kg 98. m s ( 65. m) = J U B = The change in potential energy when the car moves rom A to B is 5 5 UB UA = 59. J = 59. J P7.8 (a) We take the zero coniguration o system potential energy with the child at the lowest point o the arc. When the string is held horizontal initially, the initial position is. m above the zero level. Thus, U g = mgy=( 4 N )(. m ) = 8 J From the sketch, we see that at an angle o. the child is at a vertical height o (. m) ( cos. ) above the lowest point o the arc. Thus, U g = mgy=( 4 N )(. m ) ( cos. ) = 7 J FIG. P7.8 (c) The zero level has been selected at the lowest point o the arc. Thereore, U g = at this location. Section 7.7 Conservative and Nonconservative Forces P7.9 F mg g = = = 4. kg 9. 8 m s 9. N y (a) Work along OAC = work along OA + work along AC = F ( OA) cos 9. + F ( AC) cos8 g = ( 9. N) ( 5. m) + ( 9. N)( 5. m) ( ) g B C (5., 5.) m = 96 J W along OBC = W along OB + W along BC = ( 9. N)( 5. m) cos 8 + ( 9. N)( 5. m) cos 9. = 96 J O A FIG. P7.9 (c) Work along OC = F g ( OC) cos5 = 9. N 5. m = 96 J The results should all be the same, since gravitational orces are conservative. 794_7_ch7_p5-74.indd 64 //6 :46:54 PM

13 P7.4 (a) W = F dr and i the orce is constant, this can be written as W = F dr= F r r i, which depends only on end points, not path. 5. m 5. m W = F dr= ( ˆi+ 4ˆj) ( dˆi+ dyˆ j) = (. N) d + ( 4. N) dy 5. m 5. m W = (. N) + ( 4. N) y = 5. J+. J= 5. J The same calculation applies or all paths. P7.4 (a) The work done on the particle in its irst section o motion is and since along this path, y = W OA = = 5. m 5. m W = dˆ OA yˆ + i i ˆj yd In the net part o its path W = dyˆ AC yˆ + ˆ j i j dy For = 5. m, W AC = 5 J = 5. m 5. m and W OAC = + 5 = 5 J Following the same steps, W = dyˆ OB yˆ + ˆ j i j dy since along this path, =, W OB = since y = 5. m, W BC = 5. J = 5. m 5. m = 5. m 5. m W = dˆ BC yˆ + i i ˆj yd W OBC = + 5. = 5. J Energy o a System 65 (c) W = OC ( d ˆ+ dy ˆ ) ( y ˆ + ˆ i j i j) = yd + dy (d) = Since = y along OC, W = + OC d J F is nonconservative since the work done is path dependent. P7.4 Along each step o motion, the rictional orce is opposite in direction to the incremental displacement, so in the work cos 8 =. 5. m (a) W = ( N)(5 m)() + ( N)(5 m)() =. J The distance CO is (5 + 5 ) m = 7.7 m W = ( N)(5 m)() + ( N)(5 m)() + ( N)(7.7 m)() = 5. J (c) (d) W = ( N)(7.7 m)() + ( N)(7.7 m)() = 4.4 J The orce o riction is a nonconservative orce. 794_7_ch7_p5-74.indd 65 //6 :46:55 PM

14 66 Chapter 7 Section 7.8 Relationship Between Conservative Forces and Potential Energy 5. m 5. m P7.4 (a) W = F d = ( + 4) d = + 4 K + U = U = K = W = 4. J m (c) K = K v P7.44 (a) U = A+ B d P7.45 U( r) = K K m v = + = A B = A U = Fd =... m 6. 5 J. m B K = A+ B A r = = 4. (. ) (. ) = A B U d A A Fr = r = dr r = r. I A is positive, the positive value o radial orce indicates a orce o repulsion. U ( y P7.46 F = y y = 7 ) =( 9 7) = 79 U ( y F = y y = 7 ) =( ) = y J Thus, the orce acting at the point, y is F = Fˆi+ Fˆ j= y ˆ 7 9 i ˆj y. Section 7.9 Energy Diagrams and the Equilibrium o a System P7.47 (a) F is zero at points A, C and E; F is positive at point B and negative at point D. (c) A and E are unstable, and C is stable. F B A C E (m) D FIG. P _7_ch7_p5-74.indd 66 //6 :46:55 PM

15 Energy o a System 67 P7.48 stable unstable neutral FIG. P7.48 P7.49 (a) The new length o each spring is + L, so its etension is + L L and the orce it eerts is k( + L L) toward its ied end. The y components o the two spring orces add to zero. Their components add to + F= ˆ ik + L L = kˆi L L + L FIG. P7.49 Choose U = at =. Then at any point the potential energy o the system is U = Fd = k+ U = k + kl L + L kl + L (c) U = For negative, U has the same value as or positive. The only equilibrium point (i.e., where F = ) is =. (d) K + U + E = K + U i i mech J+ = (. kg) v + v =. 8 ms d k = d kl L d + FIG. P7.49(c) Additional Problems P7.5 The work done by the applied orce is W = F d = k + k d applied = i ma ma ma ma ma kd k d k k + = + ma = + ma k k 794_7_ch7_p5-74.indd 67 //6 :46:56 PM

16 68 Chapter 7 P7.5 At start, v 4. mscos. ˆ i 4. mssin. ˆj = + = ( 4 ms) + =( 46 ms) At ape, v. cos. ˆi ˆ j. ˆi And K = mv = (. 5 kg)( 4. 6 m s) = 9. J P7.5 (a) We write b F = a N= a(. 9 m) 5 N= a(. 5 m) 5 5 = = ln 5= b ln. 44 b ln 5 b = = 8. = b ln 4. 4 N a = 8 = m b b b 4 Nm.8 = a 5. m 4 N 8. W = Fd = 4. d.8 m 5. m. 5. m 8 4 N = 4. = 4..8 m 8. = 94 J 4 N. m 8 ( 5. m) *P7.5 (a) We assume the spring is in the horizontal plane o the motion. The radius o the puck s motion is.55 m + The spring orce causes the puck s centripetal acceleration: (4. Nm) = F = mv r = m(πr T) r = 4π m r(. s) (4. kgs ) = (.4s ) m (.55 m + ) 4. kg =.6 m m +.4 m 4. kg.6 m =.68 m m = (.6 m)(4. kg.4 m) meters (c) (d) (e) () = (.6 m.7 kg)(4. kg.4 [.7 kg]) =.95 m a nice reasonable etension We double the puck mass and ind = (.68 m.4 kg)/(4. kg.6 [.4 kg]) =.49 m more than twice as big! = (.6 m.8 kg)(4. kg.4 [.8 kg]) = 6.85 m We have to get a bigger table!! When the denominator o the raction goes to zero, the etension becomes ininite. This happens or 4. kg.4 m = ; that is or m =.84 kg. For any larger mass, the spring cannot constrain the motion. The situation is impossible. The etension is directly proportional to m when m is only a ew grams. Then it grows aster and aster, diverging to ininity or m =.84 kg. 794_7_ch7_p5-74.indd 68 //6 :46:57 PM

17 Energy o a System 69 *P7.54 (a) A time interval. I the interaction occupied no time, the orce eerted by each ball on the other would be ininite, and that cannot happen. k = F = 6 N. m = 8 MNm (c) We assume that steel has the density o its main constituent, iron, shown in Table 4.. Then its mass is ρv = ρ (4)πr = (4π)(786 kgm )(.54 m) =.674 kg and K = () mv = () (.674 kg)(5 ms) =.84 J.8 J (d) Imagine one ball running into an ininitely hard wall and bouncing o elastically. The original kinetic energy becomes elastic potential energy.84 J = () (8 7 Nm) =.45 mm.5 mm (e) The ball does not really stop with constant acceleration, but imagine it moving.45 mm orward with average speed (5 ms + ) =.5 ms. The time interval over which it stops is then.45 mm(.5 ms) = 6 5 s 4 s *P7.55 The potential energy at point is given by 5 plus the negative o the work the orce does as a particle eeling the orce is carried rom = to location. U du = Fd du = e 8 d U 5 =8/ e d 5 ( ) U = 5 8/ e = 5+ 4e 4 = + 4e The orce must be conservative because the work the orce does on the object on which it acts depends only on the original and inal positions o the object, not on the path between them. P7.56 (a) d F= ( + + ) ˆi = ( 4) ˆi d F = when =. 87 and. 55 (c) The stable point is at =. 55 point o minimum U The unstable point is at FIG. P7.56 = 87. maimum in U 794_7_ch7_p5-74.indd 69 //6 :46:58 PM

18 7 Chapter 7 P7.57 Ki + Ws + Wg = K mvi + ki k + mg cosθ = mv + ki + mgi cos = mv Ncm. cm. m. ( kg) (. 8 ms)(. 5 m) s i n.. = ( kg) v. 5 J 8. 5 J=(. 5 kg) v. 4 v = = 68. m s. 5 P7.58 (a) F = ( 5. N ) cos 5. ˆi+ sin 5. ˆ j. 5ˆi 4. ĵ (c) (d) (e) = ( + ) N F = ( 4. N ) cos5 ˆi+ sin5 ˆ j 6. 4ˆi. ĵ = ( + ) N F= F+ F = 5. 9ˆ i+ 5. ˆj N F a = = 8. ˆ i+ 77. ˆj ms m v = v + a = 4. ˆ i+ 5. ˆ j ms 8. ˆ i 77. ˆ j ms. s i t + + v = 554. ˆ i+ 7. ˆj ms r = r + i v t+ i a t r = + ( 4. ˆ i+. 5ˆ j)( ms)(. s ) + 8. ˆi+ 77. ˆ j ms. s r = r =. ˆ i+ 9. ĵ m () K = mv = ( 5. kg) ( 5. 54) + (. 7) m s = 48. kj FIG. P7.57 (g) K = mvi + F r K = ( 5. kg) ( 4. ) + ( 5. ) ( m s ) + ( 5. 9 N) (. m) + ( 5. N)( 9. m) K = J + 46 J=. 48 kj [ ] (h) The work-kinetic energy theorem is consistent with Newton s second law, used in deriving it. 794_7_ch7_p5-74.indd 7 //6 :46:58 PM

19 Energy o a System d P7.59 We evaluate by calculating (. ) 75. (. 8) +. 75(. 8) + (. 9) (. ) (. 6) +. 75(. 6) =. 86 and 75(. ) 75. (. 9) +. 75(. 9) + (. ) (. ) 75.. (. 7) +. 75(. 7) =. 79. The answer must be between these two values. We may ind it more precisely by using a value or smaller than.. Thus, we ind the integral to be. 799 N m. P7.6 (a) F N L mm F N L mm FIG. P7.6 To draw the straight line we use all the points listed and also the origin. I the coils o the spring touched each other, a bend or nonlinearity could show up at the bottom end o the graph. I the spring were stretched too ar, a nonlinearity could show up at the top end. But there is no visible evidence or a bend in the graph near either end. By least-square itting, its slope is. 5 Nmm± % = 5 Nm ± % F In F = k, the spring constant is k =, the same as the slope o the F-versus- graph. (c) F = k =( 5 Nm)(. 5 m) =. N ANSWERS TO EVEN PROBLEMS P7. (a). 8 J 8. J P7.4 Yes. It eerts a orce o equal magnitude in the opposite direction that acts over the same distance. 5. J P P Js 794_7_ch7_p5-74.indd 7 //6 :46:59 PM

20 7 Chapter 7 P7. 6. P7. (a) see the solution. J P J P7.6 (a) 575 N m 46. J P7. 8 (a). knm 5.8 cm P7. (a). ms and +.66 ms.5 ms and (c) and P7. (a) 9. kj.7 kj, larger by 9.6% P7.4 (a) mg mg + + k k k k P7.6 kgs P7. 8 I the weight o the irst tray stretches all our springs by a distance equal to the thickness o the tray, then the proportionality epressed by Hooke s law guarantees that each additional tray will have the same eect, so that the top surace o the top tray will always have the same elevation. 6 Nm. We do not need to know the length and width o the tray. P7. (a).8 J 5 J P7. (a). 94 m s. 5 m s (c). 87 m s P7. 4 (a) yes. Its kinetic energy as it enters the sand is suicient to do all o the work it must do in plowing through the pile..6 ms P7.6 (a) J N (c) ms (d).94 ns P7.8 (a) 8 J 7 J (c) P7.4 (a) see the solution 5. J P7. 4 (a). J 5. J (c) 4.4 J (d) The orce o riction is a nonconservative orce. P7.44 (a) A B U =.5A 6.B; K =.5A + 6.B P yˆi ˆj P7.48 see the solution P7.5 k ma + k ma P7.5 (a) a = 4. 8 m.kn ; b = J P7. 54 (a) A time interval. I the interaction occupied no time, the orce eerted by each ball on the other would be ininite, and that cannot happen. 8 MNm (c).8 J. We assume that steel has the same density as iron. (d).5 mm (e) 4 s 794_7_ch7_p5-74.indd 7 //6 :47: PM

21 Energy o a System 7 P7.56 (a) F = ( 4 ).87 and.55 (c) see the solution..55 is stable and.87 is unstable. ( 8. ˆ 77. ˆ ) ( ) (. ˆ i 9. ˆ ) P7.58 (a) F =. 5ˆ i+ 4. ˆj N; F = 6. 4ˆ i+. ˆj N 5. 9 ˆ i+ 5. ˆ j N (c) i+ j ms (d) i+ j ms (e) +.48 kj (g).48 kj (h) The work-kinetic energy theorem is consistent with Newton s second law, used in deriving it. P7. 6 (a) See the solution. We use all the points listed and also the origin. There is no visible evidence or a bend in the graph or nonlinearity near either end. 5 Nm± % (c). N j m 794_7_ch7_p5-74.indd 7 //6 :47: PM

22 794_7_ch7_p5-74.indd 74 //6 :47: PM